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The Statistics of Causal Inference in the Social Sciences 1 Political Science 236A Statistics 239A Jasjeet S. Sekhon UC Berkeley October 29, 2014 1 c Copyright 2014 1 / 765

The Statistics of Causal Inference in the Social Sciences

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The Statistics of Causal Inference in theSocial Sciences1

Political Science 236AStatistics 239A

Jasjeet S. Sekhon

UC Berkeley

October 29, 2014

1 c© Copyright 20141 / 765

Background

• We are concerned with causal inference in the socialsciences.

• We discuss the potential outcomes framework of causalinference in detail.

• This framework originates with Jerzy Neyman (1894-1981),the founder of the Berkeley Statistics Department.

• A key insight of the framework is that causal inference is amissing data problem.

• The framework applies regardless of the method used toestimate causal effects, whether it be quantitative orqualitative.

2 / 765

Background

• We are concerned with causal inference in the socialsciences.

• We discuss the potential outcomes framework of causalinference in detail.

• This framework originates with Jerzy Neyman (1894-1981),the founder of the Berkeley Statistics Department.

• A key insight of the framework is that causal inference is amissing data problem.

• The framework applies regardless of the method used toestimate causal effects, whether it be quantitative orqualitative.

3 / 765

The Problem

• Many of the great minds of the 18th and 19th centuriescontributed to the development of social statistics: DeMoivre, several Bernoullis, Gauss, Laplace, Quetelet,Galton, Pearson, and Yule.

• They searched for a method of statistical calculus thatwould do for social studies what Leibniz’s and Newton’scalculus did for physics.

• It quickly came apparent this is would be most difficult.• For example: deficits crowding out money versus

chemotherapy

4 / 765

The Problem

• Many of the great minds of the 18th and 19th centuriescontributed to the development of social statistics: DeMoivre, several Bernoullis, Gauss, Laplace, Quetelet,Galton, Pearson, and Yule.

• They searched for a method of statistical calculus thatwould do for social studies what Leibniz’s and Newton’scalculus did for physics.

• It quickly came apparent this is would be most difficult.• For example: deficits crowding out money versus

chemotherapy

5 / 765

The Experimental Model

• In the early 20th Century, Sir Ronald Fisher (1890-1962)helped to establish randomization as the “reasoned basisfor inference” (Fisher, 1935)

• Randomization: method by which systematic sources ofbias are made random

• Permutation (Fisherian) Inference

6 / 765

The Experimental Model

• In the early 20th Century, Sir Ronald Fisher (1890-1962)helped to establish randomization as the “reasoned basisfor inference” (Fisher, 1935)

• Randomization: method by which systematic sources ofbias are made random

• Permutation (Fisherian) Inference

7 / 765

Historical Note: Charles SandersPeirce

• Charles Sanders Peirce (1839–1914) independently, andbefore Fisher, developed permutation inference andrandomized experiments

• He introduced terms “confidence” and “likelihood”• Work was not well known until later. Betrand Russell wrote

(1959): “Beyond doubt [...] he was one of the most originalminds of the later nineteenth century, and certainly thegreatest American thinker ever.”

• Karl Popper (1972): “[He is] one of the greatestphilosophers of all times”

8 / 765

Historical Note: Charles SandersPeirce

• Charles Sanders Peirce (1839–1914) independently, andbefore Fisher, developed permutation inference andrandomized experiments

• He introduced terms “confidence” and “likelihood”• Work was not well known until later. Betrand Russell wrote

(1959): “Beyond doubt [...] he was one of the most originalminds of the later nineteenth century, and certainly thegreatest American thinker ever.”

• Karl Popper (1972): “[He is] one of the greatestphilosophers of all times”

9 / 765

There are Models and then there areModels

• Inference based on observational data remains a difficultchallenge. Especially, without rigorous mathematicaltheories such as Newtonian physics

Note the difference between the following two equations:• F = m × a

usually measured as F (force) Newtons, N; m (mass) kg; a(acceleration) as m/s2

• Y = Xβ

10 / 765

Regression is Evil

• It was hoped that statistical inference through the use ofmultiple regression would be able to provide to the socialscientist what experiments and rigorous mathematicaltheories provide, respectively, to the micro-biologist andastronomer.

• OLS has unfortunately become a black box which peoplethink solves hard problems it actually doesn’t solve. It is inthis sense Evil.

• Let’s consider a simple sample to test intuition

11 / 765

Regression is Evil

• It was hoped that statistical inference through the use ofmultiple regression would be able to provide to the socialscientist what experiments and rigorous mathematicaltheories provide, respectively, to the micro-biologist andastronomer.

• OLS has unfortunately become a black box which peoplethink solves hard problems it actually doesn’t solve. It is inthis sense Evil.

• Let’s consider a simple sample to test intuition

12 / 765

A Simple Example

• Let Y be a IID standard normal random variable, indexedby t

• Define ∆Yt = Yt − Yt−1

• Let’s estimate via OLS:

∆Yt = α + β1∆Yt−1 + β2∆Yt−2 + β3∆Yt−3

• Question: What are the values of the betas as n→∞?

13 / 765

A Simple Example II

• What about for:

∆Yt = α + β1∆Yt−1 + β2∆Yt−2 + · · ·+ βk ∆Yt−k

?

14 / 765

OLS: Sometimes Mostly Harmless

Notwithstanding the forgoing, as we shall see• OLS, and some other estimators, have some nice

properties under certain identification assumptions• These assumptions are distinct from the usual statistical

assumptions (e.g., those required for Gauss-Markov)• Relevant theorems do not assume that OLS is correct, that

the errors are IID.• e.g., what happens when we don’t assume E(ε|X ) = 0?

15 / 765

Early and Influential Methods

• John Stuart Mill (in his A System of Logic) devised a set offive methods (or canons) of inference.

• They were outlined in Book III, Chapter 8 of his book.• Unfortunately, people rarely read the very next chapter

entitled “Of Plurality of Causes: and of the Intermixture ofEffects.”

16 / 765

Mill’s Methods of Inductive Inference

• Of Agreement: “If two or more instances of thephenomenon under investigation have only onecircumstance in common, the circumstance in which aloneall the instances agree is the cause (or effect) of the givenphenomenon.”

• Of Difference: “If an instance in which the phenomenonunder investigation occurs, and an instance in which itdoes not occur, have every circumstance in common saveone, that one occurring only in the former; thecircumstance in which alone the two instances differ is theeffect, or the cause, or an indispensable part of the cause,of the phenomenon.”

17 / 765

Mill’s Methods of Inductive Inference

• Of Agreement: “If two or more instances of thephenomenon under investigation have only onecircumstance in common, the circumstance in which aloneall the instances agree is the cause (or effect) of the givenphenomenon.”

• Of Difference: “If an instance in which the phenomenonunder investigation occurs, and an instance in which itdoes not occur, have every circumstance in common saveone, that one occurring only in the former; thecircumstance in which alone the two instances differ is theeffect, or the cause, or an indispensable part of the cause,of the phenomenon.”

18 / 765

Three Other Methods

• Joint Method of Agreement and Difference: “If anantecedent circumstance is invariably present when, butonly when, a phenomenon occurs, it may be inferred to bethe cause of that phenomenon.”

• Method of Residues: “If portions of a complexphenomenon can be explained by reference to parts of acomplex antecedent circumstance, whatever remains ofthat circumstance may be inferred to be the cause of theremainder that phenomenon.”

• Method of Concomitant Variation: “If an antecedentcircumstance is observed to change proportionally with theoccurrence of a phenomenon, it is probably the cause ofthat phenomenon.”

19 / 765

Uses of Mills Methods

These methods have been used by a vast number ofresearchers, including such famous ones as Durkheim andWeber. They are known as the “most similar” and “mostdifferent” research designs in some fields (Przeworski andTeune, 1970):Here are some examples:• The Protestant Ethic• Deficits and interest rates• Health care systems and life expectancy• Gun control• Three strikes• The list goes on, and on....

20 / 765

Uses of Mills Methods

These methods have been used by a vast number ofresearchers, including such famous ones as Durkheim andWeber. They are known as the “most similar” and “mostdifferent” research designs in some fields (Przeworski andTeune, 1970):Here are some examples:• The Protestant Ethic• Deficits and interest rates• Health care systems and life expectancy• Gun control• Three strikes• The list goes on, and on....

21 / 765

Uses of Mills Methods

These methods have been used by a vast number ofresearchers, including such famous ones as Durkheim andWeber. They are known as the “most similar” and “mostdifferent” research designs in some fields (Przeworski andTeune, 1970):Here are some examples:• The Protestant Ethic• Deficits and interest rates• Health care systems and life expectancy• Gun control• Three strikes• The list goes on, and on....

22 / 765

Uses of Mills Methods

These methods have been used by a vast number ofresearchers, including such famous ones as Durkheim andWeber. They are known as the “most similar” and “mostdifferent” research designs in some fields (Przeworski andTeune, 1970):Here are some examples:• The Protestant Ethic• Deficits and interest rates• Health care systems and life expectancy• Gun control• Three strikes• The list goes on, and on....

23 / 765

Uses of Mills Methods

Mill himself thought they were inappropriate for the study ofsocial questions (Sekhon, 2004).

“Nothing can be more ludicrous than the sort ofparodies on experimental reasoning which one isaccustomed to meet with, not in popular discussiononly, but in grave treatises, when the affairs of nationsare the theme. “How,” it is asked, “can an institution bebad, when the country has prospered under it?” “Howcan such or such causes have contributed to theprosperity of one country, when another hasprospered without them?” Whoever makes use of anargument of this kind, not intending to deceive, shouldbe sent back to learn the elements of some one of themore easy physical sciences” (Mill, 1873, pp. 346–7).

24 / 765

Basic Statistical Inference

Let’s look at an example Mill himself brought up:

“In England, westerly winds blow during about twiceas great a portion of the year as easterly. If, therefore,it rains only twice as often with a westerly as with aneasterly wind, we have no reason to infer that any lawof nature is concerned in the coincidence. If it rainsmore than twice as often, we may be sure that somelaw is concerned; either there is some cause in naturewhich, in this climate, tends to produce both rain and awesterly wind, or a westerly wind has itself sometendency to produce rain.”

25 / 765

Conditional Probability

H :P(rain|westerly wind, Ω) >

P(rain|not westerly wind, Ω),

where Ω is a set of background conditions we considernecessary for a valid comparison.

A lot is hidden in the Ω. The issue becomes clearer with thepotential outcomes framework.

26 / 765

Potential Outcomes

• Fundamental problem: not observing all of the potentialoutcomes or counterfactuals

• This is the Neyman-Rubin Causal Model (Neyman 1923,Rubin 1974, Holland 1986).

• Can be used to describe problems of causal inference forboth experimental work and observational studies.

27 / 765

Potential Outcomes

• Fundamental problem: not observing all of the potentialoutcomes or counterfactuals

• This is the Neyman-Rubin Causal Model (Neyman 1923,Rubin 1974, Holland 1986).

• Can be used to describe problems of causal inference forboth experimental work and observational studies.

28 / 765

Observational Study

An observational study concerns• cause-and-effect relationships• treatments, interventions or policies and• the effects they cause

The design stage of estimating the causal effect for treatment Tis common for all Y .

29 / 765

Observational Study

An observational study concerns• cause-and-effect relationships• treatments, interventions or policies and• the effects they cause

The design stage of estimating the causal effect for treatment Tis common for all Y .

30 / 765

A Thought Experiment

An observational study could in principle have been anexperiment but for ethical concerns or logistical issues.

You are probably not estimating a causal effect if you can’tanswer Dorn’s (1953) Question: “what experiment would youhave run if you were dictator and has infinite resources?”

E.G.: Can we estimate the causal effect of race on SAT scores?

Descriptive and predictive work is something else and can beinteresting.

31 / 765

A Thought Experiment

An observational study could in principle have been anexperiment but for ethical concerns or logistical issues.

You are probably not estimating a causal effect if you can’tanswer Dorn’s (1953) Question: “what experiment would youhave run if you were dictator and has infinite resources?”

E.G.: Can we estimate the causal effect of race on SAT scores?

Descriptive and predictive work is something else and can beinteresting.

32 / 765

A Thought Experiment

An observational study could in principle have been anexperiment but for ethical concerns or logistical issues.

You are probably not estimating a causal effect if you can’tanswer Dorn’s (1953) Question: “what experiment would youhave run if you were dictator and has infinite resources?”

E.G.: Can we estimate the causal effect of race on SAT scores?

Descriptive and predictive work is something else and can beinteresting.

33 / 765

A Thought Experiment

An observational study could in principle have been anexperiment but for ethical concerns or logistical issues.

You are probably not estimating a causal effect if you can’tanswer Dorn’s (1953) Question: “what experiment would youhave run if you were dictator and has infinite resources?”

E.G.: Can we estimate the causal effect of race on SAT scores?

Descriptive and predictive work is something else and can beinteresting.

34 / 765

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A Modest Experiment

Figure 2: Brand Keyword Click Substitution

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MSN Paid MSN Natural Goog Natural

(a) MSN Test

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Goog Paid Goog Natural

(b) Google Test

MSN and Google click trac is shown for two events where paid search was suspended (Left)

and suspended and resumed (Right).

To quantify this substitution, Table 1 shows estimates from a simple pre-post comparison

as well as a simple di↵erence-in-di↵erences across search platforms. In the pre-post analysis

we regress the log of total daily clicks from MSN to eBay on an indicator for whether days

were in the period with ads turned o↵. Column 1 shows the results which suggest that

click volume was only 5.6 percent lower in the period after advertising was suspended.

This approach lacks any reasonable control group. It is apparent from Figure 2a that

trac increases in daily volatility and begins to decline after the advertising is turned

o↵. Both of these can be attributed to the seasonal nature of e-commerce. We look to

another search platform which serves as a source for eBay trac, Google, as a control

group to account for seasonal factors. During the test period on MSN, eBay continued to

purchase brand keyword advertising on Google which can serve as a control group. With

this data, we calculate the impact of brand keyword advertising on total click trac. In

the di↵erence-in-di↵erences approach, we add observations of daily trac from Google

and Yahoo! and include in the specification search engine dummies and trends.15 The

variable of interest is thus the interaction between a dummy for the MSN platform and a

dummy for treatment (ad o↵) period. Column 2 of Table 1 show a much smaller impact

15The estimates presented include date fixed e↵ects and platform specific trends but the results arevery similar without these controls.

9

From: Blake, Nosko, and Tadelis, 2014

37 / 765

General Problems with ObservationalStudies

• Prediction accuracy is uninformative• No randomization, the “reasoned basis for inference”• Selection problems abound• Data/Model Mining—e.g., not like vision• Found data, usually retrospective

Note: it is far easier to estimate the effect of a cause than thecause of an effect. Why?

38 / 765

General Problems with ObservationalStudies

• Prediction accuracy is uninformative• No randomization, the “reasoned basis for inference”• Selection problems abound• Data/Model Mining—e.g., not like vision• Found data, usually retrospective

Note: it is far easier to estimate the effect of a cause than thecause of an effect. Why?

39 / 765

General Problems with ObservationalStudies

• Prediction accuracy is uninformative• No randomization, the “reasoned basis for inference”• Selection problems abound• Data/Model Mining—e.g., not like vision• Found data, usually retrospective

Note: it is far easier to estimate the effect of a cause than thecause of an effect. Why?

40 / 765

General Problems with ObservationalStudies

• Prediction accuracy is uninformative• No randomization, the “reasoned basis for inference”• Selection problems abound• Data/Model Mining—e.g., not like vision• Found data, usually retrospective

Note: it is far easier to estimate the effect of a cause than thecause of an effect. Why?

41 / 765

General Problems with ObservationalStudies

• Prediction accuracy is uninformative• No randomization, the “reasoned basis for inference”• Selection problems abound• Data/Model Mining—e.g., not like vision• Found data, usually retrospective

Note: it is far easier to estimate the effect of a cause than thecause of an effect. Why?

42 / 765

Difficult Example: Does InformationMatter?

• We want to estimate the effect on voting behavior of payingattention to an election campaign.

• Survey research is strikingly uniform regarding theignorance of the public (e.g., Berelson, Lazarsfeld, andMcPhee, 1954; Campbell et al., 1960; J. R. Zaller, 1992).

• Although the fact of public ignorance has not beenforcefully challenged, the meaning of this observation hasbeen (Sniderman, 1993).

• Can voters use information such as polls, interest groupendorsements and partisan labels to vote like their betterinformed compatriots (e.g., Lupia, 2004; McKelvey andOrdeshook, 1985a; McKelvey and Ordeshook, 1985b;McKelvey and Ordeshook, 1986)?

43 / 765

Difficult Example: Does InformationMatter?

• We want to estimate the effect on voting behavior of payingattention to an election campaign.

• Survey research is strikingly uniform regarding theignorance of the public (e.g., Berelson, Lazarsfeld, andMcPhee, 1954; Campbell et al., 1960; J. R. Zaller, 1992).

• Although the fact of public ignorance has not beenforcefully challenged, the meaning of this observation hasbeen (Sniderman, 1993).

• Can voters use information such as polls, interest groupendorsements and partisan labels to vote like their betterinformed compatriots (e.g., Lupia, 2004; McKelvey andOrdeshook, 1985a; McKelvey and Ordeshook, 1985b;McKelvey and Ordeshook, 1986)?

44 / 765

Difficult Example: Does InformationMatter?

• We want to estimate the effect on voting behavior of payingattention to an election campaign.

• Survey research is strikingly uniform regarding theignorance of the public (e.g., Berelson, Lazarsfeld, andMcPhee, 1954; Campbell et al., 1960; J. R. Zaller, 1992).

• Although the fact of public ignorance has not beenforcefully challenged, the meaning of this observation hasbeen (Sniderman, 1993).

• Can voters use information such as polls, interest groupendorsements and partisan labels to vote like their betterinformed compatriots (e.g., Lupia, 2004; McKelvey andOrdeshook, 1985a; McKelvey and Ordeshook, 1985b;McKelvey and Ordeshook, 1986)?

45 / 765

Difficult Example: Does InformationMatter?

• We want to estimate the effect on voting behavior of payingattention to an election campaign.

• Survey research is strikingly uniform regarding theignorance of the public (e.g., Berelson, Lazarsfeld, andMcPhee, 1954; Campbell et al., 1960; J. R. Zaller, 1992).

• Although the fact of public ignorance has not beenforcefully challenged, the meaning of this observation hasbeen (Sniderman, 1993).

• Can voters use information such as polls, interest groupendorsements and partisan labels to vote like their betterinformed compatriots (e.g., Lupia, 2004; McKelvey andOrdeshook, 1985a; McKelvey and Ordeshook, 1985b;McKelvey and Ordeshook, 1986)?

46 / 765

Fundamental Problem of CausalInference

• Fundamental problem: not observing all of the potentialoutcomes or counterfactuals

• Let Yi1 denote i ’s vote intention when voter i learns duringthe campaign (i.e., is in the treatment regime).

• Let Yi0 denote i ’s vote intention when voter i does not learnduring the campaign (i.e., is in the control regime).

• Let Ti be a treatment indicator: 1 when i is in the treatmentregime and 0 otherwise.

• The observed outcome for observation i isYi = TiYi1 + (1− Ti)Yi0.

• The treatment effect for i isτi = Yi1 − Yi0.

47 / 765

Fundamental Problem of CausalInference

• Fundamental problem: not observing all of the potentialoutcomes or counterfactuals

• Let Yi1 denote i ’s vote intention when voter i learns duringthe campaign (i.e., is in the treatment regime).

• Let Yi0 denote i ’s vote intention when voter i does not learnduring the campaign (i.e., is in the control regime).

• Let Ti be a treatment indicator: 1 when i is in the treatmentregime and 0 otherwise.

• The observed outcome for observation i isYi = TiYi1 + (1− Ti)Yi0.

• The treatment effect for i isτi = Yi1 − Yi0.

48 / 765

Fundamental Problem of CausalInference

• Fundamental problem: not observing all of the potentialoutcomes or counterfactuals

• Let Yi1 denote i ’s vote intention when voter i learns duringthe campaign (i.e., is in the treatment regime).

• Let Yi0 denote i ’s vote intention when voter i does not learnduring the campaign (i.e., is in the control regime).

• Let Ti be a treatment indicator: 1 when i is in the treatmentregime and 0 otherwise.

• The observed outcome for observation i isYi = TiYi1 + (1− Ti)Yi0.

• The treatment effect for i isτi = Yi1 − Yi0.

49 / 765

Fundamental Problem of CausalInference

• Fundamental problem: not observing all of the potentialoutcomes or counterfactuals

• Let Yi1 denote i ’s vote intention when voter i learns duringthe campaign (i.e., is in the treatment regime).

• Let Yi0 denote i ’s vote intention when voter i does not learnduring the campaign (i.e., is in the control regime).

• Let Ti be a treatment indicator: 1 when i is in the treatmentregime and 0 otherwise.

• The observed outcome for observation i isYi = TiYi1 + (1− Ti)Yi0.

• The treatment effect for i isτi = Yi1 − Yi0.

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Fundamental Problem of CausalInference

• Fundamental problem: not observing all of the potentialoutcomes or counterfactuals

• Let Yi1 denote i ’s vote intention when voter i learns duringthe campaign (i.e., is in the treatment regime).

• Let Yi0 denote i ’s vote intention when voter i does not learnduring the campaign (i.e., is in the control regime).

• Let Ti be a treatment indicator: 1 when i is in the treatmentregime and 0 otherwise.

• The observed outcome for observation i isYi = TiYi1 + (1− Ti)Yi0.

• The treatment effect for i isτi = Yi1 − Yi0.

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Fundamental Problem of CausalInference

• Fundamental problem: not observing all of the potentialoutcomes or counterfactuals

• Let Yi1 denote i ’s vote intention when voter i learns duringthe campaign (i.e., is in the treatment regime).

• Let Yi0 denote i ’s vote intention when voter i does not learnduring the campaign (i.e., is in the control regime).

• Let Ti be a treatment indicator: 1 when i is in the treatmentregime and 0 otherwise.

• The observed outcome for observation i isYi = TiYi1 + (1− Ti)Yi0.

• The treatment effect for i isτi = Yi1 − Yi0.

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Probability Model I: RepeatedSampling

• Assume that Yi(t) is a simple random sample from aninfinite population, were Y is the outcome of unit i undertreatment condition t .

• Assume the treat T is randomly assigned following either:• Bernoulli assignment to each i• Complete randomization

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Probability Model II: Fixed Population

• Assume that Yi(t) is a fixed population

• Assume the treat T is randomly assigned following either:• Bernoulli assignment to each i• Complete randomization

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Assignment Mechanism, ClassicalExperiment

• individualistic• probabilistic• unconfounded: given covariates, not be dependent on any

potential outcomes

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Notation

Let:• N denote the number of units [in the sample], indexed by i• number of control units: Nc =

∑Ni=1(1− Ti); treated units:

Nt =∑N

i=1(Ti), Nc + Nt = N• Xi is a vector of K covariates, where X is a N × K matrix• Y (0) and Y (1) denote N-component vectors

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Notation

• Let T be the N-dimensional vector with elements Ti ∈ 0,1with positive probability

• Let all possible values be T = 0,1N , with cardinality 2N

• 0,1N denotes the set of all N vectors with all elementsequal to 0 or 1

• Let the subset of values for T with positive probability bedenoted by T+

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Assignment: Bernoulli Trials

• Pr(T |X ,Y (0),Y (1)) = 0.5N ,

where T+ = 0,1N = T

• Pr(T |X ,Y (0),Y (1)) = qNt · (1− q)Nc

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Notation Note

• Pr(T |X ,Y (0),Y (1)) is not the probability of a particularunit receiving the treatment

• it reflects a measure across the full population of N units ofa particular assignment vector occurring

• The unit-level assignment for unit i is:

pi(X ,Y (0),Y (1)) =∑

T:Ti =1

Pr(T |X ,Y (0),Y (1))

were we sum the probabilities across all possibleassignment vectors T for which Ti = 1

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Assignment: Complete Randomization

• An assignment mechanism that satisfies:

T+ =

T ∈ T

∣∣∣∣∣N∑

i=1

Ti = Nt

,

for some preset Nt ∈ 1,2, . . . ,N − 1

• Number of assignment vectors in this design:(

NNt

),

q =Nt

N∀ i

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Assignment: Complete Randomization

Pr (T |X ,Y (0)),Y (1)) =(

Nc! · N t !

N!

)·(

Nt

N

)Nt

·(

Nc

N

)Nc

if∑N

i=1 Ti = Nt

0 otherwise

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Experimental Data

• Under classic randomization, the inference problem isstraightforward because: T ⊥⊥ (Y (1),Y (0))

• Observations in the treatment and control groups are notexactly alike, but they are comparable—i.e., they areexchangeable

• With exchangeability plus noninterference between units,for j = 0,1 we have:

E(Y (j)|T = 1) = E(Y (j)|T = 0)

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Experimental Data

• Under classic randomization, the inference problem isstraightforward because: T ⊥⊥ (Y (1),Y (0))

• Observations in the treatment and control groups are notexactly alike, but they are comparable—i.e., they areexchangeable

• With exchangeability plus noninterference between units,for j = 0,1 we have:

E(Y (j)|T = 1) = E(Y (j)|T = 0)

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Experimental Data

• Under classic randomization, the inference problem isstraightforward because: T ⊥⊥ (Y (1),Y (0))

• Observations in the treatment and control groups are notexactly alike, but they are comparable—i.e., they areexchangeable

• With exchangeability plus noninterference between units,for j = 0,1 we have:

E(Y (j)|T = 1) = E(Y (j)|T = 0)

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Average Treatment Effect

• The Average Treatment Effect (ATE) can be estimatedsimply:

τ = Mean Outcome for the treated−Mean Outcome for the control

• In notation:τ ≡ Y1 − Y0

τ = E(Y (1)− Y (0))

= E(Y |T = 1)− E(Y |T = 0)

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Average Treatment Effect

• The Average Treatment Effect (ATE) can be estimatedsimply:

τ = Mean Outcome for the treated−Mean Outcome for the control

• In notation:τ ≡ Y1 − Y0

τ = E(Y (1)− Y (0))

= E(Y |T = 1)− E(Y |T = 0)

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Observational Data

• With observational data, the treatment and control groupsare not drawn from the same population

• Progress can be made if we assume that the two groupsare comparable once we condition on observablecovariates denoted by X

• This is the conditional independence assumption:Y (1),Y (0) ⊥⊥ T |X ,

the reasonableness of this assumption depends on thesection process

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Average Treatment Effect for theTreated

• With observational data, the treatment and control groupsare not drawn from the same population.

• Thus, we often want to estimate the average treatmenteffect for the treated (ATT):

τ |(T = 1) = E(Y (1)|T = 1)− E(Y (0)|T = 1)

• Progress can be made if we assume that the selectionprocess is the result of only observable covariates denotedby X

• We could alternative estimate the Average TreatmentEffect for the Controls (ATC).

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Average Treatment Effect for theTreated

• With observational data, the treatment and control groupsare not drawn from the same population.

• Thus, we often want to estimate the average treatmenteffect for the treated (ATT):

τ |(T = 1) = E(Y (1)|T = 1)− E(Y (0)|T = 1)

• Progress can be made if we assume that the selectionprocess is the result of only observable covariates denotedby X

• We could alternative estimate the Average TreatmentEffect for the Controls (ATC).

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Average Treatment Effect for theTreated

• With observational data, the treatment and control groupsare not drawn from the same population.

• Thus, we often want to estimate the average treatmenteffect for the treated (ATT):

τ |(T = 1) = E(Y (1)|T = 1)− E(Y (0)|T = 1)

• Progress can be made if we assume that the selectionprocess is the result of only observable covariates denotedby X

• We could alternative estimate the Average TreatmentEffect for the Controls (ATC).

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ATE=ATT

• Under random assignment

ATT = E[Y (1)−Y (0)|T = 1] = E [Y (1)−Y (0)] = ATE

• Note:E[Y |T = 1] = E[Y (0) + T (Y (1)− Y (0))|T = 1]

= E[Y (1)|T = 1],by ⊥⊥= E[Y (1)]

• Same holds for E[Y |T = 0]

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ATE=ATT

• Under random assignment

ATT = E[Y (1)−Y (0)|T = 1] = E [Y (1)−Y (0)] = ATE

• Note:E[Y |T = 1] = E[Y (0) + T (Y (1)− Y (0))|T = 1]

= E[Y (1)|T = 1],by ⊥⊥= E[Y (1)]

• Same holds for E[Y |T = 0]

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ATE=ATT

• Under random assignment

ATT = E[Y (1)−Y (0)|T = 1] = E [Y (1)−Y (0)] = ATE

• Note:E[Y |T = 1] = E[Y (0) + T (Y (1)− Y (0))|T = 1]

= E[Y (1)|T = 1],by ⊥⊥= E[Y (1)]

• Same holds for E[Y |T = 0]

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ATE=ATT

• SinceE[Y (1)− Y (0)|T = 1] = E[Y (1)− Y (0)]

= E[Y (1)]− E[Y (0)]

= E[Y |T = 1]− E[Y |T = 0]

• Then,

ATE = ATT = E[Y |T = 1]− E[Y |T = 0]

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Review and Details

• More details on ATE, ATT, and potential outcomes: [LINK]

• Some review of probability: [LINK]

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Selection on Observables

• Then, we obtain the usual exchangeability results:E(Y (j)|X ,T = 1) = E(Y (j)|X ,T = 0)

• ATT can then be estimated byτ |(T = 1) = EE(Y |X ,T = 1)−

E(Y |X ,T = 0) | T = 1

where the outer expectation is taken over the distribution ofX |(T = 1)

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Selection on Observables

• Then, we obtain the usual exchangeability results:E(Y (j)|X ,T = 1) = E(Y (j)|X ,T = 0)

• ATT can then be estimated byτ |(T = 1) = EE(Y |X ,T = 1)−

E(Y |X ,T = 0) | T = 1

where the outer expectation is taken over the distribution ofX |(T = 1)

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Observational Data: Matching

0.0 0.2 0.4 0.6 0.8

010

2030

Before Matching

X VARIABLE

Den

sity

DID NOT LEARN

DID LEARN

0.0 0.2 0.4 0.6 0.8

010

2030

After Matching

X VARIABLE

Den

sity

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Estimands• Sample Average Treatment Effect (SATE):

τS =1N

N∑i=1

[Yi(1)− Yi(0)]

• Sample Average Treatment Effect on the Treated (SATT):

τSt =

1Nt

∑i:Ti =1

[Yi(1)− Yi(0)]

• Population Average Treatment Effect (PATE):

τP = E [Y (1)− Y (0)]

• Population Average Treatment Effect on the Treated(PATT):

τPt = E [Y (1)− Y (0)|T = 1]

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CATE

Condiational ATE (CATE): conditional on the sampledistribution of sample covariates• CATE:

τ(X ) =1N

N∑i=1

E [Yi(1)− Yi(0)|Xi ]

• Condiation on the treated (CATT):

τ(X )t =1Nt

∑i:Ti =1

E [Yi(1)− Yi(0)|Xi ]

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Other Estimands

Condiational on whatever:1 Potential outcomes2 Potential response3 Other latent attributes–e.g., probability of selection4 (2) and (3) are probably related in practice

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SATE, unbiased?

• Assume completely randomized experiment. FollowNeyman (1923/1990)

• τ = Yt − Yc

• The estimator τ is unbiased for τ• Expectation taken only over the randomization distribution

• Note: E [Ti |Y (0),Y (1)] =Nt

N

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SATE, unbiased?, proof

• Rerwrite τ :

τ =1N

N∑i=1

(Ti · Yi(1)

Nt/N− (1− Ti) · Yi(0)

Nc/N

)• E [ τ |Y (0),Y (1)] =

1N∑N

i=1

(E[Ti ] · Yi(1)

Nt/N− E[1− Ti ] · Yi(0)

Nc/N

)=

1N∑N

i=1 (Yi(1)− Yi(0))

= τ

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Sample Variance

• V(Yt − Yc

)=

S2c

Nc+

S2t

Nt− S2

tcN

• S2c and S2

t are the sample variances of Y (0) and Y (1):

S2c =

1N − 1

N∑i=1

(Yi(0)− Y (0)

)2

S2t =

1N − 1

N∑i=1

(Yi(1)− Y (1)

)2

S2tc =

1N − 1

∑Ni=1[Yi(1)− Yi(0)− (Y (1)− Y (0))

]2=

1N − 1

∑Ni=1 (Yi(1)− Yi(0)− τ)2

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Alternative From

• S2tc = S2

c + S2t − 2ρtc · Sc · St

where

ρtc =1

(N − 1) · Sc · St

N∑i=1

(Yi(1)− Y (1)

)·(Yi(0)− Y (0)

)• V

(Yt − Yc

)=

Nt

N · Nc· S2

c +Nc

N · Nt· S2

t +2Nρtc · Sc · St

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Sample Variance

• When is V(Yt − Yc

)small [large]?

• An alternative: under the assumption of a constanttreatment effect, S2

c = S2t , therefore:

Vconst(Yt − Yc

)=

s2

N,

where s2c is the estimated version of S2

c

• Most used is the conservative Neyman variance estimator:

VN(Yt − Yc

)=

s2c

Nc+

s2t

Nt

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Correlation version

• Vρ =Nt

N · Nc· s2

c +Nc

N · Nt· s2

t +2Nρtc · sc · st

• By Cauchy-Schwarz inequality, |ρtc | ≤ 1, therefore

Vρ=1 = s2c ·

Nt

N · Nc+ s2

1 ·Nc

N · Nt+ sc · st ×

2N

=s2

cNc

+s2

tNt− (st − sc)2

N• If s2

c 6= s2t , then Vρ=1 < VN

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Improving the Bound

• In recent work, this bound been improved (the bound issharp, but the estimator is still conservative): Aronow,Green, and D. K. Lee (2014)

• They use Hoeffding’s inequality instead• The new bound is sharp in the sense that it is the smallest

interval containing all values of the variance that arecompatible with the observable information

• For nice results linking Neyman with the commonsandwich estimator, see: Samii and Aronow (2012)

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Infinite Sample Case

• Let E represent the expectation over the sample and thetreatment assignment; ESP over the sampling; and ET overthe treatment assignment

• τP = ESP [Yi(1)− Yi(0)](note the i)

ESP [τS] = ESP[Y (1)− Y (0)

]=

1N

N∑i=1

ESP [Yi(1)− Yi(0)]

= τP

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Infinite Sample Case

• Let σ2 = ESP [S2]. Note:

σ2tc = VSP [Yi(1)− Yi(0)] = ESP

[(Yi(1)− Yi(0)− τP)2

]• Definition of the variance of the unit-level treatment in the

population implies:

VSP(τS) = VSP[Y (1)− Y (0)

]=σ2

tcN

(1)

• What is V(τ), where τ = Yt − Yc?

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Infinite Sample Case

V(τ) = E[(

Yt − Yc − E[Yt − Yc

])2]

= E[(

Yt − Yc − ESP[Y (1)− Y (0)

])2],

With some algebra, and noting that

ET[Yt − Yc −

(Y (1)− Y (0)

)]= 0

We obtain

V(τ) = E[(

Yt − Yc − Y (1)− Y (0))2]

+ ESP

[(Y (1)− Y (0)− ESP [Y (1)− Y (0)]

)2]

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Infinite Sample Case

V(τ) = E[(

Yt − Yc − Y (1)− Y (0))2]

+ ESP

[(Y (1)− Y (0)− ESP [Y (1)− Y (0)]

)2]

Recall that

ET

[(Yt − Yc − Y (1)− Y (0)

)2]

=S2

cNc

+S2

tNt− S2

tcN

And by simple results from sampling,

ESP

[S2

cNc

+S2

tNt− S2

tcN

]=σ2

cNc

+σ2

tNt− σ2

tcN

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Infinite Sample Case

Also recall that

ESP

[(Y (1)− Y (0)− ESP [Y (1)− Y (0)]

)2]

=σ2

tcN

Therefore,

V(τ) = E[(

Yt − Yc − Y (1)− Y (0))2]

+ ESP

[(Y (1)− Y (0)− ESP [Y (1)− Y (0)]

)2]

=σ2

cNc

+σ2

tNt

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Comments

• ESP [VN ] = VSP , under the SP model• This does not hold for the other finite-sample variance

estimators we have considered, although they are sharperfor the finite-sample model

• VN is what is almost university used

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0

100

200

300

400

2011-01 2011-07 2012-01 2012-07 2013-01date

num

ber o

f pro

xim

al c

rimes

per

wee

k

CategoryASSAULT

DISORDERLY CONDUCT

DRUG/NARCOTIC

DRUNKENNESS

LARCENY/THEFT

SUSPICIOUS OCC

VANDALISM

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A Comment on Design

• What would a hypothesis test look like here? What would itmean?

• What are the assumptions to make this a causalstatement?

• Compare with Semmelweis and puerperal fever: see[Freedman, 2010, Chapter 20]

• Compare with the usual regression methods, such as thisreport: [LINK]

• Credit to: [Eytan Bakshy]

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Kepler in a Stats Department

“I sometimes have a nightmare about Kepler. Suppose a few ofus were transported back in time to the year 1600, and wereinvited by the Emperor Rudolph II to set up an ImperialDepartment of Statistics in the court at Prague. Despairing ofthose circular orbits, Kepler enrolls in our department. We teachhim the general linear model, least squares, dummy variables,everything. He goes back to work, fits the best circular orbit forMars by least squares, puts in a dummy variable for theexceptional observation - and publishes. And that’s the end,right there in Prague at the beginning of the 17th century.”

Freedman, D.A. (1985). Statistics and the scientific method. In W.M. Mason & S.E. Fienberg (Eds.), Cohort analysisin social research: Beyond the identification problem (pp. 343-366). New York: Springer-Verlag.

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Making Inference with AssumptionsFitting the Question

• The Fisher/Neyman approach required precise control overinterventions

• Neyman relied on what because the dominant way to doinference

• Fisher, and before him Pierce, Charles Sanders Peirce,outlined the way that is probably closest to the data

• With the growth of computing, a renaissance inpermutation inference

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Lady Tasting Tea

• A canonical experiment which can be used to show thepower of randomization inference.

• The example is a special case of trying to figure out if twoproportions are different, which is a very common question.

• A lady (at Cambridge) claimed that by tasting a cup of teamade with milk she can determine whether the tea infusionor milk was added first (Fisher, 1935, pp. 11–25).

• The lady was B. Muriel Bristol-Roach, who was an algabiologist. [google scholar]

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On Making Tea

• The precise way to make an optimal cup of tea has longbeen a contentious issue in China, India and Britain. e.g.,George Orwell had 11 rules for a perfect cup of tea.

• On the 100th anniversary of Orwell’s birth, the RoyalSociety of Chemistry decided to review Orwell’s rules.

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100 Years Later: On Making Tea

• The Society sternly challenged Orwell’s advice that milk bepoured in after the tea. They noted that adding milk intohot water denatures the proteins—i.e., the proteins beginto unfold and clump together.

• The Society recommended that it is better to have thechilled milk at the bottom of the cup, so it can cool the teaas it is slowly poured in (BBC, 2003).

• In India, the milk is heated together with the water and thetea is infused into the milk and water mix.

• In the west, this is generally known as chai—the genericword for tea in Hindi which comes from the Chinesecharacter (pronounced chá in Mandarin) which is alsothe source word for tea in Punjabi, Russian, Swahili andmany other languages.

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Fisher’s Experiment

The Lady:• is given eight cups of tea. Four of the cups have milk put in

first and four have tea put in first.• is presented the cups in random order without

replacement.• is told of this design so she, even if she has no ability to

discern the different types of cups, will select four cups tobe milk first and four to be tea first.

The Fisher exact test follows from examining the mechanics ofthe randomization which is done in this experiment.

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Fisher’s Randomization Test I

There are 70 ways of choosing four cups out of eight:• If one chooses four cups in succession, one has 8, 7, 6, 5

cups to choose from. There are8!

4!= 8× 7× 6× 5 = 1680

ways of choosing four cups.• But this calculation has considered it a different selection if

we choose cups in merely a different order, whichobviously doesn’t matter for the experiment.

• Since four cups can be arranged in 4! = 4× 3× 2× 1 = 24

ways, the number of possible choices is1680

24= 70.

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Fisher’s Randomization Test I

There are 70 ways of choosing four cups out of eight:• If one chooses four cups in succession, one has 8, 7, 6, 5

cups to choose from. There are8!

4!= 8× 7× 6× 5 = 1680

ways of choosing four cups.• But this calculation has considered it a different selection if

we choose cups in merely a different order, whichobviously doesn’t matter for the experiment.

• Since four cups can be arranged in 4! = 4× 3× 2× 1 = 24

ways, the number of possible choices is1680

24= 70.

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Fisher’s Randomization Test I

There are 70 ways of choosing four cups out of eight:• If one chooses four cups in succession, one has 8, 7, 6, 5

cups to choose from. There are8!

4!= 8× 7× 6× 5 = 1680

ways of choosing four cups.• But this calculation has considered it a different selection if

we choose cups in merely a different order, whichobviously doesn’t matter for the experiment.

• Since four cups can be arranged in 4! = 4× 3× 2× 1 = 24

ways, the number of possible choices is1680

24= 70.

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Fisher’s Randomization Test I

There are 70 ways of choosing four cups out of eight:• If one chooses four cups in succession, one has 8, 7, 6, 5

cups to choose from. There are8!

4!= 8× 7× 6× 5 = 1680

ways of choosing four cups.• But this calculation has considered it a different selection if

we choose cups in merely a different order, whichobviously doesn’t matter for the experiment.

• Since four cups can be arranged in 4! = 4× 3× 2× 1 = 24

ways, the number of possible choices is1680

24= 70.

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Fisher’s Randomization Test II

The Lady has:• 1 way of making 4 correct and 0 incorrect choices• 16 ways of making 3 correct and 1 incorrect choices• 36 ways of making 2 correct and 2 incorrect choices• 16 ways of making 1 correct and 3 incorrect choices• 1 way of making 0 correct and 4 incorrect choices• These five different possible outcomes taken together add

up to 70 cases out of 70.

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Fisher’s Randomization Test II

The Lady has:• 1 way of making 4 correct and 0 incorrect choices• 16 ways of making 3 correct and 1 incorrect choices• 36 ways of making 2 correct and 2 incorrect choices• 16 ways of making 1 correct and 3 incorrect choices• 1 way of making 0 correct and 4 incorrect choices• These five different possible outcomes taken together add

up to 70 cases out of 70.

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Fisher’s Randomization Test II

The Lady has:• 1 way of making 4 correct and 0 incorrect choices• 16 ways of making 3 correct and 1 incorrect choices• 36 ways of making 2 correct and 2 incorrect choices• 16 ways of making 1 correct and 3 incorrect choices• 1 way of making 0 correct and 4 incorrect choices• These five different possible outcomes taken together add

up to 70 cases out of 70.

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Fisher’s Randomization Test II

The Lady has:• 1 way of making 4 correct and 0 incorrect choices• 16 ways of making 3 correct and 1 incorrect choices• 36 ways of making 2 correct and 2 incorrect choices• 16 ways of making 1 correct and 3 incorrect choices• 1 way of making 0 correct and 4 incorrect choices• These five different possible outcomes taken together add

up to 70 cases out of 70.

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Fisher’s Randomization Test II

The Lady has:• 1 way of making 4 correct and 0 incorrect choices• 16 ways of making 3 correct and 1 incorrect choices• 36 ways of making 2 correct and 2 incorrect choices• 16 ways of making 1 correct and 3 incorrect choices• 1 way of making 0 correct and 4 incorrect choices• These five different possible outcomes taken together add

up to 70 cases out of 70.

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Fisher’s Randomization Test II

The Lady has:• 1 way of making 4 correct and 0 incorrect choices• 16 ways of making 3 correct and 1 incorrect choices• 36 ways of making 2 correct and 2 incorrect choices• 16 ways of making 1 correct and 3 incorrect choices• 1 way of making 0 correct and 4 incorrect choices• These five different possible outcomes taken together add

up to 70 cases out of 70.

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Fisher’s Randomization TestProbability I

• The Lady has p =1

70of correctly selecting the cups by

chance.• The probability of her randomly selecting 3 correct and 1

incorrect cup is1670

• But this proportion does not itself test the null hypothesis ofno ability.

• For that we want to know the chances of observing theoutcome observed (3 correct and 1 wrong), plus thechances of observing better results (4 correct cups).

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Fisher’s Randomization TestProbability I

• The Lady has p =1

70of correctly selecting the cups by

chance.• The probability of her randomly selecting 3 correct and 1

incorrect cup is1670

• But this proportion does not itself test the null hypothesis ofno ability.

• For that we want to know the chances of observing theoutcome observed (3 correct and 1 wrong), plus thechances of observing better results (4 correct cups).

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Fisher’s Randomization TestProbability I

• The Lady has p =1

70of correctly selecting the cups by

chance.• The probability of her randomly selecting 3 correct and 1

incorrect cup is1670

• But this proportion does not itself test the null hypothesis ofno ability.

• For that we want to know the chances of observing theoutcome observed (3 correct and 1 wrong), plus thechances of observing better results (4 correct cups).

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Fisher’s Randomization TestProbability I

• The Lady has p =1

70of correctly selecting the cups by

chance.• The probability of her randomly selecting 3 correct and 1

incorrect cup is1670

• But this proportion does not itself test the null hypothesis ofno ability.

• For that we want to know the chances of observing theoutcome observed (3 correct and 1 wrong), plus thechances of observing better results (4 correct cups).

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Fisher’s Randomization TestProbability II

• Otherwise, we could be rejecting the null hypothesis of noability just because the outcome observed was itself rarealthough better results could have been frequentlyobserved by chance.

• In the case of the lady correctly identifying three cupscorrectly and one incorrectly, the p-value of the Fisher

randomization tests is1670

+1

70=

1770∼= 0.24.

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Fisher’s Randomization TestProbability II

• Otherwise, we could be rejecting the null hypothesis of noability just because the outcome observed was itself rarealthough better results could have been frequentlyobserved by chance.

• In the case of the lady correctly identifying three cupscorrectly and one incorrectly, the p-value of the Fisher

randomization tests is1670

+1

70=

1770∼= 0.24.

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Fisher Exact Test

• This test is test is distribution (and model) free.• If the conditional randomization model is not correct (we

will see an example in a bit), the level will still be stillcorrect relative to the unconditional model.

• A test has level α if whenever the null is true, the chance ofrejection is no greater than α.

• But the size of the test is often less than its level if therandomization is not correct. The size of a test is thechance of rejection when the null is true.

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Fisher Exact Test

• This test is test is distribution (and model) free.• If the conditional randomization model is not correct (we

will see an example in a bit), the level will still be stillcorrect relative to the unconditional model.

• A test has level α if whenever the null is true, the chance ofrejection is no greater than α.

• But the size of the test is often less than its level if therandomization is not correct. The size of a test is thechance of rejection when the null is true.

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Fisher Exact Test

• This test is test is distribution (and model) free.• If the conditional randomization model is not correct (we

will see an example in a bit), the level will still be stillcorrect relative to the unconditional model.

• A test has level α if whenever the null is true, the chance ofrejection is no greater than α.

• But the size of the test is often less than its level if therandomization is not correct. The size of a test is thechance of rejection when the null is true.

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Binomial Randomization

• This experimental design is not very sensitive. E.g., what if

three cups are correctly identified? p =1770∼= 0.24.

• A binomial randomized experimental design would bemore sensitive.

• Follow binomial sampling: with the number ofobservations, n = 8, and the probability of having milk firstbe p = 0.5 for each cup.

• The observations are independent.

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Binomial Randomization

• This experimental design is not very sensitive. E.g., what if

three cups are correctly identified? p =1770∼= 0.24.

• A binomial randomized experimental design would bemore sensitive.

• Follow binomial sampling: with the number ofobservations, n = 8, and the probability of having milk firstbe p = 0.5 for each cup.

• The observations are independent.

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Binomial Randomization

• This experimental design is not very sensitive. E.g., what if

three cups are correctly identified? p =1770∼= 0.24.

• A binomial randomized experimental design would bemore sensitive.

• Follow binomial sampling: with the number ofobservations, n = 8, and the probability of having milk firstbe p = 0.5 for each cup.

• The observations are independent.

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Binomial Randomization

• This experimental design is not very sensitive. E.g., what if

three cups are correctly identified? p =1770∼= 0.24.

• A binomial randomized experimental design would bemore sensitive.

• Follow binomial sampling: with the number ofobservations, n = 8, and the probability of having milk firstbe p = 0.5 for each cup.

• The observations are independent.

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Binomial Randomization II

• The chance of classifying correctly eight cups binomialrandomized is 1 in 28 = 256, which is significantly lowerthan 1 in 70.

• There are 8 ways in 256 (p=0.032) of incorrectly classifyingonly one cup.

• Unlike in the canonical experiment, case of having bothmargins fixed, in the case of the binomial design, theexperiment is sensitive enough to reject the null hypothesisif the lady makes a single mistake

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Binomial Randomization II

• The chance of classifying correctly eight cups binomialrandomized is 1 in 28 = 256, which is significantly lowerthan 1 in 70.

• There are 8 ways in 256 (p=0.032) of incorrectly classifyingonly one cup.

• Unlike in the canonical experiment, case of having bothmargins fixed, in the case of the binomial design, theexperiment is sensitive enough to reject the null hypothesisif the lady makes a single mistake

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Binomial Randomization II

• The chance of classifying correctly eight cups binomialrandomized is 1 in 28 = 256, which is significantly lowerthan 1 in 70.

• There are 8 ways in 256 (p=0.032) of incorrectly classifyingonly one cup.

• Unlike in the canonical experiment, case of having bothmargins fixed, in the case of the binomial design, theexperiment is sensitive enough to reject the null hypothesisif the lady makes a single mistake

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Comments

• The extra sensitively comes purely from the design, andnot from more data.

• The binomial design is more sensitive, but there may besome reasons to prefer the design with fixed margins. E.g.,there is a chance with the binomial design that all cupswould be treated alike.

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Comments

• The extra sensitively comes purely from the design, andnot from more data.

• The binomial design is more sensitive, but there may besome reasons to prefer the design with fixed margins. E.g.,there is a chance with the binomial design that all cupswould be treated alike.

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Sharp Null

• Fisherian inference proceeds using a sharp null—i.e., all ofthe potential outcomes under the null are specified.

• Is the above true for the usuall null hypothesis that τ = 0?Under the sharp null the equivilant is: τi = 0 ∀ i .

• But note that we can pick any sharp null we wish—e.g.,τi = −1 ∀ i < 10 and τi = 20 ∀ i > 10.

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Sharp Null

• Fisherian inference proceeds using a sharp null—i.e., all ofthe potential outcomes under the null are specified.

• Is the above true for the usuall null hypothesis that τ = 0?Under the sharp null the equivilant is: τi = 0 ∀ i .

• But note that we can pick any sharp null we wish—e.g.,τi = −1 ∀ i < 10 and τi = 20 ∀ i > 10.

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Sharp Null and Potential Outcomes

• The null hypothesis is: Yi1 = Yi0

• The observed data is: Yi = TiYi1 + (1− Ti)Yi0

• Therefore, under the null:Yi = TiYi1 + (1− Ti)Yi0 = Yi0

• But other models of the PO under the null arepossible—e.g.,

Yi = Yi0 + Tiτ0,

But his could be general: Y = f (Yi0, τi), where τ is someknown vector that may vary with i .

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General Procedure

Test statistic: t(T , r), a quantity computed from treatmentassignment T ∈ Ω and the response r . For example, the mean.For a given t(T , r), we compute a significance level:

1 sharp null allows us to fix r , say at the observed value.2 treatment assignment T follows a known randomization

mechanism which we can simulate or exhaustively list.3 given (i), (ii), the observated value of the test statistic is

known for all realizations of the random treatmentassignment.

4 we seek the probability of a value of the test statistic aslarge or larger than observed

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General Procedure

Test statistic: t(T , r), a quantity computed from treatmentassignment T ∈ Ω and the response r . For example, the mean.For a given t(T , r), we compute a significance level:

1 sharp null allows us to fix r , say at the observed value.2 treatment assignment T follows a known randomization

mechanism which we can simulate or exhaustively list.3 given (i), (ii), the observated value of the test statistic is

known for all realizations of the random treatmentassignment.

4 we seek the probability of a value of the test statistic aslarge or larger than observed

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General Procedure

Test statistic: t(T , r), a quantity computed from treatmentassignment T ∈ Ω and the response r . For example, the mean.For a given t(T , r), we compute a significance level:

1 sharp null allows us to fix r , say at the observed value.2 treatment assignment T follows a known randomization

mechanism which we can simulate or exhaustively list.3 given (i), (ii), the observated value of the test statistic is

known for all realizations of the random treatmentassignment.

4 we seek the probability of a value of the test statistic aslarge or larger than observed

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General Procedure

Test statistic: t(T , r), a quantity computed from treatmentassignment T ∈ Ω and the response r . For example, the mean.For a given t(T , r), we compute a significance level:

1 sharp null allows us to fix r , say at the observed value.2 treatment assignment T follows a known randomization

mechanism which we can simulate or exhaustively list.3 given (i), (ii), the observated value of the test statistic is

known for all realizations of the random treatmentassignment.

4 we seek the probability of a value of the test statistic aslarge or larger than observed

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Signifance Level

Significance level is simply the sum of the randomizationprobabilities that lead to values of t(T , r) greater than or equalto the observed value T . So,

P[t(T , r)] ≥ T =∑t∈Ω

[t(t , r) ≥ T

]× P(T = t),

where P(T = t) is determined by the known randomizationmechanism.

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Further Reading

See Rosenbaum (2002). Observational Studies, chp2 and

Rosenbaum, P. R. (2002). “Covariance adjustment inrandomized experiments and observational studies.” StatisticalScience 17 286–327 (with discussion).

Simple introduction: Michael D. Ernst (2004). “PermutationMethods: A Basis for Exact Inference.” Statistical Science 19:4676–685.

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SUTVA: Stable Unit Treatment ValueAssumption

• We require that “the [potential outcome] observation onone unit should be unaffected by the particular assignmentof treatments to the other units” (D. R. Cox, 1958, §2.4).This is called the stable unit treatment value assumption(SUTVA) (Rubin, 1978).

• SUTVA implies that Yi1 and Yi0 (the potential outcomes forperson i) in no way depend on the treatment status of anyother person in the dataset.

• SUTVA is not just statistical independence between units!

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SUTVA: Stable Unit Treatment ValueAssumption

• We require that “the [potential outcome] observation onone unit should be unaffected by the particular assignmentof treatments to the other units” (D. R. Cox, 1958, §2.4).This is called the stable unit treatment value assumption(SUTVA) (Rubin, 1978).

• SUTVA implies that Yi1 and Yi0 (the potential outcomes forperson i) in no way depend on the treatment status of anyother person in the dataset.

• SUTVA is not just statistical independence between units!

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SUTVA: Stable Unit Treatment ValueAssumption

• We require that “the [potential outcome] observation onone unit should be unaffected by the particular assignmentof treatments to the other units” (D. R. Cox, 1958, §2.4).This is called the stable unit treatment value assumption(SUTVA) (Rubin, 1978).

• SUTVA implies that Yi1 and Yi0 (the potential outcomes forperson i) in no way depend on the treatment status of anyother person in the dataset.

• SUTVA is not just statistical independence between units!

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SUTVA: Stable Unit Treatment ValueAssumption

No-interference implies:

Y Liit = Y Lj

it ∀j 6= i

where Lj is the treatment assignment for unit i , and t ∈ 0,1denotes the potential outcomes under treatment and control.

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SUTVA: Stable Unit Treatment ValueAssumption

• Causal inference relies on a counterfactual of interest(Sekhon, 2004), and the one which is most obviouslyrelevant for political information is “how would Jane havevoted if she were better informed?”.

• There are other theoretically interesting counterfactualswhich, because of SUTVA, I do not know how toempirically answer such as “who would have won the lastelection if everyone were well informed?”.

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SUTVA: Stable Unit Treatment ValueAssumption

• Causal inference relies on a counterfactual of interest(Sekhon, 2004), and the one which is most obviouslyrelevant for political information is “how would Jane havevoted if she were better informed?”.

• There are other theoretically interesting counterfactualswhich, because of SUTVA, I do not know how toempirically answer such as “who would have won the lastelection if everyone were well informed?”.

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Another Example: Florida 2004 VotingTechnology

• In the aftermath of the 2004 Presidential election, manyresearchers argued that the optical voting machines thatare used in a majority of Florida counties caused JohnKerry to receive fewer votes than “Direct RecordingElectronic” (DRE) voting machines (Hout et al.).

• Hout et al. used a regression model to arrive at thisconclusion.

• Problem: The distributions of counties were are profoundlyimbalanced

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Another Example: Florida 2004 VotingTechnology

• In the aftermath of the 2004 Presidential election, manyresearchers argued that the optical voting machines thatare used in a majority of Florida counties caused JohnKerry to receive fewer votes than “Direct RecordingElectronic” (DRE) voting machines (Hout et al.).

• Hout et al. used a regression model to arrive at thisconclusion.

• Problem: The distributions of counties were are profoundlyimbalanced

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Another Example: Florida 2004 VotingTechnology

• In the aftermath of the 2004 Presidential election, manyresearchers argued that the optical voting machines thatare used in a majority of Florida counties caused JohnKerry to receive fewer votes than “Direct RecordingElectronic” (DRE) voting machines (Hout et al.).

• Hout et al. used a regression model to arrive at thisconclusion.

• Problem: The distributions of counties were are profoundlyimbalanced

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Dem Registration Pre/Post Matching

0.0 0.2 0.4 0.6 0.8 1.0

02

46

810

density(x = dtam$reg04p.dem[Tr], from = 0)

N = 52 Bandwidth = 0.06893

Den

sity

0.0 0.2 0.4 0.6 0.8 1.00

24

68

10

density(x = dtam$reg04p.dem[io], from = 0)

N = 8 Bandwidth = 0.02701

Den

sity

Densities of Democratic Registration proportions by county.Green lines are DREs and red lines optical counties.

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Florida 2004

Average Treatment Effect for the TreatedEstimate 0.00540SE 0.0211p-value 0.798

balance on all other covariates, example:optical dre pvalue (t-test)

pre : Dem Reg .551 .361 .000post: Dem Reg .369 .365 .556

See http://sekhon.berkeley.edu/papers/SekhonOpticalMatch.pdf for details

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Matching

• The nonparametric way to condition on X is to exactlymatch on the covariates.

• This approach fails in finite samples if the dimensionality ofX is large.

• If X consists of more than one continuous variable, exactmatching is inefficient: matching estimators with a fixednumber of matches do not reach the semi-parametricefficiency bound for average treatment effects (Abadie andG. Imbens, 2006).

• An alternative way to condition on X is to match on theprobability of being assigned to treatment—i.e., thepropensity score (P. R. Rosenbaum and Rubin, 1983). Thepropensity score is just one dimensional.

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Matching

• The nonparametric way to condition on X is to exactlymatch on the covariates.

• This approach fails in finite samples if the dimensionality ofX is large.

• If X consists of more than one continuous variable, exactmatching is inefficient: matching estimators with a fixednumber of matches do not reach the semi-parametricefficiency bound for average treatment effects (Abadie andG. Imbens, 2006).

• An alternative way to condition on X is to match on theprobability of being assigned to treatment—i.e., thepropensity score (P. R. Rosenbaum and Rubin, 1983). Thepropensity score is just one dimensional.

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Matching

• The nonparametric way to condition on X is to exactlymatch on the covariates.

• This approach fails in finite samples if the dimensionality ofX is large.

• If X consists of more than one continuous variable, exactmatching is inefficient: matching estimators with a fixednumber of matches do not reach the semi-parametricefficiency bound for average treatment effects (Abadie andG. Imbens, 2006).

• An alternative way to condition on X is to match on theprobability of being assigned to treatment—i.e., thepropensity score (P. R. Rosenbaum and Rubin, 1983). Thepropensity score is just one dimensional.

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Matching

• The nonparametric way to condition on X is to exactlymatch on the covariates.

• This approach fails in finite samples if the dimensionality ofX is large.

• If X consists of more than one continuous variable, exactmatching is inefficient: matching estimators with a fixednumber of matches do not reach the semi-parametricefficiency bound for average treatment effects (Abadie andG. Imbens, 2006).

• An alternative way to condition on X is to match on theprobability of being assigned to treatment—i.e., thepropensity score (P. R. Rosenbaum and Rubin, 1983). Thepropensity score is just one dimensional.

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Propensity Score (pscore)

• More formally the propensity score is:Pr(T = 1|X ) = E(T |X )

• It is a one-dimensional balancing score• It helps to reduces the difficulty of matching• If one balances the propensity score, one balances on the

confounders X• But if the pscore is not known, it must be estimated.• How do we know if we estimate the correct propensity

score?• It is a tautology—but we can observe some implications

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Neyman/Fisher versus OLS

• Compare the classical OLS assumptions with those from acanonical experiment described by Fisher (1935): “TheLady Tasting Tea.”

• Compare the classical OLS assumptions with those fromthe Neyman model

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Classical OLS Assumptions

A1. Yt =∑

k Xktβk + εt , t = 1,2,3, · · · ,n k = 1, · · · ,K , wheret indexes the observations and k the variables.

A2. All of the X variables are nonstochastic.A3. There is no deterministic linear relationship between any of

the X variables. More precisely, the k × k matrix∑

XtX ′t isnon-singular for every n > k .

A4. E[εt ] = 0 for every t , t = 1,2,3, · · · ,n. Since every X isassumed to be nonstochastic (A2), (A4) implies thatE[Xtεt ] = 0. (A4) always holds if there is an intercept and if(A1)–(A3) hold.

A5. The variance of the random error, ε is equal to a constant,σ2, for all values of every X (i.e., var [εt ] = σ2), and ε isnormally distributed. This assumption implies that theerrors are independent and identically distributed.

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Back to Basics

All of the assumptions which are referred to are listed on theassumptions slide.The existence of the least squares estimator is guaranteed byA1-A3. These assumptions guarantee that β exists and that it isunique.The unbiasedness of OLS is guaranteed by assumptions 1-4.So, E(β) = β.For hypothesis testing, we need all of the assumptions: A1-A5.These allow us to assume β ∼ N(β, σ2(X ′X )−1).For efficiency we require assumptions A1-A5.

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Correct Specification Assumption

The correct specification assumption implies that E(εt |Xt ) = 0.Why?Because we are modeling the conditional mean.

Yt = E(Yt |Xt ) + εt

Then

εt = Yt − E(Yt |Xt )

and

E(εt |Xt ) = E [Yt − E(Yt |Xt )|Xt ]

= E(Yt |Xt )− E [E(Yt |Xt )|Xt ]

= E(Yt |Xt )− E(Yt |Xt )

= 0

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Remarks

• The regression function E(Yt |Xt ) is used to predict Yt fromknowledge of Xt .

• The term εt is called the “regression disturbance.” The factE(εt |Xt ) = 0 implies that εt contains no systematicinformation of Xt in predicted Yt . In other words, allinformation of Xt that is useful to predict Yt has beensummarized by E(Yt |Xt ).

• The assumption that E(ε|X ) = 0 is crucial. If E(ε|X ) 6= 0, βis biased.

• Situations in which E(ε|X ) 6= 0 can arise easily. Forexample, Xt may contain errors of measurement.

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Recall Post-Treatment Bias

The lesson of the lagged differences example is to not ask toomuch from regression. See:http://sekhon.berkeley.edu/causalinf/R/difference1.R.

Many lessons follow from this. One of the most important is theissue of post-treatment bias. Post-treatment variables are thosewhich are a consequence of the treatment of interest. Hence,they will be correlated with the treatment unless their values arethe same under treatment and control.

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Recall Post-Treatment Bias

The lesson of the lagged differences example is to not ask toomuch from regression. See:http://sekhon.berkeley.edu/causalinf/R/difference1.R.

Many lessons follow from this. One of the most important is theissue of post-treatment bias. Post-treatment variables are thosewhich are a consequence of the treatment of interest. Hence,they will be correlated with the treatment unless their values arethe same under treatment and control.

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Some Intuition: Let’s derive OLS withtwo covars

• See my Government 1000 lecture notes from Harvard orany intro text book for more details

• Let’s restrict ourselves to two covariates to help with theintuition

• Our goal is to minimize∑n

i (Yi − Yi)2, where

Yi = α + β1X1i + β2X2i .• We can do this by calculating the partial derivaties with

respect to the three unknown parameters α, β1, β2,equating each to 0 and solving.

• To simplify: deviate the observed variables by their means.These mean deviated variables are denoted by yi , x1i andx2i .

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ESS =n∑i

(Yi − Yi)2 (2)

Therefore,

∂ESS∂β1

= β1

n∑i

x21i + β2

n∑i

x1ix2i −n∑i

x1iyi (3)

∂ESS∂β2

= β1

n∑i

x1ix2i + β2

n∑i

x22i −

n∑i

x2iyi (4)

These can be rewritten as:n∑i

x1iyi = β1

n∑i

x21i + β2

n∑i

x1ix2i (5)

n∑i

x2iyi = β1

n∑i

x1ix2i + β2

n∑i

x22i (6)

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Recall:n∑i

x1iyi = β1

n∑i

x21i + β2

n∑i

x1ix2i (7)

n∑i

x2iyi = β1

n∑i

x1ix2i + β2

n∑i

x22i (8)

To solve, we multiply Equation 7 by∑n

i x22i and Equation 8 by∑n

i x1ix2i and subtract the latter from the former.

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Thus,

β1 =(∑n

i x1iyi)(∑n

i x2x2i)− (

∑ni x2iyi)(

∑ni x1ix2i)

(∑n

i x21i)(∑n

i x22i)− (

∑ni x1ix2i)2

(9)

And

β2 =(∑n

i x2iyi)(∑n

i x2x1i)− (

∑ni x1iyi)(

∑ni x1ix2i)

(∑n

i x21i)(∑n

i x22i)− (

∑ni x1ix2i)2

(10)

If we do the same for α we find that:

α = Y − β1X1 − β2X2 (11)

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The equations for the estimates of β1 and β2 can be rewrittenas:

β1 =cov(X1i ,Yi) var(X2i) − cov(X2i ,Yi) cov(X1i ,X2i)

var(X1i) var(X2i) − [cov(X1i ,X2i)]2

And,

β2 =cov(X2i ,Yi) var(X1i) − cov(X1i ,Yi) cov(X1i ,X2i)

var(X1i) var(X2i) − [cov(X1i ,X2i)]2

To understand the post-treatment problem in the motivatingexample see:http://sekhon.berkeley.edu/causalinf/R/part1.R

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Post-Treatment Bias Revisited

• The discussion about post-treatment bias is related to whycertain causal questions are difficult if not impossible toanswer (Holland, 1986).

• For example, what is the causal effect of a education onvoting? How do we answer this using a cross-sectionalsurvey?

• A more charged examples: race and sex. What ispost-treatment in gender discrimination cases?

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Post-Treatment Bias Revisited

• The discussion about post-treatment bias is related to whycertain causal questions are difficult if not impossible toanswer (Holland, 1986).

• For example, what is the causal effect of a education onvoting? How do we answer this using a cross-sectionalsurvey?

• A more charged examples: race and sex. What ispost-treatment in gender discrimination cases?

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UnbiasednessSuppose (A1)-(A4) hold. Then E(β) = β—i.e., β is an unbiasedestimator of β. Proof:

β = (X ′X )−1X ′Y

= (n∑

t=1

XtX ′t )−1n∑

t=1

XtYt

By (A1)

= (n∑

t=1

XtX ′t )−1n∑

t=1

Xt (X ′t β + εt )

= (n∑

t=1

XtX ′t )−1(n∑

t=1

XtX ′t )β + (n∑

t=1

XtX ′t )−1n∑

t=1

Xtεt

= β + (n∑

t=1

XtX ′t )−1n∑

t=1

Xtεt

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Unbiasedness

β = β + (n∑

t=1

XtX ′t )−1n∑

t=1

Xtεt

Given (A2) and (A4)

E

[(

n∑t=1

XtX ′t )−1n∑

t=1

Xtεt

]= (

n∑t=1

XtX ′t )−1n∑

t=1

XtE [εt ] = 0

E(β) = β

Unbiasedness may be considered a necessary condition for agood estimator, but is certainly isn’t sufficient. Moreover, it isalso doubtful whether it is even necessary. For example, ingeneral, Maximum Likelihood Estimators are notunbiased—e.g., Logit, Probit, Tobit.

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Variance-Covariance Matrix

Suppose Assumptions (A1)-(A5) hold. Then, thevariance-covariance matrix of β is:

E[(β − β)(β − β)′

]= σ2(X ′X )−1 (12)

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Proof: Variance-Covariance Matrix

β = β +

(n∑

t=1

XtX ′t

)−1 n∑t=1

Xtεt

β − β = (X ′X )−1X ′ε

(β − β)(β − β)′ = (X ′X )−1X ′ε[(X ′X )−1X ′ε

]′= (X ′X )−1X ′ε

[ε′X (X ′X )−1

]= (X ′X )−1X ′(εε′)X (X ′X )−1

E(β − β)(β − β)′ = (X ′X )−1X ′E(εε′)X (X ′X )−1

= (X ′X )−1X ′σ2IX (X ′X )−1

move σ2 and drop I = σ2(X ′X )−1X ′X (X ′X )−1

= σ2(X ′X )−1

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The sigma-squared Matrix

Note that σ2I is an n × n matrix which look like:

σ2 0 0 · · · 0 0 00 σ2 0 · · · 0 0 0...

......

......

...0 0 0 · · · 0 σ2 00 0 0 · · · 0 0 σ2

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Estimating Sigma-squared

Recall that:Let εi = Yi − Yi , then

σ2 =

∑ε2i

n − k, (13)

where k is the number of parameters.

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Heteroscedasticity I

Assumption A5 on the assumptions slide makes thehomogeneous or constant variance assumption. Thisassumption is clearly wrong in many cases, this is particulartrue with cross national data.

We can either use an estimator which directly models theheteroscedasticity and weights each observationsappropriately—e.g, Weighted Least Squares (WLS)—or we canuse robust standard errors.

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Heteroscedasticity II

The OLS estimator is unbiased (and consistent) when there isheteroscedasticity, but it is not efficient. We say that α is anefficient unbiased estimator if for a given sample size thevariance of α is smaller than (or equal to) the variance of anyother unbiased estimator.

Why is OLS still unbiased? Recall that when we proved thatOLS is unbiased, the variance terms played no part in theproof—i.e., Assumption A5 played no part.

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Univariate Matching

• Assume that we obtain conditional independence if wecondition on one confounder, XiG, where i indexes theobservation and G ∈ T ,C denotes the treatment orcontrol group.

• Say we wish to compare two matched pairs to estimateATT, τi ′ = Yi1 − Yi ′0, where ’ denotes the matchedobservation.

• Matched pairs can be used to estimate τ because thetreatment observation reveals Yi1 and the controlobservation Yi0 under conditional exchangeability.

• Bias in the estimate of τi ′ is assumed to be some unknownincreasing function of |XiT − Xi ′C |.

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Univariate Matching

• Assume that we obtain conditional independence if wecondition on one confounder, XiG, where i indexes theobservation and G ∈ T ,C denotes the treatment orcontrol group.

• Say we wish to compare two matched pairs to estimateATT, τi ′ = Yi1 − Yi ′0, where ’ denotes the matchedobservation.

• Matched pairs can be used to estimate τ because thetreatment observation reveals Yi1 and the controlobservation Yi0 under conditional exchangeability.

• Bias in the estimate of τi ′ is assumed to be some unknownincreasing function of |XiT − Xi ′C |.

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Univariate Matching

• Assume that we obtain conditional independence if wecondition on one confounder, XiG, where i indexes theobservation and G ∈ T ,C denotes the treatment orcontrol group.

• Say we wish to compare two matched pairs to estimateATT, τi ′ = Yi1 − Yi ′0, where ’ denotes the matchedobservation.

• Matched pairs can be used to estimate τ because thetreatment observation reveals Yi1 and the controlobservation Yi0 under conditional exchangeability.

• Bias in the estimate of τi ′ is assumed to be some unknownincreasing function of |XiT − Xi ′C |.

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Univariate Matching

• Assume that we obtain conditional independence if wecondition on one confounder, XiG, where i indexes theobservation and G ∈ T ,C denotes the treatment orcontrol group.

• Say we wish to compare two matched pairs to estimateATT, τi ′ = Yi1 − Yi ′0, where ’ denotes the matchedobservation.

• Matched pairs can be used to estimate τ because thetreatment observation reveals Yi1 and the controlobservation Yi0 under conditional exchangeability.

• Bias in the estimate of τi ′ is assumed to be some unknownincreasing function of |XiT − Xi ′C |.

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Univariate Matching

• We can then minimize our bias in τi ′ and hence ˆτ byminimizing |XiT − Xi ′C | for each matched pair.

• Can be minimized non-parametrically by “nearest neighbormatching” (NN). The case with replacement is simple.

• With NN matching we just match the nearest Xi ′C to agiven XiT .

• For ATT, we match every XiT with the nearest Xi ′C .• For ATC, we match every XiC with the nearest Xi ′T .• For ATE, we match every XiT with the nearest Xi ′C AND

match every XiC with the nearest Xi ′T .• Let’s walk through some examples: nn1.R

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Univariate Matching

• We can then minimize our bias in τi ′ and hence ˆτ byminimizing |XiT − Xi ′C | for each matched pair.

• Can be minimized non-parametrically by “nearest neighbormatching” (NN). The case with replacement is simple.

• With NN matching we just match the nearest Xi ′C to agiven XiT .

• For ATT, we match every XiT with the nearest Xi ′C .• For ATC, we match every XiC with the nearest Xi ′T .• For ATE, we match every XiT with the nearest Xi ′C AND

match every XiC with the nearest Xi ′T .• Let’s walk through some examples: nn1.R

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Univariate Matching

• We can then minimize our bias in τi ′ and hence ˆτ byminimizing |XiT − Xi ′C | for each matched pair.

• Can be minimized non-parametrically by “nearest neighbormatching” (NN). The case with replacement is simple.

• With NN matching we just match the nearest Xi ′C to agiven XiT .

• For ATT, we match every XiT with the nearest Xi ′C .• For ATC, we match every XiC with the nearest Xi ′T .• For ATE, we match every XiT with the nearest Xi ′C AND

match every XiC with the nearest Xi ′T .• Let’s walk through some examples: nn1.R

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Univariate Matching

• We can then minimize our bias in τi ′ and hence ˆτ byminimizing |XiT − Xi ′C | for each matched pair.

• Can be minimized non-parametrically by “nearest neighbormatching” (NN). The case with replacement is simple.

• With NN matching we just match the nearest Xi ′C to agiven XiT .

• For ATT, we match every XiT with the nearest Xi ′C .• For ATC, we match every XiC with the nearest Xi ′T .• For ATE, we match every XiT with the nearest Xi ′C AND

match every XiC with the nearest Xi ′T .• Let’s walk through some examples: nn1.R

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Univariate Matching

• We can then minimize our bias in τi ′ and hence ˆτ byminimizing |XiT − Xi ′C | for each matched pair.

• Can be minimized non-parametrically by “nearest neighbormatching” (NN). The case with replacement is simple.

• With NN matching we just match the nearest Xi ′C to agiven XiT .

• For ATT, we match every XiT with the nearest Xi ′C .• For ATC, we match every XiC with the nearest Xi ′T .• For ATE, we match every XiT with the nearest Xi ′C AND

match every XiC with the nearest Xi ′T .• Let’s walk through some examples: nn1.R

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Univariate Matching

• We can then minimize our bias in τi ′ and hence ˆτ byminimizing |XiT − Xi ′C | for each matched pair.

• Can be minimized non-parametrically by “nearest neighbormatching” (NN). The case with replacement is simple.

• With NN matching we just match the nearest Xi ′C to agiven XiT .

• For ATT, we match every XiT with the nearest Xi ′C .• For ATC, we match every XiC with the nearest Xi ′T .• For ATE, we match every XiT with the nearest Xi ′C AND

match every XiC with the nearest Xi ′T .• Let’s walk through some examples: nn1.R

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Generalized Linear Models

Generalized linear models (GLMs) extend linear models toaccommodate both non-normal distributions andtransformations to linearity. GLMs allow unified treatment ofstatistical methodology for several important classes of models.

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GLMS

This class of models can be described by two components:stochastic and systematic.• Stochastic component:

• Normal distribution for LS: εi ∼ N(0, σ2)

• For logistic regression:

YBern(yi |πi ) = πyii (1− πi )

1−yi

πi for y = 1

1− πi for y = 0

• Systematic Component:

• For LS: E(Yi |Xi ) = Xβ• For logistic regression:

Pr(Yi = 1|Xi ) ≡ E(Yi ) ≡ πi =exp(Xiβ)

1 + exp(Xiβ)

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Logit Link vs. Identity Link

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Logit Link vs. Identity Link

The black line is the logistic fitted values and the red line is fromOLS (the identity link). For R code see, logit1.R

OLS goes unbounded, which we know cannot happen!

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Defining Assumptions

A generalized linear model may be described by the followingassumptions:

• There is a response Y observed independently at fixedvalues of stimulus variables X1, · · · ,Xk .

• The stimulus variables may only influence the distributionof Y through a single linear function called the linearpredictor η = β1X1 + · · ·+ βkXk .

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Defining Assumptions

• The distribution of Y has density of the form

f (Yi ; θi , φ) = exp[AiYiθi − γ(θi)/φ+ τ(Yi , φ/Ai)], (14)

where φ is a scale parameter (possibly known). Ai is aknown prior weight and parameter θi depends upon thelinear predictor.

This is the exponential family of distributions. Most of thedistributions you know are part of this family including thenormal, binomial and Poisson.

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Models of the Mean

The mean µ = E(Y ) is a smooth invertible function of the linearpredictor:

µ = m(η), (15)η = m−1(µ) = l(µ), (16)

where the inverse function l(·) is called the link function

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Models of the Mean: examples

For LS we use the identity link (also called the canonical link):l(µ) = µ. Thus,

Xiβ = ηi = m(µi) = µi = E(Yi)

For logistic regression we use the logit link:

l(µ) = log(

π

1− π

). Thus,

Xiβ = ηi = m(µi) =exp(ηi)

1 + exp(ηi)= µi = E(Yi) = Pr(Yi = 1|Xi),

note thatexp(ηi)

1 + exp(ηi)is the inverse logit link.

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GLM: Gaussian

GLMs allow unified treatment of many models—e.g., normal:The probability distribution function in the form one usuallysees the distribution is:

f (y ;µ) =1

(2πσ2)1/2 exp[− 1

2σ2 (y − µ)2],

where µ is the parameter of interest and σ2 is regarded, in thissetting, as a nuisance parameter.The following is the canonical form of the distributionθ = µ, γ(θ) = θ2/2 and φ = σ2 so we can write:

log f (y) =1φyµ− 1

2µ2 − 1

2y2 − 1

2log(2πφ).

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Notes on the Gaussian

• If φ, the variance or what is called the dispersion in theGLM framework, were known the distribution of y would bea one-parameter canonical exponential family.

• An unknown φ is handled as a nuisance parameter bymoment methods. For example, note how σ2 is estimatedin the least squares case.

• We can do this because θ and φ are orthogonalparameters. This means that we can estimate θ and thenconditioning on this estimate, calculate φ. We don’t have tojointly estimate both.

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GLM: Poisson Example

For a Poisson distribution with mean µ we have

log f (y) = y log(µ)− µ− log(y !),

where θ = log(µ), φ = 1, and φ(θ) = µ = eθ.

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GLM: Binomial Example

For a binomial distribution with fixed number of trials a andparameter p we take the response to be y = s/a where s is thenumber of “successes”. The density is

log f (y) = a[y log

p1− p

+ log(1− p)

]+ log

(a

ay

)where we take Ai = ai , φ = 1, θ to be the logit transform of pand γ(θ) = − log(1− p) = log(1 + eθ).

The functions supplied with R for handling generalized linearmodeling distributions include glm, gaussian, binomial,poisson, inverse.gaussian and Gamma.

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Logistic Regression

Let Y ∗ be a continuous unobserved variable such as thepropensity to vote for a particular political party, say theRepublican Party, or the propensity to be assigned to treatment.

Assume that Y ∗i has some distribution whose mean is µi . Andthat Y ∗i and Y ∗j are independent for all i 6= j , conditional on X.

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Logistic Regression

We observe Yi such that:

Yi =

1 if Y ∗i ≥ τ if i is treated or votes Republican0 if Y ∗i < τ if i is NOT treated or does NOT vote Republican

• Since Y ∗ is unobserved we usually define the threshold, τ ,to be zero.

• Note that if Y ∗i is observed and that it is normallydistributed conditional on X , we have LS regression.If only Yi is observed and that Y ∗ is distributedstandardized logistic (which is very similar to the normaldistribution), we obtain the logistic regression model.

• Note that our guess of Y ∗ is µi–or, equivalently, ηi .

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Logistic Regression Coefficients

• The estimated coefficients, β, in a logistic regression canbe interpreted in a manner very similar to that thecoefficients in LS. But there are important differenceswhich complicate the issue.

• We should interpret β as the regression coefficients of Y ∗

on X . So, β1 is what happens to Y ∗i when X1 goes up byone unit when all of the other explanatory variables remainthe same. ICK! Causal

• The interpretation is complicated by the fact that the linkfunction is nonlinear. So the effect of a one unit change inX1 on the predicted probability is different if the otherexplanatory variables are held at one given constant valueversus another. Why is this?

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Logistic Regression Assumptions andProperties

• The assumptions of logistic regression are very similar tothose of least squares—see the assumptions slide. Butthere are some important difference. The most important isthat in order to obtain consistency, logistic regressionrequires the homoscedasticity assumption while LS doesnot.

• We also need the no perfect separation assumption for theexistence of the estimator.

• Another important difference is that logistic regression isNOT unbiased. But it is consistent.

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Logistic Regression Loss Function andEstimation

The loss function for logistic regression is:

n∑i=1

− [Yi log(πi) + (1− Yi) log(1− πi)] ,

recall that πi =exp(Xiβ)

1 + exp(Xiβ).

The reasons for this loss function are made clear in a MLcourse. As will how this loss function is usuallyminimized—iterated weighted least squares.See Venables and Ripley (2002) for more estimation details.

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The Propensity Score

• The propensity score allows us to match on X withouthaving to match all of the k -variables in X . We may simplymatch on the probability of being assigned totreatment—i.e., the propensity score. The propensity scoreis just one dimensional.

• More formally the propensity score (also called thebalancing score) is:

Pr(Ti = 1|Xi) = E(Ti |Xi)

The following theory is from Rosenbaum and Rubin (1983)

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The Propensity Score

• The propensity score allows us to match on X withouthaving to match all of the k -variables in X . We may simplymatch on the probability of being assigned totreatment—i.e., the propensity score. The propensity scoreis just one dimensional.

• More formally the propensity score (also called thebalancing score) is:

Pr(Ti = 1|Xi) = E(Ti |Xi)

The following theory is from Rosenbaum and Rubin (1983)

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A1: Ignorability

Assumption 1: differences between groups are explained byobservables (ignorability)

(Y0,Y1) ⊥ T | X (A1)

Treatment assignment is “strictly ignorable” given X . This isequivalenty to:

Pr(T = 1 | Y0,Y1,X ) = Pr(T = 1 | X ), or

E(T | Y0,Y1,X ) = E(T | X )

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(A2) Overlap Assumption

Assumption 2: there exists common support for individuals inboth groups (overlap)

0 < P(T = 1|X = x) < 1 ∀x ∈ X (A2)

A1 and A2 together define strong ignorability (Rosenbaum andRubin 1983).

A2 is also known as the Experimental Treatment Assignment (ETA)assumption

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(A2) Overlap Assumption

Assumption 2: there exists common support for individuals inboth groups (overlap)

0 < P(T = 1|X = x) < 1 ∀x ∈ X (A2)

A1 and A2 together define strong ignorability (Rosenbaum andRubin 1983).

A2 is also known as the Experimental Treatment Assignment (ETA)assumption

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Observabled Quantities

We observe the conditional distributions:• F (Y1 | X ,T = 1) and• F (Y0 | X ,T = 0)

But not the joint distributions:• F (Y0,Y1 | X ,T = 1)

• F (Y0,Y1 | X )

Nor the impact distrubtion:• F (Y1 − Y0 | X ,T = 1)

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Weaker Assumptions

For ATT, weaker assumptions can be made:• A weaker conditional mean independence assumption on

Y0 is sufficient (Heckman, Ichimura, and Todd 1998).

E(Y0 | X ,T = 1) = E(Y0 | X ,T = 0) = E(Y0 | X )

• A weaker support condition:

Pr(T = 1 | X ) < 1

This is sufficient because we only need to guarantee acontrol analogue for every treatment, and not the reverse

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Pscore as a Balancing Score

Let e(X ) be the pscore: Pr(T = 1 | X ).

F (X | e(X ),T ) = F (X | e(X )),X ⊥ T | e(X )

That is, the distribution of covariates at differenttreatment/exposure levels is the same once one conditions onthe propensity score

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Balancing Score: intuition

We wish to show that if one matches on the true propensity score,asymptotically, the observed covariates, X , will be balanced betweentreatment and control groups. Assume that conditioning on the truepropensity score does not balance the underlying X covariates. Thenthere exists x1 and x2 such that e(X1) = e(X2), so that conditioning one(X ), still results in imbalance of x1 and x2. Imbalance by definitionimplies that Pr(Ti = 1 | x1) 6= Pr(Ti = 1 | x2), but this contradictse(x1) = e(x2) since by the definition of the propensity score:

e(x1) = Pr(Ti = 1 | x1), ande(x2) = Pr(Ti = 1 | x2).

Therefore, the observed covariates X must be balanced afterconditioning on the true propensity score. For details:P. R. Rosenbaum and Rubin (1983), especially theorems 1–3.

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Pscore as a Balancing Score

Proof:

Pr(T = 1 | X ,e(X )) = Pr(T = 1 | X ) = e(X )

Pr(T = 1 | e(X )) = E E(T | X ) | e(X ) =

E e(X ) | e(X ) = e(X )

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Balancing Score, finerb(X ) is a balancing score, that is,

T ⊥ X | b(X ),

if and only if b(X ) is finer than e(X )—i.e., e(X ) = f b(X ) forsome function f . Rosenbaum and Rubin Theorem 2.Suppose b(X ) is finer than e(X ). Then it is sufficient to show:

Pr(T = 1 | b(X )) = e(X )

Note:

Pr(T = 1 | b(X )) = E e(X ) | b(X )

But since b(X ) is finer than e(X ):

E e(X ) | b(X ) = e(X )

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Balancing Score, coarser

Now assume b(X ) is not finer than e(X ). So, there exists x1 and x2 s.te(x1) 6= e(x2) but b(x1) = b(x2). But, by definitionPr(T = 1 | x1) 6= Pr(T = 1 | x2), so that T and X are not conditionallyindepdnent given b(x), and thus b(X ) is not a balancing score.Therefore, to be a balancing score b(X ) must be finer than to e(X ).

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Pscore and Strong Ignorability

Theorem 3: If treatment assignment is strongly ignorable givenX , it is strongly ignorable given any balancing score b(x)

• It is sufficient to show:

Pr(T = 1 | Y0,Y1,b(x)) = Pr(T = 1 | b(x))

• By Theorem 2, this is equiv to showing:

Pr(T = 1 | Y0,Y1,b(x)) = e(x)

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Pscore and Strong Ignorability II

Note:

Pr(T = 1 | Y0,Y1,b(x)) = E Pr(T = 1 | Y0,Y1,X ) | Y0,Y1,b(X )

By assumption equals:

E Pr(T = 1 | X ) | Y0,Y1,b(x)

Which by definition equals:

E e(X ) | Y0,Y1,b(X ) ,

which since b(X ) is finer than e(X ), equals e(X ).

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Pscore as Balancing Score III

We have shown (where, Y0,1 is a compact way of writing Y0,Y1):

E(T | Y0,1,Pr(T = 1 | X )) =

E(E(T | Y0,1,X ) | Y0,1, pr(T = 1 | X )),

So that

E(T | Y0,1,X ) = E(T | X ) =⇒

E(T | Y0,1,Pr(T = 1 | X ) = E(T | Pr(T = 1 | X ))

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Correct Specification of the PropensityScore?

• It is unclear how one should correctly specify thepropensity score.

• The good news is that the propensity score is serving avery clear purpose: matching on the propensity scoreshould induce balance on the baseline covariates.

• This can be directly tested by a variety of tests such asdifference of means. But difference of means are not bythemselves sufficient.

• THE BAD NEWS: this only works for the observables.

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Estimating the Propensity Score

• We we may estimate the propensity score using logisticregression where the treatment indicator is the dependentvariable.

• We then match on either the predicted probabilities πi orthe linear predictor ηi . It is in fact preferable to match on thelinear predictor because the probabilities are compressed.

• For matching, we do not care about the parameters, β. Weonly care about getting good estimates of πi .

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Estimating the Propensity Score

• We we may estimate the propensity score using logisticregression where the treatment indicator is the dependentvariable.

• We then match on either the predicted probabilities πi orthe linear predictor ηi . It is in fact preferable to match on thelinear predictor because the probabilities are compressed.

• For matching, we do not care about the parameters, β. Weonly care about getting good estimates of πi .

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Theory of Matching, revised

Both of the proceeding assumptions can be weakend if we aresimply estimating ATTAssumption 1: differences between groups are explained byobservables

Y0⊥T |X (A1)

Assumption 2: there exists common support for individuals inboth groups

P(T = 1|X = x) < 1 ∀x ∈ X (A2)

Obvious parallel changes for ATC.

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Theory of Propensity ScoresPROBLEM: Simple matching approach fails in finite samples ifthe dimensionality of X is large.

An alternative way to condition on X is to match on theprobability of being assigned to treatment—i.e., the propensityscore. The propensity score is just one dimensional.

Defn: Propensity score is the probability that a person istreated/manipulated.

b(x) = P(T = 1|X = x)

Rosenbaum and Rubin (1983) show that we can reducematching to this single dimension, by showing that theassumptions of matching (A1 and A2) lead to,

Y0⊥T |P(X )

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Propensity Score MatchingMatching on b(X) similar to matching on X:Step 1: for each Y1i find Y0j which has similar (or proximate)b(X).

One-one matching:– Nearest-neighbor: choose closest control on P(Xi)– Caliper: same, but drop cases which are too farMany-one: use weighted average of neighbors nearP(Xi)

Y0i =∑

j

WijYj

E.g., using a weighted method where

W ∝ K (Pi − Pj

h)

Step 2: same, take average over matched values of Y1i and Y0i 228 / 765

Properties of Pscore Matching I

• The modeling portion of the estimator is limited to themodel of p(Xi). Estimation of this model requires noknowledge of the outcome.

• Unlike in the regression case, there is a clear standard forchoosing an optimal model; it is the model which balancesthe covariates, X .

• The key assumption required is that no variable has beenleft unobserved which is correlated with treatmentassignment and with the outcome. This is called theselection on observables assumption.

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Properties of Pscore Matching I

• The modeling portion of the estimator is limited to themodel of p(Xi). Estimation of this model requires noknowledge of the outcome.

• Unlike in the regression case, there is a clear standard forchoosing an optimal model; it is the model which balancesthe covariates, X .

• The key assumption required is that no variable has beenleft unobserved which is correlated with treatmentassignment and with the outcome. This is called theselection on observables assumption.

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Properties of Pscore Matching I

• The modeling portion of the estimator is limited to themodel of p(Xi). Estimation of this model requires noknowledge of the outcome.

• Unlike in the regression case, there is a clear standard forchoosing an optimal model; it is the model which balancesthe covariates, X .

• The key assumption required is that no variable has beenleft unobserved which is correlated with treatmentassignment and with the outcome. This is called theselection on observables assumption.

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Properties of Pscore Matching II

• No functional form is implied for the relationship betweentreatment and outcome. No homogeneous causal effectassumption has been made.

• Since we are interested in a lower dimensionalrepresentation of Xi , in particular p(Xi), we do not need toestimate consistently any of the individual parameters inour propensity model—they don’t even need to beidentified.

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Properties of Pscore Matching II

• No functional form is implied for the relationship betweentreatment and outcome. No homogeneous causal effectassumption has been made.

• Since we are interested in a lower dimensionalrepresentation of Xi , in particular p(Xi), we do not need toestimate consistently any of the individual parameters inour propensity model—they don’t even need to beidentified.

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Properties of Pscore Matching III

• Because we are taking a conditional expectation, we needto decide what to condition on. If we condition too little, wedo not obtain strong ignorability. If we condition onvariables which are not baseline variables, we also obtainbiased estimates.

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Practice of Matching (revisited)

Step 1: for each individual i in the treated group (T=1) find oneor more individuals i ′ from the untreated group (T=0) who isequivalent (or proximate) in terms of X.

(a) If more than one observation matches, take average

(b) If no exact matches exist, either drop case (because ofnon-common support) or take closest / next best match(es)

Step 2:

τ =1N

∑(Y1i |X )− 1

N ′∑

(Y0i ′ |X )

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A simple example

i Ti Xi Yi J Yi (0) Yi (1) KM(i)1 0 2 7 5 7 8 32 0 4 8 4,6 8 7.5 13 0 5 6 4,6 6 7.5 0

i Ti Xi Yi J Yi (0) Yi (1) KM(i)4 1 3 9 1,2 7.5 9 15 1 2 8 1 7 8 16 1 3 6 1,2 7.5 6 17 1 1 5 1 7 5 0

where observed outcome is Y , covariate to match on is X , J indexesmatches, KM(i) is number of times used as a match divided by #J ;Yi = 1/#J (i)

∑l∈J (i) Yl .

The estimate of ATE is .1428. ATT is −0.25. And ATC is23

.

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Alternative test statisticsTreatment may be multiplicative, rather than constant additive.So the ratio of treated/non-treated should be constant: e.g.,Y1 = aY0

log(Y1)− log(Y0) = log(Y1/Y0) = log(a)

so,

τ2 =1N

∑log(Y1i |X )− 1

N ′∑

log(Y0i ′ |X )

Robust way to discern differences (e.g., if you fear outliersdistorting means) transform to ranks:

Let Ri be the rank of the i observation, pooling over treated andnon-treated.

τ3 =1N

∑(Ri |X )− 1

N ′∑

(Ri ′ |X )

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Comments

• If you think the variances are changed by treatment, butnot the location, then test differences in variances.

• The point: your test statistic should be motivated by yourtheory of what is changing when manipulation / treatmentoccurs!

• One could in theory estimate any model one wishes on thematched data. Such as regression, logistic regression etc.

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RDD Readings• Lee, D., 2008. “Randomized experiments from non-random

selection in U.S. house elections.” Journal of Econometrics142:2 675–697.

• Devin Caughey and Jasjeet S. Sekhon. “Elections and theRegression-Discontinuity Design: Lessons from Close U.S.House Races, 1942–2008.” Political Analysis. 19 (4): 385–408.2011.

• Thistlethwaite, D. and D. Campbell. 1960.“Regression-Discontinuity Analysis: An alternative to the ex postfacto experiment.” Journal of Educational Psychology 51 (6):309–317.

• Gerber, Alan S. and Donald P. Green. “Field Experiments andNatural Experiments.” Oxford Handbook of PoliticalMethodology.

• Hahn J., P. Todd, and W.V. der Klaauw, 2001. “Identification andEstimation of Treatment Effects with a Regression-DiscontinuityDesign.” Econometrica.

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Regression Discontinuity Design

Context: Individual receives treatment only iff observedcovariate Zi crosses known threshold z0.

Logic: compare individuals just above and below threshold

Key assumptions:1. Probability of receiving treatment jumps discontinuously at z02. Continuity of outcome at threshold

E [Y0|Z = z] and E [Y1|Z = z] are continuous in z at z0

3. Confounders vary smoothly with Z—i.e., there are no jumpsat the treatment threshold z0

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Regression Discontinuity Design

Context: Individual receives treatment only iff observedcovariate Zi crosses known threshold z0.

Logic: compare individuals just above and below threshold

Key assumptions:1. Probability of receiving treatment jumps discontinuously at z02. Continuity of outcome at threshold

E [Y0|Z = z] and E [Y1|Z = z] are continuous in z at z0

3. Confounders vary smoothly with Z—i.e., there are no jumpsat the treatment threshold z0

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Regression Discontinuity Design

Context: Individual receives treatment only iff observedcovariate Zi crosses known threshold z0.

Logic: compare individuals just above and below threshold

Key assumptions:1. Probability of receiving treatment jumps discontinuously at z02. Continuity of outcome at threshold

E [Y0|Z = z] and E [Y1|Z = z] are continuous in z at z0

3. Confounders vary smoothly with Z—i.e., there are no jumpsat the treatment threshold z0

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Regression Discontinuity Design

If the potential outcomes are distributed smoothly at thecut-point, the RD design estimates the average causal effect oftreatment at the cut-point, Zi = c:

τRD ≡ E[Yi(1)− Yi(0)|Zi = c] =

limZi↓c

E[Yi(1)|Zi = c]− limZi↑c

E[Yi(0)|Zi = c]

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Two cases of RD

Constant treatment effect: Zi does not independently impactoutcome except through treatment

Variable treatment effect: Zi also impacts outcome

Issue: We are estimating the Local Average Treatment Effect(LATE)

244 / 765

RDD with Constant Treatment Effects

0 20 40 60 80 100

020

4060

8010

0

Z

Y

Non−Treated Treated

245 / 765

RDD with Variable Treatment Effects

0 20 40 60 80 100

020

4060

8010

0

Z

Y

Non−Treated Treated

246 / 765

Example 1: Scholarship Winners

• From Kenya to the U.S. RDD has been used to estimatethe effect of school scholarships.

• In these cases, z is the measure of academic performancewhich determines if one receives a scholarship—e.g., testscores, GPA.

• Of course, comparing everyone who wins a scholarshipwith everyone who loses is not a good way to proceed.

• Focus attention on people who just won and who just lost.• This reduces omitted variable/selection bais, why?

247 / 765

RDD and Omitted Variable Bias I

• Let’s assume that there is a homogeneous causal effect:Yi = α + βTi + γXi + εi ,

where T is the treatment indicator and X is a confounder.• Then:

E(Yi | Ti = 1)− E(Yi | Ti = 0)

= [α + β + γE(Xi | Ti = 1) + E(εi | Ti = 1)]

− [α + 0 + γE(Xi | Ti = 0) + E(εi | Ti = 0)]

= β + γ [E(Xi | Ti = 1)− E(Xi | Ti = 0)]

True Effect + Bias

248 / 765

RDD and Omitted Variable Bias I

• Let’s assume that there is a homogeneous causal effect:Yi = α + βTi + γXi + εi ,

where T is the treatment indicator and X is a confounder.• Then:

E(Yi | Ti = 1)− E(Yi | Ti = 0)

= [α + β + γE(Xi | Ti = 1) + E(εi | Ti = 1)]

− [α + 0 + γE(Xi | Ti = 0) + E(εi | Ti = 0)]

= β + γ [E(Xi | Ti = 1)− E(Xi | Ti = 0)]

True Effect + Bias

249 / 765

RDD and Omitted Variable Bias I

• Let’s assume that there is a homogeneous causal effect:Yi = α + βTi + γXi + εi ,

where T is the treatment indicator and X is a confounder.• Then:

E(Yi | Ti = 1)− E(Yi | Ti = 0)

= [α + β + γE(Xi | Ti = 1) + E(εi | Ti = 1)]

− [α + 0 + γE(Xi | Ti = 0) + E(εi | Ti = 0)]

= β + γ [E(Xi | Ti = 1)− E(Xi | Ti = 0)]

True Effect + Bias

250 / 765

RDD and Omitted Variable Bias I

• Let’s assume that there is a homogeneous causal effect:Yi = α + βTi + γXi + εi ,

where T is the treatment indicator and X is a confounder.• Then:

E(Yi | Ti = 1)− E(Yi | Ti = 0)

= [α + β + γE(Xi | Ti = 1) + E(εi | Ti = 1)]

− [α + 0 + γE(Xi | Ti = 0) + E(εi | Ti = 0)]

= β + γ [E(Xi | Ti = 1)− E(Xi | Ti = 0)]

True Effect + Bias

251 / 765

RDD and Omitted Variable Bias II

• RecallE(Yi | Ti = 1)− E(Yi | Ti = 0)

= β + γ [E(Xi | Ti = 1)− E(Xi | Ti = 0)]

True Effect + Bias• Since we assume that X varies smoothly with respect to Z

(i.e., there are no jumps at the treatment threshold z0):= β + γ [(x∗ + δ)− x∗]= β + γ∗δ

≈ β

252 / 765

RDD and Omitted Variable Bias II

• RecallE(Yi | Ti = 1)− E(Yi | Ti = 0)

= β + γ [E(Xi | Ti = 1)− E(Xi | Ti = 0)]

True Effect + Bias• Since we assume that X varies smoothly with respect to Z

(i.e., there are no jumps at the treatment threshold z0):= β + γ [(x∗ + δ)− x∗]= β + γ∗δ

≈ β

253 / 765

RDD and Omitted Variable Bias II

• RecallE(Yi | Ti = 1)− E(Yi | Ti = 0)

= β + γ [E(Xi | Ti = 1)− E(Xi | Ti = 0)]

True Effect + Bias• Since we assume that X varies smoothly with respect to Z

(i.e., there are no jumps at the treatment threshold z0):= β + γ [(x∗ + δ)− x∗]= β + γ∗δ

≈ β

254 / 765

Sharp vs. Fuzzy Design

• Sharp design: Ti is a deterministic function of Zi

• Fuzzy design: Ti is a function of Zi and an exogenousrandom variable:

• Fuzzy design requires observing Zi separately from therandom component

• We will cover this after instrumental variables (J. D. Angrist,G. W. Imbens, and Rubin, 1996)

• because intention-to-treat and IV for compliance correctioncan be used

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Incumbency Advantage (IA)

• The Personal Incumbency Advantage is the candidatespecific advantage that results from the benefits of officeholding—e.g:• name recognition; incumbency cue• constituency service• public position taking• providing pork

• The Incumbent Party Advantage:• ceteris paribus, voters have a preference for remaining with

the same party they had before• parties have organizational advantages in some districts

256 / 765

Incumbency Advantage (IA)

• The Personal Incumbency Advantage is the candidatespecific advantage that results from the benefits of officeholding—e.g:• name recognition; incumbency cue• constituency service• public position taking• providing pork

• The Incumbent Party Advantage:• ceteris paribus, voters have a preference for remaining with

the same party they had before• parties have organizational advantages in some districts

257 / 765

Problems with Estimating the IA

Empirical work• agrees there is a significant incumbency advantage• debates over source and magnitude

Theoretical work• argues positive estimates of IA are spurious:

Cox and Katz 2002; Zaller 1998; Rivers 1988• selection and survivor bias• strategic candidate entry and exit

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Estimates of Incumbency Advantage

Early empirical strategies:• Regression model (Gelman-King)∼10%

• Sophomore-surge (Erikson; Levitt-Wolfram)∼7%

Design-based approaches• Surprise loss of incumbents through death (Cox-Katz)

2%• Redistricting and personal vote (Sekhon-Titiunik)

0%, Texas; 2%, California• Regression-Discontinuity with close election (Lee)

10% increase in vote shares for incumbent party40% increase in win probability for incumbent party

259 / 765

Estimates of Incumbency Advantage

Early empirical strategies:• Regression model (Gelman-King)∼10%

• Sophomore-surge (Erikson; Levitt-Wolfram)∼7%

Design-based approaches• Surprise loss of incumbents through death (Cox-Katz)

2%• Redistricting and personal vote (Sekhon-Titiunik)

0%, Texas; 2%, California• Regression-Discontinuity with close election (Lee)

10% increase in vote shares for incumbent party40% increase in win probability for incumbent party

260 / 765

Regression-Discontinuity (RD) andElections

Logic of applying Regression-Discontinuity to elections• treatment: being elected (incumbent)• assigned by getting 50%+1 of vote• candidates near 50% randomly assigned to treatment• first applications: D. S. Lee, 2001, Pettersson-Lidbom,

2001

We show• RD doesn’t work for U.S. House elections• there are substantive implications

261 / 765

Regression-Discontinuity (RD) andElections

Logic of applying Regression-Discontinuity to elections• treatment: being elected (incumbent)• assigned by getting 50%+1 of vote• candidates near 50% randomly assigned to treatment• first applications: D. S. Lee, 2001, Pettersson-Lidbom,

2001

We show• RD doesn’t work for U.S. House elections• there are substantive implications

262 / 765

Regression-Discontinuity (RD) andElections

Logic of applying Regression-Discontinuity to elections• treatment: being elected (incumbent)• assigned by getting 50%+1 of vote• candidates near 50% randomly assigned to treatment• first applications: D. S. Lee, 2001, Pettersson-Lidbom,

2001

We show• RD doesn’t work for U.S. House elections• there are substantive implications

263 / 765

More Information and Cleaner Data

We created a new version of U.S. House elections dataset

Key features:• Corrected vote (and outcome) data, 1942–2008

validation of election results for every close race• Incorporates 30 additional covariates

e.g., campaign expenditures, CQ scores• Recounts, fraud, and other oddities:

random sample of 50% of close elections

264 / 765

Non-incumbent at T+1

Incumbent at T+1

0 50 1000

0.5

1

Democratic Vote ProportionElection T

Pro

port

ion

Dem

ocra

ticW

inE

lect

ion

T+1

265 / 765

Non-incumbent at T−1Incumbent at T−1

∆ = 0

0 50 1000

0.5

1

Democratic Vote ProportionElection T

Pro

port

ion

Dem

ocra

ticW

inE

lect

ion

T−1

266 / 765

Covariate Balance

VariableName

ValidCases

D WinMean

D LossMean

Dem Win t − 1 85 0.58 0.19

Dem % t − 1 85 51 45

Dem Spending % 47 54 45

Dem Donation % 34 56 45

CQ Rating −1, 0, 1 69 0.23 −0.29

Dem Inc in Race 85 0.49 0.14

Rep Inc in Race 85 0.28 0.62

Partisan Swing 85 −1.7 4.0

Dem Win t + 1 85 0.74 0.33

Dem % t + 1 85 53 43

0 .05 .1 1p−value

267 / 765

Covariate Balance

VariableName

ValidCases

D WinMean

D LossMean

Dem Sec of State 85 0.47 0.31

Dem Governor 85 0.40 0.48

Dem Pres % Margin 79 −0.09 −0.10

Voter Turnout % 85 37 34

Pct Gov't Worker 73 5.1 4.4

Pct Urban 73 70 65

Pct Black 73 4.9 5.0

Pct Foreign Born 73 4.0 4.1

0 .05 .1 1p−value

268 / 765

Incumbent Party’s Margin in CloseElections

Incumbent Party Margin (%)

Fre

quen

cy C

ount

in 0

.5%

Bin

s

−4 −2 0 2 4

0

10

20

30

40

50

60

37

30

43

29

43

3734

26

59

49

33

41

53 52

57

52

269 / 765

Democratic Margin in Close Elections

Democratic Margin (%)

Fre

quen

cy C

ount

in 0

.5%

Bin

s

−4 −2 0 2 4

0

10

20

30

40

50

60

45

37 38

51

46

4143 42 43

39

28

38

31

5047

40

270 / 765

Broken Down by Incumbent Party

Democrat−Held Seats

Democratic Margin (%)

Fre

quen

cy C

ount

in 0

.5%

Bin

s

−4 −2 0 2 4

0

10

20

30

40

18

14 15 15

2220

15

8

25

20

12

17 17

23

31

22

271 / 765

Broken Down by Incumbent Party

Republican−Held Seats

Democratic Margin (%)

−4 −2 0 2 4

0

10

20

30

40

27

23 23

36

24

21

28

34

18 19

16

21

14

27

1618

272 / 765

CQ Rating and Democratic Victory inElections Decided by Less Than 0.5%

Dem Win t − 1 Dem Win tRep Favored (23) 17% 30%Tossup (25) 24% 52%Dem Favored (21) 90% 71%

273 / 765

CQ Rating and Democratic Victory inElections Decided by Less Than 0.5%

Dem Win t − 1 Dem Win tRep Favored (23) 17% 30%Tossup (25) 24% 52%Dem Favored (21) 90% 71%

Even in tossup elections, incumbent party won about two-thirdsof elections

274 / 765

Democratic Win Proportion Next Election (T+1)

Democratic Margin (%), Election T

Pro

port

ion

Dem

ocra

tic W

in, E

lect

ion

T+

1

0.2

0.4

0.6

0.8

1.0

−20 −10 0 10 20

275 / 765

Democratic Win Proportion Previous Election (T−1)

Democratic Margin (%), Election T

Pro

port

ion

Dem

ocra

tic W

in, E

lect

ion

T−

1

0.0

0.2

0.4

0.6

0.8

1.0

−20 −10 0 10 20

276 / 765

Democratic Campaign Spending %

Democratic Margin (%)

Dem

ocra

tic S

pend

ing

%

30

40

50

60

70

−20 −10 0 10 20

277 / 765

Donations to Democratic %

Democratic Margin (%)

Don

atio

ns to

Dem

ocra

tic C

andi

ate

%

20

30

40

50

60

70

80

−20 −10 0 10 20

278 / 765

Divergence Figures

The following two figures plot the divergence of covariatedistributions in the immediate neighborhood of the cut-point.Covariate differences are plotted against the mid-point of thedisjoint interval tested (e.g., 1.5 for the interval (−1.75,−1.25),(1.75,1.25). Loess lines highlight the trends in the imbalance.

279 / 765

Divergence Figure 1Balance Tests of Ten Covariates in Disjoint 0.5% Intervals

Absolute Value of Democratic Margin (%), Election t

Min

imum

p−

Val

ue

0.0

0.1

0.2

0.3

0.4

0.5

0 1 2 3 4 5 6 7 8 9 10

280 / 765

Divergence Figure 2Treated−Control Difference in Proportion of Democratic Victories, t+1 vs. t−1

Absolute Value of Democratic Margin (%), Election t

Diff

eren

ce in

Pro

port

ion

0.0

0.2

0.4

0.6

0.8 t+1

t−1

0 1 2 3 4 5 6 7 8 9 10

281 / 765

Lagged Democratic Win in RDD: 1st and 2nd-order polynomial

282 / 765

Lagged Democratic Win in RDD: 3rd and 4th-order polynomial

283 / 765

Lagged Democratic Win in RDD: 8th-order polynomial

284 / 765

National Partisan Swings

• Partisan swings are imbalanced:• 1958 (pro-Democratic tide): all 6 close elections occurred

in Republican-held seats• 1994 (pro-Republican tide): all 5 close elections occurred

in Democratic-held seats• Close elections do not generally occur in 50/50 districts

285 / 765

National Democratic Swing

Democratic Margin (%)

Dem

ocra

tic S

win

g (%

)

−3

−2

−1

0

1

2

3

−20 −10 0 10 20

286 / 765

There is Strategic Exit

Strategic exit among incumbents• 20% who barely win retire prior to next election• but all candidates who barely won first election

ran for reelection• even outside of the RD window, previous election margin is

predictive of retirements not due to running for higheroffice:• e.g., 1994: Equal percentage of Republicans and

Democrats retired, but Republicans left to run for higheroffice (13 of 20) Democrats did not (7 of 27)

• 2006: 18 Republican open seats versus 9 Democratic openseats

• 2010: In non-safe seats, 6 Republicans retired, but 15Democrats retired

287 / 765

There is Strategic Exit

Strategic exit among incumbents• 20% who barely win retire prior to next election• but all candidates who barely won first election

ran for reelection• even outside of the RD window, previous election margin is

predictive of retirements not due to running for higheroffice:• e.g., 1994: Equal percentage of Republicans and

Democrats retired, but Republicans left to run for higheroffice (13 of 20) Democrats did not (7 of 27)

• 2006: 18 Republican open seats versus 9 Democratic openseats

• 2010: In non-safe seats, 6 Republicans retired, but 15Democrats retired

288 / 765

There is Strategic Exit

Strategic exit among incumbents• 20% who barely win retire prior to next election• but all candidates who barely won first election

ran for reelection• even outside of the RD window, previous election margin is

predictive of retirements not due to running for higheroffice:• e.g., 1994: Equal percentage of Republicans and

Democrats retired, but Republicans left to run for higheroffice (13 of 20) Democrats did not (7 of 27)

• 2006: 18 Republican open seats versus 9 Democratic openseats

• 2010: In non-safe seats, 6 Republicans retired, but 15Democrats retired

289 / 765

Democratic Incumbent in Race

Democratic Margin (%)

Pro

port

ion

of R

aces

with

Dem

ocra

tic In

cum

bent

0.0

0.2

0.4

0.6

0.8

−20 −10 0 10 20

290 / 765

Money and Incumbents in CloseElections

The imbalances in campaign resources occur because:

• When incumbent candidates run, they have more money:62% of donations and 58% of expenditures

Money isn’t predictive of winning, although they usually win

• In open seats, the incumbent party’s candidates raise 50%of the money but spend 54%

Winners have more money:

winners raise 54% versus 41% (p=0.01)spend 56% versus 49% (p=0.02)

291 / 765

Money and Incumbents in CloseElections

The imbalances in campaign resources occur because:

• When incumbent candidates run, they have more money:62% of donations and 58% of expenditures

Money isn’t predictive of winning, although they usually win

• In open seats, the incumbent party’s candidates raise 50%of the money but spend 54%

Winners have more money:

winners raise 54% versus 41% (p=0.01)spend 56% versus 49% (p=0.02)

292 / 765

Recounts?

New data on close races• for random sample of half of all 130 elections in the

50% + /− 0.75 window• detailed information from news coverage and

administration records on each electionRecounts found in half of the elections (35 recounts)• Imbalance on previous Democratic victory is greater in

recounted elections: a treated-control difference of 0.56versus 0.12

• In the 35 recounted elections, the first election recountwinner was only changed three times

• In each of the three cases, however, the incumbentrepresentative was the ultimate winner

293 / 765

Recounts?

New data on close races• for random sample of half of all 130 elections in the

50% + /− 0.75 window• detailed information from news coverage and

administration records on each electionRecounts found in half of the elections (35 recounts)• Imbalance on previous Democratic victory is greater in

recounted elections: a treated-control difference of 0.56versus 0.12

• In the 35 recounted elections, the first election recountwinner was only changed three times

• In each of the three cases, however, the incumbentrepresentative was the ultimate winner

294 / 765

Recounts?

New data on close races• for random sample of half of all 130 elections in the

50% + /− 0.75 window• detailed information from news coverage and

administration records on each electionRecounts found in half of the elections (35 recounts)• Imbalance on previous Democratic victory is greater in

recounted elections: a treated-control difference of 0.56versus 0.12

• In the 35 recounted elections, the first election recountwinner was only changed three times

• In each of the three cases, however, the incumbentrepresentative was the ultimate winner

295 / 765

Benefits of RD

• Observable implications• Still making weaker assumptions than model based

estimators—e.g., Gelman-King

• Conjecture: RD will work better for elections with lessprofessionalization and where cut point is less known

• Examples of smoothness with RD:• UK: Eggers and Hainmueller (2009)• Brazil: Hidalgo (2010)• California assembly elections• Canadian Parliamentary elections

296 / 765

Benefits of RD

• Observable implications• Still making weaker assumptions than model based

estimators—e.g., Gelman-King

• Conjecture: RD will work better for elections with lessprofessionalization and where cut point is less known

• Examples of smoothness with RD:• UK: Eggers and Hainmueller (2009)• Brazil: Hidalgo (2010)• California assembly elections• Canadian Parliamentary elections

297 / 765

Gelman–King Estimator

E[Vt+1] = β0 + β1Pt + β2(Pt × Rt+1) + β3Vt ,

where Vt ∈ [0,1] is the Dem Share in election t

Pt ∈ −1,1 is the Winning Party in election t

and Rt+1 ∈ 0,1 is a dummy variable indicating whether theIncumbent Runs in election t + 1

298 / 765

Modified Gelman–King Estimator

E[Vt+1] = δ0 + δ1Pt + δ2(Pt × Rt+1) + δ3Mt

+ δ4M2t + δ5M3

t + δ6M4t

+ δ7(Mt × Pt ) + δ8(M2t × Pt ) + δ9(M3

t × Pt )

+ δ10(M4t × Pt )

Democratic Margin in election t , denoted Mt , is substituted inplace of Democratic Share (Vt )

299 / 765

Lee’s Model

Compare with Lee’s specification, trivially modified so thatPt ∈ −1,1 is substituted for Democratic Victory ∈ 0,1:

E[Vt+1] = γ0 + γ1Pt + γ2Mt + γ3M2t + γ4M3

t + γ5M4t

+ γ6(Mt × Pt ) + γ7(M2t × Pt ) + γ8(M3

t × Pt )

+ γ9(M4t × Pt )

300 / 765

Comments

• Models of campaign finance fail to accurately predict in thecase they all agree on: e.g., Baron (1989), Synder(1989,1990), Erikson and Palfrey (2000)

• Why should elections be random?e.g., close votes in legislatures

• Other designs may be better: redistricting (Sekhon andTitiunik 2012)

• Even when RD works, averaging over elections is delicate

301 / 765

Comments

• Models of campaign finance fail to accurately predict in thecase they all agree on: e.g., Baron (1989), Synder(1989,1990), Erikson and Palfrey (2000)

• Why should elections be random?e.g., close votes in legislatures

• Other designs may be better: redistricting (Sekhon andTitiunik 2012)

• Even when RD works, averaging over elections is delicate

302 / 765

Comments

• Models of campaign finance fail to accurately predict in thecase they all agree on: e.g., Baron (1989), Synder(1989,1990), Erikson and Palfrey (2000)

• Why should elections be random?e.g., close votes in legislatures

• Other designs may be better: redistricting (Sekhon andTitiunik 2012)

• Even when RD works, averaging over elections is delicate

303 / 765

Comments

• Models of campaign finance fail to accurately predict in thecase they all agree on: e.g., Baron (1989), Synder(1989,1990), Erikson and Palfrey (2000)

• Why should elections be random?e.g., close votes in legislatures

• Other designs may be better: redistricting (Sekhon andTitiunik 2012)

• Even when RD works, averaging over elections is delicate

304 / 765

Comments

• Similar results with open seats; but not much data• Imbalance is so bad there is no support near the cutpoint• Fuzzy RD will not work here: cannot observed random

component separately from the endogenous one

305 / 765

Fraud?

• Only few cases of fraud described in secondary literature(e.g., Campbell 2005) also in RD window

• Given the imbalances of resources a priori,fraud isn’t needed as an explanation

306 / 765

Other RD examples to read:

Make sure to read:

Eggers, Andy and Jens Hainmueller. 2009. “The Value ofPolitical Power: Estimating Returns to Office in Post-War BritishPolitics.” American Political Science Review 104(4):513–533.

307 / 765

Mahalanobis Distance

• The most common method of multivariate matching isbased on the Mahalanobis distance. The Mahalanobisdistance measure between any two column vectors isdefined as:

md(Xi ,Xj) =

(Xi − Xj)′S−1(Xi − Xj)

12

where Xi and Xj are two different observations and S isthe sample covariance matrix of X .

• Mahalanobis distance is an appropriate distance measureif each covariate has an elliptic distribution whose shape iscommon between treatment and control groups (Mitchelland Krzanowski, 1985; Mitchell and Krzanowski, 1989).

• In finite samples, Mahalanobis distance will not be optimal.

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Mahalanobis Distance

• The most common method of multivariate matching isbased on the Mahalanobis distance. The Mahalanobisdistance measure between any two column vectors isdefined as:

md(Xi ,Xj) =

(Xi − Xj)′S−1(Xi − Xj)

12

where Xi and Xj are two different observations and S isthe sample covariance matrix of X .

• Mahalanobis distance is an appropriate distance measureif each covariate has an elliptic distribution whose shape iscommon between treatment and control groups (Mitchelland Krzanowski, 1985; Mitchell and Krzanowski, 1989).

• In finite samples, Mahalanobis distance will not be optimal.

309 / 765

Affine Invariance

• When can matching confounders make bias worse? e.g.,what if the propensity score model is incorrect?

• An affinely invariant matching method is a matchingmethod which produces the same matches if thecovariates X are affinely transformed. An affinetransformation is any transformation that preservescollinearity (i.e., all points lying on a line initially still lie on aline after transformation) and ratios of distances (e.g., themidpoint of a line segment remains the midpoint aftertransformation).

310 / 765

Affine Invariance

• When can matching confounders make bias worse? e.g.,what if the propensity score model is incorrect?

• An affinely invariant matching method is a matchingmethod which produces the same matches if thecovariates X are affinely transformed. An affinetransformation is any transformation that preservescollinearity (i.e., all points lying on a line initially still lie on aline after transformation) and ratios of distances (e.g., themidpoint of a line segment remains the midpoint aftertransformation).

311 / 765

Properties of Matching Algorithms

• All affinely invariant matching methods have the EqualPercent Bias Reduction (EPBR) property under someconditions.

• If X are distributed with ellipsoidal distributions, then theEPBR property holds for affinely invariant matchingmethods (Rubin and Thomas, 1992).

• There is an extension to a restricted class of mixtures(Rubin and Stuart, 2006): discriminant mixtures ofproportional ellipsoidally symmetric distributions.

312 / 765

Properties of Matching Algorithms

• All affinely invariant matching methods have the EqualPercent Bias Reduction (EPBR) property under someconditions.

• If X are distributed with ellipsoidal distributions, then theEPBR property holds for affinely invariant matchingmethods (Rubin and Thomas, 1992).

• There is an extension to a restricted class of mixtures(Rubin and Stuart, 2006): discriminant mixtures ofproportional ellipsoidally symmetric distributions.

313 / 765

Properties of Matching Algorithms

• All affinely invariant matching methods have the EqualPercent Bias Reduction (EPBR) property under someconditions.

• If X are distributed with ellipsoidal distributions, then theEPBR property holds for affinely invariant matchingmethods (Rubin and Thomas, 1992).

• There is an extension to a restricted class of mixtures(Rubin and Stuart, 2006): discriminant mixtures ofproportional ellipsoidally symmetric distributions.

314 / 765

Equal Percent Bias Reduction (EPBR)

• Let Z be the expected value of X in the matched controlgroup. Then we say that a matching procedure is EPBR if

E(X |T = 1)− Z = γ E(X |T = 1)− E(X |T = 0)for a scalar 0 ≤ γ ≤ 1.

• We say that a matching method is EPBR for X becausethe percent reduction in the mean biases for each of thematching variables is the same.

• In general, if a matching method is not EPBR, then thebias for some linear function of X is increased.

• If the mapping between X and Y is not linear, EPBR isonly of limited value.

315 / 765

Equal Percent Bias Reduction (EPBR)

• Let Z be the expected value of X in the matched controlgroup. Then we say that a matching procedure is EPBR if

E(X |T = 1)− Z = γ E(X |T = 1)− E(X |T = 0)for a scalar 0 ≤ γ ≤ 1.

• We say that a matching method is EPBR for X becausethe percent reduction in the mean biases for each of thematching variables is the same.

• In general, if a matching method is not EPBR, then thebias for some linear function of X is increased.

• If the mapping between X and Y is not linear, EPBR isonly of limited value.

316 / 765

Equal Percent Bias Reduction (EPBR)

• Let Z be the expected value of X in the matched controlgroup. Then we say that a matching procedure is EPBR if

E(X |T = 1)− Z = γ E(X |T = 1)− E(X |T = 0)for a scalar 0 ≤ γ ≤ 1.

• We say that a matching method is EPBR for X becausethe percent reduction in the mean biases for each of thematching variables is the same.

• In general, if a matching method is not EPBR, then thebias for some linear function of X is increased.

• If the mapping between X and Y is not linear, EPBR isonly of limited value.

317 / 765

When is EPBR a Good Property?

• In general, if a matching method is not EPBR, than thebias for a particular linear function of X is increased.

• We may not want EPBR if we have some specificknowledge that one covariate is more important thananother. For example,

Y = α T + X 41 + 2X2,

where X > 1. In this case we should be generally moreconcerned with X1 than X2.

• In finite samples, Mahalanobis distance and propensityscore matching will not be optimal because X will not beellipsoidally distributed in a finite sample even if that is itstrue distribution.

318 / 765

Can an Observational Study Recoverthe Experimental Benchmark?

• LaLonde (1986) examined a randomized job trainingexperiment: National Supported Work DemonstrationProgram (NSW).

• We know the experimental benchmark from the NSW.• LaLonde replaced the experimental controls with

observational controls from the Current Population Survey• Could standard econometric techniques recover the

experimental estimate? All failed (e.g., regression and IV).

319 / 765

Can an Observational Study Recoverthe Experimental Benchmark?

• LaLonde (1986) examined a randomized job trainingexperiment: National Supported Work DemonstrationProgram (NSW).

• We know the experimental benchmark from the NSW.• LaLonde replaced the experimental controls with

observational controls from the Current Population Survey• Could standard econometric techniques recover the

experimental estimate? All failed (e.g., regression and IV).

320 / 765

Can an Observational Study Recoverthe Experimental Benchmark?

• LaLonde (1986) examined a randomized job trainingexperiment: National Supported Work DemonstrationProgram (NSW).

• We know the experimental benchmark from the NSW.• LaLonde replaced the experimental controls with

observational controls from the Current Population Survey• Could standard econometric techniques recover the

experimental estimate? All failed (e.g., regression and IV).

321 / 765

Can an Observational Study Recoverthe Experimental Benchmark?

• LaLonde (1986) examined a randomized job trainingexperiment: National Supported Work DemonstrationProgram (NSW).

• We know the experimental benchmark from the NSW.• LaLonde replaced the experimental controls with

observational controls from the Current Population Survey• Could standard econometric techniques recover the

experimental estimate? All failed (e.g., regression and IV).

322 / 765

The Controversy

• Dehejia & Wahba used a subset for which 2-yearspre-treatment earnings available, and claimpropensity-score matching succeeds (1997; 1999).

• Smith & Todd dispute Dehejia & Wahba’s claims (2001).• The original question remains unresolved. Both sides have

basically agreed to disagree: R. H. Dehejia and Wahba(2002), R. Dehejia (2005), J. Smith and P. Todd (2005a),and J. Smith and P. Todd (2005b).

323 / 765

The Controversy

• Dehejia & Wahba used a subset for which 2-yearspre-treatment earnings available, and claimpropensity-score matching succeeds (1997; 1999).

• Smith & Todd dispute Dehejia & Wahba’s claims (2001).• The original question remains unresolved. Both sides have

basically agreed to disagree: R. H. Dehejia and Wahba(2002), R. Dehejia (2005), J. Smith and P. Todd (2005a),and J. Smith and P. Todd (2005b).

324 / 765

The Controversy

• Dehejia & Wahba used a subset for which 2-yearspre-treatment earnings available, and claimpropensity-score matching succeeds (1997; 1999).

• Smith & Todd dispute Dehejia & Wahba’s claims (2001).• The original question remains unresolved. Both sides have

basically agreed to disagree: R. H. Dehejia and Wahba(2002), R. Dehejia (2005), J. Smith and P. Todd (2005a),and J. Smith and P. Todd (2005b).

325 / 765

Genetic Matching (GenMatch)

Genetic matching is a new general method for performingmultivariate matching. GenMatch:• maximizes the balance of observed potential confounders

across matched treated and control units• uses an evolutionary search algorithm to determine the

weight each covariate is given• a genetic algorithm is used (Mebane and Sekhon, 2011)

326 / 765

Genetic Matching (GenMatch)

Genetic matching is a new general method for performingmultivariate matching. GenMatch:• maximizes the balance of observed potential confounders

across matched treated and control units• uses an evolutionary search algorithm to determine the

weight each covariate is given• a genetic algorithm is used (Mebane and Sekhon, 2011)

327 / 765

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328 / 765

GenMatch

• Debate arises because existing matching methods fail toobtain reliable levels of balance in this dataset.

• GenMatch is able to reliably estimate the causal effectswhen other methods fail because it achieves substantiallybetter balance.

• GenMatch in an hour produces vastly better balance thanhuman researchers working away for ten years.

329 / 765

GenMatch

• No information about any outcome is used• The method is nonparametric and does not depend on a

propensity score (pscore can be included)• Genetic matching can reduce the bias in X covariates in

cases where conventional methods of matching increasebias. Who general is this?

• If selection on observables holds, but EPBR does not, haslower bias and MSE in the estimand in the usualsimulations (will generally depend on loss function)

330 / 765

More General Method of MeasuringDistance

• A more general way to measure distance is defined by:

d(Xi ,Xj) =

(Xi − Xj)′ (S−1/2)′WS−1/2(Xi − Xj)

12

where W is a k × k positive definite weight matrix andS1/2 is the Cholesky decomposition of S which is thevariance-covariance matrix of X .

• All elements of W are zero except down the main diagonal.The main diagonal consists of k parameters which must bechosen.

• This leaves the problem of choosing the free elements ofW . For identification, there are only k − 1 free parameters.

331 / 765

More General Method of MeasuringDistance

• A more general way to measure distance is defined by:

d(Xi ,Xj) =

(Xi − Xj)′ (S−1/2)′WS−1/2(Xi − Xj)

12

where W is a k × k positive definite weight matrix andS1/2 is the Cholesky decomposition of S which is thevariance-covariance matrix of X .

• All elements of W are zero except down the main diagonal.The main diagonal consists of k parameters which must bechosen.

• This leaves the problem of choosing the free elements ofW . For identification, there are only k − 1 free parameters.

332 / 765

Parameterization

• GenMatch uses the propensity score if it is known or if itcan be estimated.

• The propensity score is estimated and its linear predictor,µ, is matched upon along with the covariates X once theyhave been adjusted so as to be uncorrelated with the linearpredictor.

• Combining is good because:• Propensity score matching is good at minimizing the

discrepancy along the propensity score• Mahalanobis distance is good at minimizing the distance

between individual coordinates of X (orthogonal to thepropensity score) (P. R. Rosenbaum and Rubin, 1985).

333 / 765

Parameterization

• GenMatch uses the propensity score if it is known or if itcan be estimated.

• The propensity score is estimated and its linear predictor,µ, is matched upon along with the covariates X once theyhave been adjusted so as to be uncorrelated with the linearpredictor.

• Combining is good because:• Propensity score matching is good at minimizing the

discrepancy along the propensity score• Mahalanobis distance is good at minimizing the distance

between individual coordinates of X (orthogonal to thepropensity score) (P. R. Rosenbaum and Rubin, 1985).

334 / 765

Parameterization

• GenMatch uses the propensity score if it is known or if itcan be estimated.

• The propensity score is estimated and its linear predictor,µ, is matched upon along with the covariates X once theyhave been adjusted so as to be uncorrelated with the linearpredictor.

• Combining is good because:• Propensity score matching is good at minimizing the

discrepancy along the propensity score• Mahalanobis distance is good at minimizing the distance

between individual coordinates of X (orthogonal to thepropensity score) (P. R. Rosenbaum and Rubin, 1985).

335 / 765

Optimization

• Many loss functions are possible. Such as:• minimize the largest discrepancy• minimize the mean or median discrepancy• minimize some other quantile• restrict the above to only uniformly improving moves

• Our recommended algorithm attempts to minimize thelargest discrepancy at every step (minimizing the infinitynorm).

• For a given set of matches resulting from a given W , theloss is defined as the minimum p-value observed across aseries of balance tests.

336 / 765

Optimization

• Many loss functions are possible. Such as:• minimize the largest discrepancy• minimize the mean or median discrepancy• minimize some other quantile• restrict the above to only uniformly improving moves

• Our recommended algorithm attempts to minimize thelargest discrepancy at every step (minimizing the infinitynorm).

• For a given set of matches resulting from a given W , theloss is defined as the minimum p-value observed across aseries of balance tests.

337 / 765

Optimization

• Many loss functions are possible. Such as:• minimize the largest discrepancy• minimize the mean or median discrepancy• minimize some other quantile• restrict the above to only uniformly improving moves

• Our recommended algorithm attempts to minimize thelargest discrepancy at every step (minimizing the infinitynorm).

• For a given set of matches resulting from a given W , theloss is defined as the minimum p-value observed across aseries of balance tests.

338 / 765

Optimization

• Many loss functions are possible. Such as:• minimize the largest discrepancy• minimize the mean or median discrepancy• minimize some other quantile• restrict the above to only uniformly improving moves

• Our recommended algorithm attempts to minimize thelargest discrepancy at every step (minimizing the infinitynorm).

• For a given set of matches resulting from a given W , theloss is defined as the minimum p-value observed across aseries of balance tests.

339 / 765

Optimization

• Many loss functions are possible. Such as:• minimize the largest discrepancy• minimize the mean or median discrepancy• minimize some other quantile• restrict the above to only uniformly improving moves

• Our recommended algorithm attempts to minimize thelargest discrepancy at every step (minimizing the infinitynorm).

• For a given set of matches resulting from a given W , theloss is defined as the minimum p-value observed across aseries of balance tests.

340 / 765

Optimization

• Many loss functions are possible. Such as:• minimize the largest discrepancy• minimize the mean or median discrepancy• minimize some other quantile• restrict the above to only uniformly improving moves

• Our recommended algorithm attempts to minimize thelargest discrepancy at every step (minimizing the infinitynorm).

• For a given set of matches resulting from a given W , theloss is defined as the minimum p-value observed across aseries of balance tests.

341 / 765

Measuring Balance

• There are many different ways of testing for balance andwe cannot summarize the vast literature here.

• The choice of balance will be domain specific. E.g., userandomization inference if you can as Bowers and Hansen(2006) do.

• But the tests should be powerful and there should be manyoff them because different tests are sensitive to differentdepartures from balance.

• It is important the maximum discrepancy be small.p-values conventionally understood to signal balance (e.g.,0.10) are often too low to produce reliable estimates.

342 / 765

Measuring Balance

• There are many different ways of testing for balance andwe cannot summarize the vast literature here.

• The choice of balance will be domain specific. E.g., userandomization inference if you can as Bowers and Hansen(2006) do.

• But the tests should be powerful and there should be manyoff them because different tests are sensitive to differentdepartures from balance.

• It is important the maximum discrepancy be small.p-values conventionally understood to signal balance (e.g.,0.10) are often too low to produce reliable estimates.

343 / 765

Measuring Balance

• There are many different ways of testing for balance andwe cannot summarize the vast literature here.

• The choice of balance will be domain specific. E.g., userandomization inference if you can as Bowers and Hansen(2006) do.

• But the tests should be powerful and there should be manyoff them because different tests are sensitive to differentdepartures from balance.

• It is important the maximum discrepancy be small.p-values conventionally understood to signal balance (e.g.,0.10) are often too low to produce reliable estimates.

344 / 765

Measuring Balance

• There are many different ways of testing for balance andwe cannot summarize the vast literature here.

• The choice of balance will be domain specific. E.g., userandomization inference if you can as Bowers and Hansen(2006) do.

• But the tests should be powerful and there should be manyoff them because different tests are sensitive to differentdepartures from balance.

• It is important the maximum discrepancy be small.p-values conventionally understood to signal balance (e.g.,0.10) are often too low to produce reliable estimates.

345 / 765

Balance Tests

• the p-values from these balance tests cannot beinterpreted as true probabilities because of standardpre-test problems, but they remain useful measures ofbalance

• By default, tests are conducted for all univariate baselinecovariates, as well as their first-order interactions andquadratic terms.

• The analyst may add tests of any function of X desired,including additional nonlinear functions and higher orderinteractions.

• The tests conducted are t-tests for the difference of meansand nonparametric bootstrap Kolmogorov-Smirnovdistributional tests.

346 / 765

Balance Tests

• the p-values from these balance tests cannot beinterpreted as true probabilities because of standardpre-test problems, but they remain useful measures ofbalance

• By default, tests are conducted for all univariate baselinecovariates, as well as their first-order interactions andquadratic terms.

• The analyst may add tests of any function of X desired,including additional nonlinear functions and higher orderinteractions.

• The tests conducted are t-tests for the difference of meansand nonparametric bootstrap Kolmogorov-Smirnovdistributional tests.

347 / 765

Balance Tests

• the p-values from these balance tests cannot beinterpreted as true probabilities because of standardpre-test problems, but they remain useful measures ofbalance

• By default, tests are conducted for all univariate baselinecovariates, as well as their first-order interactions andquadratic terms.

• The analyst may add tests of any function of X desired,including additional nonlinear functions and higher orderinteractions.

• The tests conducted are t-tests for the difference of meansand nonparametric bootstrap Kolmogorov-Smirnovdistributional tests.

348 / 765

Balance Tests

• the p-values from these balance tests cannot beinterpreted as true probabilities because of standardpre-test problems, but they remain useful measures ofbalance

• By default, tests are conducted for all univariate baselinecovariates, as well as their first-order interactions andquadratic terms.

• The analyst may add tests of any function of X desired,including additional nonlinear functions and higher orderinteractions.

• The tests conducted are t-tests for the difference of meansand nonparametric bootstrap Kolmogorov-Smirnovdistributional tests.

349 / 765

Genetic Optimization

• The optimization problem described above is difficult andirregular, and we utilize an evolutionary algorithm calledGENOUD (Mebane and Sekhon, 2011)

• Random search also works better than the usual matchingmethods, but is less efficient than GENOUD.

350 / 765

Genetic Optimization

• The optimization problem described above is difficult andirregular, and we utilize an evolutionary algorithm calledGENOUD (Mebane and Sekhon, 2011)

• Random search also works better than the usual matchingmethods, but is less efficient than GENOUD.

351 / 765

Monte Carlo Experiments

• Two Monte Carlos are presented.• MC 1: the experimental conditions satisfy assumptions for

EPBR1 X covariates are distributed multivariate normal2 propensity score is correctly specified3 linear mapping from X to Y

• MC 2: the assumptions required for EPBR are notsatisfied:

1 X covariates are discrete and others are skewed and havepoint masses: they have the same distributions as thecovariates of the LaLonde (1986) data.

2 propensity score is incorrectly specified3 mapping from X to Y is nonlinear

352 / 765

Monte Carlo Experiments

• Two Monte Carlos are presented.• MC 1: the experimental conditions satisfy assumptions for

EPBR1 X covariates are distributed multivariate normal2 propensity score is correctly specified3 linear mapping from X to Y

• MC 2: the assumptions required for EPBR are notsatisfied:

1 X covariates are discrete and others are skewed and havepoint masses: they have the same distributions as thecovariates of the LaLonde (1986) data.

2 propensity score is incorrectly specified3 mapping from X to Y is nonlinear

353 / 765

Experimental Condition 1: MultivariateNormal Distribution of Covariates

Estimator Bias RMSEBias

Bias GMMSE

MSE GM

Raw −604 .686 24.6 27.4Mahalanobis (MH) −8.63 .173 3.50 1.75Pscore −2.45 .210 .993 2.57Pscore + MH −5.96 .160 2.41 1.49GenMatch −2.47 .130

354 / 765

Experimental Condition 2: Distributionof Lalonde Covariates

Estimator Bias RMSEBias

Bias GMMSE

MSE GM

Raw 485 1611 19.0 18.2Mahalanobis (MH) −717 959 28.0 6.45Pscore 512 1294 20.0 11.7Pscore + MH 428 743 16.8 3.87GenMatch 25.6 378Random Search 112 493 4.38 1.30

355 / 765

Experimental Condition 2: Distributionof Lalonde Covariates

Estimator Bias RMSEBias

Bias GMMSE

MSE GM

Raw 485 1611 19.0 18.2Mahalanobis (MH) −717 959 28.0 6.45Pscore 512 1294 20.0 11.7Pscore + MH 428 743 16.8 3.87GenMatch 25.6 378Random Search 112 493 4.38 1.30

356 / 765

Summary of Monte Carlos

• Genetic matching reliably reduces both the bias and theMSE of the estimated causal effect even whenconventional methods of matching increase bias.

• When the assumptions of the usual methods are satisfied,GenMatch still has lower MSE.

• Caution 1: selection on observables holds in both MonteCarlos

• Caution 2: loss function seemed to work. What about adifferent one?

357 / 765

Summary of Monte Carlos

• Genetic matching reliably reduces both the bias and theMSE of the estimated causal effect even whenconventional methods of matching increase bias.

• When the assumptions of the usual methods are satisfied,GenMatch still has lower MSE.

• Caution 1: selection on observables holds in both MonteCarlos

• Caution 2: loss function seemed to work. What about adifferent one?

358 / 765

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359 / 765

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360 / 765

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361 / 765

0.00 0.05 0.10 0.15 0.20 0.25

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$)

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362 / 765

Mahalanobis Distance

• Distributions of Xs are not ellipsoidal, so there’s no reasonto think MD will work• Dehejia Wahba sample: balance is poor, but estimates are

right on target!• LaLonde sample: balance exceeds conventional standards,

but estimates are off the mark

• Lesson: Don’t count on getting lucky. Instead, achieve bestpossible balance, and look at outcomes only once, afterbalance is attained.

363 / 765

Mahalanobis Distance

• Distributions of Xs are not ellipsoidal, so there’s no reasonto think MD will work• Dehejia Wahba sample: balance is poor, but estimates are

right on target!• LaLonde sample: balance exceeds conventional standards,

but estimates are off the mark

• Lesson: Don’t count on getting lucky. Instead, achieve bestpossible balance, and look at outcomes only once, afterbalance is attained.

364 / 765

Mahalanobis Distance

• Distributions of Xs are not ellipsoidal, so there’s no reasonto think MD will work• Dehejia Wahba sample: balance is poor, but estimates are

right on target!• LaLonde sample: balance exceeds conventional standards,

but estimates are off the mark

• Lesson: Don’t count on getting lucky. Instead, achieve bestpossible balance, and look at outcomes only once, afterbalance is attained.

365 / 765

GenMatch Summary

• GenMatch in an hour produces vastly better balance thanhuman researchers working away for ten years.

• Machine learning can come to the rescue• But without the RCT, we would have a debate of how much

to adjust• Caution is warranted• No amount of statistical modeling or algorithmic wizardry

can resolve these questions

366 / 765

GenMatch Summary

• GenMatch in an hour produces vastly better balance thanhuman researchers working away for ten years.

• Machine learning can come to the rescue• But without the RCT, we would have a debate of how much

to adjust• Caution is warranted• No amount of statistical modeling or algorithmic wizardry

can resolve these questions

367 / 765

Opiates for the Matches

• Rubin: design trumps analysisbut design cannot be mass produced

• Selection on observables assumed too readily:• no design• difficult questions about inference ignored• balance checks are rudimentary• no placebo tests

• ties to Neyman-Rubin model weakening• how well does it work in practice?• statistical theory is of little guidance

368 / 765

Opiates for the Matches

• Rubin: design trumps analysisbut design cannot be mass produced

• Selection on observables assumed too readily:• no design• difficult questions about inference ignored• balance checks are rudimentary• no placebo tests

• ties to Neyman-Rubin model weakening• how well does it work in practice?• statistical theory is of little guidance

369 / 765

Opiates for the Matches

• Rubin: design trumps analysisbut design cannot be mass produced

• Selection on observables assumed too readily:• no design• difficult questions about inference ignored• balance checks are rudimentary• no placebo tests

• ties to Neyman-Rubin model weakening• how well does it work in practice?• statistical theory is of little guidance

370 / 765

Public Health and Medical Literature

• Use of matching has a longer history• Many more experiments so a hope of calibration• The interventions are actually done: another source of

calibration• But resistance to comparing experiments with

observational studies:• external validity• heterogeneous causal effect: casemix• private data (e.g., Hormone Replacement Therapy)

371 / 765

Public Health and Medical Literature

• Use of matching has a longer history• Many more experiments so a hope of calibration• The interventions are actually done: another source of

calibration• But resistance to comparing experiments with

observational studies:• external validity• heterogeneous causal effect: casemix• private data (e.g., Hormone Replacement Therapy)

372 / 765

Non-Randomized Studies (NRS)

• Randomized Controlled Trials (RCTs) are the used toevaluate clinical interventions.

• But RCTs are often unavailable and the only evidence isfrom NRS

• For cost effectiveness some prefer observational studies• Increasingly common to use propensity score matching,

but:• selection on observables assumption may be false• parametric adjustment may be inadequate

373 / 765

Non-Randomized Studies (NRS)

• Randomized Controlled Trials (RCTs) are the used toevaluate clinical interventions.

• But RCTs are often unavailable and the only evidence isfrom NRS

• For cost effectiveness some prefer observational studies• Increasingly common to use propensity score matching,

but:• selection on observables assumption may be false• parametric adjustment may be inadequate

374 / 765

Summary

• Example: evaluation of Pulmonary Artery Catheterization(PAC) versus no-PAC. NRS found that PAC significantlyincreases mortality while the RCT found no difference inmortality. Sekhon and Grieve 2008

• Genetic Matching gives similar results to the RCT butusing the NRS

• People failed to check balance after spending million ofdollars collecting the data

• They didn’t even plot the data

375 / 765

Pulmonary Artery Catheterization(PAC)

• PAC is an invasive cardiac monitoring device for critical illpatients (ICU)—e.g., myocardial infarction (ischaemicheart disease)

• Widely used for the past 30 years: spend $2 billion in U.S.per year

• RCT find no effect; seven NRS find that PAC increasesmortality (e.g., Connors et al. JAMA 1996)

• RCT: causal effect really is zero• Why do seven NRS disagree with RCTs?

376 / 765

Pulmonary Artery Catheterization(PAC)

• PAC is an invasive cardiac monitoring device for critical illpatients (ICU)—e.g., myocardial infarction (ischaemicheart disease)

• Widely used for the past 30 years: spend $2 billion in U.S.per year

• RCT find no effect; seven NRS find that PAC increasesmortality (e.g., Connors et al. JAMA 1996)

• RCT: causal effect really is zero• Why do seven NRS disagree with RCTs?

377 / 765

PAC

• NRS all ICU admissions to 57 UK ICUs 2003-4• 1052 cases with PAC; 32,499 controls (from a database of

N≈2M)• Propensity score model: Age, reasons for admission,

organ failure, probability of death, LOS, teaching hospital,size of unit

• logistic regression, predict p (PAC)• match on propensity score• GenMatch: p score and the same underlying covariates

multivariate matching algorithm, to max balance

378 / 765

PAC

• NRS all ICU admissions to 57 UK ICUs 2003-4• 1052 cases with PAC; 32,499 controls (from a database of

N≈2M)• Propensity score model: Age, reasons for admission,

organ failure, probability of death, LOS, teaching hospital,size of unit

• logistic regression, predict p (PAC)• match on propensity score• GenMatch: p score and the same underlying covariates

multivariate matching algorithm, to max balance

379 / 765

PAC

• NRS all ICU admissions to 57 UK ICUs 2003-4• 1052 cases with PAC; 32,499 controls (from a database of

N≈2M)• Propensity score model: Age, reasons for admission,

organ failure, probability of death, LOS, teaching hospital,size of unit

• logistic regression, predict p (PAC)• match on propensity score• GenMatch: p score and the same underlying covariates

multivariate matching algorithm, to max balance

380 / 765

PAC

• NRS all ICU admissions to 57 UK ICUs 2003-4• 1052 cases with PAC; 32,499 controls (from a database of

N≈2M)• Propensity score model: Age, reasons for admission,

organ failure, probability of death, LOS, teaching hospital,size of unit

• logistic regression, predict p (PAC)• match on propensity score• GenMatch: p score and the same underlying covariates

multivariate matching algorithm, to max balance

381 / 765

PAC

• NRS all ICU admissions to 57 UK ICUs 2003-4• 1052 cases with PAC; 32,499 controls (from a database of

N≈2M)• Propensity score model: Age, reasons for admission,

organ failure, probability of death, LOS, teaching hospital,size of unit

• logistic regression, predict p (PAC)• match on propensity score• GenMatch: p score and the same underlying covariates

multivariate matching algorithm, to max balance

382 / 765

Balance of baseline probability ofDeath, PSCORE

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.2

0.4

0.6

0.8

1.0

no−PAC

PA

C

383 / 765

Balance of baseline probability ofDeath, GenMatch

0.0 0.2 0.4 0.6 0.8 1.0

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PA

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384 / 765

Balance Measures

Variable No PAC PAC T-test KS P-value

IMProbUnmatched 30.8 56.2 0.00 0.00Pscore 54.8 0.25 0.06GenMatch 56.1 0.80 0.87

% admitted to hospital in LondonUnmatched 9.2 8.7 0.55Pscore 10.0 0.26GenMatch 8.7 1.00

385 / 765

Balance Measures

Variable No PAC PAC KS P-value

AgeUnmatched 30.8 56.2 0.00Pscore 54.8 0.06GenMatch 56.1 0.87

% admitted for emergency surgeryUnmatched 20.2 23.1 0.03Pscore 21.3 0.32GenMatch 23.7 0.66

% admitted to teaching hospitalUnmatched 37.7 42.6 0.00Pscore 44.5 0.38GenMatch 42.6 1.00

386 / 765

Comparison of outcomes acrossmethods

PAC vs No PAC

Method Odds Ratio 95% CI

RCT 1.13 (0.87 to 1.47)GenMatch 1.10 (0.93 to 1.31)Pscore 1.22 (1.03 to 1.45)Unmatched 3.51 (3.09 to 3.97)

387 / 765

Comparison of outcomes acrossmethods

PAC vs No PAC

Method Odds Ratio 95% CI

RCT 1.13 (0.87 to 1.47)GenMatch 1.10 (0.93 to 1.31)Pscore 1.22 (1.03 to 1.45)Unmatched 3.51 (3.09 to 3.97)

388 / 765

Monte Carlo Setup

• Use data from PAC-Man RCT (Harvey et al. 2005)• Select treatment via a “true” propensity score• mapping from X to Y is nonlinear• Costs and QALYs generated assuming gamma and zero

inflated Poisson distributions• Net benefits valued at £30,000 pounds per QALY• True INB=0• 1000 MCs

• For matching assume the propensity score is incorrect, butuse the one used by Connors 1997.

389 / 765

Propensity Score Setup

Pscore based on previous study (Connors 1997):

pscore = α + α1probmort + α2misspmort + α3emersurg +

α4age + α5elecsurg + α6unit + α7rate + α8history +

α9age× probmort + α10age× emersurg +

α11age× elecsurg

Using the same information what is relative performance ofdifferent matching methods? bias, RMSE compared to true INB

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MC ResultsIncremental Net Benefit

Method Bias RMSEBias

Bias GMRMSE

RMSE GM

GenMatch −147 565Pscore 936 3039 6.4 5.4

True value: 0 pounds

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Conclusion

• NICE (National Institute of Clinical Medicine) is coming tothe U.S. in some form

• Cost effectiveness uses NRS more than FDA typeregulators

• The biases are profound, even when selection onobservables holds

• Variance estimates profoundly underestimate the trueuncertainty

• Most extant objections are moral: I have a scientificconcern

• What does this say about our field?

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Conclusion

• NICE (National Institute of Clinical Medicine) is coming tothe U.S. in some form

• Cost effectiveness uses NRS more than FDA typeregulators

• The biases are profound, even when selection onobservables holds

• Variance estimates profoundly underestimate the trueuncertainty

• Most extant objections are moral: I have a scientificconcern

• What does this say about our field?

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Conclusion

• NICE (National Institute of Clinical Medicine) is coming tothe U.S. in some form

• Cost effectiveness uses NRS more than FDA typeregulators

• The biases are profound, even when selection onobservables holds

• Variance estimates profoundly underestimate the trueuncertainty

• Most extant objections are moral: I have a scientificconcern

• What does this say about our field?

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Conclusion

• Machine learning can help solve the easy problem ofbalance

• GenMatch achieves excellent covariate balance• It gives similar results to the RCT• But without the RCT, we would have a debate of how much

to adjust• Caution is warranted• Talking NICE into releasing observational data before

parallel RCT results

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Conclusion

• Machine learning can help solve the easy problem ofbalance

• GenMatch achieves excellent covariate balance• It gives similar results to the RCT• But without the RCT, we would have a debate of how much

to adjust• Caution is warranted• Talking NICE into releasing observational data before

parallel RCT results

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The Problem

• When overcoming identification problems, we often turn torandomized controlled trials (RCTs)

• Often, though, RCTs are not a feasible, so we turn toobservational studies, or non-random studies (NRSs)

• The problem, then, is how to combine this information inorder to provide evidence for treatment effects in the fullpopulation of interest.• RCTs raise issues of Randomization Bias (Heckman and

J. A. Smith, 1995): poor external validity• NRSs raise issues of Selection Bias, or non random

assignment to treatment: poor internal validity

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The Opportunity

• Explosion of data sources: administrative, electronicmedical records (EMR), online behavior

• Population data is becoming more common and moreprecise

• How can it be used?

• Policy makers/firms:“let’s just use the big data to make causal inference”

• Tension between identification vs. machinelearning/prediction

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The Problem

• Population effects cannot be estimated withoutassumptions:• Target population difficult to describe• Target population usually dynamic• General equilibrium effects as treatment is widely deployed

(SUTVA violations)

• Our results can be used for the easier case of comparingtwo different experiments:

1 Experiment 1: A vs. B and2 Experiment 2: A vs. C3 We wish to compare: B vs. C

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Example

• A growing interest in the cost of health care• Interest in system-wide issues—e.g., Dartmouth Atlas of

Health Care• Comparing the cost of UCLA versus Mayo Clinic is a

causal question• Comparing the cost and health outcomes in the USA

versus UK is also a causal question, but a less clear one• We examine a simple case: conduct a cost effectiveness

analysis (CEA) for one medical procedure

400 / 765

Example

• A growing interest in the cost of health care• Interest in system-wide issues—e.g., Dartmouth Atlas of

Health Care• Comparing the cost of UCLA versus Mayo Clinic is a

causal question• Comparing the cost and health outcomes in the USA

versus UK is also a causal question, but a less clear one• We examine a simple case: conduct a cost effectiveness

analysis (CEA) for one medical procedure

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Estimands

• RCTs allow us to identify the Sample Average TreatmentEffect (SATE), which is asymptotically equivalent to theSample Average Treatment Effect on the Treated (SATT)

• We are often interested in the treatment effect for thosewho would receive treatment in practice, or the PopulationAverage Treatment Effect on the Treated (PATT).

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Our Method

We:• develop a theoretical decomposition of the bias of going

from SATT to PATT• introduce a new method to combine RCTs and NRSs

• Using Genetic Matching to maximize the internal validity• SATE→ SATT

• Using Maximum Entropy Weighting to maximize theexternal validity• SATT→ PATT

• most importantly, provide placebo tests to validate theidentifying assumptions.

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Pulmonary Artery Catheterization(PAC)

• PAC is an invasive cardiac monitoring device for critical illpatients (ICU)—e.g., myocardial infarction (ischaemicheart disease)

• Widely used for the past 30 years: spend $2 billion in U.S.per year

• RCT find no effect; seven NRS find that PAC increasesmortality (e.g., Connors et al. JAMA 1996)

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Pulmonary Artery Catheterization

• RCT: a publicly funded, pragmatic experiment done in 65UK ICUs in 2000-2004.• 1014 subjects, 506 who received PAC• No difference in hospital mortality (p = 0.39)

• NRS: all ICU admissions to 57 UK ICUs in 2003-2004• 1052 cases with PAC and 32,499 controls• One observational study was able to find no difference in

hospital mortality (p = 0.29)

• However, the populations between the two studies differ,and we are interested in identifying PATT.

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Schematic of SATE to PATT

In the next figure: double arrows indicate exchangeability ofpotential outcomes and dashed arrows indicate adjustment ofthe covariate distribution

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We can think of the bias in our estimate of the PATT as acombination of bias due to a possible lack of internal validity ofthe estimate of SATT and bias due to poor external validity ofthe estimate from the RCT.

B = BI + BE

We can decompose of these biases in more detail as anextension of the Heckman, Ichimura, et al. (1998)decomposition of bias.

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Some Definitions

• Let X denote a set of conditioning covariates in the samplepopulation and W denote a set of conditioning covariatesin the target population

• Let Y1 and Y0 denote the potential outcomes for a subject i• Let T ∈ (0,1) be an indicator for whether or not subject i

was in the treatment or control group• Let I ∈ (0,1) be an indicator for whether or not subject i

was in the sample population

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BI = ES11\SI1E(Y0|X ,T = 1, I = 1)

− ES10\SI1E(Y0|X ,T = 0, I = 1)

+ EdF (T =1)−dF (T =0),SI1E(Y0|X ,T = 0, I = 1)

+ EdF (T =1),SI1E(Y0|X ,T = 1, I = 1)− E(Y0|X ,T = 0, I = 1)

The internal validity problem is due to bias in the RCT estimateof SATT.

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Example: Propensity Score Distributions

Propensity Score P(X)

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P(X

)

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BI = ES11\SI1E(Y0|X ,T = 1, I = 1)

−ES10\SI1E(Y0|X ,T = 0, I = 1)

+EdF (T =1)−dF (T =0),SI1E(Y0|X ,T = 0, I = 1)

+EdF (T =1),SI1E(Y0|X ,T = 1, I = 1)− E(Y0|X ,T = 0, I = 1)

This is bias due to treated units who are not in the overlapregion of the sample treated and sample controls.

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Example: Propensity Score Distributions

Propensity Score P(X)

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050

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trol P

(X)

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P(X

)

413 / 765

BI = ES11\SI1E(Y0|X ,T = 1, I = 1)

−ES10\SI1E(Y0|X ,T = 0, I = 1)

+EdF (T =1)−dF (T =0),SI1E(Y0|X ,T = 0, I = 1)

+EdF (T =1),SI1E(Y0|X ,T = 1, I = 1)− E(Y0|X ,T = 0, I = 1)

This is bias due to controls units who are not in the overlapregion of the sample treated and sample controls.

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Example: Propensity Score Distributions

Propensity Score P(X)

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050

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trol P

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P(X

)

415 / 765

BI = ES11\SI1E(Y0|X ,T = 1, I = 1)

−ES10\SI1E(Y0|X ,T = 0, I = 1)

+EdF (T =1)−dF (T =0),SI1E(Y0|X ,T = 0, I = 1)

+EdF (T =1),SI1E(Y0|X ,T = 1, I = 1)− E(Y0|X ,T = 0, I = 1)

This is bias due to imbalance of the sample treated andsample control units in the overlap region.

416 / 765

Example: Propensity Score Distributions

Propensity Score P(X)

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050

100

150

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050

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-50

050

100

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trol P

(X)

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P(X

)

417 / 765

Example: Propensity Score Distributions

Propensity Score P(X)

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050

100

150

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050

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trol P

(X)

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P(X

)

418 / 765

BI = ES11\SI1E(Y0|X ,T = 1, I = 1)

−ES10\SI1E(Y0|X ,T = 0, I = 1)

+EdF (T =1)−dF (T =0),SI1E(Y0|X ,T = 0, I = 1)

+EdF (T =1),SI1E(Y0|X ,T = 1, I = 1)− E(Y0|X ,T = 0, I = 1)

This is the usual definition of bias, or the selection issue.This should be minimized because we have an RCT, so thereshouldn’t be confounding of treatment.

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BE = ES11\ST1E(Yi |W ,T = 1, I = 1)

− ES01\ST1E(Yi |W ,T = 1, I = 0)

+ EdF (I=1)−dF (I=0),ST1E(Yi |W ,T = 1, I = 0)

+ EdF (I=1),ST1E(Yi |W ,T = 1, I = 1)− E(Yi |W ,T = 1, I = 0)

for i ∈ (0,1)

External validity problem:• RCT is not a random sample of the population.• Treatment can mean something different between the RCT

and NRS.

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Example: Propensity Score Distributions

Propensity Score P(X)

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050

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050

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ulat

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ted

P(W

)S

ampl

e Tr

eate

d P

(W)

421 / 765

BE = ES11\ST1E(Yi |W ,T = 1, I = 1)

−ES01\ST1E(Yi |W ,T = 1, I = 0)

+EdF (I=1)−dF (I=0),ST 1E(Yi |W ,T = 1, I = 0)

+EdF (I=1),ST1E(Yi |W ,T = 1, I = 1)− E(Yi |W ,T = 1, I = 0)

for i ∈ (0,1)

This is bias due to sample treated units who are not in theoverlap region of the sample treated and population treated.

422 / 765

Example: Propensity Score Distributions

Propensity Score P(X)

-150

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050

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150

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050

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050

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ulat

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ted

P(W

)S

ampl

e Tr

eate

d P

(W)

423 / 765

BE = ES11\ST1E(Yi |W ,T = 1, I = 1)

−ES01\ST1E(Yi |W ,T = 1, I = 0)

+EdF (I=1)−dF (I=0),ST 1E(Yi |W ,T = 1, I = 0)

+EdF (I=1),ST1E(Yi |W ,T = 1, I = 1)− E(Yi |W ,T = 1, I = 0)

for i ∈ (0,1)

This is bias due to population treated units who are not inthe overlap region of the sample treated and population treated.

424 / 765

Example: Propensity Score Distributions

Propensity Score P(X)

-150

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050

100

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050

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050

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ulat

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Trea

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P(W

)S

ampl

e Tr

eate

d P

(W)

425 / 765

BE = ES11\ST1E(Yi |W ,T = 1, I = 1)

−ES01\ST1E(Yi |W ,T = 1, I = 0)

+EdF (I=1)−dF (I=0),ST 1E(Yi |W ,T = 1, I = 0)

+EdF (I=1),ST1E(Yi |W ,T = 1, I = 1)− E(Yi |W ,T = 1, I = 0)

for i ∈ (0,1)

This is bias due to imbalance of the sample treated andpopulation treated units in the overlap region.

426 / 765

Example: Propensity Score Distributions

Propensity Score P(X)

-150

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050

100

150

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050

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P(W

)S

ampl

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d P

(W)

427 / 765

Example: Propensity Score Distributions

Propensity Score P(X)

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050

100

150

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050

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P(W

)S

ampl

e Tr

eate

d P

(W)

428 / 765

BE = ES11\ST1E(Yi |W ,T = 1, I = 1)

−ES01\ST1E(Yi |W ,T = 1, I = 0)

+EdF (I=1)−dF (I=0),ST 1E(Yi |W ,T = 1, I = 0)

+EdF (I=1),ST1E(Yi |W ,T = 1, I = 1)− E(Yi |W ,T = 1, I = 0)

for i ∈ (0,1)

This is the usual definition of bias, or the sample selectionbias. There is nothing we can do to fix this, and the bias can beof opposite sign and any magnitude.

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The Method

Let θs be a stratified treatment effect from the randomized trial,θws be the treatment effect of reweighted strata, and θ be thetrue population treatment effect.

θs → θws → θ

• Estimate θs using Genetic Matching• Reweight the strata using Maximum Entropy weighting to

estimate θws

• Run a placebo test to validate the identifying assumptionsand provide evidence for how close θws is to θ.

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Some Definitions

• Let W denote a set of conditioning covariates, with thedistribution of the population treated observation

• Let Ys,t denote the potential outcomes for a given subjectin sample s and treatment t

• Let T ∈ (0,1) be an indicator for whether or not subject iwas in the treatment (T = 1) or control (T = 0) group

• Let S ∈ (0,1) be an indicator for whether or not subject iwas in the RCT (S = 1) or target population (S = 0)

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Model

• Assume that both the RCT and the target population dataare simple random samples from two different infinitepopulations

• Unlike the paper, we assume here that W = W t = W c

• The expectation E01· is a weighted mean of the Wspecific means, E(Ys1|W ,S = 1,T = 1), with weightsaccording to the distribution of W in the treated targetpopulation, Pr(W |S = 0,T = 1)

• SATE is defined as:

τSATE = E(Y11 − Y10|S = 1),

where the expectation is over the (random) units in S = 1(the RCT)

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Assumptions

A.1 Treatment is consistent across studies:Yi01 = Yi11 and Yi00 = Yi10

A.2 Strong Ignorability of Sample Assignment for Treated:

(Yi01,Yi11)⊥⊥Si |(Wi ,Ti = 1) 0 < Pr(Si = 1|Wi ,Ti = 1) < 1

A.3 Strong Ignorability of Sample Assignment forTreated-Controls:

(Yi00,Yi10)⊥⊥Si |(Wi ,Ti = 1) 0 < Pr(Si = 1|Wi ,Ti = 1) < 1

A.4 SUTVA: no interference between units

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PATT

Sufficient to identify:

τPATT = E01E(Yi |Wi ,Si = 1,Ti = 1)−E01E(Yi |Wi ,Si = 1,Ti = 0)

But the assumptions also imply:

E(Yi |Si = 0,Ti = 1)− E01E(Yi |Wi ,Si = 1,Ti = 1) = 0

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Placebo Test

E(Yi |Si = 0,Ti = 1)− E01E(Yi |Wi ,Si = 1,Ti = 1) = 0

• The difference between the mean outcome of the NRStreated and mean outcome of the reweighted RCT treatedshould be zero

• If not 0, at least one assumptions has failed

• There is a similar placebo test for controls, however, it doesnot provide as much information

• Could fail due to lack of overlap, for example

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Placebo Test

E(Yi |Si = 0,Ti = 1)− E01E(Yi |Wi ,Si = 1,Ti = 1) = 0

• The difference between the mean outcome of the NRStreated and mean outcome of the reweighted RCT treatedshould be zero

• If not 0, at least one assumptions has failed

• There is a similar placebo test for controls, however, it doesnot provide as much information

• Could fail due to lack of overlap, for example

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Difference-in-Difference

An alternative design:

τPATTDID = E01E(Y |W ,S = 1,T = 1)− E(Y |W ,S = 1,T = 0)−[E01E(Y |W ,S = 1,T = 1) − E(Y |S = 0,T = 1)]

• The first difference is the adjusted experimental estimandand is intuitively a measure of the adjusted average effect

• The second difference is defined as the difference betweenthe RCT treated and NRS treated

• Required that A3, A4, and part of A1 hold (Yi00 = Yi10)

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IV Alternative• Our first method didn’t used outcomes observed in NRS• “Diff-in-Diff” estimator uses outcomes from NRS, and

estimates PATT under relaxed sample ignorabilityassumption

• Alternative approach: use sample indicator S as IV undernon-ignorable treatment assignment

• Need additional monotonicity assumption: T is increasingin S (conditional on W ), with prob 1. Treatment incidenceis (weakly) higher for everyone in the RCT sample.Reasonable?

• Then, we can identify the following LATE:

E [Y |S = 1] E [Y |S = 0]

E [T |S = 1] E [T |S = 0]

which is consistent estimator for E [Y1 − Y0|Ts=1 > Ts=0]

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Statistical Methods

GenMatch:• Matching method with automated balance optimization.

Maximum Entropy weighting:• Weighting method that assigns weights such that they

simultaneously meet a set of consistency constraints whilemaximizing Shannon’s measure of entropy.

• Consistency constraints are based on moments of thepopulation based on the NRS.

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Statistical Methods

GenMatch:• Matching method with automated balance optimization.

Maximum Entropy weighting:• Weighting method that assigns weights such that they

simultaneously meet a set of consistency constraints whilemaximizing Shannon’s measure of entropy.

• Consistency constraints are based on moments of thepopulation based on the NRS.

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Covariate Balance in RCT

Balance Plots

0.00

0.05

0.10

1.00

MeanTreated

MeanControl P−valuesSubgroup

Elective Surgery 0.06 0.06

Emergency Surgery 0.28 0.27

Non−Surgical 0.66 0.67

Teaching 0.22 0.22

Non−teaching 0.78 0.78

IMProb 0.53 0.53

Age 64.19 64.37

Female 0.43 0.43

Ventilation 0.89 0.91

Unit size 2.24 2.23

BM t−testAM t−test

BM KS−testAM KS−test

441 / 765

Balance Before and After Adjustment

0.00

0.05

0.10

1.00

MeanRCT

MeanNRS

Mean RCTAfter MaxentAdjustment

Mean RCTAfter IPSWAdjustment

P−valuesSubgroup

ElectiveSurgery

0.06 0.09 0.09 0.07

EmergencySurgery

0.28 0.23 0.23 0.24

Non−Surgical

0.66 0.68 0.68 0.69

Teaching 0.22 0.43 0.43 0.28

Non−teaching

0.78 0.57 0.57 0.72

IMProb 0.53 0.52 0.52 0.53

Age 64.19 61.86 61.86 63.49

Female 1.57 1.61 1.61 1.6

Ventilation 0.89 0.86 0.86 0.88

Unit size 2.24 2.66 2.45 2.28

Before AdjustmentAfter Maxent Adjustment

After IPSW Adjustment

442 / 765

Placebo Tests

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Strata

Mortality PlacebosNaive P.value FDR P.value

Placebo Test P−Values

Male

Female

Non−TeachingHospital

TeachingHospital

Non−surgical

EmergencySurgery

ElectiveSurgery

Overall

443 / 765

Placebo Tests

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Stratum PowerNaive p−value FDR p−value

Placebo Test P−Values

Non−TeachingHospital 0.85

−3917net benefit−1635cost−0.03survival

TeachingHospital 0.27

5765net benefit3934cost−0.04survival

Non−Surgical 0.832566net benefit747cost−0.04survival

EmergencySurgery 0.28

−1821net benefit2226cost−0.07survival

ElectiveSurgery 0.081

−11917net benefit−3069cost0.08survival

Overall 0.96201net benefit733cost−0.03survival

OutcomeDifferenceObs − Adj

444 / 765

Population Treatment Effects on Hospital Survival Rates

−0.4 −0.2 0 0.2 0.4 0.6

Strata PATT SATT

Treatment Effect EstimatesEffect on Survival Rate (% points)

Non−TeachingHospital

TeachingHospital

Non−Surgical

EmergencySurgery

ElectiveSurgery

Overall

445 / 765

Population Treatment Effects on Costs

−25000 −15000 −5000 0 5000 10000 15000

Strata PATT SATT

Treatment Effect EstimatesEffect on Costs in £

Non−TeachingHospital

TeachingHospital

Non−Surgical

EmergencySurgery

ElectiveSurgery

Overall

446 / 765

Population Treatment Effects on Cost-Effectiveness

−60000 −20000 0 20000 40000 60000 80000 1e+05

Strata PATT SATT

Treatment Effect EstimatesEffect on Incremental Net Benefit (Valuing £ 20K per Quality Adjusted Life Year (QALY))

Non−TeachingHospital

TeachingHospital

Non−Surgical

EmergencySurgery

ElectiveSurgery

Overall

447 / 765

Conclusions and Implications

• We pass placebo tests for both costs and hospital mortality,thus validating our identifying assumptions for the PATT.

• This has implications for Cost-Effectiveness Analysessince CEAs are often based on observational studies

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• See Hartman et al., forthcoming: [LINK]• See the references in that paper on Maximum Entropy

weighting methods. These are also used in SyntheticMatching—i.e., Abadie, Diamond, and Hainmueller, 2010;Hainmueller, 2012

449 / 765

Maximum EntropyJaynes defined the principle of maximum entropy as:

maxp

S(p) = −n∑

i=1

pi ln pi

s.t .

∑n

i=1 pi = 1∑ni=1 pigr (xi) =

∑ni=1 pigri = ar r = 1, . . . ,m

pi ≥ 0 i = 1,2, . . . ,n

• Equation (1) is referred to as a natural constraint, stating that allprobabilities must sum to unity.

• Equation (2), the m moment constraints, are referred to as theconsistency constraints. Each ar represents an r -th ordermoment, or characteristic moment, of the probability distribution.

• Equation (3) the final constraint ensures that all probabilities arenon-negative. This is always met.

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Balance Tests and the PropensityScore

Any given propensity score model may not be very good atbalancing the X covariates. After matching on the pscore weneed to test if the underlying X covariates have actually beenbalanced. You need to test more than just balance of thepscore.The MatchBalance() function in Matching provides a numberof tests, the two main ones being the t-test and theKolmogorov-Smirnov Test—the KS test.The KS test tries to determine if two distributions differsignificantly. The KS-test has the advantage of making noassumption about the distribution of data—i.e., it is distributionfree and non-parametric. However, because of this generality,other tests, such as the t-test, are more sensitive to certaindifferences—e.g., mean differences.

451 / 765

Balance Tests and the PropensityScore

Any given propensity score model may not be very good atbalancing the X covariates. After matching on the pscore weneed to test if the underlying X covariates have actually beenbalanced. You need to test more than just balance of thepscore.The MatchBalance() function in Matching provides a numberof tests, the two main ones being the t-test and theKolmogorov-Smirnov Test—the KS test.The KS test tries to determine if two distributions differsignificantly. The KS-test has the advantage of making noassumption about the distribution of data—i.e., it is distributionfree and non-parametric. However, because of this generality,other tests, such as the t-test, are more sensitive to certaindifferences—e.g., mean differences.

452 / 765

Balance Tests and the PropensityScore

Any given propensity score model may not be very good atbalancing the X covariates. After matching on the pscore weneed to test if the underlying X covariates have actually beenbalanced. You need to test more than just balance of thepscore.The MatchBalance() function in Matching provides a numberof tests, the two main ones being the t-test and theKolmogorov-Smirnov Test—the KS test.The KS test tries to determine if two distributions differsignificantly. The KS-test has the advantage of making noassumption about the distribution of data—i.e., it is distributionfree and non-parametric. However, because of this generality,other tests, such as the t-test, are more sensitive to certaindifferences—e.g., mean differences.

453 / 765

KS Test Details and Example

A great description of the standard KS test of offered at thiswebpage: http://www.physics.csbsju.edu/stats/KS-test.html.But the standard KS test does not provide the correct p-valuefor noncontinous variables. But the bootstrap KS does providethe correct p-values. And additional problem arises if we wantto test if distribution from estimated propensity scores differ. Wethen need to do another bootstrap to take into account thedistributions of the parameters in the propensity model.

Also seehttp://sekhon.berkeley.edu/causalinf/R/ks1.R

454 / 765

KS Test Details and Example

A great description of the standard KS test of offered at thiswebpage: http://www.physics.csbsju.edu/stats/KS-test.html.But the standard KS test does not provide the correct p-valuefor noncontinous variables. But the bootstrap KS does providethe correct p-values. And additional problem arises if we wantto test if distribution from estimated propensity scores differ. Wethen need to do another bootstrap to take into account thedistributions of the parameters in the propensity model.

Also seehttp://sekhon.berkeley.edu/causalinf/R/ks1.R

455 / 765

The Bootstrap

Bootstrap methods are a good way of obtaining confidenceintervals and other statistical estimates which require generallyweaker assumptions than the usual analytical approaches.The bootstrap is also useful when we don’t know of ananalytical solution. We lack such solutions when:

1 we are using complicated statistics for which a samplingdistribution is difficult to solve, such as two step estimators

2 we don’t want to make population assumptions

456 / 765

The Bootstrap

Bootstrap methods are a good way of obtaining confidenceintervals and other statistical estimates which require generallyweaker assumptions than the usual analytical approaches.The bootstrap is also useful when we don’t know of ananalytical solution. We lack such solutions when:

1 we are using complicated statistics for which a samplingdistribution is difficult to solve, such as two step estimators

2 we don’t want to make population assumptions

457 / 765

The Bootstrap

Bootstrap methods are a good way of obtaining confidenceintervals and other statistical estimates which require generallyweaker assumptions than the usual analytical approaches.The bootstrap is also useful when we don’t know of ananalytical solution. We lack such solutions when:

1 we are using complicated statistics for which a samplingdistribution is difficult to solve, such as two step estimators

2 we don’t want to make population assumptions

458 / 765

The Bootstrap

Bootstrap methods are a good way of obtaining confidenceintervals and other statistical estimates which require generallyweaker assumptions than the usual analytical approaches.The bootstrap is also useful when we don’t know of ananalytical solution. We lack such solutions when:

1 we are using complicated statistics for which a samplingdistribution is difficult to solve, such as two step estimators

2 we don’t want to make population assumptions

459 / 765

Percentile Bootstrap IntervalsThe basic setup is as follows:• We have a sample of size n from a given from a probability

distribution F

F → (x1, x2, x3, · · · , xn)

• The empirical distribution function F is defined to be the

discrete distribution that puts equal probability,1n

, on eachvalue xi .

• A set X which is made up of various xi has a probabilityassigned by F :

ˆprobX = #xi ∈ X/n

• We can make program via the plug-in principle. Theplug-in estimate of a parameter θ = t(F ) is θ = t(F )

460 / 765

Percentile Bootstrap IntervalsThe basic setup is as follows:• We have a sample of size n from a given from a probability

distribution F

F → (x1, x2, x3, · · · , xn)

• The empirical distribution function F is defined to be the

discrete distribution that puts equal probability,1n

, on eachvalue xi .

• A set X which is made up of various xi has a probabilityassigned by F :

ˆprobX = #xi ∈ X/n

• We can make program via the plug-in principle. Theplug-in estimate of a parameter θ = t(F ) is θ = t(F )

461 / 765

Percentile Bootstrap IntervalsThe basic setup is as follows:• We have a sample of size n from a given from a probability

distribution F

F → (x1, x2, x3, · · · , xn)

• The empirical distribution function F is defined to be the

discrete distribution that puts equal probability,1n

, on eachvalue xi .

• A set X which is made up of various xi has a probabilityassigned by F :

ˆprobX = #xi ∈ X/n

• We can make program via the plug-in principle. Theplug-in estimate of a parameter θ = t(F ) is θ = t(F )

462 / 765

Percentile Bootstrap IntervalsThe basic setup is as follows:• We have a sample of size n from a given from a probability

distribution F

F → (x1, x2, x3, · · · , xn)

• The empirical distribution function F is defined to be the

discrete distribution that puts equal probability,1n

, on eachvalue xi .

• A set X which is made up of various xi has a probabilityassigned by F :

ˆprobX = #xi ∈ X/n

• We can make program via the plug-in principle. Theplug-in estimate of a parameter θ = t(F ) is θ = t(F )

463 / 765

Bootstrap Intervals Algorithm

• For a dataset (denoted “full sample”) iterate the following Btimes where B is a large number such as 1,000.

1 Generate a random sample of size n from x1, · · · , xn withreplacement. This is called a bootstrap sample.

2 calculate the statistic of interest using this bootstrapsample, denoted θb.

• The confidence interval for θ (our full sample estimate) isobtained by taking the quantiles from the samplingdistribution in our bootstrap samples: θb.

• For example, for 95% CI, calculate the 2.5% and 97.5%quantiles of the distribution of θb.

464 / 765

Bootstrap Intervals Algorithm

• For a dataset (denoted “full sample”) iterate the following Btimes where B is a large number such as 1,000.

1 Generate a random sample of size n from x1, · · · , xn withreplacement. This is called a bootstrap sample.

2 calculate the statistic of interest using this bootstrapsample, denoted θb.

• The confidence interval for θ (our full sample estimate) isobtained by taking the quantiles from the samplingdistribution in our bootstrap samples: θb.

• For example, for 95% CI, calculate the 2.5% and 97.5%quantiles of the distribution of θb.

465 / 765

Bootstrap Intervals Algorithm

• For a dataset (denoted “full sample”) iterate the following Btimes where B is a large number such as 1,000.

1 Generate a random sample of size n from x1, · · · , xn withreplacement. This is called a bootstrap sample.

2 calculate the statistic of interest using this bootstrapsample, denoted θb.

• The confidence interval for θ (our full sample estimate) isobtained by taking the quantiles from the samplingdistribution in our bootstrap samples: θb.

• For example, for 95% CI, calculate the 2.5% and 97.5%quantiles of the distribution of θb.

466 / 765

Bootstrap Intervals Algorithm

• For a dataset (denoted “full sample”) iterate the following Btimes where B is a large number such as 1,000.

1 Generate a random sample of size n from x1, · · · , xn withreplacement. This is called a bootstrap sample.

2 calculate the statistic of interest using this bootstrapsample, denoted θb.

• The confidence interval for θ (our full sample estimate) isobtained by taking the quantiles from the samplingdistribution in our bootstrap samples: θb.

• For example, for 95% CI, calculate the 2.5% and 97.5%quantiles of the distribution of θb.

467 / 765

Bootstrap Intervals Algorithm

• For a dataset (denoted “full sample”) iterate the following Btimes where B is a large number such as 1,000.

1 Generate a random sample of size n from x1, · · · , xn withreplacement. This is called a bootstrap sample.

2 calculate the statistic of interest using this bootstrapsample, denoted θb.

• The confidence interval for θ (our full sample estimate) isobtained by taking the quantiles from the samplingdistribution in our bootstrap samples: θb.

• For example, for 95% CI, calculate the 2.5% and 97.5%quantiles of the distribution of θb.

468 / 765

Full Sample Estimate

• An estimate for the full sample quantity is provided by:

θ∗ =B∑

b=1

θb/B

This is an alternative to whatever we usually do in the fullsample, whether it be the usual least squares estimates,the mean, median or whatever.

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Bootstrapping Regression Coefficients

• For a dataset (denoted “full sample”) iterate the following Btimes.

• Generate a random sample of size n from Y ,X, withreplacement. Denote this sample of data Sb.

• Estimate your model using the dataset Sb. This results in avector of bootstrap sample estimates of β denoted βb.

Y b = βbX b + εb

• Create CIs and SEs in the usual bootstrap fashion basedon the distribution βb.

470 / 765

Bootstrapping Regression Coefficients

• For a dataset (denoted “full sample”) iterate the following Btimes.

• Generate a random sample of size n from Y ,X, withreplacement. Denote this sample of data Sb.

• Estimate your model using the dataset Sb. This results in avector of bootstrap sample estimates of β denoted βb.

Y b = βbX b + εb

• Create CIs and SEs in the usual bootstrap fashion basedon the distribution βb.

471 / 765

Bootstrapping Regression Coefficients

• For a dataset (denoted “full sample”) iterate the following Btimes.

• Generate a random sample of size n from Y ,X, withreplacement. Denote this sample of data Sb.

• Estimate your model using the dataset Sb. This results in avector of bootstrap sample estimates of β denoted βb.

Y b = βbX b + εb

• Create CIs and SEs in the usual bootstrap fashion basedon the distribution βb.

472 / 765

Bootstrapping Regression Coefficients

• For a dataset (denoted “full sample”) iterate the following Btimes.

• Generate a random sample of size n from Y ,X, withreplacement. Denote this sample of data Sb.

• Estimate your model using the dataset Sb. This results in avector of bootstrap sample estimates of β denoted βb.

Y b = βbX b + εb

• Create CIs and SEs in the usual bootstrap fashion basedon the distribution βb.

473 / 765

CommentsSo far we’ve talked about non-parametric bootstraps. And:• We are treating the sample as a population and sampling

from it.• does makes that assumption that the observations xi are

independent. It does not assume that they arehomoscedastic.

• can be applied to any function of the data which is smooth(this is a technical assumption) such as least squaresregression.

• the resulting estimates, point estimates, CIs and standard

errors have1√n

asymptotics just like the usual normal

theory estimates. There are better bootstraps with, for

example,1n

asymptotics. Examples of such bootstraps are:calibrated bootstrap, percentile-t bootstrap and the biascorrected, BCa, bootstrap.

474 / 765

CommentsSo far we’ve talked about non-parametric bootstraps. And:• We are treating the sample as a population and sampling

from it.• does makes that assumption that the observations xi are

independent. It does not assume that they arehomoscedastic.

• can be applied to any function of the data which is smooth(this is a technical assumption) such as least squaresregression.

• the resulting estimates, point estimates, CIs and standard

errors have1√n

asymptotics just like the usual normal

theory estimates. There are better bootstraps with, for

example,1n

asymptotics. Examples of such bootstraps are:calibrated bootstrap, percentile-t bootstrap and the biascorrected, BCa, bootstrap.

475 / 765

CommentsSo far we’ve talked about non-parametric bootstraps. And:• We are treating the sample as a population and sampling

from it.• does makes that assumption that the observations xi are

independent. It does not assume that they arehomoscedastic.

• can be applied to any function of the data which is smooth(this is a technical assumption) such as least squaresregression.

• the resulting estimates, point estimates, CIs and standard

errors have1√n

asymptotics just like the usual normal

theory estimates. There are better bootstraps with, for

example,1n

asymptotics. Examples of such bootstraps are:calibrated bootstrap, percentile-t bootstrap and the biascorrected, BCa, bootstrap.

476 / 765

CommentsSo far we’ve talked about non-parametric bootstraps. And:• We are treating the sample as a population and sampling

from it.• does makes that assumption that the observations xi are

independent. It does not assume that they arehomoscedastic.

• can be applied to any function of the data which is smooth(this is a technical assumption) such as least squaresregression.

• the resulting estimates, point estimates, CIs and standard

errors have1√n

asymptotics just like the usual normal

theory estimates. There are better bootstraps with, for

example,1n

asymptotics. Examples of such bootstraps are:calibrated bootstrap, percentile-t bootstrap and the biascorrected, BCa, bootstrap.

477 / 765

Distributions of Difficult to HandleQuantities

The bootstrap provides an easy way to to obtain the samplingdistribution of quantities for which it is often not known ordifficult to obtain sampling distributions. For example:

• CIs for the median or other quantiles• CIs and SEs for LMS coefficients• In a regression, is β1 > β2

• In a regression, isβ1

β2<β3

β4• Two stage regression estimation. Say a logit in the first

stage and a linear model in the second.• Goodness-of-fit for two model which may not be nested.

478 / 765

Distributions of Difficult to HandleQuantities

The bootstrap provides an easy way to to obtain the samplingdistribution of quantities for which it is often not known ordifficult to obtain sampling distributions. For example:

• CIs for the median or other quantiles• CIs and SEs for LMS coefficients• In a regression, is β1 > β2

• In a regression, isβ1

β2<β3

β4• Two stage regression estimation. Say a logit in the first

stage and a linear model in the second.• Goodness-of-fit for two model which may not be nested.

479 / 765

Distributions of Difficult to HandleQuantities

The bootstrap provides an easy way to to obtain the samplingdistribution of quantities for which it is often not known ordifficult to obtain sampling distributions. For example:

• CIs for the median or other quantiles• CIs and SEs for LMS coefficients• In a regression, is β1 > β2

• In a regression, isβ1

β2<β3

β4• Two stage regression estimation. Say a logit in the first

stage and a linear model in the second.• Goodness-of-fit for two model which may not be nested.

480 / 765

Distributions of Difficult to HandleQuantities

The bootstrap provides an easy way to to obtain the samplingdistribution of quantities for which it is often not known ordifficult to obtain sampling distributions. For example:

• CIs for the median or other quantiles• CIs and SEs for LMS coefficients• In a regression, is β1 > β2

• In a regression, isβ1

β2<β3

β4• Two stage regression estimation. Say a logit in the first

stage and a linear model in the second.• Goodness-of-fit for two model which may not be nested.

481 / 765

Distributions of Difficult to HandleQuantities

The bootstrap provides an easy way to to obtain the samplingdistribution of quantities for which it is often not known ordifficult to obtain sampling distributions. For example:

• CIs for the median or other quantiles• CIs and SEs for LMS coefficients• In a regression, is β1 > β2

• In a regression, isβ1

β2<β3

β4• Two stage regression estimation. Say a logit in the first

stage and a linear model in the second.• Goodness-of-fit for two model which may not be nested.

482 / 765

Distributions of Difficult to HandleQuantities

The bootstrap provides an easy way to to obtain the samplingdistribution of quantities for which it is often not known ordifficult to obtain sampling distributions. For example:

• CIs for the median or other quantiles• CIs and SEs for LMS coefficients• In a regression, is β1 > β2

• In a regression, isβ1

β2<β3

β4• Two stage regression estimation. Say a logit in the first

stage and a linear model in the second.• Goodness-of-fit for two model which may not be nested.

483 / 765

Bootstrap Hypothesis Tests from CIs

This is the simplest way:• Given test statistic θ, we are interested in obtaining the

achieved significance level (ASL) of the test which is:

ASL = ProbHo

θb ≥ θ

• Smaller the value of ASL, the stronger the evidenceagainst Ho.

• A confidence interval (θlo, θup) is the set of plausible valuesof θ having observed θ. Values not ruled out as beingunlikely.

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Difference of Means (case 1)

• For a dataset iterate the following B times.1 Generate a random sample of size n from x1, · · · , xn with

replacement. Do the same for the m obs in z.2 calculate the test statistic of interest using this bootstrap

sample, denoted θb = mean(xb)−mean(zb).

• The ASL under Ho = 0 is

ˆprob(Ho) = #θb ≥ 0

/B,

485 / 765

Difference of Means (case 1)

• For a dataset iterate the following B times.1 Generate a random sample of size n from x1, · · · , xn with

replacement. Do the same for the m obs in z.2 calculate the test statistic of interest using this bootstrap

sample, denoted θb = mean(xb)−mean(zb).

• The ASL under Ho = 0 is

ˆprob(Ho) = #θb ≥ 0

/B,

486 / 765

Difference of Means (case 1)

• For a dataset iterate the following B times.1 Generate a random sample of size n from x1, · · · , xn with

replacement. Do the same for the m obs in z.2 calculate the test statistic of interest using this bootstrap

sample, denoted θb = mean(xb)−mean(zb).

• The ASL under Ho = 0 is

ˆprob(Ho) = #θb ≥ 0

/B,

487 / 765

Difference of Means (case 1)

• For a dataset iterate the following B times.1 Generate a random sample of size n from x1, · · · , xn with

replacement. Do the same for the m obs in z.2 calculate the test statistic of interest using this bootstrap

sample, denoted θb = mean(xb)−mean(zb).

• The ASL under Ho = 0 is

ˆprob(Ho) = #θb ≥ 0

/B,

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Which Side?

• Note that the previous ASL is based on the proportion ofthe empiric above 0. Or the portion which is most relevantif mean(x) < mean(z)—i.e., (mean(x) - mean(z)) < 0

• The portion of the empiric below zero would be of interestin rejecting the null if mean(x) > mean(z). If we decidewhich side to look at based on the full sample statistic, weneed to adjust our ASL by the dividing by the probability ofchoosing the side (generally .5). For details see the[bs1mc1.R], [bs1mc1b.R] files.

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The forging algorithm is presented in the [bs1.R] file as case 1.The following cases are also presented in that R file. Pleaseexamine it closely.

The proceeding algorithm only works when the test statisticunder the null is centered around zero. If not, we need tosubtract the null from θb.

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Difference of Means (case 2)

Here’s another way to test the equivalent hypothesis.• Bootstrap as in case 1, but now difference the full sample

estimate which is θ• The bootstrap ASL of interest if mean(x) < mean(z),

ˆprob(Ho) = #θb − θ < θ

/B,

else

ˆprob(Ho) = #θb − θ ≥ θ

/B,

491 / 765

Difference of Means (case 2)

Here’s another way to test the equivalent hypothesis.• Bootstrap as in case 1, but now difference the full sample

estimate which is θ• The bootstrap ASL of interest if mean(x) < mean(z),

ˆprob(Ho) = #θb − θ < θ

/B,

else

ˆprob(Ho) = #θb − θ ≥ θ

/B,

492 / 765

Difference of Means (case 3)The studentized version• Bootstrap as in case 2, but let’s studentize the quantities:

θ =mean(x)−mean(z)√

var(x) + var(z)

θb is analogously defined• The bootstrap ASL of interest if mean(x) < mean(z)

ˆprob(Ho) = #θb − θ < θ

/B,

and for mean(x) > mean(z)

ˆprob(Ho) = #θb − θ ≥ θ

/B,

493 / 765

Difference of Means (case 4)

Here we can test if the two distributions F and G are equal. Wewill, however, use this just for the means though. The KS test isbased on this bootstrap. See the code for [ks.boot]• Draw B samples of size n + m with replacement from

c(x , z). Call the first n observations xb and the remainingm observations zb.

• Construct θ and calculate ASL as in case 2.This version is much like a permutation test except that we are(1) sampling with replacement and (2) we are learning aboutthe distribution under the null by sampling instead of byconstruction.

494 / 765

Difference of Means (case 4)

Here we can test if the two distributions F and G are equal. Wewill, however, use this just for the means though. The KS test isbased on this bootstrap. See the code for [ks.boot]• Draw B samples of size n + m with replacement from

c(x , z). Call the first n observations xb and the remainingm observations zb.

• Construct θ and calculate ASL as in case 2.This version is much like a permutation test except that we are(1) sampling with replacement and (2) we are learning aboutthe distribution under the null by sampling instead of byconstruction.

495 / 765

Different Null Hypothesizes

• Code which adapts case four to use the KS test is providedat: [bs2.R]

• All four cases are based on the null hypothesis that themeans in the two samples are equal. The null hypothesisneed not be so restrictive for the first three cases.

• Why in the fourth case doesn’t non-equality make sense?We could generalize (much like a sharp null).

• The following cases generalize the first three cases for thenon-zero difference setting. For R code see: [bs1_null.R].

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Difference of Means Redux (case 1∗)

• Bootstrap as before. Note that the null hypothesis is H0

• The bootstrap ASL of interest if[mean(x)−mean(z)] < H0, is

ˆprob(Ho) = #θb ≥ H0

/B,

else

ˆprob(Ho) = #θb < H0

/B.

497 / 765

Difference of Means Redux (case 1∗)

• Bootstrap as before. Note that the null hypothesis is H0

• The bootstrap ASL of interest if[mean(x)−mean(z)] < H0, is

ˆprob(Ho) = #θb ≥ H0

/B,

else

ˆprob(Ho) = #θb < H0

/B.

498 / 765

Difference of Means Redux (case 2∗)

• Bootstrap as before.• The bootstrap ASL of interest if

[mean(x)−mean(z)] < H0, is

ˆprob(Ho) = #θb − θ < θ − H0

/B,

else

ˆprob(Ho) = #θb − θ ≥ θ − H0

/B,

499 / 765

Difference of Means Redux (case 3∗)The studentized version• Bootstrap as before and let’s studentize the quantities:

θ =mean(x)−mean(z)√

var(x) + var(z)

θb is analogously defined. And the studentized H0 is:

H ′0 =mean(x)−mean(z)− H0√

var(x) + var(z)

• The bootstrap ASL of interest if[mean(x)−mean(z)] < H0, is

ˆprob(Ho) = #θb − θ < H ′0

/B,

and for mean(x) > mean(z)

ˆprob(Ho) = #θb − θ ≥ H ′0

/B,

500 / 765

Difference of Means Redux (case 3∗)The studentized version• Bootstrap as before and let’s studentize the quantities:

θ =mean(x)−mean(z)√

var(x) + var(z)

θb is analogously defined. And the studentized H0 is:

H ′0 =mean(x)−mean(z)− H0√

var(x) + var(z)

• The bootstrap ASL of interest if[mean(x)−mean(z)] < H0, is

ˆprob(Ho) = #θb − θ < H ′0

/B,

and for mean(x) > mean(z)

ˆprob(Ho) = #θb − θ ≥ H ′0

/B,

501 / 765

Parametric Bootstrap

The previous bootstraps assume that the observations areindependent. But this may not be the case. Also we may wantto assume that X is fixed. In these cases, one may estimate aparametric bootstrap if it can be assumed that the residuals of aparametric model are independent even if the original data isnot. This often occurs with time-series models.

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Parametric Bootstrap Algorithm• Estimate the model in the full-sample:

Y = βX + ε

• Iterate the following B times.• Generate a random sample of size n from ε, with

replacement. Denote this εb. Create the bootstrap sampleby adding the new εb to the full sample X and β (there isNO estimation in this step, just addition):

Y b = βX + εb

• Estimate your model using the full sample X but thebootstrap sample Y b. This results in a vector of bootstrapestimates of β estimates denoted βb.

Y b = βbX + εb

• Create CIs and SEs in the usual bootstrap fashionbased on the distribution βb.

503 / 765

Parametric Bootstrap Algorithm• Estimate the model in the full-sample:

Y = βX + ε

• Iterate the following B times.• Generate a random sample of size n from ε, with

replacement. Denote this εb. Create the bootstrap sampleby adding the new εb to the full sample X and β (there isNO estimation in this step, just addition):

Y b = βX + εb

• Estimate your model using the full sample X but thebootstrap sample Y b. This results in a vector of bootstrapestimates of β estimates denoted βb.

Y b = βbX + εb

• Create CIs and SEs in the usual bootstrap fashionbased on the distribution βb.

504 / 765

Parametric Bootstrap Algorithm• Estimate the model in the full-sample:

Y = βX + ε

• Iterate the following B times.• Generate a random sample of size n from ε, with

replacement. Denote this εb. Create the bootstrap sampleby adding the new εb to the full sample X and β (there isNO estimation in this step, just addition):

Y b = βX + εb

• Estimate your model using the full sample X but thebootstrap sample Y b. This results in a vector of bootstrapestimates of β estimates denoted βb.

Y b = βbX + εb

• Create CIs and SEs in the usual bootstrap fashionbased on the distribution βb.

505 / 765

Parametric Bootstrap Algorithm• Estimate the model in the full-sample:

Y = βX + ε

• Iterate the following B times.• Generate a random sample of size n from ε, with

replacement. Denote this εb. Create the bootstrap sampleby adding the new εb to the full sample X and β (there isNO estimation in this step, just addition):

Y b = βX + εb

• Estimate your model using the full sample X but thebootstrap sample Y b. This results in a vector of bootstrapestimates of β estimates denoted βb.

Y b = βbX + εb

• Create CIs and SEs in the usual bootstrap fashionbased on the distribution βb.

506 / 765

Parametric Bootstrap Algorithm• Estimate the model in the full-sample:

Y = βX + ε

• Iterate the following B times.• Generate a random sample of size n from ε, with

replacement. Denote this εb. Create the bootstrap sampleby adding the new εb to the full sample X and β (there isNO estimation in this step, just addition):

Y b = βX + εb

• Estimate your model using the full sample X but thebootstrap sample Y b. This results in a vector of bootstrapestimates of β estimates denoted βb.

Y b = βbX + εb

• Create CIs and SEs in the usual bootstrap fashionbased on the distribution βb.

507 / 765

Two Nice Bootstrap PropertiesBootstrap CI estimates are:

1 Range Preserving. The confidence interval cannot extendbeyond the range which estimates of θ range in bootstrapsamples. For example, if we are bootstrapping estimates ofa proportion, our confidence interval cannot be less than 0or greater than 1. The usual normal theory confidenceintervals are not range persevering.

2 Take into account more general uncertainty. Manyestimators, such as LMS, have the usual statisticaluncertainty (due to sampling) but also have uncertaintybecause the algorithm used to obtain the estimates isstochastic or otherwise suboptimal. Bootstrap CIs will takethis uncertainty into account. See the R help on the lqsfunction for more information about how the LMS estimatesare obtained.

508 / 765

Two Nice Bootstrap PropertiesBootstrap CI estimates are:

1 Range Preserving. The confidence interval cannot extendbeyond the range which estimates of θ range in bootstrapsamples. For example, if we are bootstrapping estimates ofa proportion, our confidence interval cannot be less than 0or greater than 1. The usual normal theory confidenceintervals are not range persevering.

2 Take into account more general uncertainty. Manyestimators, such as LMS, have the usual statisticaluncertainty (due to sampling) but also have uncertaintybecause the algorithm used to obtain the estimates isstochastic or otherwise suboptimal. Bootstrap CIs will takethis uncertainty into account. See the R help on the lqsfunction for more information about how the LMS estimatesare obtained.

509 / 765

Bootstrap Readings

For bootstrap algorithms see Venables and Ripley (2002).Especially, p.133-138 (section 5.7) and p.163-165 (section 6.6).The R "boot" command is of particular interest.For a statistical discussion see Chapter 16 (p.493ff) of Fox(2002).For additional readings see:

• Efron, Bradley and Tibshirani, Robert J. 1994. AnIntroduction to the Bootstrap. Chapman & Hall. ISBN:0412042312.

• Davison, A. C. and Hinkely, D. V. 1997. Bootstrap Methodsand their Applications. Cambridge University Press.

• Go to jstor and search for Efron and/or Tibshirani.

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Bootstrap Readings

For bootstrap algorithms see Venables and Ripley (2002).Especially, p.133-138 (section 5.7) and p.163-165 (section 6.6).The R "boot" command is of particular interest.For a statistical discussion see Chapter 16 (p.493ff) of Fox(2002).For additional readings see:

• Efron, Bradley and Tibshirani, Robert J. 1994. AnIntroduction to the Bootstrap. Chapman & Hall. ISBN:0412042312.

• Davison, A. C. and Hinkely, D. V. 1997. Bootstrap Methodsand their Applications. Cambridge University Press.

• Go to jstor and search for Efron and/or Tibshirani.

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Bootstrap Readings

For bootstrap algorithms see Venables and Ripley (2002).Especially, p.133-138 (section 5.7) and p.163-165 (section 6.6).The R "boot" command is of particular interest.For a statistical discussion see Chapter 16 (p.493ff) of Fox(2002).For additional readings see:

• Efron, Bradley and Tibshirani, Robert J. 1994. AnIntroduction to the Bootstrap. Chapman & Hall. ISBN:0412042312.

• Davison, A. C. and Hinkely, D. V. 1997. Bootstrap Methodsand their Applications. Cambridge University Press.

• Go to jstor and search for Efron and/or Tibshirani.

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The Search Problem

• Many search problems have an exponential asymptoticorder: O(cN), c > 1 where N is the sample size.

• Finding the optimal set of matches is such a problem.E.G., we want to estimate ATT and do matching withoutreplacement:• with 10 treated and 20 control obs: 184,756 possible

matches• with 20 treated and 40 control obs: 13,784,652,8820

possible matches• with 40 treated and 80 control obs: 1.075072e+23• in the LaLonde data with 185 treated and 260 control:

1.633347e+69• with 185 treated and 4000 control: computer infinity

Matching with replacement makes the search problemexplode even more quickly.

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The Search Problem

• Many search problems have an exponential asymptoticorder: O(cN), c > 1 where N is the sample size.

• Finding the optimal set of matches is such a problem.E.G., we want to estimate ATT and do matching withoutreplacement:• with 10 treated and 20 control obs: 184,756 possible

matches• with 20 treated and 40 control obs: 13,784,652,8820

possible matches• with 40 treated and 80 control obs: 1.075072e+23• in the LaLonde data with 185 treated and 260 control:

1.633347e+69• with 185 treated and 4000 control: computer infinity

Matching with replacement makes the search problemexplode even more quickly.

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The Search Problem

• Many search problems have an exponential asymptoticorder: O(cN), c > 1 where N is the sample size.

• Finding the optimal set of matches is such a problem.E.G., we want to estimate ATT and do matching withoutreplacement:• with 10 treated and 20 control obs: 184,756 possible

matches• with 20 treated and 40 control obs: 13,784,652,8820

possible matches• with 40 treated and 80 control obs: 1.075072e+23• in the LaLonde data with 185 treated and 260 control:

1.633347e+69• with 185 treated and 4000 control: computer infinity

Matching with replacement makes the search problemexplode even more quickly.

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The Search Problem

• Many search problems have an exponential asymptoticorder: O(cN), c > 1 where N is the sample size.

• Finding the optimal set of matches is such a problem.E.G., we want to estimate ATT and do matching withoutreplacement:• with 10 treated and 20 control obs: 184,756 possible

matches• with 20 treated and 40 control obs: 13,784,652,8820

possible matches• with 40 treated and 80 control obs: 1.075072e+23• in the LaLonde data with 185 treated and 260 control:

1.633347e+69• with 185 treated and 4000 control: computer infinity

Matching with replacement makes the search problemexplode even more quickly.

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The Search Problem

• Many search problems have an exponential asymptoticorder: O(cN), c > 1 where N is the sample size.

• Finding the optimal set of matches is such a problem.E.G., we want to estimate ATT and do matching withoutreplacement:• with 10 treated and 20 control obs: 184,756 possible

matches• with 20 treated and 40 control obs: 13,784,652,8820

possible matches• with 40 treated and 80 control obs: 1.075072e+23• in the LaLonde data with 185 treated and 260 control:

1.633347e+69• with 185 treated and 4000 control: computer infinity

Matching with replacement makes the search problemexplode even more quickly.

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The Search Problem

• Many search problems have an exponential asymptoticorder: O(cN), c > 1 where N is the sample size.

• Finding the optimal set of matches is such a problem.E.G., we want to estimate ATT and do matching withoutreplacement:• with 10 treated and 20 control obs: 184,756 possible

matches• with 20 treated and 40 control obs: 13,784,652,8820

possible matches• with 40 treated and 80 control obs: 1.075072e+23• in the LaLonde data with 185 treated and 260 control:

1.633347e+69• with 185 treated and 4000 control: computer infinity

Matching with replacement makes the search problemexplode even more quickly.

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The Search Problem

• Many search problems have an exponential asymptoticorder: O(cN), c > 1 where N is the sample size.

• Finding the optimal set of matches is such a problem.E.G., we want to estimate ATT and do matching withoutreplacement:• with 10 treated and 20 control obs: 184,756 possible

matches• with 20 treated and 40 control obs: 13,784,652,8820

possible matches• with 40 treated and 80 control obs: 1.075072e+23• in the LaLonde data with 185 treated and 260 control:

1.633347e+69• with 185 treated and 4000 control: computer infinity

Matching with replacement makes the search problemexplode even more quickly.

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The Solution

• It is impossible to search all possible matches in finite time.• We need to make the problem tractable• This is done by finding a solution which is “good enough”• A variety of global search algorithms can do this including

simulated annealing and genetic optimization• Both are variants of random search

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Difficult Optimization Problems

• Loss functions of statistical models are often irregular• ML models are generally not globally concave• As such, deriviative methods fail to find the global

maximum• And Taylor series methods for SEs do not provide correct

coverage

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Normal Mixtures

Mixtures of Normal Densities provides a nice way to createdifficult optimization surfaces.

The Claw:

12

N(0,1) +4∑

m=0

110

N(m2− 1, (

110

))

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The Claw

-3 -2 -1 0 1 2 3

0.0

0.1

0.2

0.3

0.4

0.5

0.6

••

0

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Normal Mixtures

Asymmetric Double Claw:

1∑m=0

46100

N(2m − 1,23

) +

3∑m=1

1300

N(−m2,

1100

) +

3∑m=1

7300

N(m2,

7100

)

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Asymetric Double Claw

-3 -2 -1 0 1 2 3

0.0

0.1

0.2

0.3

0.4

••

•••

0

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Normal Mixtures

Discrete Comb Density:

2∑m=0

27

N((12m − 15)

7,27

) +

2∑m=0

121

N(2m + 16

7,

121

)

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Discrete Comb Density

-3 -2 -1 0 1 2 3

0.0

0.1

0.2

0.3

0.4 •• ••••

0 2431

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Discrete Comb Density: Close Up

-3 -2 -1 0 1 2 3

0.39

60.

398

0.40

00.

402

•• ••••

0 24

31

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Evolutionary Search (EA)

• An EA uses a collection of heuristic rules to modify apopulation of trial solutions in such a way that eachgeneration of trial values tends to be on average betterthan its predecessor.

• The EA in GENOUD assumes a solution is a vector of realnumbers, each number being a value for a scalarparameter of a function to be optimized.

• parameter==gene• Each heuristic rule, or operator, acts on one or more trial

solutions from the current population to produce one ormore trial solutions to be included in the new population.

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Evolutionary Search (EA)

• An EA uses a collection of heuristic rules to modify apopulation of trial solutions in such a way that eachgeneration of trial values tends to be on average betterthan its predecessor.

• The EA in GENOUD assumes a solution is a vector of realnumbers, each number being a value for a scalarparameter of a function to be optimized.

• parameter==gene• Each heuristic rule, or operator, acts on one or more trial

solutions from the current population to produce one ormore trial solutions to be included in the new population.

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Description

• A GA uses a set of randomized genetic operators to evolvea finite population of finite code-strings over a series ofgenerations (Holland 1975; Goldberg 1989; Grefenstetteand Baker 1989)

• The operators used in GA implementations vary, but thebasic set of operators can be defined as reproduction,mutation and crossover (Davis 1991; Filho, Treleaven andAlippi 1994).

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General Description of Operators

• reproduction: a copy of an individual in the currentpopulation is reproduced into the new one. Analogous toasexual reproduction.

• mutation: randomly alter parameter values (genes) so asto search the entire space

• crossover: Create a new individual from one or moreindividuals from the current population. Splice someparameters from individual A with the parameter valuesfrom another individual. Analogous to sexual reproduction.

• All three can be combined to create mixed operators

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General Description of Operators

• reproduction: a copy of an individual in the currentpopulation is reproduced into the new one. Analogous toasexual reproduction.

• mutation: randomly alter parameter values (genes) so asto search the entire space

• crossover: Create a new individual from one or moreindividuals from the current population. Splice someparameters from individual A with the parameter valuesfrom another individual. Analogous to sexual reproduction.

• All three can be combined to create mixed operators

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General Description of Operators

• reproduction: a copy of an individual in the currentpopulation is reproduced into the new one. Analogous toasexual reproduction.

• mutation: randomly alter parameter values (genes) so asto search the entire space

• crossover: Create a new individual from one or moreindividuals from the current population. Splice someparameters from individual A with the parameter valuesfrom another individual. Analogous to sexual reproduction.

• All three can be combined to create mixed operators

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General Description of Operators

• reproduction: a copy of an individual in the currentpopulation is reproduced into the new one. Analogous toasexual reproduction.

• mutation: randomly alter parameter values (genes) so asto search the entire space

• crossover: Create a new individual from one or moreindividuals from the current population. Splice someparameters from individual A with the parameter valuesfrom another individual. Analogous to sexual reproduction.

• All three can be combined to create mixed operators

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The Actual Operators

Some notation:

X =[X1, . . . ,Xn

]is the vector of n parameters Xi . x i is the

lower bound and x i is the upper bound on values for Xi . xi isthe current value of Xi , and x is the current value of X.N = 1, . . . ,n. p ∼ U(0,1) means that p is drawn from theuniform distribution on the [0,1] interval.

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Operators 1-4

1 Cloning. Copy Xt into the next generation, Xt+1.2 Uniform Mutation. At random choose i ∈ N. Select a value

xi ∼ U(x i , x i). Set Xi = xi .3 Boundary Mutation. At random choose i ∈ N. Set either

Xi = x i or Xi = x i , with probability 1/2 of using each value.4 Non-uniform Mutation. At random choose i ∈ N. Compute

p = (1− t/T )Bu, where t is the current generation number,T is the maximum number of generations, B > 0 is atuning parameter and u ∼ U(0,1). Set eitherXi = (1− p)xi + px i or Xi = (1− p)xi + px i , with probability1/2 of using each value.

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Operators 1-4

1 Cloning. Copy Xt into the next generation, Xt+1.2 Uniform Mutation. At random choose i ∈ N. Select a value

xi ∼ U(x i , x i). Set Xi = xi .3 Boundary Mutation. At random choose i ∈ N. Set either

Xi = x i or Xi = x i , with probability 1/2 of using each value.4 Non-uniform Mutation. At random choose i ∈ N. Compute

p = (1− t/T )Bu, where t is the current generation number,T is the maximum number of generations, B > 0 is atuning parameter and u ∼ U(0,1). Set eitherXi = (1− p)xi + px i or Xi = (1− p)xi + px i , with probability1/2 of using each value.

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Operators 1-4

1 Cloning. Copy Xt into the next generation, Xt+1.2 Uniform Mutation. At random choose i ∈ N. Select a value

xi ∼ U(x i , x i). Set Xi = xi .3 Boundary Mutation. At random choose i ∈ N. Set either

Xi = x i or Xi = x i , with probability 1/2 of using each value.4 Non-uniform Mutation. At random choose i ∈ N. Compute

p = (1− t/T )Bu, where t is the current generation number,T is the maximum number of generations, B > 0 is atuning parameter and u ∼ U(0,1). Set eitherXi = (1− p)xi + px i or Xi = (1− p)xi + px i , with probability1/2 of using each value.

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Operators 1-4

1 Cloning. Copy Xt into the next generation, Xt+1.2 Uniform Mutation. At random choose i ∈ N. Select a value

xi ∼ U(x i , x i). Set Xi = xi .3 Boundary Mutation. At random choose i ∈ N. Set either

Xi = x i or Xi = x i , with probability 1/2 of using each value.4 Non-uniform Mutation. At random choose i ∈ N. Compute

p = (1− t/T )Bu, where t is the current generation number,T is the maximum number of generations, B > 0 is atuning parameter and u ∼ U(0,1). Set eitherXi = (1− p)xi + px i or Xi = (1− p)xi + px i , with probability1/2 of using each value.

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Operators 5-7

5 Polytope Crossover. Using m = max(2,n) vectors x fromthe current population and m random numbers pj ∈ (0,1)such that

∑mj=1 pj = 1, set X =

∑mj=1 pjxj .

6 Simple Crossover. Choose a single integer i from N. Usingtwo parameter vectors, x and y, for the i setXi = pxi + (1− p)yi and Yi = pyi + (1− p)xi , wherep ∈ (0,1) is a fixed number.

7 Whole Non-uniform Mutation. Do non-uniform mutation forall the elements of X.

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Operators 5-7

5 Polytope Crossover. Using m = max(2,n) vectors x fromthe current population and m random numbers pj ∈ (0,1)such that

∑mj=1 pj = 1, set X =

∑mj=1 pjxj .

6 Simple Crossover. Choose a single integer i from N. Usingtwo parameter vectors, x and y, for the i setXi = pxi + (1− p)yi and Yi = pyi + (1− p)xi , wherep ∈ (0,1) is a fixed number.

7 Whole Non-uniform Mutation. Do non-uniform mutation forall the elements of X.

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Operators 5-7

5 Polytope Crossover. Using m = max(2,n) vectors x fromthe current population and m random numbers pj ∈ (0,1)such that

∑mj=1 pj = 1, set X =

∑mj=1 pjxj .

6 Simple Crossover. Choose a single integer i from N. Usingtwo parameter vectors, x and y, for the i setXi = pxi + (1− p)yi and Yi = pyi + (1− p)xi , wherep ∈ (0,1) is a fixed number.

7 Whole Non-uniform Mutation. Do non-uniform mutation forall the elements of X.

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Operators 8

8 Heuristic Crossover. Choose p ∼ U(0,1). Using twoparameter vectors, x and y, compute z = p(x− y) + x. If zsatisfies all constraints, use it. Otherwise choose another pvalue and repeat. Set z equal to the better of x and y if asatisfactory mixed z is not found by a preset number ofattempts. In this fashion produce two z vectors.

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Operators 9

9 Local-minimum Crossover. NOT USED IN GENMATCH.Choose p ∼ U(0,1). Starting with x, run BFGSoptimization up to a preset number of iterations to produce~x. Compute z = p~x + (1− p)x. If z satisfies boundaryconstraints, use it. Otherwise shrink p by setting p = p/2and recompute z. If a satisfactory z is not found by apreset number of attempts, return x. This operators isextremely computationally intensive, use sparingly.

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Markov Chains

GAs are Markov chains:

A Markov chain describes at successive times the states of a system.At these times the system may have changed from the state it was inthe moment before to another or stayed in the same state. Thechanges of state are called transitions. The Markov property meansthat the conditional probability distribution of the state in the future,given the state of the process currently and in the past, depends onlyon its current state and not on its state in the past.

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Markov Chains II

A Markov chain is a sequence of random variablesX1,X2,X3, . . . with the Markov property, that is, given thepresent state, the future state and past states are independent.Formally:

Pr (Xn+1 = x | Xn = xn, . . . ,X1 = x1,X0 = x0) =

Pr (Xn+1 = x | Xn = xn)

The possible values of Xi form a countable set S which is thestate space of the chain. Continuous time chains exist.

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• For more on Markov chains cites in GENOUD article orhttp://random.mat.sbg.ac.at/~ste/diss/node6.html

• A finite GA with random reproduction and mutation is anaperiodic and irreducible Markov chain.

• Such a chain converges at an exponential rate to a uniquestationary distribution (Billingsley 1986, 128).

• the probability that each population occurs rapidlyconverges to a constant, positive value.

• Nix and Vose (1992; Vose 1993) show that in a GA wherethe probability that each code-string is selected toreproduce is proportional to its observed fitness, thestationary distribution strongly emphasizes populationsthat contain code-strings that have high fitness values.

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Population Size

• asymptotically in the population size i.e., in the limit for aseries of GAs with successively larger populationspopulations that have suboptimal average fitness haveprobabilities approaching zero in the stationary distribution,while the probability for the population that has optimalaverage fitness approaches one.

• The crucial practical implication from the theoretical resultsof Nix and Vose is that a GA’s success as an optimizerdepends on having a sufficiently large population.

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Robustness of RandomizationInference

• Randomization inference is often more robust thanalternatives

• An example of this is when method of moment estimatorsbecome close to unidentified in finite samples

• An illustration is provide by the Wald (1940) estimator,which is often used to estimate treatment effects withinstruments

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What is IV Used For?

• IV used for causal inference• The most compelling common example: estimate the

average treatment effect when there is one-waynon-compliance in an experiment.

• Assumptions are weak in this case. In most otherexamples, the behavioral assumptions are strong.

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One Way Non-Compliance

Only two types of units: compliers and never takersWe have five different parameters:

(i) the proportion of compliers in the experimental population(α)

(ii) the average response of compliers assigned to treatment(W )

(iii) the average response of compliers assigned to control (C)(iv) the difference between W and C, which is the average

effect of treatment on the compliers (R)(v) the average response of never-treat units assigned to

control (Z )

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Estimation I

• α can be estimated by calculating the proportion ofcompliers observed in the treatment group

• The average response of compliers to treatment, W , is theaverage response of compliers in the treatment group

• Z , the average response of never-treat units to control, isestimated by the average response among units in thetreatment group who refused treatment

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Estimation II

This leaves C and R.For R, the control group contains a mix of compliers andnever-treat units. We know (in expectation) the proportion ofeach type there must be in control because we can estimatethis proportion in the treated group.

α denotes the proportion of compliers in the experimentalpopulation, and assume that α > 0. Under the model, theproportion of never-treat units must be 1− α.

Denote the average observed responses in treatment andcontrol by Y t , Y c , these are sample quantities which aredirectly observed.

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Estimation III

E(Y c) = αC + (1− α)Z .

Therefore,

C =E(Y c)− (1− α)Z

α.

An obvious estimator for C is

ˆC =Y c − (1− α) ˆZ

α.

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ETT

Then the only remaining quantity is R, the average effect oftreatment on the compliers—i.e., the Effect of Treatment on theTreated (ETT). This can be estimated by

ˆW − ˆC =Y t − Y c

α. (17)

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Instrumental Variables (IV)

• IV methods solve the problem of missing or unknowncontrol IF the instruments are valid

• Simple example under the unnecessary assumption ofconstant effects. This assumption is only used to simplythe presentation here:

α = Y1i − Y0i

Y0i = β + εi ,

where β ≡ E [Y0i ]

• When we do permutation inference, we will assume underthe null, unlike the Wald estimator, that the potentialoutcomes are fixed.

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Towards the Wald Estimator I

• The potential outcomes model can now be written:Yi = β + αTi + εi , (18)

But Ti is likely correlated with εi .• Suppose: a third variable Zi which is correlated with Ti , but

is unrelated to Y for any other reason—i.e., Y0i ⊥⊥ Zi andE [εi | Zi ] = 0.

• Zi is said to be an IV or “an instrument” for the causaleffect of T on Y

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Towards the Wald Estimator II

• Suppose that Zi is dichotomous (0,1). Then

α =E [Yi | Z = 1]− E [Yi | Z = 0]

E [Ti | Z = 1]− E [Ti | Z = 0](19)

• The sample analog of this equation is called the Waldestimator, since it first appear in Wald (1940) onerrors-in-variables problems.

• More general versions for continuous, multi-valued, ormultiple instruments.

• Problem: what if in finite samples the denominator in Eq 19is close to zero?

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Conditions for a Valid Instrument

• Instrument Relevance:cov(Z ,T ) 6= 0

• Instrument Exogeneity:cov(Z , ε) = 0

• These conditions ensure that the part of X that iscorrelated with Z only contains exogenous variation

• Instrument relevance is testable• Instrument exogeneity is NOT testable. It must be true by

design

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Weak Instruments

• A weak instrument is one where the denominator in Eq 19is close to zero.

• This poses two distinct problems:• if the instrument is extremely weak, it may provide little or

no useful information• commonly used statistical methods for IV do not accurately

report this lack of information

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Two Stage Least Squares Estimator(TSLS)

Y = βT + ε (20)T = Zγ + υ, (21)

where Y , ε, υ are N × 1, and Z is N × K , where K is thenumber of instruments• Note we do not assume that E(T , ε) = 0, which is the

central problem• We assume instead E(υ | Z ) = 0, E(υ, ε) = 0, E(Z , ε) = 0,E(Z ,T ) 6= 0

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TSLS Estimator

βiv = (T ′PzT )−1T ′PzY , (22)

where Pz = Z (Z ′Z )−1Z ′, the projection matrix for Z . It can beshown that:

• plim βols = β +σT ,ε

σ2T

• plim βiv = β +σT ,ε

σ2T

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Quarter of Birth and Returns toSchooling

• J. Angrist and Krueger (1991) want to estimate the causaleffect of education

• This is difficult, so they propose a way to estimate thecausal effect of compulsory school attendance on earnings

• Every state has a minimum number δ of years of schoolingthat all students must have

• But, the laws are written in terms of the age at which astudent can leave school

• Because birth dates vary, individual students are requiredto attend between δ and δ + 1 years of schooling

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Quarter of Birth

• The treatment T is years of education• The instrument Z is quarter of birth (exact birth date to be

precise)• The outcome Y are earnings

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The Data

• Census data is used• 329,509 men born between 1930 and 1939• Observe: years of schooling, birth date, earnings in 1980• mean number of years of education is 12.75• Example instrument: being born in the fourth quarter of the

year, which is 1 for 24.5% of sample

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How Much Information Is There?

• If the number of years of education is regressed on thisquarter-of-birth indicator, the least squares regressioncoefficient is 0.092 with standard error 0.013

• if log-earnings are regressed on the quarter-of-birthindicator, the coefficient is 0.0068 with standard error0.0027, so being born in the fourth quarter is associatedwith about 2

3% higher earnings.

• Wald estimator:0.00680.092

≈ 0.074

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Estimates Using the Wald Estimator

Estimate 95% lower 95% upper

Simple Wald Estimator

0.074 0.019 0.129

Multivariate TSLS Estimator

0.074 0.058 0.090

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Problem

• replace the actual quarter-of-birth variable by a randomlygenerated instrument that carries no information because itis unrelated to years of education.

• As first reported by Bound, Jaeger, and Baker (1995), theTSLS estimate incorrectly suggests that the data areinformative, indeed, very informative when there are manyinstruments.

• The 95% confidence interval: 0.042, 0.078• But the true estimate is 0!

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Method

• There are S strata with ns units in stratum s and Nsubjects in total

• YCsi is the control (T = 0) potential outcome for unit i instrata s

• Ytsi is the potential outcome for unit i in strata s if the unitreceived treatment t 6= 0

• Simplifying assumption (not necessary), effect isproportional:Ytsi − YCsi = βt .

• t is years of education beyond the minimum that arerequired by law

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Instruments

• In stratum s there is a preset, sorted, fixed list of nsinstrument settings hsj , j = 1, · · · ,ns, where hsj ≤ hs,j+1 ∀s, j

• h = (h11,h12, · · · ,h1,n1 ,h2,1, · · · ,hS,ns )T

• Instrument settings in h are randomly permuted withinstrata

• Assignment of instrument settings, z, is z=ph where p is astratified permutation matrix—i.e., an N × N block diagonalmatrix with S blocks, p1, · · · ,pS.

• Block ps is an ns × ns permutation matrix—i.e., ps is amatrix of 0s and 1s s.t. each row and column sum to 1

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Permutations

• Let Ω be the set of all stratified permutation matrices p, soΩ is a set containing |Ω| =

∏Ss=1 ns! matrices, where |Ω|

denotes the number of elements of the set Ω

• Pick a random P from Ω where Pr(P = p) =1|Ω| for each

p ∈ Ω

• Then Z = Ph is a random permutation of h within strata, sothe i th unit in stratum s receives instrument setting Zsi

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Outcomes

• For each z there is a tsiz for each unit who then has anoutcome YCsi + βtsiz

• Tsi is the dose, treatment value, for unit i in stratum s soTsi = tsiz

• Let Ysi be the response for this unit, so Ysi = YCsi + βTsi

• Write T = (T11, · · · ,T TS,ns

) andY = (Y11, · · · ,Y T

S,ns)

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Hypothesis Testing I

• We wish to test H0 : β = β0

• Let q(·) be a method of scoring response such as theirranks within strata

• Let ρ(Z) be some way of scoring the instrument settingssuch that ρ(ph) = pρ(h) for each p ∈ Ω

• The test statistic is U = q(Y− β0(T ))Tρ(Z)

• For appropriate scores, U can be Wilcoxon’s stratified ranksum statistic, the Hodges-Lehmann aligned rank statistic,the stratified Spearman rank correlation, etc

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Hypothesis Testing II

• If H0 were true, Y− β0T = YC would be fixed, not varyingwith Z: q(Y− β0T) = q(YC) = q would also be fixed

• If the null is false, Y− β0T = YC + (β − β0)T will be relatedto the dose T and related to Z

• Our test amounts to looking for an absence of arelationship between Y− β0T and Z.

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Exact Test

• An exact test computes q(Y− β0T), which is the fixedvalue q = q(YC), in which case U = qT Pρ(H)

• The chance that U ≤ u under H0 is the proportion of p ∈ Ωthat qT Pρ(H) ≤ u

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Comparison of instrumental variableestimates with uninformative data

Procedure 95% lower 95% upper

TSLS 0.042 0.078Permute ranks -1 1

Permute log-earnings -1 1

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Results with Uninformative Instrument

Dotted line is TSLS; solid line is randomization test using ranks; anddashed line is randomization test using the full observed data

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References

This treatment is based on:• G. W. Imbens and P. Rosenbaum (2005): “Robust,

Accurate Confidence Intervals with a Weak Instrument:Quarter of Birth and Education,” Journal of the RoyalStatistical Society, Series A, vol 168(1), 109–126.

• J. Angrist and Krueger (1991): “Does compulsory schoolattendance affect earnings?” QJE 1991; 106: 979–1019.

• Bound, Jaeger, and Baker (1995): “Problems withInstrumental Variables Estimation when the CorrelationBetween the Instruments and the Endogenous Regressorsis Weak,” JASA 90, June 1995, 443–450.

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Background Reading

• Angrist, J.D. & Krueger, A.B. 2001. Instrumental Variablesand the Search for Identification: From Supply andDemand to Natural Experiments. Journal of EconomicPerspectives, 15(4), 69-85.

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Example: Incumbency Advantage

• The purpose is to show that different identificationstrategies lead to different estimands.

• These differences are often subtle• Some research designs are so flawed, that even random

assignment would not identify the treatment effect ofinterest

• For details, see: Sekhon and Titiunik “Redistricting and thePersonal Vote: When Natural Experiments are NeitherNatural nor Experiments” http://sekhon.berkeley.edu/papers/incumb.pdf

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Incumbency Advantage (IA)

• Extensive literature on the Incumbency Advantage (IA) inthe U.S.

• One of the most studied topics in political science• Senior scholars e.g.:

Ansolabehere, Brady, Cain, Cox, Erikson, Ferejohn,Fiorina, Gelman, Jacobson, Nagler, Katz, Kernell, King,Levitt, Mayhew, Niemi, Snyder, Stewart

• Junior/junior senior scholars e.g.:Ashworth, Bueno de Mesquita, Carson, Desposato,Engstrom, Gordon, Huber, Hirano, Landa, Lee, Prior,Roberts

582 / 765

Incumbency Advantage (IA)

• Extensive literature on the Incumbency Advantage (IA) inthe U.S.

• One of the most studied topics in political science• Senior scholars e.g.:

Ansolabehere, Brady, Cain, Cox, Erikson, Ferejohn,Fiorina, Gelman, Jacobson, Nagler, Katz, Kernell, King,Levitt, Mayhew, Niemi, Snyder, Stewart

• Junior/junior senior scholars e.g.:Ashworth, Bueno de Mesquita, Carson, Desposato,Engstrom, Gordon, Huber, Hirano, Landa, Lee, Prior,Roberts

583 / 765

So much work

“I am convinced that one more article demonstratingthat House incumbents tend to win reelection willinduce a spontaneous primal scream among allcongressional scholars across the nation.”Charles O. Jones (1981)

A Cautionary Tale. This should be an easy problem: clearquestion, a lot of data, and what people thought was a cleanidentification strategy.

584 / 765

So much work

“I am convinced that one more article demonstratingthat House incumbents tend to win reelection willinduce a spontaneous primal scream among allcongressional scholars across the nation.”Charles O. Jones (1981)

A Cautionary Tale. This should be an easy problem: clearquestion, a lot of data, and what people thought was a cleanidentification strategy.

585 / 765

So much work

“I am convinced that one more article demonstratingthat House incumbents tend to win reelection willinduce a spontaneous primal scream among allcongressional scholars across the nation.”Charles O. Jones (1981)

A Cautionary Tale. This should be an easy problem: clearquestion, a lot of data, and what people thought was a cleanidentification strategy.

586 / 765

Incumbency Advantage (IA)

• The Personal Incumbency Advantage is the candidatespecific advantage that results from the benefits of officeholding—e.g:• name recognition• constituency service• public position taking• providing pork

• Different from the Incumbent Party Advantage:• marginally, voters have a preference for remaining with the

same party they had before (D. S. Lee, 2008b)• parties have organizational advantages in some districts

587 / 765

Incumbency Advantage (IA)

• The Personal Incumbency Advantage is the candidatespecific advantage that results from the benefits of officeholding—e.g:• name recognition• constituency service• public position taking• providing pork

• Different from the Incumbent Party Advantage:• marginally, voters have a preference for remaining with the

same party they had before (D. S. Lee, 2008b)• parties have organizational advantages in some districts

588 / 765

Problems with Estimating the IA

• Empirical work agrees on there being an advantage, butmuch disagreement about the sources of the IA

• Theoretical work argues that selection issues are thecause of the positive estimates of IA(e.g., G. W. Cox and Katz, 2002; J. Zaller, 1998):

Main concerns: strategic candidate entry and exit, andsurvival bias

589 / 765

Exploiting Redistricting

• Ansolabehere, Snyder, and Stewart (2000) use thevariation brought about by decennial redistricting plans toidentify the causal effect of the personal IA

• After redistricting, most incumbents face districts thatcontain a combination of old and new voters

• They compare an incumbent’s vote share in the new part ofthe district with her vote share in the old part of the district.

• They hope to avoid some of the selection issues becausethe electoral environment is constant

• They analyze U.S. House elections at the county level from1872 to 1990

• Others have also used this design and found similar resultsDesposato and Petrocik (2003); Carson, Engstrom andRoberts (2007).

590 / 765

Exploiting Redistricting

• Ansolabehere, Snyder, and Stewart (2000) use thevariation brought about by decennial redistricting plans toidentify the causal effect of the personal IA

• After redistricting, most incumbents face districts thatcontain a combination of old and new voters

• They compare an incumbent’s vote share in the new part ofthe district with her vote share in the old part of the district.

• They hope to avoid some of the selection issues becausethe electoral environment is constant

• They analyze U.S. House elections at the county level from1872 to 1990

• Others have also used this design and found similar resultsDesposato and Petrocik (2003); Carson, Engstrom andRoberts (2007).

591 / 765

Exploiting Redistricting

• Ansolabehere, Snyder, and Stewart (2000) use thevariation brought about by decennial redistricting plans toidentify the causal effect of the personal IA

• After redistricting, most incumbents face districts thatcontain a combination of old and new voters

• They compare an incumbent’s vote share in the new part ofthe district with her vote share in the old part of the district.

• They hope to avoid some of the selection issues becausethe electoral environment is constant

• They analyze U.S. House elections at the county level from1872 to 1990

• Others have also used this design and found similar resultsDesposato and Petrocik (2003); Carson, Engstrom andRoberts (2007).

592 / 765

Exploiting Redistricting

• Ansolabehere, Snyder, and Stewart (2000) use thevariation brought about by decennial redistricting plans toidentify the causal effect of the personal IA

• After redistricting, most incumbents face districts thatcontain a combination of old and new voters

• They compare an incumbent’s vote share in the new part ofthe district with her vote share in the old part of the district.

• They hope to avoid some of the selection issues becausethe electoral environment is constant

• They analyze U.S. House elections at the county level from1872 to 1990

• Others have also used this design and found similar resultsDesposato and Petrocik (2003); Carson, Engstrom andRoberts (2007).

593 / 765

Using Redistricting to EstimateIncumbency

• We propose a new method for estimating incumbencyadvantage (IA)

• Previous redistricting designs result in biased estimates ofincumbency

• The wrong potential outcomes were used• Leads to bias even if voters were redistricted randomly!• In practice, a selection on observables assumption must

be made• The selection on observables assumption in prior work is

theoretically implausible and fails a placebo test• Use redistricting to answer other questions as well—e.g.,

how voters respond to candidates’ race and ethnicity

594 / 765

Using Redistricting to EstimateIncumbency

• We propose a new method for estimating incumbencyadvantage (IA)

• Previous redistricting designs result in biased estimates ofincumbency

• The wrong potential outcomes were used• Leads to bias even if voters were redistricted randomly!• In practice, a selection on observables assumption must

be made• The selection on observables assumption in prior work is

theoretically implausible and fails a placebo test• Use redistricting to answer other questions as well—e.g.,

how voters respond to candidates’ race and ethnicity

595 / 765

Using Redistricting to EstimateIncumbency

• We propose a new method for estimating incumbencyadvantage (IA)

• Previous redistricting designs result in biased estimates ofincumbency

• The wrong potential outcomes were used• Leads to bias even if voters were redistricted randomly!• In practice, a selection on observables assumption must

be made• The selection on observables assumption in prior work is

theoretically implausible and fails a placebo test• Use redistricting to answer other questions as well—e.g.,

how voters respond to candidates’ race and ethnicity

596 / 765

Using Redistricting to EstimateIncumbency

• We propose a new method for estimating incumbencyadvantage (IA)

• Previous redistricting designs result in biased estimates ofincumbency

• The wrong potential outcomes were used• Leads to bias even if voters were redistricted randomly!• In practice, a selection on observables assumption must

be made• The selection on observables assumption in prior work is

theoretically implausible and fails a placebo test• Use redistricting to answer other questions as well—e.g.,

how voters respond to candidates’ race and ethnicity

597 / 765

Where we are Going

One interpretation of our results:• No (TX) or little (CA) personal incumbency advantage in

U.S. House elections• Significant and large incumbent party effects

Results are more consistent with:• Voters learn the type of new incumbent very quickly• Unless there is a mismatch between the partisanship of

voters and their incumbent. Then, voters take more time tolearn the type of their incumbent

598 / 765

Where we are Going

One interpretation of our results:• No (TX) or little (CA) personal incumbency advantage in

U.S. House elections• Significant and large incumbent party effects

Results are more consistent with:• Voters learn the type of new incumbent very quickly• Unless there is a mismatch between the partisanship of

voters and their incumbent. Then, voters take more time tolearn the type of their incumbent

599 / 765

Where we are Going

One interpretation of our results:• No (TX) or little (CA) personal incumbency advantage in

U.S. House elections• Significant and large incumbent party effects

Results are more consistent with:• Voters learn the type of new incumbent very quickly• Unless there is a mismatch between the partisanship of

voters and their incumbent. Then, voters take more time tolearn the type of their incumbent

600 / 765

Baseline Vote Share for Incumbent House Member, Texas

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.2

0.4

0.6

0.8

1.0

Redistricted

Non

red

istr

icte

d

601 / 765

Baseline Vote Share for Incumbent House Member, California

602 / 765

Lagged Democratic Win in RDD: 1st and 2nd-order polynomial

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Lagged Democratic Win in RDD: 3rd and 4th-order polynomial

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Lagged Democratic Win in RDD: 8th-order polynomial

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Lagged Democratic Win in RDD: local regression

606 / 765

The Basic Setup

• Redistricting induces variation in two dimensions:• time: voters vote both before and after redistricting• cross-sectional: some voters are moved to a different

district while others stay in the district

• Can estimate IA by comparing the behavior of voters whoare:• moved to a new district (new voters)• to the behavior of voters whose district remains unchanged

(old voters).

607 / 765

The Basic Setup

• Redistricting induces variation in two dimensions:• time: voters vote both before and after redistricting• cross-sectional: some voters are moved to a different

district while others stay in the district

• Can estimate IA by comparing the behavior of voters whoare:• moved to a new district (new voters)• to the behavior of voters whose district remains unchanged

(old voters).

608 / 765

New Voters vs. Old Voters

Although this comparison seems intuitive, there are importantcomplications:• New voters are naturally defined as the voters whose

district changes between one election and another• BUT there is an ambiguity in the way in which old voters

are defined• Old voters could be either:

• the electorate of the district to which new voters are moved(new neighbors), or

• the electorate of the district to which new voters belongedbefore redistricting occurred (old neighbors)

609 / 765

New Voters vs. Old Voters

Although this comparison seems intuitive, there are importantcomplications:• New voters are naturally defined as the voters whose

district changes between one election and another• BUT there is an ambiguity in the way in which old voters

are defined• Old voters could be either:

• the electorate of the district to which new voters are moved(new neighbors), or

• the electorate of the district to which new voters belongedbefore redistricting occurred (old neighbors)

610 / 765

Before one-time redistricting

District A

District B

B's original voters

A's original voters

611 / 765

One-time redistricting

District A

District B

Voters to be redistricted

at election t B's original voters

A's original voters

612 / 765

After one-time redistricting

District A

District B

Old neighbors

New voters

New neighbors

613 / 765

First identification strategy:old-neighbors design

District B

New neighbors

Not used

District A

Control Treated

District B

New voters

Old neighbors

614 / 765

Second identification strategy:new neighbors design

District B

New neighbors

Control

District A

Not used Treated

New voters

Old neighbors

615 / 765

Two Obvious Designs

There are then two obvious designs under the randomizationwhere everyone is in a certain district at t − 1 and someprecincts are randomly moved to another district at t .• Compare new voters with old neighbors• Compare new voters with new neighbors—this is the

Ansolabehere, Snyder, and Stewart (2000) design• Randomization ensures exchangeability for the first design,

but not the second• This occurs because the history of new voters with new

neighbors is not balanced by randomization

616 / 765

Two Obvious Designs

There are then two obvious designs under the randomizationwhere everyone is in a certain district at t − 1 and someprecincts are randomly moved to another district at t .• Compare new voters with old neighbors• Compare new voters with new neighbors—this is the

Ansolabehere, Snyder, and Stewart (2000) design• Randomization ensures exchangeability for the first design,

but not the second• This occurs because the history of new voters with new

neighbors is not balanced by randomization

617 / 765

Two Obvious Designs

There are then two obvious designs under the randomizationwhere everyone is in a certain district at t − 1 and someprecincts are randomly moved to another district at t .• Compare new voters with old neighbors• Compare new voters with new neighbors—this is the

Ansolabehere, Snyder, and Stewart (2000) design• Randomization ensures exchangeability for the first design,

but not the second• This occurs because the history of new voters with new

neighbors is not balanced by randomization

618 / 765

Before two-time redistricting(election t-1)

District A

District B

B's original voters

A's original voters

619 / 765

District A

District B

Voters to be redistricted

at election t B's original voters

A's original voters

Before two-time redistricting(election t-1)

620 / 765

Two-time redistricting(election t)

District A

District B

B's original voters

A's original votersEarly

new voters

(arrived at t)

Voters to be redistricted at

election t+1

621 / 765

Two-time redistricting(election t+1)

District A

District B

Earlynew voters

(arrived at t)

B's original voters

A's original votersLate

new voters

(arrived at t+1)

622 / 765

After two-time redistricting(election t+1)

District A

District B

Earlynew voters

(arrived at t)

B's original voters

A's original votersLate

new voters

(arrived at t+1)

Control

Not used

Treated

Not used

623 / 765

More Formally

• Let Ti be equal to 1 if precinct i is moved from one districtto another before election t and equal to 0 if it is not moved

• Let Di be equal to 1 if precinct i has new voters in itsdistrict at t and equal to 0 otherwise

• Let Y00 (i , t) be precinct i ’s outcome Ti = 0 and Di = 0 it isnot moved and does not have new neighbors

• Let Y01 (i , t) be precinct i ’s outcome if Ti = 0 and Di = 1 theprecinct is not moved and has new neighbors

• Let Y11 (i , t) be precinct i ’s outcome if Ti = 1 and Di = 1 theprecinct is moved and has new neighbors

624 / 765

More Formally

• Let Ti be equal to 1 if precinct i is moved from one districtto another before election t and equal to 0 if it is not moved

• Let Di be equal to 1 if precinct i has new voters in itsdistrict at t and equal to 0 otherwise

• Let Y00 (i , t) be precinct i ’s outcome Ti = 0 and Di = 0 it isnot moved and does not have new neighbors

• Let Y01 (i , t) be precinct i ’s outcome if Ti = 0 and Di = 1 theprecinct is not moved and has new neighbors

• Let Y11 (i , t) be precinct i ’s outcome if Ti = 1 and Di = 1 theprecinct is moved and has new neighbors

625 / 765

More Formally

• Let Ti be equal to 1 if precinct i is moved from one districtto another before election t and equal to 0 if it is not moved

• Let Di be equal to 1 if precinct i has new voters in itsdistrict at t and equal to 0 otherwise

• Let Y00 (i , t) be precinct i ’s outcome Ti = 0 and Di = 0 it isnot moved and does not have new neighbors

• Let Y01 (i , t) be precinct i ’s outcome if Ti = 0 and Di = 1 theprecinct is not moved and has new neighbors

• Let Y11 (i , t) be precinct i ’s outcome if Ti = 1 and Di = 1 theprecinct is moved and has new neighbors

626 / 765

More Formally

• Let Ti be equal to 1 if precinct i is moved from one districtto another before election t and equal to 0 if it is not moved

• Let Di be equal to 1 if precinct i has new voters in itsdistrict at t and equal to 0 otherwise

• Let Y00 (i , t) be precinct i ’s outcome Ti = 0 and Di = 0 it isnot moved and does not have new neighbors

• Let Y01 (i , t) be precinct i ’s outcome if Ti = 0 and Di = 1 theprecinct is not moved and has new neighbors

• Let Y11 (i , t) be precinct i ’s outcome if Ti = 1 and Di = 1 theprecinct is moved and has new neighbors

627 / 765

More Formally

• Let Ti be equal to 1 if precinct i is moved from one districtto another before election t and equal to 0 if it is not moved

• Let Di be equal to 1 if precinct i has new voters in itsdistrict at t and equal to 0 otherwise

• Let Y00 (i , t) be precinct i ’s outcome Ti = 0 and Di = 0 it isnot moved and does not have new neighbors

• Let Y01 (i , t) be precinct i ’s outcome if Ti = 0 and Di = 1 theprecinct is not moved and has new neighbors

• Let Y11 (i , t) be precinct i ’s outcome if Ti = 1 and Di = 1 theprecinct is moved and has new neighbors

628 / 765

Fundamental Problem of CausalInference

For each precinct, we observe only one of its three potentialoutcomes:

Y (i , t) = Y00 (i , t) · (1− Ti) · (1− Di) +

Y01 (i , t) · (1− Ti) · Di +

Y11 (i , t) · Ti · Di

We can estimate two different ATT’s:

ATT0 ≡ E [Y11 (i , t)− Y00 (i , t) | Ti = 1,Di = 1]

ATT1 ≡ E [Y11 (i , t)− Y01 (i , t) | Ti = 1,Di = 1]

629 / 765

Fundamental Problem of CausalInference

For each precinct, we observe only one of its three potentialoutcomes:

Y (i , t) = Y00 (i , t) · (1− Ti) · (1− Di) +

Y01 (i , t) · (1− Ti) · Di +

Y11 (i , t) · Ti · Di

We can estimate two different ATT’s:

ATT0 ≡ E [Y11 (i , t)− Y00 (i , t) | Ti = 1,Di = 1]

ATT1 ≡ E [Y11 (i , t)− Y01 (i , t) | Ti = 1,Di = 1]

630 / 765

Fundamental Problem of CausalInference

For each precinct, we observe only one of its three potentialoutcomes:

Y (i , t) = Y00 (i , t) · (1− Ti) · (1− Di) +

Y01 (i , t) · (1− Ti) · Di +

Y11 (i , t) · Ti · Di

We can estimate two different ATT’s:

ATT0 ≡ E [Y11 (i , t)− Y00 (i , t) | Ti = 1,Di = 1]

ATT1 ≡ E [Y11 (i , t)− Y01 (i , t) | Ti = 1,Di = 1]

631 / 765

Identification of ATT0

ATT0 ≡ E [Y11 (i , t)− Y00 (i , t) | Ti = 1,Di = 1]

Is identified if:

E [Y00 (i , t) | Ti = 1,Di = 1] = E [Y00 (i , t) | Ti = 0,Di = 0]

ATT0 requires that voters who stay in A and voters who aremoved from A to B would have the same average outcomes ifthey hadn’t been moved.

632 / 765

Identification of ATT0

ATT0 ≡ E [Y11 (i , t)− Y00 (i , t) | Ti = 1,Di = 1]

Is identified if:

E [Y00 (i , t) | Ti = 1,Di = 1] = E [Y00 (i , t) | Ti = 0,Di = 0]

ATT0 requires that voters who stay in A and voters who aremoved from A to B would have the same average outcomes ifthey hadn’t been moved.

633 / 765

Identification of ATT1

ATT1 ≡ E [Y11 (i , t)− Y01 (i , t) | Ti = 1,Di = 1]

Is identified if:

E [Y01 (i , t) | Ti = 1,Di = 1] = E [Y01 (i , t) | Ti = 0,Di = 1]

ATT1 requires that voters who are originally in B and voters whoare moved from A to B would have the same average outcomesif A’s voters would not have been moved even though theywould be in different districts.

Randomization does not imply that B’s old voters are a validcounterfactual for B’s new voters

634 / 765

Identification of ATT1

ATT1 ≡ E [Y11 (i , t)− Y01 (i , t) | Ti = 1,Di = 1]

Is identified if:

E [Y01 (i , t) | Ti = 1,Di = 1] = E [Y01 (i , t) | Ti = 0,Di = 1]

ATT1 requires that voters who are originally in B and voters whoare moved from A to B would have the same average outcomesif A’s voters would not have been moved even though theywould be in different districts.

Randomization does not imply that B’s old voters are a validcounterfactual for B’s new voters

635 / 765

Identification of ATT1

Required for identification:

E [Y01 (i , t) | Ti = 1,Di = 1] = E [Y01 (i , t) | Ti = 0,Di = 1]

Since this is not guaranteed to hold under randomization, onemust make the following assumption (even with randomization):

E [Y1 (i , t) | Ti = 1,Di = 1,X ] = E [Y1 (i , t) | Ti = 0,Di = 1,X ]

where X is a vector of observable characteristics which correctfor imbalance on district specific history.

636 / 765

The Best Design: MultipleRedistrictings

Let Wi,t+1 = 1 if precinct i is moved from district A to district Bat election t + 1.

Let Wi,t+1 = 0 if precinct i is moved from A to B at election t .

Let Y0 (i , t + 1) denote the outcome of i at election t + 1 ifWi,t+1 = 0

Let Y1 (i , t + 1) denote the outcome of i at election t + 1 ifWi,t+1 = 1

The parameter of interest ATTB is

ATTB ≡ E[Y1 (i , t + 1)− Y0 (i , t + 1) |Wi,t+1 = 1

]

637 / 765

The Best Design: MultipleRedistrictings

Let Wi,t+1 = 1 if precinct i is moved from district A to district Bat election t + 1.

Let Wi,t+1 = 0 if precinct i is moved from A to B at election t .

Let Y0 (i , t + 1) denote the outcome of i at election t + 1 ifWi,t+1 = 0

Let Y1 (i , t + 1) denote the outcome of i at election t + 1 ifWi,t+1 = 1

The parameter of interest ATTB is

ATTB ≡ E[Y1 (i , t + 1)− Y0 (i , t + 1) |Wi,t+1 = 1

]

638 / 765

The Best Design: MultipleRedistrictings

Let Wi,t+1 = 1 if precinct i is moved from district A to district Bat election t + 1.

Let Wi,t+1 = 0 if precinct i is moved from A to B at election t .

Let Y0 (i , t + 1) denote the outcome of i at election t + 1 ifWi,t+1 = 0

Let Y1 (i , t + 1) denote the outcome of i at election t + 1 ifWi,t+1 = 1

The parameter of interest ATTB is

ATTB ≡ E[Y1 (i , t + 1)− Y0 (i , t + 1) |Wi,t+1 = 1

]

639 / 765

The Best Design: MultipleRedistrictings

Let Wi,t+1 = 1 if precinct i is moved from district A to district Bat election t + 1.

Let Wi,t+1 = 0 if precinct i is moved from A to B at election t .

Let Y0 (i , t + 1) denote the outcome of i at election t + 1 ifWi,t+1 = 0

Let Y1 (i , t + 1) denote the outcome of i at election t + 1 ifWi,t+1 = 1

The parameter of interest ATTB is

ATTB ≡ E[Y1 (i , t + 1)− Y0 (i , t + 1) |Wi,t+1 = 1

]

640 / 765

The Best Design: MultipleRedistrictings

Let Wi,t+1 = 1 if precinct i is moved from district A to district Bat election t + 1.

Let Wi,t+1 = 0 if precinct i is moved from A to B at election t .

Let Y0 (i , t + 1) denote the outcome of i at election t + 1 ifWi,t+1 = 0

Let Y1 (i , t + 1) denote the outcome of i at election t + 1 ifWi,t+1 = 1

The parameter of interest ATTB is

ATTB ≡ E[Y1 (i , t + 1)− Y0 (i , t + 1) |Wi,t+1 = 1

]

641 / 765

The parameter of interest ATTB is

ATTB ≡ E[Y1 (i , t + 1)− Y0 (i , t + 1) |Wi,t+1 = 1

]which is identified by

E[Y0 (i , t + 1) |Wi,t+1 = 1

]= E

[Y0 (i , t + 1) |Wi,t+1 = 0

]By randomization we have

E[Y0 (i , t − 1) |Wi,t+1 = 1

]= E

[Y0 (i , t − 1) |Wi,t+1 = 0

]Randomization along with the following stability assumptionprovides identification:

E[Y0 (i , t + 1)− Y0 (i , t − 1) |Wi,t+1 = 1

]=

E[Y0 (i , t + 1)− Y0 (i , t − 1) |Wi,t+1 = 0

]

642 / 765

The parameter of interest ATTB is

ATTB ≡ E[Y1 (i , t + 1)− Y0 (i , t + 1) |Wi,t+1 = 1

]which is identified by

E[Y0 (i , t + 1) |Wi,t+1 = 1

]= E

[Y0 (i , t + 1) |Wi,t+1 = 0

]By randomization we have

E[Y0 (i , t − 1) |Wi,t+1 = 1

]= E

[Y0 (i , t − 1) |Wi,t+1 = 0

]Randomization along with the following stability assumptionprovides identification:

E[Y0 (i , t + 1)− Y0 (i , t − 1) |Wi,t+1 = 1

]=

E[Y0 (i , t + 1)− Y0 (i , t − 1) |Wi,t+1 = 0

]

643 / 765

The parameter of interest ATTB is

ATTB ≡ E[Y1 (i , t + 1)− Y0 (i , t + 1) |Wi,t+1 = 1

]which is identified by

E[Y0 (i , t + 1) |Wi,t+1 = 1

]= E

[Y0 (i , t + 1) |Wi,t+1 = 0

]By randomization we have

E[Y0 (i , t − 1) |Wi,t+1 = 1

]= E

[Y0 (i , t − 1) |Wi,t+1 = 0

]Randomization along with the following stability assumptionprovides identification:

E[Y0 (i , t + 1)− Y0 (i , t − 1) |Wi,t+1 = 1

]=

E[Y0 (i , t + 1)− Y0 (i , t − 1) |Wi,t+1 = 0

]

644 / 765

Selection on Observables

• Redistricting was not actually randomized• We, like previous work, make a selection on observables

assumption:

E[Y0 (i , t − 1) |Wi,t+1 = 1,X

]= E

[Y0 (i , t − 1) |Wi,t+1 = 0,X

]• We conduct a placebo test which is enabled by our

two-redistricting research design• We use the multiple redistrictings which were done in

Texas between 2002 and 2006. We have mergedVTD-level election, registration and census data withcandidate data such as quality measures and Nominate.

645 / 765

Selection on Observables

• Redistricting was not actually randomized• We, like previous work, make a selection on observables

assumption:

E[Y0 (i , t − 1) |Wi,t+1 = 1,X

]= E

[Y0 (i , t − 1) |Wi,t+1 = 0,X

]• We conduct a placebo test which is enabled by our

two-redistricting research design• We use the multiple redistrictings which were done in

Texas between 2002 and 2006. We have mergedVTD-level election, registration and census data withcandidate data such as quality measures and Nominate.

646 / 765

Selection on Observables

• Redistricting was not actually randomized• We, like previous work, make a selection on observables

assumption:

E[Y0 (i , t − 1) |Wi,t+1 = 1,X

]= E

[Y0 (i , t − 1) |Wi,t+1 = 0,X

]• We conduct a placebo test which is enabled by our

two-redistricting research design• We use the multiple redistrictings which were done in

Texas between 2002 and 2006. We have mergedVTD-level election, registration and census data withcandidate data such as quality measures and Nominate.

647 / 765

Placebo Test: Key Covariates

• Treated: VTDs to be redistricted in 2004• Control: VTDs to remain in same district in 2004• Placebo test:

• Match in 2000 Congressional district• Match in 2000 VTD-level covariates: presidential vote,

senate vote, house vote, governor vote, turnout,registration, etc.

• In 2002 there should be no significant difference betweenour treated and control groups in outcomes

• Our set of covariates pass the test• “Presidential vote” fails the test

648 / 765

Placebo Test: Key Covariates

• Treated: VTDs to be redistricted in 2004• Control: VTDs to remain in same district in 2004• Placebo test:

• Match in 2000 Congressional district• Match in 2000 VTD-level covariates: presidential vote,

senate vote, house vote, governor vote, turnout,registration, etc.

• In 2002 there should be no significant difference betweenour treated and control groups in outcomes

• Our set of covariates pass the test• “Presidential vote” fails the test

649 / 765

Placebo Test: Key Covariates

• Treated: VTDs to be redistricted in 2004• Control: VTDs to remain in same district in 2004• Placebo test:

• Match in 2000 Congressional district• Match in 2000 VTD-level covariates: presidential vote,

senate vote, house vote, governor vote, turnout,registration, etc.

• In 2002 there should be no significant difference betweenour treated and control groups in outcomes

• Our set of covariates pass the test• “Presidential vote” fails the test

650 / 765

Placebo Test: Key Covariates

• Treated: VTDs to be redistricted in 2004• Control: VTDs to remain in same district in 2004• Placebo test:

• Match in 2000 Congressional district• Match in 2000 VTD-level covariates: presidential vote,

senate vote, house vote, governor vote, turnout,registration, etc.

• In 2002 there should be no significant difference betweenour treated and control groups in outcomes

• Our set of covariates pass the test• “Presidential vote” fails the test

651 / 765

Placebo Test: Key Covariates

• Treated: VTDs to be redistricted in 2004• Control: VTDs to remain in same district in 2004• Placebo test:

• Match in 2000 Congressional district• Match in 2000 VTD-level covariates: presidential vote,

senate vote, house vote, governor vote, turnout,registration, etc.

• In 2002 there should be no significant difference betweenour treated and control groups in outcomes

• Our set of covariates pass the test• “Presidential vote” fails the test

652 / 765

Placebo Test: Key Covariates

• Treated: VTDs to be redistricted in 2004• Control: VTDs to remain in same district in 2004• Placebo test:

• Match in 2000 Congressional district• Match in 2000 VTD-level covariates: presidential vote,

senate vote, house vote, governor vote, turnout,registration, etc.

• In 2002 there should be no significant difference betweenour treated and control groups in outcomes

• Our set of covariates pass the test• “Presidential vote” fails the test

653 / 765

Placebo Test: Key Covariates

• Treated: VTDs to be redistricted in 2004• Control: VTDs to remain in same district in 2004• Placebo test:

• Match in 2000 Congressional district• Match in 2000 VTD-level covariates: presidential vote,

senate vote, house vote, governor vote, turnout,registration, etc.

• In 2002 there should be no significant difference betweenour treated and control groups in outcomes

• Our set of covariates pass the test• “Presidential vote” fails the test

654 / 765

Estimation

• Genetic Matching (GenMatch) is used to achieve covariatebalance

• Hodges-Lehmann interval estimation, bivariateoverdispersed GLM estimation and Abadie-Imbens CIs arecalculated post-matching. Substantive inferences remainunchanged.

• Hodges-Lehmann intervals are presented in the following.

655 / 765

Estimation

• Genetic Matching (GenMatch) is used to achieve covariatebalance

• Hodges-Lehmann interval estimation, bivariateoverdispersed GLM estimation and Abadie-Imbens CIs arecalculated post-matching. Substantive inferences remainunchanged.

• Hodges-Lehmann intervals are presented in the following.

656 / 765

Balance Tests, TX

Before Matching After MatchingVariable mean diff D-stat KS-pval mean diff D-stat KS-pval

Dem Pres. vote ’00 .0447 .100 0.00 .00459 .0337 0.953Dem House vote ’00 .159 .305 0.00 .00693 .0344 0.678Dem House vote ’98 .127 .340 0.00 .00585 .0368 0.996Dem Senate vote ’00 .0426 .120 0.00 .00576 .0317 0.846Dem Governor vote ’98 .0305 .0974 0.00 .00510 .0241 0.942Dem Att. Gen. vote ’98 .0353 .141 0.00 .00683 .0358 0.868Dem Compt. vote ’98 .0304 .208 0.00 .00499 .0373 0.994Voter turnout ’00 .0331 .102 0.00 .00607 .0327 0.943Voter turnout ’98 .028 .199 0.00 .0111 .0378 0.235Registration ’00 .0308 .157 0.00 .00736 .0608 0.601

The mean difference are the the simple differences between treatment and control, the D-statistic is the largestdifference in the empirical QQ-plot on the scale of the variable, and the KS-pvalue is from the bootstrappedKolmogrov-Smirnov test.

657 / 765

Placebo Tests with All Key Covariates,TX

Estimate 95% CI p.value

Incumbent vote ’02 0.00245 −0.00488 0.00954 0.513

Turnout ’02 0.00334 −0.00443 0.0112 0.412

658 / 765

Placebo Tests with All Key Covariates,TX

Estimate 95% CI p.value

Incumbent vote ’02 0.00245 −0.00488 0.00954 0.513

Turnout ’02 0.00334 −0.00443 0.0112 0.412

659 / 765

Placebo Test TX: Presidential vote only

What happens if placebo test is done matching on presidentialvote only?• Achieve excellent balance on 2000 presidential vote• But there is still a significant effect in 2002, before

redistricting occurs• Past presidential vote is not sufficient to satisfy this

placebo test.• The significant effect estimated in the placebo test includes

the actual estimate of Ansolabehere, Snyder, and Stewart(2000) in its confidence interval.

660 / 765

Placebo Test TX: Presidential vote only

What happens if placebo test is done matching on presidentialvote only?• Achieve excellent balance on 2000 presidential vote• But there is still a significant effect in 2002, before

redistricting occurs• Past presidential vote is not sufficient to satisfy this

placebo test.• The significant effect estimated in the placebo test includes

the actual estimate of Ansolabehere, Snyder, and Stewart(2000) in its confidence interval.

661 / 765

Placebo Test TX: Presidential vote only

What happens if placebo test is done matching on presidentialvote only?• Achieve excellent balance on 2000 presidential vote• But there is still a significant effect in 2002, before

redistricting occurs• Past presidential vote is not sufficient to satisfy this

placebo test.• The significant effect estimated in the placebo test includes

the actual estimate of Ansolabehere, Snyder, and Stewart(2000) in its confidence interval.

662 / 765

Placebo Test TX: Presidential vote only

What happens if placebo test is done matching on presidentialvote only?• Achieve excellent balance on 2000 presidential vote• But there is still a significant effect in 2002, before

redistricting occurs• Past presidential vote is not sufficient to satisfy this

placebo test.• The significant effect estimated in the placebo test includes

the actual estimate of Ansolabehere, Snyder, and Stewart(2000) in its confidence interval.

663 / 765

cc

QQ plot:2000 Presidential Vote

0.2 0.3 0.4 0.5 0.6 0.7 0.8

0.2

0.3

0.4

0.5

0.6

0.7

0.8

placebo control

plac

ebo

trea

ted

QQ Plot:2002 Incumbent Vote

0.4 0.5 0.6 0.7 0.8 0.9 1.0

0.4

0.5

0.6

0.7

0.8

0.9

1.0

placebo control

plac

ebo

trea

ted

664 / 765

Placebo Test TX: Presidential vote only

Estimate 95% CI p.value

Incumbent vote ’02 0.0285 0.0160 0.0413 0.00

Turnout ’02 0.00246 −0.0180 0.0218 0.831

665 / 765

Placebo Test TX: Presidential vote only

Estimate 95% CI p.value

Incumbent vote ’02 0.0285 0.0160 0.0413 0.00

Turnout ’02 0.00246 −0.0180 0.0218 0.831

666 / 765

Incumbency Advantage TX: SameParty

Estimate 95% CI p.value

Two-Time Redistricting DesignIncumbent vote ’04 0.00637 −0.00428 0.0177 0.254Incumbent vote ’06 0.00843 −0.00938 0.0258 0.457

One-Time Redistricting DesignIncumbent vote ’04 0.00214 −0.00807 0.0124 0.690Incumbent vote ’06 0.00472 −0.00539 0.0149 0.378

667 / 765

Incumbency Advantage TX: SameParty

Estimate 95% CI p.value

Two-Time Redistricting DesignIncumbent vote ’04 0.00637 −0.00428 0.0177 0.254Incumbent vote ’06 0.00843 −0.00938 0.0258 0.457

One-Time Redistricting DesignIncumbent vote ’04 0.00214 −0.00807 0.0124 0.690Incumbent vote ’06 0.00472 −0.00539 0.0149 0.378

668 / 765

Incumbency Advantage TX: DifferentParty

Estimate 95% CI p.value

Incumbent vote ’04 0.119 0.0595 0.191 0.000

Incumbent vote ’06 0.0389 0.00973 0.0692 0.0106

669 / 765

Incumbency Advantage CA

Estimate 95% CI p.value

Same-Party, One-Time Redistricting DesignIncumb vote ’02 0.0219 0.0171 0.0266 0.000Incumb vote ’04 0.0240 0.0195 0.0284 0.000Incumb vote ’06 −0.0072 −0.0129 −0.0015 0.0125

Different-Party, One-Time Redistricting DesignIncumb vote ’02 0.1025 0.0925 0.1122 0.000Incumb vote ’04 0.1020 0.0932 0.1109 0.000Incumb vote ’06 0.0307 0.0186 0.0428 0.000

670 / 765

Summary

• Our results are consistent with theoretical arguments thatexisting positive estimates are plagued by selectionproblems (Ashworth and Bueno de Mesquita, 2007;G. W. Cox and Katz, 2002; J. Zaller, 1998).

• Voters learn the type of new incumbent very quickly• Find a significant incumbent party effect. Consistent with

the results of Lee (2008).• Finding a positive incumbency effect is much less common

outside of the U.S. Some work even documents a negativeincumbency effect (Linden, 2004; Uppal, 2005).

671 / 765

Summary

• Our results are consistent with theoretical arguments thatexisting positive estimates are plagued by selectionproblems (Ashworth and Bueno de Mesquita, 2007;G. W. Cox and Katz, 2002; J. Zaller, 1998).

• Voters learn the type of new incumbent very quickly• Find a significant incumbent party effect. Consistent with

the results of Lee (2008).• Finding a positive incumbency effect is much less common

outside of the U.S. Some work even documents a negativeincumbency effect (Linden, 2004; Uppal, 2005).

672 / 765

Summary

• Our results are consistent with theoretical arguments thatexisting positive estimates are plagued by selectionproblems (Ashworth and Bueno de Mesquita, 2007;G. W. Cox and Katz, 2002; J. Zaller, 1998).

• Voters learn the type of new incumbent very quickly• Find a significant incumbent party effect. Consistent with

the results of Lee (2008).• Finding a positive incumbency effect is much less common

outside of the U.S. Some work even documents a negativeincumbency effect (Linden, 2004; Uppal, 2005).

673 / 765

A Cautionary Tale

• This should be an easy problem: clear question, a lot ofdata, and what people thought was a clean identificationstrategy

• “Natural experiments” require careful theoretical andstatistical work to make valid inferences

• Real experiments require that researchers a priori designthe study so that randomization will ensure theidentification of the causal effect

• Different ideal experiments imply different identifyingassumptions and hence different experimental designs

674 / 765

A Cautionary Tale

• This should be an easy problem: clear question, a lot ofdata, and what people thought was a clean identificationstrategy

• “Natural experiments” require careful theoretical andstatistical work to make valid inferences

• Real experiments require that researchers a priori designthe study so that randomization will ensure theidentification of the causal effect

• Different ideal experiments imply different identifyingassumptions and hence different experimental designs

675 / 765

A Cautionary Tale

• This should be an easy problem: clear question, a lot ofdata, and what people thought was a clean identificationstrategy

• “Natural experiments” require careful theoretical andstatistical work to make valid inferences

• Real experiments require that researchers a priori designthe study so that randomization will ensure theidentification of the causal effect

• Different ideal experiments imply different identifyingassumptions and hence different experimental designs

676 / 765

Hodges-Lehmann estimator

• Assume additive treatment effect• Subtract the hypothesized treatment effect from the treated

responses, and test hypothesis using these adjustedresponses

• Hodges-Lehmann (HL) estimate is the value that whensubtracted from the responses yields adjusted responsesthat are free of treatment effect, in the sense that the(Wilcoxon Signed Rank) test statistic equals its expectationin the absence of treatment effect

• Wilcoxon Signed Rank test statistic: form matchedtreated-control pairs, rank absolute value of treated-controldifferences within matched pairs, sum ranks for pairs inwhich treated response is higher than control response

677 / 765

Hodges-Lehmann estimator

• Assume additive treatment effect• Subtract the hypothesized treatment effect from the treated

responses, and test hypothesis using these adjustedresponses

• Hodges-Lehmann (HL) estimate is the value that whensubtracted from the responses yields adjusted responsesthat are free of treatment effect, in the sense that the(Wilcoxon Signed Rank) test statistic equals its expectationin the absence of treatment effect

• Wilcoxon Signed Rank test statistic: form matchedtreated-control pairs, rank absolute value of treated-controldifferences within matched pairs, sum ranks for pairs inwhich treated response is higher than control response

678 / 765

Hodges-Lehmann estimator

• Assume additive treatment effect• Subtract the hypothesized treatment effect from the treated

responses, and test hypothesis using these adjustedresponses

• Hodges-Lehmann (HL) estimate is the value that whensubtracted from the responses yields adjusted responsesthat are free of treatment effect, in the sense that the(Wilcoxon Signed Rank) test statistic equals its expectationin the absence of treatment effect

• Wilcoxon Signed Rank test statistic: form matchedtreated-control pairs, rank absolute value of treated-controldifferences within matched pairs, sum ranks for pairs inwhich treated response is higher than control response

679 / 765

Hodges-Lehmann estimator

• Assume additive treatment effect• Subtract the hypothesized treatment effect from the treated

responses, and test hypothesis using these adjustedresponses

• Hodges-Lehmann (HL) estimate is the value that whensubtracted from the responses yields adjusted responsesthat are free of treatment effect, in the sense that the(Wilcoxon Signed Rank) test statistic equals its expectationin the absence of treatment effect

• Wilcoxon Signed Rank test statistic: form matchedtreated-control pairs, rank absolute value of treated-controldifferences within matched pairs, sum ranks for pairs inwhich treated response is higher than control response

680 / 765

Effect of Incumbent Ethnicity onTurnout: White to Hispanic Incumbent

Estimate 95% CI p.value

Incumbent vote ’04 −0.00674 −0.0165 0.00551 0.233Incumbent vote ’06 0.0167 −0.0075 0.0391 0.177

Turnout ’04 −0.0406 −0.0699 −0.00696 0.0187Turnout ’06 −0.0139 −0.0413 0.0151 0.340

681 / 765

Estimating the Effect of IncumbentEthnicity on Registration: White to

Hispanic Incumbent

Estimate 95% CI p.value

Hispanic Reg ’04 −0.0197 −0.0338 −0.00692 0.00335Hispanic Reg ’06 −0.0279 −0.0429 −0.0138 0.000143

Non-Hispanic Reg ’06 −0.00233 −0.0393 0.0332 0.912Non-Hispanic Reg ’04 −0.00449 −0.0381 0.0256 0.784

682 / 765

Effect of Incumbent Race on Turnout:White to White Incumbent

Estimate 95% CI p.value

Incumbent vote ’04 0.00555 −0.00503 0.0166 0.268

Incumbent vote ’06 0.012 −0.000877 0.0267 0.071

683 / 765

Inverse Probability Weighting

• Inverse probability weighting: Treated observations are

weighted by1

P(Ti = 1 | Xi)and control observations by

11− P(Ti = 1 | Xi)

• This is a property of nature, not simply our creation.• Example: imagine propensity score is 0.2, for every treated

subject there are 9 control subjects.• What would happen if everyone received treatment? What

happens to the other four control subjects once they areassigned treatment?

684 / 765

Three Parts, PRE-Intervention

The PRE-INTERVENTION distribution can be decomposed into3 conditional probabilities:

P(Y = 1,T = 1,X = x) = P(Y = 1 | T = 1,X = x)×P(T = 1 | X = x)× P(X = x)

P(Y = 1 | T = 1,X = x) describes how the outcome dependson treatment and X .

P(T = 1 | X = x) describes how subjects choose treatmentprior to intervention

P(X = x) describes the prior distribution of the covariates

685 / 765

Three Parts, PRE-Intervention

The PRE-INTERVENTION distribution can be decomposed into3 conditional probabilities:

P(Y = 1,T = 1,X = x) = P(Y = 1 | T = 1,X = x)×P(T = 1 | X = x)× P(X = x)

P(Y = 1 | T = 1,X = x) describes how the outcome dependson treatment and X .

P(T = 1 | X = x) describes how subjects choose treatmentprior to intervention

P(X = x) describes the prior distribution of the covariates

686 / 765

Three Parts, PRE-Intervention

The PRE-INTERVENTION distribution can be decomposed into3 conditional probabilities:

P(Y = 1,T = 1,X = x) = P(Y = 1 | T = 1,X = x)×P(T = 1 | X = x)× P(X = x)

P(Y = 1 | T = 1,X = x) describes how the outcome dependson treatment and X .

P(T = 1 | X = x) describes how subjects choose treatmentprior to intervention

P(X = x) describes the prior distribution of the covariates

687 / 765

Three Parts, PRE-Intervention

The PRE-INTERVENTION distribution can be decomposed into3 conditional probabilities:

P(Y = 1,T = 1,X = x) = P(Y = 1 | T = 1,X = x)×P(T = 1 | X = x)× P(X = x)

P(Y = 1 | T = 1,X = x) describes how the outcome dependson treatment and X .

P(T = 1 | X = x) describes how subjects choose treatmentprior to intervention

P(X = x) describes the prior distribution of the covariates

688 / 765

Three Parts, POST-Intervention

The POST-Intervention distribution is denoted by P ′::

P ′(Y = 1,T = 1,X = x) = P ′(Y = 1 | T = 1,X = x)×P ′(T = 1 | X = x)× P ′(X = x)

P ′(X = x) = P(X = x) because X is pretreatment

P ′(Y = 1 | T = 1,X = x) = P(Y = 1 | T = 1,X = x) becauseof the selection on observables assumption.

P ′(T = 1 | X = x) 6= P(T = 1 | X = x). Why?

689 / 765

Three Parts, POST-Intervention

The POST-Intervention distribution is denoted by P ′::

P ′(Y = 1,T = 1,X = x) = P ′(Y = 1 | T = 1,X = x)×P ′(T = 1 | X = x)× P ′(X = x)

P ′(X = x) = P(X = x) because X is pretreatment

P ′(Y = 1 | T = 1,X = x) = P(Y = 1 | T = 1,X = x) becauseof the selection on observables assumption.

P ′(T = 1 | X = x) 6= P(T = 1 | X = x). Why?

690 / 765

Three Parts, POST-Intervention

The POST-Intervention distribution is denoted by P ′::

P ′(Y = 1,T = 1,X = x) = P ′(Y = 1 | T = 1,X = x)×P ′(T = 1 | X = x)× P ′(X = x)

P ′(X = x) = P(X = x) because X is pretreatment

P ′(Y = 1 | T = 1,X = x) = P(Y = 1 | T = 1,X = x) becauseof the selection on observables assumption.

P ′(T = 1 | X = x) 6= P(T = 1 | X = x). Why?

691 / 765

Three Parts, POST-Intervention

The POST-Intervention distribution is denoted by P ′::

P ′(Y = 1,T = 1,X = x) = P ′(Y = 1 | T = 1,X = x)×P ′(T = 1 | X = x)× P ′(X = x)

P ′(X = x) = P(X = x) because X is pretreatment

P ′(Y = 1 | T = 1,X = x) = P(Y = 1 | T = 1,X = x) becauseof the selection on observables assumption.

P ′(T = 1 | X = x) 6= P(T = 1 | X = x). Why?

692 / 765

Treatment distribution

After the intervention, P ′(T = 1 | X = X ) is either 0 or 1.Therefore, the post-intervention distribution is:

P ′(Y = 1,T = 1,X = x) = (23)P(Y = 1 | T = 1,X = x)× P(X = x)

Equivalently:

P ′(Y = 1,T = 1,X = x) =P(Y = 1,T = 1,X = x)

P(T = 1 | X = x)(24)

Eq. 23 leads to regression. Eq. 24 leads to IPW. Since (23) and(24) are asymptotically equivalent, they must by identified bythe same assumptions.

693 / 765

Treatment distribution

After the intervention, P ′(T = 1 | X = X ) is either 0 or 1.Therefore, the post-intervention distribution is:

P ′(Y = 1,T = 1,X = x) = (23)P(Y = 1 | T = 1,X = x)× P(X = x)

Equivalently:

P ′(Y = 1,T = 1,X = x) =P(Y = 1,T = 1,X = x)

P(T = 1 | X = x)(24)

Eq. 23 leads to regression. Eq. 24 leads to IPW. Since (23) and(24) are asymptotically equivalent, they must by identified bythe same assumptions.

694 / 765

Treatment distribution

After the intervention, P ′(T = 1 | X = X ) is either 0 or 1.Therefore, the post-intervention distribution is:

P ′(Y = 1,T = 1,X = x) = (23)P(Y = 1 | T = 1,X = x)× P(X = x)

Equivalently:

P ′(Y = 1,T = 1,X = x) =P(Y = 1,T = 1,X = x)

P(T = 1 | X = x)(24)

Eq. 23 leads to regression. Eq. 24 leads to IPW. Since (23) and(24) are asymptotically equivalent, they must by identified bythe same assumptions.

695 / 765

IPW Readings

• Freedman, D.A. and Berk, R.A. (2008). WeightingRegressions by Propensity Scores. Evaluation Review32,4 392-409.

• Kang, J.D.Y. and Schafer, J.L. (2007). Demystifying DoubleRobustness: A Comparison of Alternative Strategies forEstimating a Population Mean from Incomplete Data.Statistical Science 22,4 523-539.

• Robins, J., Sued, M., Lei-Gomez, Q. and Rotnitzky (2007).Comment: Performance of Double-Robust EstimatorsWhen “Inverse Probability” Weights are Highly Variable.Statistical Science 22,4 544-559.

696 / 765

Collaborative Targeted MaximumLikelihood

• Double robust estimator use knowledge of Y , but theyshow poor properties when the pscore is close to 0 or 1.

• Even if the pscore model is correct, one is often better offnot adjusted by the pscore and just modelling Y—e.g.,Freedman and Berk 2008

• Even true for bounded doubly robust estimators• Influence curve may be bounded, but MSE is still

increased by using the pscore• C-TMLE helps (van der Laan and Gruber 2009); extends

targeted ML• One can extend C-TMLE to trim data as matching

does—adjusted collaborative targeted maximum likelihood• Incorporate various ways of trimming including that of

Crump, Hotz, Imbens, and Mitnik (2006)697 / 765

Collaborative Targeted MaximumLikelihood

• Double robust estimator use knowledge of Y , but theyshow poor properties when the pscore is close to 0 or 1.

• Even if the pscore model is correct, one is often better offnot adjusted by the pscore and just modelling Y—e.g.,Freedman and Berk 2008

• Even true for bounded doubly robust estimators• Influence curve may be bounded, but MSE is still

increased by using the pscore• C-TMLE helps (van der Laan and Gruber 2009); extends

targeted ML• One can extend C-TMLE to trim data as matching

does—adjusted collaborative targeted maximum likelihood• Incorporate various ways of trimming including that of

Crump, Hotz, Imbens, and Mitnik (2006)698 / 765

Collaborative Targeted MaximumLikelihood

• Double robust estimator use knowledge of Y , but theyshow poor properties when the pscore is close to 0 or 1.

• Even if the pscore model is correct, one is often better offnot adjusted by the pscore and just modelling Y—e.g.,Freedman and Berk 2008

• Even true for bounded doubly robust estimators• Influence curve may be bounded, but MSE is still

increased by using the pscore• C-TMLE helps (van der Laan and Gruber 2009); extends

targeted ML• One can extend C-TMLE to trim data as matching

does—adjusted collaborative targeted maximum likelihood• Incorporate various ways of trimming including that of

Crump, Hotz, Imbens, and Mitnik (2006)699 / 765

The Problem

• O =(X, T, Y)X are confoundersT is the binary treatment indicator (or censoring model)Y is the outcome

• Estimand: ψ = EX [E [Y | T = 1,X ]− E [Y | T = 0,X ]]

• Two nuisance parameters:Q(T ,X ) ≡ E [Y | T ,X ]g(T ,X ) ≡ P(T | X )

• Inverse probability weighting:

ψIPTW =1n

n∑i=1

[I(Ti = 1)− I(Ti = 0)]Yi

gn(Ti ,Xi)

700 / 765

The Problem

• O =(X, T, Y)X are confoundersT is the binary treatment indicator (or censoring model)Y is the outcome

• Estimand: ψ = EX [E [Y | T = 1,X ]− E [Y | T = 0,X ]]

• Two nuisance parameters:Q(T ,X ) ≡ E [Y | T ,X ]g(T ,X ) ≡ P(T | X )

• Inverse probability weighting:

ψIPTW =1n

n∑i=1

[I(Ti = 1)− I(Ti = 0)]Yi

gn(Ti ,Xi)

701 / 765

The Problem

• O =(X, T, Y)X are confoundersT is the binary treatment indicator (or censoring model)Y is the outcome

• Estimand: ψ = EX [E [Y | T = 1,X ]− E [Y | T = 0,X ]]

• Two nuisance parameters:Q(T ,X ) ≡ E [Y | T ,X ]g(T ,X ) ≡ P(T | X )

• Inverse probability weighting:

ψIPTW =1n

n∑i=1

[I(Ti = 1)− I(Ti = 0)]Yi

gn(Ti ,Xi)

702 / 765

Alternatives• Double Robust-IPTW

ψDR =1n

n∑i=1

[I(Ti = 1)− I(Ti = 0)]

gn(Ti | Xi)(Yi −Q(T ,X ))

+1n

n∑i=1

Qn(1,Xi)−Qn(0,Xi)

• Collaborative Targeted Maximum Likelihood

ψC−TMLE =1n

n∑i=1

Q∗n(1,Xi)−Q∗n(0,Xi)

where Q1n adjustes a preliminary non-parametric estimate

of Q0n byεkn = arg maxPnQk−1

g (Pn)(ε)

All estimation via machine learning—e.g., a super learnerthat is based on a convex combination of non-parametricestimators with

√n oracle properties.

703 / 765

Alternatives• Double Robust-IPTW

ψDR =1n

n∑i=1

[I(Ti = 1)− I(Ti = 0)]

gn(Ti | Xi)(Yi −Q(T ,X ))

+1n

n∑i=1

Qn(1,Xi)−Qn(0,Xi)

• Collaborative Targeted Maximum Likelihood

ψC−TMLE =1n

n∑i=1

Q∗n(1,Xi)−Q∗n(0,Xi)

where Q1n adjustes a preliminary non-parametric estimate

of Q0n byεkn = arg maxPnQk−1

g (Pn)(ε)

All estimation via machine learning—e.g., a super learnerthat is based on a convex combination of non-parametricestimators with

√n oracle properties.

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The Collaborative Part

Q1n adjustes a preliminary non-parmetric estimate of Q0

n by

εkn = arg maxPnQk−1g (Pn)(ε)

Define our update based on g() as:

h1 =

(I[T = 1]

g1(1 | X )− I[T = 0]

g1(0 | X )

)ε1 is fitted by regression Y on h with offset Q0

n , with theinfluence curve

IC(P0) = D∗(Q,g0, ψ0) + ICg0

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The Adjusted Part

• If the assignment part (g()) near the bounds, trim as inCrump, Hotz, Imbens, and Mitnik (2006).

• maximize the efficiency of the estimand as determined bythe influence—e.g., return the lowest MSE estimandclosest to ATE.

• estimate all parts by super learner: combinenon-parametric estimators via maximum likelihood crossvalidation

• downside: understandable balance checks are not possiblesince g() now only corrects for the residuals of Q()

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The Adjusted Part

• If the assignment part (g()) near the bounds, trim as inCrump, Hotz, Imbens, and Mitnik (2006).

• maximize the efficiency of the estimand as determined bythe influence—e.g., return the lowest MSE estimandclosest to ATE.

• estimate all parts by super learner: combinenon-parametric estimators via maximum likelihood crossvalidation

• downside: understandable balance checks are not possiblesince g() now only corrects for the residuals of Q()

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The Adjusted Part

• If the assignment part (g()) near the bounds, trim as inCrump, Hotz, Imbens, and Mitnik (2006).

• maximize the efficiency of the estimand as determined bythe influence—e.g., return the lowest MSE estimandclosest to ATE.

• estimate all parts by super learner: combinenon-parametric estimators via maximum likelihood crossvalidation

• downside: understandable balance checks are not possiblesince g() now only corrects for the residuals of Q()

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Causal Mediation

Examines the roles of intermediate variables that lie in thecausal paths between treatment and outcome variables:

YTM

T is a treatment, M is a mediator, Y is an outcome

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Mediation Analysis is Popular

• Researchers want to know how and why treatment has aneffect not simply that it does

• Very popular—e.g., Baron and Kenny (1986), cited 19,988times. Linear version goes back to Haavelmo (1943) andSimon (1943)

• New assumptions have been proposed that providenonparametric identification—e.g., Robins and Greenland(1992), Pearl (2001), Robins (2003), Petersen, Sinisi andvan der Laan (2006), Hafeman and VanderWeele (2010),Imai et al. (2010).

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Mediation Analysis is Problematic

• These assumptions are usually implausible• I provide analytical results which demonstrate that these

assumptions are fragile: when the assumptions are slightlyweakened, identification is lost

• Troubling because researchers claim it is essential toestimate causal mediation effects to test substantivetheories

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Example: Religious Practice in India

• Joint work with Pradeep Chhibber• Do Muslims political leaders have greater trust in a political

context than other religious leaders?• Can leaders use religious symbols to politically mobilize

supporters?• We are especially interested in differences between

Muslims and Hindus• Religious practice differs sharply between Muslims and

Hindus, and between Sunnis and Shiites• Conjecture: Islamic religious leaders, particularly Sunni,

have greater political role than Hindu religious leaders

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Example: Religious Practice in India

• Joint work with Pradeep Chhibber• Do Muslims political leaders have greater trust in a political

context than other religious leaders?• Can leaders use religious symbols to politically mobilize

supporters?• We are especially interested in differences between

Muslims and Hindus• Religious practice differs sharply between Muslims and

Hindus, and between Sunnis and Shiites• Conjecture: Islamic religious leaders, particularly Sunni,

have greater political role than Hindu religious leaders

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Why People Care?

Current Question:• Concern about the role of Imams in radicalization• This role may vary across different Islamic groups and

contexts

Historical Question:• Long running debate on the political role of Imams. Islamic

organization was never centralized—no Pope.• But Christians and Muslim both regularly launched

cross-continental crusades in the last millennium.

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Why People Care?

Current Question:• Concern about the role of Imams in radicalization• This role may vary across different Islamic groups and

contexts

Historical Question:• Long running debate on the political role of Imams. Islamic

organization was never centralized—no Pope.• But Christians and Muslim both regularly launched

cross-continental crusades in the last millennium.

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Simple Graph

YTM

T is a treatment, M is a mediator, Y is an outcome

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Graph with Pre-Treatment Variables

X

M YZ

T

X are possibly unobserved pre-treatment variables

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Graph with Other Moderators

XM

YZ

T

X are possibly unobserved pre-treatment variablesZ are possibly unobserved post-treatment variables

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We Still Have Simplifications

XM

YZ

T

X are possibly unobserved pre-treatment variablesZ are possibly unobserved post-treatment variables

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Some Notation

• Binary treatment: Ti ∈ 0,1• Mediator: Mi ∈M• Outcome: Yi ∈ Y• Pre-treatment covariates: Xi ∈ X• Post-treatment covariates: Zi ∈ Z

• Potential outcomes (standard case): Yi1,Yi0,Yi = Yi1T + Yi0(1− T )

• Potential mediators: Mi(t) where Mi = Mi(Ti)

• Potential outcomes: Yi(t ,m) where Yi = Y (Ti ,Mi(Ti))

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Defining Causal Mediation Effects

• Total Causal Effect:τi ≡ Yi(1,Mi(1))− Yi(0,Mi(0))

• Natural Indirect Effect:δi(t) ≡ Yi(t ,Mi(1))− Yi(t ,Mi(0))

Causal effect of the change in Mi on Yi that would beinduced by treatment while holding treatment constant

• Controlled Effect of the Mediator:controlled ≡ Yi(t ,m)− Yi(t ,m′)

m 6= m′

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Defining Causal Mediation Effects

• Total Causal Effect:τi ≡ Yi(1,Mi(1))− Yi(0,Mi(0))

• Natural Indirect Effect:δi(t) ≡ Yi(t ,Mi(1))− Yi(t ,Mi(0))

Causal effect of the change in Mi on Yi that would beinduced by treatment while holding treatment constant

• Controlled Effect of the Mediator:controlled ≡ Yi(t ,m)− Yi(t ,m′)

m 6= m′

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Defining Causal Mediation Effects

• Total Causal Effect:τi ≡ Yi(1,Mi(1))− Yi(0,Mi(0))

• Natural Indirect Effect:δi(t) ≡ Yi(t ,Mi(1))− Yi(t ,Mi(0))

Causal effect of the change in Mi on Yi that would beinduced by treatment while holding treatment constant

• Controlled Effect of the Mediator:controlled ≡ Yi(t ,m)− Yi(t ,m′)

m 6= m′

723 / 765

Defining Direct Effects

• Direct effects:ζi(t) ≡ Yi(1,Mi(t))− Yi(0,Mi(t))

• Causal effect of Ti on Yi holding mediator constant at itspotential value that would be induced when Ti = t : thenatural direct effect (Pearl 2001) or the pure/total directeffect (Robins and Greenland 1992).

• Total effect equals mediation (indirect) effect + direct effect:

τi = δi(t) + ζi(1− t) =12δi(0) + δi(1) + ζi(0) + ζi(1)

724 / 765

Defining Direct Effects

• Direct effects:ζi(t) ≡ Yi(1,Mi(t))− Yi(0,Mi(t))

• Causal effect of Ti on Yi holding mediator constant at itspotential value that would be induced when Ti = t : thenatural direct effect (Pearl 2001) or the pure/total directeffect (Robins and Greenland 1992).

• Total effect equals mediation (indirect) effect + direct effect:

τi = δi(t) + ζi(1− t) =12δi(0) + δi(1) + ζi(0) + ζi(1)

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Defining Direct Effects

• Direct effects:ζi(t) ≡ Yi(1,Mi(t))− Yi(0,Mi(t))

• Causal effect of Ti on Yi holding mediator constant at itspotential value that would be induced when Ti = t : thenatural direct effect (Pearl 2001) or the pure/total directeffect (Robins and Greenland 1992).

• Total effect equals mediation (indirect) effect + direct effect:

τi = δi(t) + ζi(1− t) =12δi(0) + δi(1) + ζi(0) + ζi(1)

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Identification of Average CausalMediation Effect

• Average Causal Mediation Effect (ACME):δ(t) ≡ E(δi(t)) = E Yi(t ,Mi(1))− Yi(t ,Mi(0))

• Issue: we can never observe Yi(t ,Mi(1− t))

• One assumption, Sequential Ignorability (e.g., Imai et al.2010):

Yi(t ′,m),Mi(t)

⊥⊥ Ti | Xi = x (25)

Yi(t ′,m) ⊥⊥ Mi(t) | Ti = t ,Xi = x , (26)

for t , t ′ = 0,1; 0 < Pr(Ti = t | Xi = x) and0 < p(Mi = m | Ti = t ,Xi = x), ∀ x ∈ X and m ∈ M

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Identification of Average CausalMediation Effect

• Average Causal Mediation Effect (ACME):δ(t) ≡ E(δi(t)) = E Yi(t ,Mi(1))− Yi(t ,Mi(0))

• Issue: we can never observe Yi(t ,Mi(1− t))

• One assumption, Sequential Ignorability (e.g., Imai et al.2010):

Yi(t ′,m),Mi(t)

⊥⊥ Ti | Xi = x (25)

Yi(t ′,m) ⊥⊥ Mi(t) | Ti = t ,Xi = x , (26)

for t , t ′ = 0,1; 0 < Pr(Ti = t | Xi = x) and0 < p(Mi = m | Ti = t ,Xi = x), ∀ x ∈ X and m ∈ M

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Sequential Ignorability Graph

X

M YZ

T

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Pearl’s (2001) Version of SequentialIgnorability

In order to identify δ(t∗):

Yi(t ,m) ⊥⊥ Mi(t∗) | Xi = x , (27)p(Y (t ,m) | Xi = x) and (28)

p(Mi(t∗) | Xi = x) are identifiable,

for all t = 0,1.

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Pearl’s (2001) Version of SequentialIgnorability

In order to identify δ(t∗):

Yi(t ,m) ⊥⊥ Mi(t∗) | Xi = x , (27)p(Y (t ,m) | Xi = x) and (28)

p(Mi(t∗) | Xi = x) are identifiable,

for all t = 0,1.

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Other Post-Treatment Variables

Yi(t ′,m),Mi(t)

⊥⊥ Ti | Xi = x

Yi(t ,m) ⊥⊥ Mi(t) | Ti = t ,Zi = z,Xi = x (29)

Equation 29 is not sufficient to identify ACME without furtherassumptions. Robins’ (2003) “no-interaction” assumption:

Yi(1,m)− Yi(0,m) = Bi ,

where Bi is a random variable independent of m

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Post-Treatment DAG

XM

YZ

T

733 / 765

Nonparametric Bounds and SensitivityTests

• Imai et al. (2010) derive sharp nonparametric bounds forACME when their particular assumption does not hold

• The nonparametric bounds are uninformative when theassumptions are weakened

• I derive nonparametric sensitivity tests to allow for partialweakening, and the bounds for ACME for the Robins andGreenland (1992) assumption.

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Nonparametric Bounds

• Assume randomization of T :Yi(t ′,m),Mi(t)

⊥⊥ Ti

• For Sequential Ignorability do not assume (26):Yi(t ′,m) ⊥⊥ Mi(t) | Ti = t ,Xi = x

• For Robins (2003), no longer assume:Yi(1,m)− Yi(0,m) = Bi ,

but still assume:Yi(t ,m) ⊥⊥ Mi(t) | Ti = t ,Zi = z,Xi = x

735 / 765

Nonparametric Bounds

In the case of binary M, Y :

max

−P001 − P011

−P000 − P001 − P100

−P011 − P010 − P110

≤ δ(1) ≤ min

P101 + P111

P000 + P100 + P101

P010 + P110 + P111

where Pymt ≡ Pr(Yi = y ,Mi = m | Ti = t), ∀ y ,m, t ∈ 0,1

the bounds will always include zero

736 / 765

Sensitivity Tests

• The identification assumptions ensure that within eachtreatment group the mediator is assigned independent ofpotential outcomes:

Pr (Yi(t ,m) | Mi = 1,Ti = t ′)Pr (Yi(t ,m) | Mi = 0,Ti = t ′)

= 1

∀ t ,m• Sensitivity test: calculate sharp bounds for a fix deviation

from this ratio

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A Series of Experiments

• Measure reported trust in political leaders, randomizereligious versus secular leaders

• Prompt people to vote, randomize religious versus secularprompt. Measure voter turnout and vote.

• Conduct experiments on different groups (Sunni/Shiite,Hindus, Sikhs) and at different locations (majority/minority)

• Natural experiment: Iranian Shiite Imams in Afghanistan• More Experiments: Governments

738 / 765

A Series of Experiments

• Measure reported trust in political leaders, randomizereligious versus secular leaders

• Prompt people to vote, randomize religious versus secularprompt. Measure voter turnout and vote.

• Conduct experiments on different groups (Sunni/Shiite,Hindus, Sikhs) and at different locations (majority/minority)

• Natural experiment: Iranian Shiite Imams in Afghanistan• More Experiments: Governments

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Example Photos

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Simple Trust Experiment

At a recent meeting celebrating India’s democracy this leadingpolitician (show photo) said:

“Politicians like me from different political parties tryhard to represent the interests of the people whosupport us and vote for us.”

Do you think he can be trusted?

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Trust in Uttar Pradesh

Muslim Hindu

UP, Aligarh

Estimate 15.1% 3.79%p-value 0.000686 0.387

UP, Kanpur

Estimate 12.6% -1.20%p-value 0.0118 0.791

742 / 765

Tamil Nadu

Muslim Hindu

Estimate 12.1% -5.09%p-value 0.02 0.21

743 / 765

Shiite Muslims

Uttar Pradesh, Lucknow

Muslim Hindu

Estimate 4.80% 4.72%p-value 0.337 0.336

744 / 765

Get-Out-The-Vote (GOTV) Experiment

• 2009 general election for the 15th Lok Sabha (April/May)• GOTV Experiment with three arms:

• control: no contact prior to the election

• religious: receive GOTV appeal with religious picture

• secular: receive GOTV appeal with secular picture

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Non-Religious Leader Appeal

This learned person has written a book. The book says thatPolitics affects whether the government looks after the interestsof people like you and the interests of your community. Heurged everyone to VOTE!

He wrote that if you as a citizen want to have your input inmaking politics and government work for your community, youneed to VOTE in order to send a message to those in power.Your interests and the interests of your community will not beattended to if you do not VOTE.

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GOTV Results: Hindu Vote Proportions

Religious Secular Control

Bharatiya Janata Party 0.44 0.32 0.36

Janata Dal–S 0.22 0.42 0.27

Congress 0.28 0.22 0.29

747 / 765

GOTV Results: Muslim VoteProportions

Religious Secular Control

Bharatiya Janata Party 0.20 0.18 0.21

Janata Dal–S 0.18 0.03 0.26

Congress 0.59 0.70 0.44

Bahujan Samaj Party 0.00 0.06 0.08

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Other Experiments: Triangulation

• Iranian Shiite Imams in Afghanistan, a natural experimentduring Muharram

• Public Safety Canada: Doing things Berkeley won’t

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Conclusion

• Triangulation will never be logically sufficient

• Many moving parts: political trust, voting, the strategicrelationship between political parties, majority/minority,Shiite vs. Sunni, levels of violence, media saturation

• The temptation of reductionism is widespread: logicalpositivism, a common language in which all scientificpropositions can be expressed

• Causality and manipulation are separable concepts

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Conclusion

• Triangulation will never be logically sufficient

• Many moving parts: political trust, voting, the strategicrelationship between political parties, majority/minority,Shiite vs. Sunni, levels of violence, media saturation

• The temptation of reductionism is widespread: logicalpositivism, a common language in which all scientificpropositions can be expressed

• Causality and manipulation are separable concepts

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Manipulation and Reductionism

• A good Mantra: “No causation without manipulation”• But this does not imply the Dogma of reductionism

common in the literature• Classic Example:

• Incoherent: “What is the causal effect of being white onincome/grades/happiness?”

• But this does not imply that race does not play a role ingenerating these outcomes

• We test theories of how different religions are organized.Do Imams have a greater political role than other religiousleaders? Are they more trusted?

752 / 765

Manipulation and Reductionism

• A good Mantra: “No causation without manipulation”• But this does not imply the Dogma of reductionism

common in the literature• Classic Example:

• Incoherent: “What is the causal effect of being white onincome/grades/happiness?”

• But this does not imply that race does not play a role ingenerating these outcomes

• We test theories of how different religions are organized.Do Imams have a greater political role than other religiousleaders? Are they more trusted?

753 / 765

Manipulation and Reductionism

• A good Mantra: “No causation without manipulation”• But this does not imply the Dogma of reductionism

common in the literature• Classic Example:

• Incoherent: “What is the causal effect of being white onincome/grades/happiness?”

• But this does not imply that race does not play a role ingenerating these outcomes

• We test theories of how different religions are organized.Do Imams have a greater political role than other religiousleaders? Are they more trusted?

754 / 765

Manipulation and Reductionism

• A good Mantra: “No causation without manipulation”• But this does not imply the Dogma of reductionism

common in the literature• Classic Example:

• Incoherent: “What is the causal effect of being white onincome/grades/happiness?”

• But this does not imply that race does not play a role ingenerating these outcomes

• We test theories of how different religions are organized.Do Imams have a greater political role than other religiousleaders? Are they more trusted?

755 / 765

What is Reductionism?

• Extreme Reductionism: every scientific sentence has a fulltranslation in sense-datum

• Moderate Reductionism: each scientific sentence has itsown separate empirical content

• Moderate Holism: a scientific sentence need not implyempirical consequences by itself. A bigger cluster isusually needed.

• Classic Example: Copernican vs. Ptolemaic system. Betterexample: Aryabhat.a (500CE)—diurnal motion of the earth,elliptic orbits calculated relative to the sun, correctexplanation of lunar and solar eclipses, but still geocentric.

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What is Reductionism?

• Extreme Reductionism: every scientific sentence has a fulltranslation in sense-datum

• Moderate Reductionism: each scientific sentence has itsown separate empirical content

• Moderate Holism: a scientific sentence need not implyempirical consequences by itself. A bigger cluster isusually needed.

• Classic Example: Copernican vs. Ptolemaic system. Betterexample: Aryabhat.a (500CE)—diurnal motion of the earth,elliptic orbits calculated relative to the sun, correctexplanation of lunar and solar eclipses, but still geocentric.

757 / 765

What is Reductionism?

• Extreme Reductionism: every scientific sentence has a fulltranslation in sense-datum

• Moderate Reductionism: each scientific sentence has itsown separate empirical content

• Moderate Holism: a scientific sentence need not implyempirical consequences by itself. A bigger cluster isusually needed.

• Classic Example: Copernican vs. Ptolemaic system. Betterexample: Aryabhat.a (500CE)—diurnal motion of the earth,elliptic orbits calculated relative to the sun, correctexplanation of lunar and solar eclipses, but still geocentric.

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What’s the Problem?

• Some want to weaken the Mantra of “No causation withoutmanipulation” in order theorize about race, gender, religion

• Others justify estimating causal mediation effects becausethey are needed to test theories—e.g., natural direct effect,pure/total direct effect.

• Assumptions needed to estimate natural mediation effectsare usually implausible.

• Good work on the many possible assumptions need toidentify the mediation effect—e.g., Robins and Greenland(1992); Pearl (2001); Robins (2003); Petersen, Sinisi, vander Laan (2006); Imai, Keele, Tingley (2010), . . .

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What’s the Problem?

• Some want to weaken the Mantra of “No causation withoutmanipulation” in order theorize about race, gender, religion

• Others justify estimating causal mediation effects becausethey are needed to test theories—e.g., natural direct effect,pure/total direct effect.

• Assumptions needed to estimate natural mediation effectsare usually implausible.

• Good work on the many possible assumptions need toidentify the mediation effect—e.g., Robins and Greenland(1992); Pearl (2001); Robins (2003); Petersen, Sinisi, vander Laan (2006); Imai, Keele, Tingley (2010), . . .

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Keep the Mantra; Drop the Dogma

• The central issue is key parts of a scientific theory may nottouch data in the way reductionism suggests.

• The Mantra should be kept: we need to be clear aboutwhat we are estimating.

• But our theories can have quantities that are notmanipulatable, even in theory.

• We make progress the way science always has: findingcases where the theories make different predictions,falsification, triangulate, etc.

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Keep the Mantra; Drop the Dogma

• The central issue is key parts of a scientific theory may nottouch data in the way reductionism suggests.

• The Mantra should be kept: we need to be clear aboutwhat we are estimating.

• But our theories can have quantities that are notmanipulatable, even in theory.

• We make progress the way science always has: findingcases where the theories make different predictions,falsification, triangulate, etc.

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Philosophy of science is about as useful to scientists asornithology is to birds.

—Richard Feynman

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Some Citations

• Baron, R. M. and Kenny, D. A. (1986). The moderator-mediator variable distinction in social psychologicalresearch: Conceptual, strategic, and statistical considerations. Journal of Personality and SocialPsychology 51 1173-1182.

• Imai, Kosuke, Luke Keele, and Teppei Yamamoto. (2010). “Identification, Inference, and Sensitivity Analysisfor Causal Mediation Effects.” Statistical Science, Vol. 25, No. 1 (February), pp. 51-71.

• Pearl, J . (2001). Direct and indirect effects. In Proceedings of the Seventeenth Conference on Uncertaintyin Artificial Intelligence (J. S. Breese and D. Koller, eds.) 411-420. Morgan Kaufman, San Francisco, CA.

• Pearl, J. (2010). An introduction to causal inference. Int. J. Biostat. 6 Article 7.

• Robins, J. M. (2003). Semantics of causal DAG models and the identification of direct and indirect effects.In Highly Structured Stochastic Systems (P. J. Green, N. L. Hjort and S. Richardson, eds.) 70-81. OxfordUniv. Press, Oxford. MR2082403

• Robins, J. M. and Greenland, S. (1992). Identifiability and exchangeability for direct and indirect effects.Epidemiology 3 143-155.

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References I

Abadie, Alberto, Alexis Diamond, and Jens Hainmueller (2010).“Synthetic control methods for comparative case studies:Estimating the effect of Californiaâs tobacco controlprogram”. In: Journal of the American Statistical Association105.490.

Abadie, Alberto and Guido Imbens (2006). “Large SampleProperties of Matching Estimators for Average TreatmentEffects”. In: Econometrica 74.1, pp. 235–267.

Angrist, J and AB Krueger (1991). “Does compulsory schoolattendance affect earnings?” In: Quarterly Journal ofEconomics 106, pp. 979–1019.

Angrist, Joshua D., Guido W. Imbens, and Donald B. Rubin(1996). “Identification of Causal Effects Using InstrumentalVariables”. In: Journal of the American StatisticalAssociation 91.434, pp. 444–455.

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References II

Ansolabehere, Stephen, James M. Snyder, andCharles Stewart (2000). “Old Voters, New Voters, and thePersonal Vote: Using Redistricting to Measure theIncumbency Advantage”. In: American Journal of PoliticalScience 44.1, pp. 17–34.

Aronow, Peter M, Donald P Green, and Donald KK Lee (2014).“Sharp bounds on the variance in randomized experiments”.In: The Annals of Statistics 42.3, pp. 850–871.

Ashworth, Scott and Ethan Bueno de Mesquita (2007).“Electoral Selection, Strategic Challenger Entry, and theIncumbency Advantage”. Working Paper.

BBC (2003). “How to make a perfect cuppa”. June 25. URL:http://news.bbc.co.uk/1/hi/uk/3016342.stm.

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References III

Berelson, Bernard R., Paul F. Lazarsfeld, andWilliam N. McPhee (1954). Voting: A Study of OpinionFormation in a Presidential Campaign. Chicago: Universityof Chicago Press.

Blake, Tom, Chris Nosko, and Steven Tadelis (2014).Consumer heterogeneity and paid search effectiveness: Alarge scale field experiment. Tech. rep. National Bureau ofEconomic Research.

Bound, J., D. Jaeger, and R. Baker (1995). “Problems withInstrumental Variables Estimation when the CorrelationBetween the Instruments and the Endogenous Regressors isWeak”. In: Journal of the American Statistical Association90, pp. 443–450.

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References IV

Bowers, Jake and Ben Hansen (2006). “Attributing Effects to ACluster Randomized Get-Out-The-Vote Campaign”.Technical Report #448, Statistics Department, University ofMichigan.http://www-personal.umich.edu/~jwbowers/PAPERS/bowershansen2006-10TechReport.pdf.

Campbell, Angus et al. (1960). The American Voter. New York:John Wiley & Sons.

Cox, David R. (1958). Planning of Experiments. New York:Wiley.

Cox, Gary W. and Jonathan N. Katz (2002). Elbridge Gerry’sSalamander: The Electoral Consequences of theReapportionment Revolution. New York: CambridgeUniversity Press.

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References VDehejia, Rajeev (2005). “Practical Propensity Score Matching:

A Reply to Smith and Todd”. In: Journal of Econometrics125.1–2, pp. 355–364.

Dehejia, Rajeev H. and Sadek Wahba (2002). “PropensityScore Matching Methods for Nonexperimental CausalStudies”. In: Review of Economics and Statistics 84.1,pp. 151–161.

Dehejia, Rajeev and Sadek Wahba (1997). “Causal Effects inNon-Experimental Studies: Re-Evaluating the Evaluation ofTraining Programs”. Rejeev Dehejia, Econometric Methodsfor Program Evaluation. Ph.D. Dissertation, HarvardUniversity, Chapter 1.

– (1999). “Causal Effects in Non-Experimental Studies:Re-Evaluating the Evaluation of Training Programs”. In:Journal of the American Statistical Association 94.448,pp. 1053–1062.

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References VI

Fisher, Ronald A. (1935). Design of Experiments. New York:Hafner.

Freedman, David A (2010). Statistical Models and CausalInference: A Dialogue with the Social Sciences. CambridgeUniversity Press.

Hainmueller, Jens (2012). “Entropy Balancing: A MultivariateReweighting Method to Produce Balanced Samples inObservational Studies”. In: Political Analysis 20.1, pp. 25–46.

Hartman, Erin et al. (forthcoming). “From SATE to PATT:Combining Experimental with Observational Studies toEstimate Population Treatment Effects”. In: Journal of theRoyal Statistical Society, Series A.

Heckman, James J., Hidehiko Ichimura, et al. (1998).“Characterizing Selection Bias Using Experimental Data”. In:Econometrica 66.5, pp. 1017–1098.

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References VII

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