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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 / 778
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.
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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
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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 @
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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”
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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
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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
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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→∞?
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A Simple Example II
• What about for:
∆Yt = α + β1∆Yt−1 + β2∆Yt−2 + · · ·+ βk ∆Yt−k
?
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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?
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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.”
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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.”
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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.”
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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....
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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).
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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.”
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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.
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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.
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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 .
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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.
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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
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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?
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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)?
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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
35 / 778
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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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 selection
process 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
• 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 by
τ |(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
44 / 778
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
49 / 778
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 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 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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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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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 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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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
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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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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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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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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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
• 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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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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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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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 149 / 778
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 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 nojumps at 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)
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RDD with Constant Treatment Effects
0 20 40 60 80 100
020
4060
8010
0
Z
Y
Non−Treated Treated
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RDD with Variable Treatment Effects
0 20 40 60 80 100
020
4060
8010
0
Z
Y
Non−Treated Treated
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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?
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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
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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
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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 withthe same party they had before
• parties have organizational advantages in some districts
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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 AdvantageEarly 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
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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
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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
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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
172 / 778
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
173 / 778
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
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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
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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
176 / 778
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
177 / 778
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
178 / 778
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
179 / 778
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%
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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
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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
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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
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Democratic Campaign Spending %
Democratic Margin (%)
Dem
ocra
tic S
pend
ing
%
30
40
50
60
70
−20 −10 0 10 20
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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
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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.
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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
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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
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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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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
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National Democratic Swing
Democratic Margin (%)
Dem
ocra
tic S
win
g (%
)
−3
−2
−1
0
1
2
3
−20 −10 0 10 20
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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 ispredictive 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
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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
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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)
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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 election@Recounts 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
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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 less
professionalization 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
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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
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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 )
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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 )
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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 and
Titiunik 2012)
@• Even when RD works, averaging over elections is delicate
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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
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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
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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.
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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.
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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).
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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.
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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.
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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.
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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 theexperimental estimate? All failed (e.g., regression and IV).
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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).
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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)
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0.00 0.05 0.10 0.15 0.20 0.25
500
1000
1500
2000
2500
Dehejia Wahba Sample
Lowest p−value (KS & paired t−tests)
Est
imat
ed A
vera
ge T
reat
men
tt E
ffect
for
Tre
ated
($)
Experimental benchmark estimate ($1794)
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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.
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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)
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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.
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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).
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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.
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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.
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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.
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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.
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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 not
satisfied:1 X covariates are discrete and others are skewed and have
point 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
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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
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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 25.6 378@Random Search 112 493 4.38 1.30
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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?
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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?
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0.00 0.05 0.10 0.15 0.20 0.25
020
0000
6000
0010
0000
0
Dehejia Wahba Sample
Lowest p−value (KS & paired t−tests)
Squ
ared
Err
or fr
om E
xper
imen
tal B
ench
mar
k
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0.00 0.05 0.10 0.15 0.20 0.25
500
1000
1500
2000
2500
Dehejia Wahba Sample
Lowest p−value (KS & paired t−tests)
Est
imat
ed A
vera
ge T
reat
men
tt E
ffect
for
Tre
ated
($)
Experimental benchmark estimate ($1794)
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0.00 0.05 0.10 0.15 0.20 0.25
5000
0010
0000
015
0000
020
0000
0
Lalonde Sample
Lowest p−value (KS & paired t−tests)
Squ
ared
Err
or fr
om E
xper
imen
tal B
ench
mar
k
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0.00 0.05 0.10 0.15 0.20 0.25
−15
00−
1000
−50
00
500
1000
Lalonde Sample
Lowest p−value (KS & paired t−tests)
Est
imat
ed A
vera
ge T
reat
men
t Effe
ct fo
r T
reat
ed (
$)
Experimental benchmark estimate ($884)Lower bound, 95% CI of experimental estimate
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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 best
possible balance, and look at outcomes only once, afterbalance is attained.
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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
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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
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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)
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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
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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
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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)
• RCT: causal effect really is zero• Why do seven NRS disagree with RCTs?
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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
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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
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Balance of baseline probability ofDeath, GenMatch
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
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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
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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
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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)
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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.
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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
• 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
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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)
-150
-100
-50
050
100
150
-150
-100
-50
050
100
150
0 0.2 0.4 0.6 0.8 1
Con
trol P
(X)
Trea
ted
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)
-150
-100
-50
050
100
150
-150
-100
-50
050
100
150
-150
-100
-50
050
100
150
0 0.2 0.4 0.6 0.8 1
Con
trol P
(X)
Trea
ted
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 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)
-150
-100
-50
050
100
150
-150
-100
-50
050
100
150
-150
-100
-50
050
100
150
0 0.2 0.4 0.6 0.8 1
Con
trol P
(X)
Trea
ted
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 imbalance of the sample treated andsample control units in the overlap region.
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Example: Propensity Score Distributions
Propensity Score P(X)
-150
-100
-50
050
100
150
-150
-100
-50
050
100
150
-150
-100
-50
050
100
150
-150
-100
-50
050
100
150
0 0.2 0.4 0.6 0.8 1
Con
trol P
(X)
Trea
ted
P(X
)
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Example: Propensity Score Distributions
Propensity Score P(X)
-150
-100
-50
050
100
150
-150
-100
-50
050
100
150
0 0.2 0.4 0.6 0.8 1
Con
trol P
(X)
Trea
ted
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 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)
-150
-100
-50
050
100
150
-150
-100
-50
050
100
150
0 0.2 0.4 0.6 0.8 1
Pop
ulat
ion
Trea
ted
P(W
)S
ampl
e Tr
eate
d P
(W)
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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),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.
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Example: Propensity Score Distributions
Propensity Score P(X)
-150
-100
-50
050
100
150
-150
-100
-50
050
100
150
-150
-100
-50
050
100
150
0 0.2 0.4 0.6 0.8 1
Pop
ulat
ion
Trea
ted
P(W
)S
ampl
e Tr
eate
d P
(W)
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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),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.
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Example: Propensity Score Distributions
Propensity Score P(X)
-150
-100
-50
050
100
150
-150
-100
-50
050
100
150
-150
-100
-50
050
100
150
0 0.2 0.4 0.6 0.8 1
Pop
ulat
ion
Trea
ted
P(W
)S
ampl
e Tr
eate
d P
(W)
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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),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.
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Example: Propensity Score Distributions
Propensity Score P(X)
-150
-100
-50
050
100
150
-150
-100
-50
050
100
150
-150
-100
-50
050
100
150
-150
-100
-50
050
100
150
0 0.2 0.4 0.6 0.8 1
Pop
ulat
ion
Trea
ted
P(W
)S
ampl
e Tr
eate
d P
(W)
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Example: Propensity Score Distributions
Propensity Score P(X)
-150
-100
-50
050
100
150
-150
-100
-50
050
100
150
0 0.2 0.4 0.6 0.8 1
Pop
ulat
ion
Trea
ted
P(W
)S
ampl
e Tr
eate
d P
(W)
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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),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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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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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
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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
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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
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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
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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
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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
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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
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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
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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.
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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
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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
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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 )
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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.
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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.
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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.
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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 firststage and a linear model in the second. @
• Goodness-of-fit for two model which may not be nested.
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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,
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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,
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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,
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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.
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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.
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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,
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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,
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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.
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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.
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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 [email protected]., 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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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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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. Computep = (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 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
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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.
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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• 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
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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
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Exploiting Redistricting• Ansolabehere, Snyder, and Stewart (2000) use the
variation 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 partof the district with her vote share in the old part of thedistrict. @
• 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).
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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
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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
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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
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Baseline Vote Share for Incumbent House Member, California
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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
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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 who
are:• moved to a new district (new voters)• to the behavior of voters whose district remains unchanged
(old voters).
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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)
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Before one-time redistricting
District A
District B
B's original voters
A's original voters
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One-time redistricting
District A
District B
Voters to be redistricted
at election t B's original voters
A's original voters
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After one-time redistricting
District A
District B
Old neighbors
New voters
New neighbors
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First identification strategy:old-neighbors design
District B
New neighbors
Not used
District A
Control Treated
District B
New voters
Old neighbors
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Second identification strategy:new neighbors design
District B
New neighbors
Control
District A
Not used Treated
New voters
Old neighbors
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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
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Before two-time redistricting(election t-1)
District A
District B
B's original voters
A's original voters
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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)
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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
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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)
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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
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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
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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]
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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.
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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
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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.
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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
]410 / 778
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
]
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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 ourtwo-redistricting research design @
• We use the multiple redistrictings which were done inTexas between 2002 and 2006. We have mergedVTD-level election, registration and census data withcandidate data such as quality measures and Nominate.
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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
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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.
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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.
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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
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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.
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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
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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
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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
420 / 778
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
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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
422 / 778
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).
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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
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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
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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
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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
427 / 778
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
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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?
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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
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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)because of the selection on observables assumption.
@P ′(T = 1 | X = x) 6= P(T = 1 | X = x). Why?
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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.
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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.
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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)434 / 778
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)
435 / 778
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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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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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
444 / 778
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
446 / 778
We Still Have Simplifications
XM
YZ
T
X are possibly unobserved pre-treatment variablesZ are possibly unobserved post-treatment variables
447 / 778
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 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
451 / 778
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.
453 / 778
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
455 / 778
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
457 / 778
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
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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
460 / 778
Example Photos
461 / 778
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
463 / 778
Tamil Nadu
Muslim Hindu
Estimate 12.1% -5.09%p-value 0.02 0.21
464 / 778
Shiite Muslims
Uttar Pradesh, Lucknow
Muslim Hindu
Estimate 4.80% 4.72%p-value 0.337 0.336
465 / 778
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
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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: logical
positivism, 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?
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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.@Better example: Aryabhat.a (500CE)—diurnal motion ofthe earth, elliptic orbits calculated relative to the sun,correct explanation of lunar and solar eclipses, but stillgeocentric.
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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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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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