Probabilistic Inference

PROBABILISTIC INFERENCE

AGENDA

Conditional probability Independence Intro to Bayesian Networks

REMEMBER: PROBABILITY NOTATION LEAVES VALUES IMPLICIT

P(AB) = P(A)+P(B)- P(AB) means

P(A=a B=b) = P(A=a) + P(B=b)- P(A=a B=b)

For all aVal(A) and bVal(B)

A and B are random variables.A=a and B=b are events.

Random variables indicate many possible combinations of events

CONDITIONAL PROBABILITY

P(A,B) = P(A|B) P(B)= P(B|A) P(A)

P(A|B) is the posterior probability of A given knowledge of B

Axiomatic definition:P(A|B) = P(A,B)/P(B)

CONDITIONAL PROBABILITY

P(A|B) is the posterior probability of A given knowledge of B

“For each value of b: given that I know B=b, what do I believe about A?”

If a new piece of information C arrives, the agent’s new belief (if it obeys the rules of probability) should be P(A|B,C)

CONDITIONAL DISTRIBUTIONS

State P(state)

C, T, P 0.108

C, T, P 0.012

C, T, P 0.072

C, T, P 0.008

C, T, P 0.016

C, T, P 0.064

C, T, P 0.144

C, T, P 0.576

P(Cavity|Toothache) = P(CavityToothache)/P(Toothache) =

(0.108+0.012)/(0.108+0.012+0.016+0.064) = 0.6

Interpretation: After observing Toothache, the patient is no longer an “average” one, and the prior probability (0.2) of Cavity is no longer valid

P(Cavity|Toothache) is calculated by keeping the ratios of the probabilities of the 4 cases of Toothache unchanged, and normalizing their sum to 1

UPDATING THE BELIEF STATE

The patient walks into the dentists door

Let D now observe evidence E: Toothache holds with probability 0.8 (e.g., “the patient says so”)

How should D update its belief state?

State P(state)

C, T, P 0.108

C, T, P 0.012

C, T, P 0.072

C, T, P 0.008

C, T, P 0.016

C, T, P 0.064

C, T, P 0.144

C, T, P 0.576


P(Toothache|E) = 0.8 We want to compute

P(CTP|E)= P(CP|T,E) P(T|E)

Since E is not directly related to the cavity or the probe catch, we consider that C and P are independent of E given T, hence:P(CP|T,E) = P(CP|T)

P(CTP|E)= P(CPT) P(T|E)/P(T)

State P(state)

C, T, P 0.108

C, T, P 0.012

C, T, P 0.072

C, T, P 0.008

C, T, P 0.016

C, T, P 0.064

C, T, P 0.144

C, T, P 0.576






State P(state)

C, T, P 0.108

C, T, P 0.012

C, T, P 0.072

C, T, P 0.008

C, T, P 0.016

C, T, P 0.064

C, T, P 0.144

C, T, P 0.576

These rows should be scaled to sum to 0.8







State P(state)

C, T, P 0.108 0.432

C, T, P 0.012 0.048

C, T, P 0.072 0.018

C, T, P 0.008 0.002

C, T, P 0.016 0.064

C, T, P 0.064 0.256

C, T, P 0.144 0.036

C, T, P 0.576 0.144



ISSUES

If a state is described by n propositions, then a belief state contains 2n states (possibly, some have probability 0)

Modeling difficulty: many numbers must be entered in the first place

Computational issue: memory size and time

INDEPENDENCE OF EVENTS

Two events A=a and B=b are independent if P(A=a B=b) = P(A=a) P(B=b)

hence P(A=a|B=b) = P(A=a) Knowing B=b doesn’t give you any

information about whether A=a is true

INDEPENDENCE OF RANDOM VARIABLES

Two random variables A and B are independent if

P(A,B) = P(A) P(B)

hence P(A|B) = P(A) Knowing B doesn’t give you any information

about A

[This equality has to hold for all combinations of values that A and B can take on, i.e., all events A=a and B=b are independent]

SIGNIFICANCE OF INDEPENDENCE

If A and B are independent, then P(A,B) = P(A) P(B)

=> The joint distribution over A and B can be defined as a product of the distribution of A and the distribution of B

Rather than storing a big probability table over all combinations of A and B, store two much smaller probability tables!

To compute P(A=a B=b), just look up P(A=a) and P(B=b) in the individual tables and multiply them together

CONDITIONAL INDEPENDENCE

Two random variables A and B are conditionally independent given C, if

P(A, B|C) = P(A|C) P(B|C)

hence P(A|B,C) = P(A|C) Once you know C, learning B doesn’t give

you any information about A

[again, this has to hold for all combinations of values that A,B,C can take on]

SIGNIFICANCE OF CONDITIONAL INDEPENDENCE

Consider Rainy, Thunder, and RoadsSlippery Ostensibly, thunder doesn’t have anything

directly to do with slippery roads… But they happen together more often when it

rains, so they are not independent… So it is reasonable to believe that Thunder

and RoadsSlippery are conditionally independent given Rainy

So if I want to estimate whether or not I will hear thunder, I don’t need to think about the state of the roads, just whether or not it’s raining!

Toothache and PCatch are independent given Cavity, but this relation is hidden in the numbers! [Quiz]

Bayesian networks explicitly represent independence among propositions to reduce the number of probabilities defining a belief state

State P(state)

C, T, P 0.108

C, T, P 0.012

C, T, P 0.072

C, T, P 0.008

C, T, P 0.016

C, T, P 0.064

C, T, P 0.144

C, T, P 0.576

BAYESIAN NETWORK Notice that Cavity is the “cause” of both Toothache

and PCatch, and represent the causality links explicitly

Give the prior probability distribution of Cavity Give the conditional probability tables of Toothache

and PCatchCavity

Toothache

P(Cavity)

0.2

P(Toothache|c)

CavityCavity

0.60.1

PCatch

P(PCatch|c)

CavityCavity

0.90.02

5 probabilities, instead of 7

P(CTP) = P(TP|C) P(C) = P(T|C) P(P|C) P(C)

CONDITIONAL PROBABILITY TABLES

Cavity

Toothache

P(Cavity)

0.2

P(Toothache|c)

CavityCavity

0.60.1

PCatch

P(PCatch|c)

CavityCavity

0.90.02

P(t|c) P(t|c)

CavityCavity

0.60.1

0.40.9

Rows sum to 1

If X takes n values, just store n-1 entries

P(CTP) = P(TP|C) P(C) = P(T|C) P(P|C) P(C)

SIGNIFICANCE OF BAYESIAN NETWORKS

If we know that variables are conditionally independent, we should be able to decompose joint distribution to take advantage of it

Bayesian networks are a way of efficiently factoring the joint distribution into conditional probabilities

And also building complex joint distributions from smaller models of probabilistic relationships

But… What knowledge does the BN encode about the

distribution? How do we use a BN to compute probabilities of

variables that we are interested in?

BAYES’ RULE AND OTHER PROBABILITY MANIPULATIONS

P(A,B) = P(A|B) P(B)= P(B|A) P(A)

P(A|B) = P(B|A) P(A) / P(B)

Gives us a way to manipulate distributions e.g. P(B) = Sa P(B|A=a) P(A=a) Can derive P(A|B), P(B) using only P(B|A) and

P(A)

A MORE COMPLEX BN

Burglary Earthquake

Alarm

MaryCallsJohnCalls

causes

effects

Directed acyclic graph

Intuitive meaning of arc from x to y:

“x has direct influence on y”

B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

Size of the CPT for a node with k parents: 2k

A MORE COMPLEX BN

10 probabilities, instead of 31

WHAT DOES THE BN ENCODE?

Each of the beliefs JohnCalls and MaryCalls is independent of Burglary and Earthquake given Alarm or Alarm

Burglary Earthquake

Alarm

MaryCallsJohnCalls

For example, John doesnot observe any burglariesdirectly

P(BJ) P(B) P(J)P(BJ|A) = P(B|A) P(J|A)

WHAT DOES THE BN ENCODE?

The beliefs JohnCalls and MaryCalls are independent given Alarm or Alarm

For instance, the reasons why John and Mary may not call if there is an alarm are unrelated

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(BJ|A) = P(B|A) P(J|A)P(JM|A) = P(J|A) P(M|A)P(BJ|A) = P(B|A) P(J|A)P(JM|A) = P(J|A) P(M|A)

A node is independent of its non-descendants given its parents


WHAT DOES THE BN ENCODE?Burglary Earthquake

Alarm

MaryCallsJohnCallsA node is

independent of its non-descendants given its parents


Burglary and Earthquake are independent

Burglary and Earthquake are independent

The beliefs JohnCalls and MaryCalls are independent given Alarm or Alarm

For instance, the reasons why John and Mary may not call if there is an alarm are unrelated

LOCALLY STRUCTURED WORLD

A world is locally structured (or sparse) if each of its components interacts directly with relatively few other components

In a sparse world, the CPTs are small and the BN contains much fewer probabilities than the full joint distribution

If the # of entries in each CPT is bounded by a constant, i.e., O(1), then the # of probabilities in a BN is linear in n – the # of propositions – instead of 2n for the joint distribution

EQUATIONS INVOLVING RANDOM VARIABLES GIVE RISE TO CAUSALITY RELATIONSHIPS

C = A B C = max(A,B) Constrains joint probability P(A,B,C) Nicely encoded as causality relationship

C

A B

Conditional probability given by equation rather than a CPT

NAÏVE BAYES MODELS

P(Cause,Effect1,…,Effectn)= P(Cause) Pi P(Effecti | Cause)

Cause

Effect1 Effect2 Effectn

NAÏVE BAYES CLASSIFIER

P(Class,Feature1,…,Featuren)= P(Class) Pi P(Featurei | Class)

Class

Feature1 Feature2 Featuren

P(C|F1,….,Fk) = P(C,F1,….,Fk)/P(F1,….,Fk)

= 1/Z P(C) Pi P(Fi|C)

Given features, what class?

Spam / Not Spam

English / French/ Latin

…

Word occurrences

BUT DOES A BN REPRESENT A BELIEF STATE?

IN OTHER WORDS, CAN WE COMPUTE THE FULL JOINT DISTRIBUTION OF THE PROPOSITIONS FROM IT?

CALCULATION OF JOINT PROBABILITY

B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

P(JMABE) = ??

P(JMABE)= P(JM|A,B,E) P(ABE)= P(J|A,B,E) P(M|A,B,E) P(ABE)(J and M are independent given A)

P(J|A,B,E) = P(J|A)(J and B and J and E are independent given A)

P(M|A,B,E) = P(M|A) P(ABE) = P(A|B,E) P(B|E) P(E)

= P(A|B,E) P(B) P(E)(B and E are independent)

P(JMABE) = P(J|A)P(M|A)P(A|B,E)P(B)P(E)

Burglary Earthquake

Alarm

MaryCallsJohnCalls


B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

P(JMABE)= P(J|A)P(M|A)P(A|B,E)P(B)P(E)= 0.9 x 0.7 x 0.001 x 0.999 x 0.998= 0.00062


B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

P(JMABE)= P(J|A)P(M|A)P(A|B,E)P(B)P(E)= 0.9 x 0.7 x 0.001 x 0.999 x 0.998= 0.00062

P(x1x2…xn) = Pi=1,…,nP(xi|parents(Xi))

full joint distribution table


B E P(A|…)

TTFF

TFTF

0.950.940.290.001

Burglary Earthquake

Alarm

MaryCallsJohnCalls

P(B)

0.001

P(E)

0.002

A P(J|…)

TF

0.900.05

A P(M|…)

TF

0.700.01

P(x1x2…xn) = Pi=1,…,nP(xi|parents(Xi))

full joint distribution table

P(jmabe)= P(j|a)P(m|a)P(a|b,e)P(b)P(e)= 0.9 x 0.7 x 0.001 x 0.999 x 0.998= 0.00062

Since a BN defines the full joint distribution of a set of propositions, it represents a belief state

Since a BN defines the full joint distribution of a set of propositions, it represents a belief state

QUERYING THE BN

The BN gives P(T|C) What about P(C|T)? P(Cavity|T=t)

= P(Cavity T=t)/P(T=t)= P(T=t|Cavity) P(Cavity) / P(T=t)[Bayes’ rule]

Querying a BN is just applying Bayes’ rule on a larger scale… algorithms next time

Cavity

Toothache

P(C)

0.1

C P(T|C)

TF

0.40.01111

MORE COMPLICATED SINGLY-CONNECTED BELIEF NET

Radio

Battery

SparkPlugs

Starts

Gas

Moves

SOME APPLICATIONS OF BN

Medical diagnosis Troubleshooting of hardware/software

systems Fraud/uncollectible debt detection Data mining Analysis of genetic sequences Data interpretation, computer vision, image

understanding

Region = {Sky, Tree, Grass, Rock}

R2

R4R3

R1

Above

BN to evaluate insurance risks

PURPOSES OF BAYESIAN NETWORKS

Efficient and intuitive modeling of complex causal interactions

Compact representation of joint distributions O(n) rather than O(2n)

Algorithms for efficient inference with given evidence (more on this next time)

Documents

Probabilistic Inference