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Probabilistic Inference

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Probabilistic Inference. Agenda. Review of probability theory Joint, conditional distributions Independence Marginalization and conditioning Intro to Bayesian Networks. Probabilistic Belief. Consider a world where a dentist agent D meets with a new patient P - PowerPoint PPT Presentation

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Page 1: Probabilistic Inference

Probabilistic Inference

Page 2: Probabilistic Inference

Agenda

• Review of probability theory

• Joint, conditional distributions

• Independence

• Marginalization and conditioning

• Intro to Bayesian Networks

Page 3: Probabilistic Inference

Probabilistic Belief Consider a world where a dentist agent D

meets with a new patient P

D is interested in only whether P has a cavity; so, a state is described with a single proposition – Cavity

Before observing P, D does not know if P has a cavity, but from years of practice, he believes Cavity with some probability p and Cavity with probability 1-p

The proposition is now a boolean random variable and (Cavity, p) is a probabilistic belief

Page 4: Probabilistic Inference

Probabilistic Belief State

The world has only two possible states, which are respectively described by Cavity and Cavity

The probabilistic belief state of an agent is a probabilistic distribution over all the states that the agent thinks possible

In the dentist example, D’s belief state is:

Cavity Cavityp 1-p

Page 5: Probabilistic Inference

Multivariate Belief State

We now represent the world of the dentist D using three propositions – Cavity, Toothache, and PCatch

D’s belief state consists of 23 = 8 states each with some probability:

{CavityToothachePCatch, CavityToothachePCatch, CavityToothachePCatch,...}

Page 6: Probabilistic Inference

The belief state is defined by the full joint probability of the

propositions

PCatch PCatch PCatch PCatch

Cavity 0.108 0.012 0.072 0.008

Cavity 0.016 0.064 0.144 0.576

Toothache Toothache

Page 7: Probabilistic Inference

Probabilistic Inference

PCatch PCatch PCatch PCatch

Cavity 0.108 0.012 0.072 0.008

Cavity 0.016 0.064 0.144 0.576

Toothache Toothache

P(Cavity Toothache) = 0.108 + 0.012 + ...= 0.28

Page 8: Probabilistic Inference

Probabilistic Inference

PCatch PCatch PCatch PCatch

Cavity 0.108 0.012 0.072 0.008

Cavity 0.016 0.064 0.144 0.576

Toothache Toothache

P(Cavity) = 0.108 + 0.012 + 0.072 + 0.008= 0.2

Page 9: Probabilistic Inference

Probabilistic Inference

PCatch PCatch PCatch PCatch

Cavity 0.108 0.012 0.072 0.008

Cavity 0.016 0.064 0.144 0.576

Toothache Toothache

Marginalization: P(c) = tpc P(ctpc) using the conventions that c = Cavity or Cavity and that t is the sum over t = {Toothache, Toothache}

Page 10: Probabilistic Inference

Komolgorov’s Probability Axioms

• 0 P(a) 1• P(true) = 1, P(false) = 0• P(a b) = P(a) + P(b) - P(a b)

Page 11: Probabilistic Inference

Decoding Probability Notation

• Capital letters A,B,C denote random variables

• Each random variable X can take one of a set of possible values xVal(X)– Confusingly, this is often left implicit

• Probabilities defined over logical statements, e.g., P(X=x)– Confusingly, this is often written P(x)…

Page 12: Probabilistic Inference

Decoding Probability Notation

• Interpretation of P(A) depends on if A treated as a logical statement, or a random variable

• As a logical statement– Just the probability that A is true– P(A) = 1-P(A)

• As a random variable– Denotes both P(A is true) and P(A is false)– For non-boolean variables, denotes P(A=a) for all a

Val(A)

• Computation requires marginalization– Belief space A, B, C, D, … often left implicit– Marginalize the joint distribution over all combinations of

B, C, D… (event space)

Page 13: Probabilistic Inference

Decoding Probability Notation

• How to interpret P(AB) = P(A,B)?• As a logical statement

– Probability that both A and B are true• As random variables

– {P(A=0,B=0), P(A=1,B=0), P(A=0,B=1), P(A=1,B=1)}

– P(A=a,B=b) for all aVal(A), bVal(B) in general

• Equal to P(A)P(B)? No!!! (except under very special conditions)

Page 14: Probabilistic Inference

Quiz: What does this mean?

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

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

For all aVal(A) and bVal(B)

Page 15: Probabilistic Inference

Marginalization

• Express P(A,B) in terms of P(A,B,C) • If C is boolean,

– P(A,B) = P(A,B,C=0)+ P(A,B,C=1)

• In general– P(A,B) = cVal(C) P(A,B,C=c)

Page 16: Probabilistic Inference

Conditional Probability

P(AB) = P(A|B) P(B)= P(B|A) P(A)

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

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

Page 17: Probabilistic Inference

PCatch PCatch PCatch PCatch

Cavity 0.108 0.012 0.072 0.008

Cavity 0.016 0.064 0.144 0.576

Toothache Toothache

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 unchanged, and normalizing their sum to 1

Page 18: Probabilistic Inference

PCatch PCatch PCatch PCatch

Cavity 0.108 0.012 0.072 0.008

Cavity 0.016 0.064 0.144 0.576

Toothache Toothache

P(Cavity|Toothache) = P(CavityToothache)/P(Toothache)= (0.108+0.012)/(0.108+0.012+0.016+0.064) = 0.6

P(Cavity|Toothache)=P(CavityToothache)/P(Toothache)= (0.016+0.064)/(0.108+0.012+0.016+0.064) = 0.4

P(c|Toochache) = (P(Cavity|Toothache), P(Cavity|Toothache))= P(c Toothache)

= pc P(c Toothache pc) = [(0.108, 0.016) + (0.012, 0.064)] = (0.12, 0.08) = (0.6, 0.4)

normalizationconstant

Page 19: Probabilistic Inference

Conditioning

• Express P(A) in terms of P(A|B), P(B)• If C is boolean,

– P(A) = P(A|B=0) P(B=0)+ P(A|B=1) P(B=1)

• In general– P(A) = bVal(B) P(A|B=b) P(B=b)

Page 20: Probabilistic Inference

Conditional Probability

P(AB) = P(A|B) P(B)= P(B|A) P(A)

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

P(Cavity) = tpc P(Cavitytpc)

= tpc P(Cavity|t,pc) P(tpc)

Page 21: Probabilistic Inference

Independence

Two random variables A and B are independent if

P(AB) = P(A) P(B) hence if P(A|B) = P(A)

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

P(AB|C) = P(A|C) P(B|C)hence if P(A|B,C) = P(A|C)

Page 22: Probabilistic Inference

Updating the Belief State

PCatch PCatch PCatch PCatch

Cavity 0.108 0.012 0.072 0.008

Cavity 0.016 0.064 0.144 0.576

Toothache Toothache

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

How should D update its belief state?

Page 23: Probabilistic Inference

Updating the Belief State

PCatch PCatch PCatch PCatch

Cavity 0.108 0.012 0.072 0.008

Cavity 0.016 0.064 0.144 0.576

Toothache Toothache

Let E be the evidence such that P(Toothache|E) = 0.8 We want to compute P(ctpc|E) = P(cpc|t,E) P(t|E) Since E is not directly related to the cavity or the probe

catch, we consider that c and pc are independent of E given t, hence: P(cpc|t,E) = P(cpc|t)

Page 24: Probabilistic Inference

Let E be the evidence such that P(Toothache|E) = 0.8 We want to compute P(ctpc|E) = P(cpc|t,E) P(t|E) Since E is not directly related to the cavity or the probe

catch, we consider that c and pc are independent of E given t, hence: P(cpc|t,E) = P(cpc|t) To get these 4 probabilities

we normalize their sum to 0.2

To get these 4 probabilities we normalize their sum to 0.8

Updating the Belief State

PCatch PCatch PCatch PCatch

Cavity 0.108 0.012 0.072 0.008

Cavity 0.016 0.064 0.144 0.576

Toothache Toothache

0.432

0.064 0.256

0.048 0.018

0.036 0.144

0.002

Page 25: Probabilistic Inference

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

Page 26: Probabilistic Inference

Bayes’ Rule and other Probability Manipulations

P(AB) = P(A|B) P(B)= P(B|A) P(A)

P(A|B) = P(B|A) P(A) / P(B) Can manipulate distributions, e.g.

P(B) = a P(B|A=a) P(A=a) Can derive P(A|B), P(B) using only P(B|A) and P(A)

So, if any variables are conditionally independent, we should be able to decompose joint distribution to take advantage of it

Page 27: Probabilistic Inference

Toothache and PCatch are independent given Cavity (or Cavity), but this relation is hidden in the numbers ! [Verify this]

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

PCatch PCatch PCatch PCatch

Cavity 0.108 0.012 0.072 0.008

Cavity 0.016 0.064 0.144 0.576

Toothache Toothache

Page 28: Probabilistic Inference

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 PCatch

Cavity

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(ctpc) = P(tpc|c) P(c) = P(t|c) P(pc|c) P(c)

Page 29: Probabilistic Inference

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(ctpc) = P(tpc|c) P(c) = P(t|c) P(pc|c) P(c)

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

Page 30: Probabilistic Inference

Naïve Bayes Models

P(Cause,Effect1,…,Effectn)

= P(Cause) i P(Effecti | Cause)

Cause

Effect1 Effect2 Effectn

Page 31: Probabilistic Inference

Naïve Bayes Classifier

P(Class,Feature1,…,Featuren)

= P(Class) i P(Featurei | Class)

Class

Feature1 Feature2 Featuren

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

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

Given features, what class?

Spam / Not Spam

English / French/ Latin

Word occurrences

Page 32: Probabilistic Inference

Equations Involving Random Variables

• 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

Page 33: Probabilistic Inference

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”

Page 34: Probabilistic Inference

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

Page 35: Probabilistic Inference

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)

Page 36: Probabilistic Inference

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

What does the BN encode?

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

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

Page 37: Probabilistic Inference

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

What does the BN encode?

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

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

Burglary and Earthquake are independent

Burglary and Earthquake are independent

Page 38: Probabilistic Inference

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

Page 39: Probabilistic Inference

But does a BN represent a belief state?

In other words, can we compute the full joint

distribution of the propositions from it?

Page 40: Probabilistic Inference

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) = ??

Page 41: Probabilistic Inference

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

Page 42: Probabilistic Inference

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(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

Page 43: Probabilistic Inference

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(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) = i=1,…,nP(xi|parents(Xi))

full joint distribution table

Page 44: Probabilistic Inference

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(x1x2…xn) = i=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

Page 45: Probabilistic Inference

The BN gives P(t|c) What about P(c|t)? P(Cavity|t)

= P(Cavity t)/P(t)= P(t|Cavity) P(Cavity) /

P(t)[Bayes’ rule]

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

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

Querying the BN

Cavity

Toothache

P(C)

0.1

C P(T|c)

TF

0.40.01111

Page 46: Probabilistic Inference

More Complicated Singly-Connected Belief Net

Radio

Battery

SparkPlugs

Starts

Gas

Moves

Page 47: Probabilistic Inference

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

Page 48: Probabilistic Inference

Region = {Sky, Tree, Grass, Rock}

R2

R4R3

R1

Above

Page 49: Probabilistic Inference

BN to evaluate insurance risks

Page 50: Probabilistic Inference

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 new evidence (more on this next time)

Page 51: Probabilistic Inference

Reminder

• HW4 due on Thursday!