Artificial Intelligence 06.01 introduction bayesian_networks

Artificial IntelligenceIntroduction to Bayesian Networks

Andres Mendez-Vazquez

March 2, 2016

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Outline1 History

The History of Bayesian Applications

2 Bayes TheoremEverything Starts at SomeplaceWhy Bayesian Networks?

3 Bayesian NetworksDefinitionMarkov ConditionExample

Using the Markok ConditionRepresenting the Joint DistributionObservations

Causality and Bayesian NetworksPrecautionary Tale

Causal DAGInference in Bayesian NetworksExampleGeneral Strategy of InferenceInference - An Overview

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Outline1 History







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HistoryHistory

‘60s The first expert systems. IF-THEN rules.1968 Attempts to use probabilities in expert systems (Gorry &Barnett).1973 Gave up - to heavy calculations! (Gorry).1976 MYCIN: Medical predicate logic expert system with certaintyfactors (Shortliffe).1976 PROSPECTOR: Predicts the likely location of mineral deposits.Uses Bayes’ rule. (Duda et al.).

Summary until mid ’80s“Pure logic will solve the AI problems!”“Probability theory is intractable to use and too complicated forcomplex models.”

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HistoryHistory



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HistoryHistory



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HistoryHistory



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HistoryHistory



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HistoryHistory



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HistoryHistory



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But...

More History1986 Bayesian networks were revived and reintroduced to expertsystems (Pearl).1988 Breakthrough for efficient calculation algorithms (Lauritzen &Spiegelhalter) tractable calculations on Bayesian Networkss.1995 In Windows95™ for printer-trouble shooting and Officeassistance (“the paper clip”).1999 Bayesian Networks are getting more and more used. Ex. Geneexpression analysis, Business strategy etc.2000 Widely used - A Bayesian Network tool will be shipped withevery Windows™ Commercial Server.

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But...


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But...


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But...


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But...


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Furtheron 2000-2015

Bayesian Networks are use inSpam Detection.Gene Dicovery.Signal Processing.Ranking.Forecasting.etc.

Something NotableWe are interested more and more on building automatically BayesianNetworks using data!!!

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Furtheron 2000-2015

Bayesian Networks are use inSpam Detection.Gene Dicovery.Signal Processing.Ranking.Forecasting.etc.

Something NotableWe are interested more and more on building automatically BayesianNetworks using data!!!

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Bayesian Network Advantages

Many of Them1 Since in a Bayesian network encodes all variables, missing data entries

can be handled successfully.2 When used for learning casual relationships, they help better

understand a problem domain as well as forecast consequences.3 it is ideal to use a Bayesian network for representing prior data and

knowledge.4 Over-fitting of data can be avoidable when using Bayesian networks

and Bayesian statistical methods.

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Outline1 History







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Bayes TheoremOne Version

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

WhereP(A) is the prior probability or marginal probability of A. It is"prior" in the sense that it does not take into account any informationabout B.P(A|B) is the conditional probability of A, given B. It is also calledthe posterior probability because it is derived from or depends uponthe specified value of B.P(B|A) is the conditional probability of B given A. It is also calledthe likelihood.P(B) is the prior or marginal probability of B, and acts as anormalizing constant.

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A Simple Example

Consider two related variables:1 Drug (D) with values y or n2 Test (T ) with values +ve or –ve

Initial ProbabilitiesP(D = y) = 0.001P(T = +ve|D = y) = 0.8P(T = +ve|D = n) = 0.01

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A Simple Example



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A Simple Example



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A Simple Example



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A Simple Example



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A Simple Example

What is the probability that a person has taken the drug?

P (D = y|T = +ve) =P (T = +ve|D = y) P (D=y)

P (T = +ve|D = y) P (D=y) + P (T = +ve|D = n) P (D=n)

Let me develop the equationUsing simply

P (A, B) = P (A|B) P (B) (Chain Rule) (1)

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A Simple Example

What is the probability that a person has taken the drug?

P (D = y|T = +ve) =P (T = +ve|D = y) P (D=y)

P (T = +ve|D = y) P (D=y) + P (T = +ve|D = n) P (D=n)

Let me develop the equationUsing simply

P (A, B) = P (A|B) P (B) (Chain Rule) (1)

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Outline1 History







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A More Complex Case

Increase ComplexitySuppose now that there is a similar link between Lung Cancer (L) anda chest X-ray (X) and that we also have the following relationships:

I History of smoking (S) has a direct influence on bronchitis (B) andlung cancer (L);

I L and B have a direct influence on fatigue (F).

QuestionWhat is the probability that someone has bronchitis given that theysmoke, have fatigue and have received a positive X-ray result?

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A More Complex Case





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A More Complex Case





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A More Complex Case





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A More Complex Case

Short Hand

P (b1|s1, f1, x1) =P (b1, s1, f1, x1)

P (s1, f1, x1)=

∑l P (b1, s1, f1, x1, l)∑b,l P (b, s1, f1, x1, l)

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Values for the Complex Case

TableFeature Value When the Feature Takes this Value

H h1 There is a history of smokingh2 There is no history of smoking

B b1 Bronchitis is presentb2 Bronchitis is absent

L l1 Lung cancer is presentl2 Lung cancer is absent

F f1 Fatigue is presentf2 Fatigue is absent

C c1 Chest X-ray is positivec2 Chest X-ray is negative

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Problem with Large Instances

The joint probability distribution P(b, s, f , x , l)For five binary variables there are 25 = 32 values in the jointdistribution (for 100 variables there are over 2100 values)How are these values to be obtained?

We can try to do inferenceTo obtain posterior distributions once some evidence is availablerequires summation over an exponential number of terms!!!

OkWe need something BETTER!!!

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Bayesian Networks

DefinitionA Bayesian network consists of

A GraphI Nodes represent the random variables.I Directed edges (arrows) between pairs of nodes.I it must be a Directed Acyclic Graph (DAG) – no directed cycles.I The graph represents independence relationships between variables.

This allows to defineConditional Probability Specifications:

I The conditional probability of each variable given its parents in theDAG.

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Bayesian Networks





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Bayesian Networks





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Bayesian Networks





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Bayesian Networks





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Bayesian Networks





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Bayesian Networks





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Example

DAG for the previous Lung Cancer ProblemH

B L

F C

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Markov Condition

DefinitionSuppose we have a joint probability distribution P of the randomvariables in some set V and a DAG G = (V , E).

I We say that (G, P) satisfies the Markov condition if for each variableX ∈ V , {X} is conditionally independent of the set of all itsnon-descendents given the set of all its parents.

NotationPAX= set of parents of X .NDX= set of non-descendants of X .

We use the following the notation

IP ({X} , NDX |PAX )

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Markov Condition






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Markov Condition






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Markov Condition






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Markov Condition






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Example

We have thatH

B L

F C

Given the previous DAG we haveNode PA Conditional Independence

C {L} IP ({C} , {H , B, F} | {L})B {H} IP ({B} , {L, C} | {H})F {B, L} IP ({F} , {H , C} | {B, L})L {H} IP ({L} , {B} | {H})

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Example

We have thatH

B L

F C

Given the previous DAG we haveNode PA Conditional Independence

C {L} IP ({C} , {H , B, F} | {L})B {H} IP ({B} , {L, C} | {H})F {B, L} IP ({F} , {H , C} | {B, L})L {H} IP ({L} , {B} | {H})

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Using the Markov Condition

First Decompose a Joint Distribution using the Chain RuleP (c, f , l, b, h) = P (c|b, s, l, f ) P (f |b, h, l) P (l|b, h) P (b|h) P (h) (2)

Using the Markov condition in the following DAG

We have the following equivalencesP (c|b, h, l, f ) = P (c|l)P (f |b, h, l) = P (f |b, l)P (l|b, h) = P (l|h)

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Using the Markov ConditionFirst Decompose a Joint Distribution using the Chain Rule

P (c, f , l, b, h) = P (c|b, s, l, f ) P (f |b, h, l) P (l|b, h) P (b|h) P (h) (2)

Using the Markov condition in the following DAGH

B L

F C


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Using the Markov ConditionFirst Decompose a Joint Distribution using the Chain Rule

P (c, f , l, b, h) = P (c|b, s, l, f ) P (f |b, h, l) P (l|b, h) P (b|h) P (h) (2)

Using the Markov condition in the following DAGH

B L

F C


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Using the Markov Condition

Finally

P (c, f , l, b, h) = P (c|l) P (f |b, l) P (l|h) P (b|h) P (h) (3)

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Representing the Joint Distribution

Theorem 1.4If (G, P) satisfies the Markov condition, then P is equal to the product ofits conditional distributions of all nodes given values of their parents,whenever these conditional distributions exist.

General RepresentationIn general, for a network with nodes X1, X2, ..., Xn ⇒

P (x1, x2, ..., xn) =n∏

i=1P (xi |PA (xi))

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Representing the Joint Distribution

Theorem 1.4If (G, P) satisfies the Markov condition, then P is equal to the product ofits conditional distributions of all nodes given values of their parents,whenever these conditional distributions exist.

General RepresentationIn general, for a network with nodes X1, X2, ..., Xn ⇒

P (x1, x2, ..., xn) =n∏

i=1P (xi |PA (xi))

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Proof of Theorem 1.4

We prove the case where P is discreteOrder the nodes so that if Y is a descendant of Z , then Y follows Z inthe ordering.

Topological Sorting.

This is calledAncestral ordering.

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Proof

For example

The ancestral ordering are

[H , L, B, C , F ] and [H , B, L, F , C ] (4)

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Proof

For example

The ancestral ordering are

[H , L, B, C , F ] and [H , B, L, F , C ] (4)

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Proof

NowLet X1, X2, ..., Xn be the resultant ordering.

For a given set of values of x1, x2, ..., xn

Let pai be the subsets of these values containing the values of X ′i s parents

Thus, we need to prove that whenever P (pai) 6= 0 for 1 ≤ i ≤ n

P (xn , xn−1, ..., x1) = P (xn |pan) P(xn−1|pan−1

)...P (x1|pa1) (5)

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Proof






)...P (x1|pa1) (5)

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Proof






)...P (x1|pa1) (5)

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ProofSomething NotableWe show this using induction on the number of variables in the network.

Assume then P (pai) 6= 0 for 1 ≤ i ≤ n for a combination of x ′isvalues.

Base Case of InductionSince pa1 is empty, then

P (x1) = P (x1|pa1) (6)

Inductive HypothesisSuppose for this combination of values of the xi ’s that

P (xi , xi−1, ..., x1) = P (xi |pai) P(xi−1|pai−1

)...P (x1|pa1) (7)

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P (x1) = P (x1|pa1) (6)



)...P (x1|pa1) (7)

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P (x1) = P (x1|pa1) (6)



)...P (x1|pa1) (7)

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P (x1) = P (x1|pa1) (6)



)...P (x1|pa1) (7)

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P (x1) = P (x1|pa1) (6)



)...P (x1|pa1) (7)

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P (x1) = P (x1|pa1) (6)



)...P (x1|pa1) (7)

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Proof

Inductive StepWe need show for this combination of values of the xi ’s that

P (xi+1, xi , ..., x1) = P(xi+1|pai+1

)P (xi |pai) ...P (x1|pa1) (8)

Case 1For this combination of values:

P (xi , xi−1, ..., x1) = 0 (9)

By Conditional Probability, we have

P (xi+1, xi , ..., x1) = P (xi+1|xi , ..., x1) P (xi , ..., x1) = 0 (10)

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Proof


P (xi+1, xi , ..., x1) = P(xi+1|pai+1

)P (xi |pai) ...P (x1|pa1) (8)


P (xi , xi−1, ..., x1) = 0 (9)



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Proof


P (xi+1, xi , ..., x1) = P(xi+1|pai+1

)P (xi |pai) ...P (x1|pa1) (8)


P (xi , xi−1, ..., x1) = 0 (9)



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Proof

Due to the previous equalities and the inductive hypothesisThere is some k, 1 ≤ k ≤ i such that P (xk |pak) = 0 because after all

P (xi |pai) P(xi−1|pai−1

)...P (x1|pa1) = 0 (11)

Thus, the equality holdsNow for the Case 2

Case 2For this combination of values P (xi , xi−1, ..., x1) 6= 0

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Proof



)...P (x1|pa1) = 0 (11)



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Proof



)...P (x1|pa1) = 0 (11)



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Proof

Thus by the Rule of Conditional Probability

P (xi+1, xi , ..., x1) = P (xi+1|xi , ..., x1) P (xi , ..., x1)

Definition Markov Condition (Remember!!!)Suppose we have a joint probability distribution P of the randomvariables in some set V and a DAG G = (V , E).


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Proof





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Proof





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Proof

Given this Markov Condition and the fact that X1, ..., Xi are allnon-descendants of Xi+1

We have that

P (xi+1, xi , ..., x1) = P(xi+1|pai+1

)P (xi , ..., x1)

= P(xi+1|pai+1

)P (xi |pai) · · ·P (x1|pa1) (IH)

Q.E.D.

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Proof

Given this Markov Condition and the fact that X1, ..., Xi are allnon-descendants of Xi+1

We have that

P (xi+1, xi , ..., x1) = P(xi+1|pai+1

)P (xi , ..., x1)

= P(xi+1|pai+1

)P (xi |pai) · · ·P (x1|pa1) (IH)

Q.E.D.

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Now

OBSERVATIONS1 An enormous saving can be made regarding the number of values

required for the joint distribution.2 To determine the joint distribution directly for n binary variables 2n

values are required.3 For a Bayesian Network with n binary variables and each node has at

most k parents then less than 2kn values are required!!!

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Now





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Now





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It is more!!!

Theorem 1.5Let a DAG G be given in which each node is a random variable, andlet a discrete conditional probability distribution of each node givenvalues of its parents in G be specified.Then, the product of these conditional distributions yields a jointprobability distribution P of the variables, and (G, P) satisfies theMarkov condition.

NoteNotice that the theorem requires that specified conditionaldistributions be discrete.Often in the case of continuous distributions it still holds.

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It is more!!!



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It is more!!!



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It is more!!!



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Causality in Bayesian Networks

Definition of a CauseThe one, such as a person, an event, or a condition, that is responsible foran action or a result.

HoweverAlthough useful, this simple definition is certainly not the last word on theconcept of causation.

Actually Philosophers are still wrangling the issue!!!

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Nevertheless, It sheds light in the issueIf the action of making variable X take some value sometimeschanges the value taken by a variable Y .

Causality

Here, we assume X is responsible for sometimes changing Y ’s valueThus, we conclude X is a cause of Y .

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Nevertheless, It sheds light in the issueIf the action of making variable X take some value sometimeschanges the value taken by a variable Y .

Causality

Here, we assume X is responsible for sometimes changing Y ’s valueThus, we conclude X is a cause of Y .

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Furthermore

FormallyWe say we manipulate X when we force X to take some value.

We say X causes Y if there is some manipulation of X that leads toa change in the probability distribution of Y .

ThusWe assume causes and their effects are statistically correlated.

HoweverVariables can be correlated without one causing the other.

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Furthermore





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Furthermore





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Furthermore





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Precautionary Tale: Causality and Bayesian Networks

ImportantNot every Bayesian Networks describes causal relationships between thevariables.

ConsiderConsider the dependence between Lung Cancer, L, and the X-raytest, X .By focusing on just these variables we might be tempted to representthem by the following Bayesian Networks.

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L X

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However, we can try the same

L X

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Remark

Be CarefulIt is tempting to think that Bayesian Networkss can be created by creatinga DAG where the edges represent direct causal relationships between thevariables.

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However

Causal DAGGiven a set of variables V , if for every X , Y ∈ V we draw an edge from Xto Y ⇐⇒ X is a direct cause of Y relative to V , we call the resultantDAG a causal DAG.

We wantIf we create a causal DAG G = (V , E) and assume the probabilitydistribution of the variables in V satisfies the Markov condition with G:

we say we are making the causal Markov assumption.

In GeneralThe Markov condition holds for a causal DAG.

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However





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However





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However





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However, we still want to know if the Markov ConditionHolds

RemarkThere are several thing that the DAG needs to have in order to have theMarkov Condition.

Examples of thoseCommon CausesCommon Effects

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How to have a Markov Assumption : Common Causes

Consider

Smoking

Bronchitis Lung Cancer

Markov condition

Ip ({B} , {L} | {S})⇒ P(b|l, s) = P(b|s) (12)

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Consider

Smoking

Bronchitis Lung Cancer

Markov condition

Ip ({B} , {L} | {S})⇒ P(b|l, s) = P(b|s) (12)

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If we know the causal relationships

S → B and S → L (13)

Now!!!If we know the person is a smoker.

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If we know the causal relationships

S → B and S → L (13)

Now!!!If we know the person is a smoker.

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Then, because of the blocking of information from smokingFinding out that he has Bronchitis will not give us any more informationabout the probability of him having Lung Cancer.

Markov conditionIt is satisfied!!!

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Then, because of the blocking of information from smokingFinding out that he has Bronchitis will not give us any more informationabout the probability of him having Lung Cancer.

Markov conditionIt is satisfied!!!

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How to have a Markov Assumption : Common Effects

Consider

Alarm

Burglary Earthquake

Markov Condition

lp (B, E)⇒ P(b|e) = P(b) (14)

ThusWe would expect Burglary and Earthquake to be independent of eachother which is in agreement with the Markov condition.

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Consider

Alarm

Burglary Earthquake

Markov Condition

lp (B, E)⇒ P(b|e) = P(b) (14)


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Consider

Alarm

Burglary Earthquake

Markov Condition

lp (B, E)⇒ P(b|e) = P(b) (14)


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HoweverWe would, however expect them to be conditionally dependent givenAlarm.

ThusIf the alarm has gone off, news that there had been an earthquake would‘explain away’ the idea that a burglary had taken place.

ThenAgain in agreement with the Markov condition.

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The Causal Markov Condition

What do we want?The basic idea is that the Markov condition holds for a causal DAG.

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Rules to construct A Causal Graph

Conditions1 There must be no hidden common causes.2 There must not be selection bias.3 There must be no feedback loops.

ObservationsEven with these there is a lot of controversy as to its validity.It seems to be false in quantum mechanical.

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Hidden Common Causes?Given the following DAG

H

X Y

Z

Something NotableIf a DAG is created on the basis of causal relationships between thevariables under consideration then X and Y would be marginallyindependent according to the Markov condition.

ThusIf H is hidden, they will normally be dependent.

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H

X Y

Z



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H

X Y

Z



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Inference in Bayesian Networks

What do we want from Bayesian Networks?The main point of Bayesian Networkss is to enable probabilistic inferenceto be performed.

Two different types of inferences1 Belief Updating.2 Abduction Inference.

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Belief updatingIt is used to obtain the posterior probability of one or more variables givenevidence concerning the values of other variables.

Abductive inferenceIt finds the most probable configuration of a set of variables (hypothesis)given certain evidence.

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Belief updatingIt is used to obtain the posterior probability of one or more variables givenevidence concerning the values of other variables.

Abductive inferenceIt finds the most probable configuration of a set of variables (hypothesis)given certain evidence.

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Using the Structure I

Consider the following Bayesian NetworksBurgalary Earthquake

Alarm

JohnCalls MaryCalls

P(B)

0.001

P(E)

0.002

P(B)

0.001

P(E)

0.002

B E P(A|B,E)

T T 0.95

T F 0.94

F T 0.29

F F 0.001

P(B)

0.001

A P(JC|A)

T 0.9

F 0.05

A P(MC|A)

T 0.7

F 0.01

Consider answering a query in a Bayesian NetworkQ= set of query variablese= evidence (set of instantiated variable-value pairs)Inference = computation of conditional distribution P(Q|e)

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Alarm

JohnCalls MaryCalls

P(B)

0.001

P(E)

0.002

P(B)

0.001

P(E)

0.002

B E P(A|B,E)

T T 0.95

T F 0.94

F T 0.29

F F 0.001

P(B)

0.001

A P(JC|A)

T 0.9

F 0.05

A P(MC|A)

T 0.7

F 0.01


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Alarm

JohnCalls MaryCalls

P(B)

0.001

P(E)

0.002

P(B)

0.001

P(E)

0.002

B E P(A|B,E)

T T 0.95

T F 0.94

F T 0.29

F F 0.001

P(B)

0.001

A P(JC|A)

T 0.9

F 0.05

A P(MC|A)

T 0.7

F 0.01


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Alarm

JohnCalls MaryCalls

P(B)

0.001

P(E)

0.002

P(B)

0.001

P(E)

0.002

B E P(A|B,E)

T T 0.95

T F 0.94

F T 0.29

F F 0.001

P(B)

0.001

A P(JC|A)

T 0.9

F 0.05

A P(MC|A)

T 0.7

F 0.01


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Using the Structure II

ExamplesP(burglary|alarm)P(earthquake|JCalls, MCalls)P(JCalls, MCalls|burglary, earthquake)

SoCan we use the structure of the Bayesian Network to answer such queriesefficiently?

AnswerYES

Note: Generally speaking, complexity is inversely proportional tosparsity of graph

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AnswerYES


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AnswerYES


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AnswerYES


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AnswerYES


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AnswerYES


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ExampleDAG

D

B E

CA F G

We have the following modelp (a, b, c, d, e, f , g) is modeled by

p (a|b) p (c|b) p (f |e) p (g|e) p (b|d) p (e|d) p (d)

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ExampleDAG

D

B E

CA F G

We have the following modelp (a, b, c, d, e, f , g) is modeled by

p (a|b) p (c|b) p (f |e) p (g|e) p (b|d) p (e|d) p (d)

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Example

DAG

D

B E

CA F G

We want to calculate the following

p (a|c, g)

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Example

DAG

D

B E

CA F G

We want to calculate the following

p (a|c, g)

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ExampleDAG

D

B E

CA F G

However, a direct calculation leads to use a demarginalization

p (a|c, g) =∑

b,d,e,fp (a, b, d, e, f |c, g)

This will require that if we fix the value of a, c and g to have a complexityof O

(m4) with m = max {|B| , |D| , |E | , |F |}

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ExampleDAG

D

B E

CA F G

However, a direct calculation leads to use a demarginalization

p (a|c, g) =∑

b,d,e,fp (a, b, d, e, f |c, g)

This will require that if we fix the value of a, c and g to have a complexityof O

(m4) with m = max {|B| , |D| , |E | , |F |}

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ExampleWe get some information about (a = ai , c = ci , g = gi)

D

B E

CA F G

However, we re-express the equation using the chain representation

p (a = ai , b, d, e, f |c = ci , g = gi) =...

p (a = ai |b) p (b|d, c = ci) p (d|e) p (e, f |g = gi)

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ExampleWe get some information about (a = ai , c = ci , g = gi)

D

B E

CA F G

However, we re-express the equation using the chain representation

p (a = ai , b, d, e, f |c = ci , g = gi) =...

p (a = ai |b) p (b|d, c = ci) p (d|e) p (e, f |g = gi)

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Example

DAG

D

B E

CA F G

Now, we re-order the sum∑b

p (a = ai |b)∑

dp (b|d, c = ci)

∑e

p (d|e)∑

fp (e, f |g = gi)

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Example

DAG

D

B E

CA F G

Now, we re-order the sum∑b

p (a = ai |b)∑

dp (b|d, c = ci)

∑e

p (d|e)∑

fp (e, f |g = gi)

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ExampleNow, using the relation about E

D

B E

CA F G

Using this information, we can reduce one of the sums bymarginalization ∑

fp (e, f |g = gi) = p (e|g = gi)

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ExampleNow, using the relation about E

D

B E

CA F G

Using this information, we can reduce one of the sums bymarginalization ∑

fp (e, f |g = gi) = p (e|g = gi)

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Example

DAG

D

B E

CA F G

Thus, we can reduce the size of our sum∑b

p (a = ai |b)∑

dp (b|d, c = ci)

∑e

p (d|e) p (e|g = gi)

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Example

DAG

D

B E

CA F G

Thus, we can reduce the size of our sum∑b

p (a = ai |b)∑

dp (b|d, c = ci)

∑e

p (d|e) p (e|g = gi)

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Example

DAG

D

B E

CA F G

Now, we can calculate the probability of D by using the chain rule

p (d|e) p (e|g = gi) = p (d|e, g = gi) p (e|g = gi) = p (d, e|g = gi)

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Example

DAG

D

B E

CA F G

Now, we can calculate the probability of D by using the chain rule

p (d|e) p (e|g = gi) = p (d|e, g = gi) p (e|g = gi) = p (d, e|g = gi)

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Example

DAG

D

B E

CA F G

Now, we can calculate the probability of D by using the chain rule∑b

p (a = ai |b)∑

dp (b|d, c = ci)

∑e

p (d, e|g = gi)

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Example

DAG

D

B E

CA F G

Now, we can calculate the probability of D by using the chain rule∑b

p (a = ai |b)∑

dp (b|d, c = ci)

∑e

p (d, e|g = gi)

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Example

DAG

D

B E

CA F G

Now, we sum over all possible values of E∑e

p (d, e|g = gi) = p (d|g = gi)

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Example

DAG

D

B E

CA F G

Now, we sum over all possible values of E∑e

p (d, e|g = gi) = p (d|g = gi)

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Example

DAG

D

B E

CA F G

We get the following∑b

p (a = ai |b)∑

dp (b|d, c = ci) p (d|g = gi)

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Example

DAG

D

B E

CA F G

We get the following∑b

p (a = ai |b)∑

dp (b|d, c = ci) p (d|g = gi)

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ExampleDAG

D

B E

CA F G

Again the chain rule for D

p (b|d, c = ci) p (d|g = gi) = p (b|d, c = ci , g = gi) p (d|c = ci , g = gi)= p (b, d|c = ci , g = gi)

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ExampleDAG

D

B E

CA F G

Again the chain rule for D

p (b|d, c = ci) p (d|g = gi) = p (b|d, c = ci , g = gi) p (d|c = ci , g = gi)= p (b, d|c = ci , g = gi)

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Example

DAG

D

B E

CA F G

Now, we sum over all possible values of D∑b

p (a = ai |b) p (b|c = ci , g = gi)

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Example

DAG

D

B E

CA F G

Now, we sum over all possible values of D∑b

p (a = ai |b) p (b|c = ci , g = gi)

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Example

DAG

D

B E

CA F G

Now, we use the chain rule for reducing again

p (a = ai |b) p (b|) = p (a = ai , b|c = ci , g = gi)

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Example

DAG

D

B E

CA F G

Now, we use the chain rule for reducing again

p (a = ai |b) p (b|) = p (a = ai , b|c = ci , g = gi)

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Example

DAG

D

B E

CA F G

Now, we use the chain rule for reducing again∑b

p (a = ai , b|c = ci , g = gi) = p (a = ai |c = ci , g = gi)

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Example

DAG

D

B E

CA F G

Now, we use the chain rule for reducing again∑b

p (a = ai , b|c = ci , g = gi) = p (a = ai |c = ci , g = gi)

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Complexity

Because this can be computed using a sequence of four for loopsThe complexity simply becomes O (m) when compared with O

(m4)

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Outline1 History







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General Strategy for Inference

QueryWant to compute P(q|e)!!!

Step 1P(q|e) = P(q,e)

P(e) = aP(q, e), since a = P(e) is constant wrt Q.

Step 2P(q, e) =

∑a..z P(q, e, a, b, . . . .z), by the law of total probability.

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Step 2P(q, e) =


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Step 2P(q, e) =


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General Strategy for inference

Step 3∑a..z P(q, e, a, b, . . . .z) =

∑a..z

∏P(variable i | parents i) (using

Bayesian network factoring)

Step 4Distribute summations across product terms for efficient computation.

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General Strategy for inference

Step 3∑a..z P(q, e, a, b, . . . .z) =

∑a..z

∏P(variable i | parents i) (using

Bayesian network factoring)

Step 4Distribute summations across product terms for efficient computation.

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Outline1 History







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Inference – An OverviewCase 1

Trees and singly connected networks – only one path between any twonodes:

I Message passing (Pearl, 1988)

Case 2Multiply connected networks:

I A range of algorithms including cut-set conditioning (Pearl, 1988),junction tree propagation (Lauritzen and Spiegelhalter, 1988), bucketelimination (Dechter, 1996) to mention a few.

I A range of algorithms for approximate inference.

NotesBoth exact and approximate inference are NP-hard in the worst case.Here the focus will be on message passing and junction treepropagation for discrete variables.

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Artificial Intelligence 06.01 introduction bayesian_networks

Engineering