baye’s rule. baye’s rule and reasoning allows use of uncertain causal knowledge knowledge: given...

Baye’s Rule

(b)

1 where)()|(

)(

)()|()|(

PPP

P

PPP aab

b

aabba

)|()|()|(

)()|()|(

acabacb

aacbcba

PPP

PPP

)|()(),...,,(

stributionability DiJoint Prob Full- s Baye'Naive

1CauseEffectCauseEffectEffectCause

iin PPP

Baye’s Rule and Reasoning

Allows use of uncertain causal knowledge Knowledge: given a cause what is the likelihood of

seeing particular effects (conditional probabilities) Reasoning: Seeing some effects, how do we infer the

likelihood of a cause.

This can be very complicated: need joint probability distribution of (k+1) variables, i.e., 2k+1 numbers.

Use conditional independence to simplify expressions.Allows sequential step by step computation

)...(

)()|...()...|(

21

2121

k

kk eee

HHeeeeeeH

P

PPP

Bayesian/Belief NetworkTo avoid problems of enumerating large joint probabilities

Use causal knowledge and independence to simplify reasoning, and draw inferences

)()|().......,....,|(),....,|(

),....,(),....,|(),....,,(

13221

22121

nnnnn

nnn

XPXXPXXXPXXXP

XXPXXXPXXXP

Bayesian NetworksAlso called Belief Network or probabilistic network Nodes – random variables, one variable per

node Directed Links between pairs of nodes. AB

A has a direct influence on B With no directed cycles

A conditional distribution for each node given its parents

))(|( ii XParentsXP

Cavity

Toothache Catch

WeatherMust determine theDomain specific topology.

Bayesian NetworksNext step is to determine the conditional probability distribution for each variable. Represented as a conditional probability

table (CPT) giving the distribution over Xi for each combination of the parent value.

Once CPT is determined, the full joint probability distribution is represented by the network.

The network provides a complete description of a domain.

Belief Networks: ExampleIf you go to college, this will effect the likelihood that you will study and the likelihood that you will party. Studying and partying effect your chances of exam success, and partying effects your chances of having fun.Variables: College, Study, Party, Exam (success), FunCausal Relations: College will affect studying College will affect parting Studying and partying will affect exam success Partying affects having fun. College

PartyStudy

FunExam

College example: CPTs

CPT

Discrete Variables only in this format

College

PartyStudy

FunExam

P(C)

0.2

C P(S)

True 0.8

False 0.2

C P(P)

True 0.6

False 0.5

S P P(E)

True True 0.6

True False 0.9

False

True 0.1

False

False 0.2

P P(F)

True 0.9

False 0.7

Belief Networks: Compactness

A CPT for Boolean variable Xi with k Boolean parents is 2k rows for combinations of parent valuesEach row requires one number p for Xi = true (the number Xi = false is 1-p) Row must sum to 1. Conditional Probability

If each variable had no more than k parents, then complete network requires O(n2k) numbers i.e., the numbers grow linearly in n vs. O(2n) for the full

joint distribution

College net has 1+2+2+4+2=11 numbers

Belief Networks: Joint Probability Distribution Calculation

Global semantics defines the full joint distribution as the product of local distributions: ))(|(),...,,(

121

n

iiin XParentsXXXX PP

)|(),|()|()|()(

)(

fun. haveor party not but exams your on successful

be andstudy will that youand college togoingofyProbabilit

PFPPSEPCPPCSPCP

FEPSCP

0.2*0.8*0.4*0.9*0.3 = 0.01728

Can use the networks to make inferences.

College

PartyStudy

FunExam Every value in a full joint probability distributioncan be calculated.


College

PartyStudy

FunExam

P(C)

0.2

C P(S)

True 0.8

False 0.2

C P(P)

True 0.6

False 0.5

S P P(E)

True True 0.6

True False 0.9

False

True 0.1

False

False 0.2

P P(F)

True 0.9

False 0.7

)|(),|()|()|()( PFPPSEPCPPCSPCP 0.2*0.8*0.4*0.9*0.3 = 0.01728

)( FEPSCP

Network ConstructionMust ensure network and distribution are good representations of the domain. Want to rely on conditional independence

relationships. First, rewrite the joint distribution in

terms of the conditional probability.

Repeat for each conjunctive probability

),...,(),...,|(),...,( 11111 xxPxxxPxxP nnnn

n

i

nnnn

xxxP

xPxxPxxPxxxPxxP

ii

11

11211111

),...,|(

)()|()...,...,(),...,|(),...,(

1 Chain Rule

Network Construction Note is equivalent to:

where the partial order is defined by the graph structure.

n

iiin xxxPxxP

1111 ),...,|(),...,(

))(|(),...,|( 11 iiii XParentsXXXX PP

},...,{)( 11 XXXParents ii

The above equation says that the network correctly represents the domainonly if each node is conditionally independent of its predecessors in the node ordering, given the node’s parents. Means: Parents of Xi needs to contain all nodes in X1,…,Xi-1 that have a direct influence on Xi.

College example:

P(F|C, S, P, E) = P(F|P)

College

PartyStudy

FunExam

P(C)

0.2

C P(S)

True 0.8

False 0.2

C P(P)

True 0.6

False 0.5

S P P(E)

True True 0.6

True False 0.9

False

True 0.1

False

False 0.2

P P(F)

True 0.9

False 0.7

Compact NetworksBayesian networks are sparse, therefore, much more compact than full joint distribution. Sparse: each subcomponent interacts directly

with a bounded number of other nodes independent of the total number of components.

Usually linearly bounded complexity. College net has 1+2+2+4+2=11 numbers Fully connected domain = full joint distribution.

Must determine the correct network topology. Add “root causes” first then the variables that

they influence.

Network ConstructionNeed a method such that a series of locally testable assertions of conditional independence guarantees the required global semantics

1. Choose an ordering of variables X1, …., Xn

2. For i = 1 to nadd Xi to networkselect parents from X1, …, Xi-1 such

thatP(Xi |Parents(Xi)) = P(Xi | X1,… Xi-1 )

The choice of parents guarantees the global semantics

onconstructibyXParentsX

rulechainXXXXX

ii

n

i

ii

n

in

))(|(

),...,|(),...,(

1

111

1

P

PP

Compact Networks

Party Study

College

Fun

Exam

College

PartyStudy

FunExam

Network Construction: Alternative

Start with topological semantics that specifies the conditional independence relationships. Defined by either:

A node is conditionally independent of its non-descendants, given its parents.

A node is conditionally independent of all other nodes given its parents, children, and children’s parents: Markov Blanket.

Then reconstruct the CPTs.

X


Each node is conditionally independent of its non-descendants given its parents

Local semantics

Global semantics

Exam is independent of College, given the values of Study and Party.


Each node is conditionally independent of its parents, children and children’s parents. – Markov Blanket

U1 Um

…

…

XZ1j Znj

Y1Yn

College is independent of fun, given Party.

Canonical DistributionCompleting a node’s CPT requires up to O(2k) numbers. (k – number of parents) If the parent child relationship is

arbitrary, than can be difficult to do.

Standard patterns can be named along with a few parameters to satisfy the CPT. Canonical distribution

Deterministic NodesSimplest form is to use deterministic nodes. A value is specified exactly by its parent’s

values. No uncertainty.

But what about relationships that are uncertain? If someone has a fever do they have a cold, the

flu, or a stomach bug?

Can you have a cold or stomach bug without a fever?

Noisy-Or RelationshipsA Noisy-or relationship permits uncertainty related to the each parent causing a child to be true. The causal relationship may be inhibited. Assumes:

All possible causes are known. Can have a miscellaneous category if necessary

(leak node)

Inhibition of a particular parent is independent of inhibiting other parents.

Can you have a cold or stomach bug without a fever? Fever is true iff cold, Flu, or Malaria is true.

ExampleGiven:

1.0),,|(

2.0),,|(

6.0),,|(

malariaflucoldfeverP



Example

Cold

Flu Malaria

P( Fever)

F F F 1.0

F F T 0.1

F T F 0.2

F T T

T F F 0.6

T F T

T T F

T T T

0.2 * 0.1 = 0.02

0.6 * 0.1 = 0.06

0.6 * 0.2 = 0.12

0.6 * 0.2 * 0.1 = 0.012

Requires O(k) parameters rather than O(2k)

Networks with Continuous Variables

How are continuous variables represented? Discretization using intervals

Can result in loss of accuracy and large CPTs Define probability density functions

specified by a finite number of parameters. i.e. Gaussian distribution

Hybrid Bayesian NetworksContains both discrete and continuous variables.

Specification of such a network requires: Conditional distribution for a continuous

variable with discrete or continuous parents.

Conditional distribution for a discrete variable with continuous parents.

Example

subsidy harvest

Cost

Buys

Discrete parent Continuous parent

Discrete parent isExplicitly enumerated.

Continuous parent is represented as a distribution. Cost c depends on the distribution function for h.

A linear Gaussian distribution can be used. Have to define the distribution forboth values of subsidy.

Continuous child with a discrete parent and a continuous parent

Example

subsidy harvest

Cost

BuysDiscrete child

Continuous parent

Discrete child with a continuous parent

Set a threshold for cost.

Can use a integral of the standard normal distribution.

Underlying decision processhas a hard threshold but the Threshold’s location moves based upon randomGaussian noise. Probit Distribution

ExampleProbit distribution Usually a better fit for real problems

Logit distribution Uses sigmoid function to determine

threshold. Can be mathematically easier to work

with.

Baye’s Networks and Exact Inference

Notation X: Query variable E: set of evidence variables E1,…Em

e: a particular observed event Y: set of nonevidence variables Y1,…Ym

Also called hidden variables. The complete set of variables: A query: P(X|e)

YEX }{X


College

PartyStudy

FunExam

P(C)

0.2

C P(S)

True 0.8

False 0.2

C P(P)

True 0.6

False 0.5

S P P(E)

True True 0.6

True False 0.9

False

True 0.1

False

False 0.2

P P(F)

True 0.9

False 0.7

Example QueryIf you succeeded on an exam and had fun, what is the probability of partying? P(Party|Exam=true, Fun=true)

Inference by EnumerationFrom Chap 13 we know:

From this Chapter we have:

P(x,b,y) in the joint distribution can be represented as products of the conditional probabilities.

y

XXX ),,(),()|( yePePeP

))(|(),...,,(1

21

n


Inference by EnumerationA query can be answered using a Baye’s Net by computing the sums of products of the conditional probabilities from the network.

Example QueryIf you succeeded on an exam and had fun, what is the probability of partying? P(Party|Exam=true, Fun=true)

What are the hidden variables?

Example QueryLet: C = College PR = Party S = Study E = Exam F =Fun

Then we have from eq. 13.6 (p.476):

C S

CSfeprPfeprfepr ),,,,(),,(),|( PP

Example QueryUsing

we can put in terms of the CPT entries.

))(|(),...,,(1

21

n


C S

CPprfPprSePCSPCprPfepr )()|(),|()|()|(),|( P

The worst case complexity of this equation is: O(n2n) for n variables.

C S

CSfeprfeprfepr ),,,,(),,(),|( PPP

Example QueryImproving the calculation P(f|pr) is a constant so it can be moved

out of the summation over C and S.

The move the elements that only involve C and not S to outside the summation over S.

C S

CPprSePCSPCprPprfPfepr )(),|()|()|()|(),|( P

C S

prSePCSPCprPCPprfPfepr ),|()|()|()()|(),|( P

College example:

College

PartyStudy

FunExam

P(C)

0.2

C P(S)

True 0.8

False 0.2

C P(PR)

True 0.6

False 0.5

S PR P(E)

True True 0.6

True False 0.9

False

True 0.1

False

False 0.2

PR P(F)

True 0.9

False 0.7

Example Query

+

+ +

P(f|pr).9

P(c).2

P(s|c).8

P(e|s,pr).6

2.

)|( csP

1.

),|( prseP

8.

)|( csP

8.

)( cP

.48 + .02 = .5.12 + .08 = .2

.06 + .08 = .14

.126Similarly for P( pr|e,f).

Still O(2n)

2.

)|( csP

P(pr|c).6 5.

)|( cprP

P(e|s,pr).6 1.

),|( prseP

c s

c s

prsePcsPcprPcPprfP

prsePcsPcprPcPprfPfePR

),|()|()|()()|(

,),|()|()|()()|(),|(

P

C S

prSePCSPCprPCPprfPfepr ),|()|()|()()|(),|( P

Variable EliminationA problem with the enumeration method is that particular products can be computed multiple times, thus reducing efficiency. Reduce the number of duplicate calculations by

doing the calculation once and saving it for later.

Variable elimination evaluates expressions from right to left, stores the intermediate results and sums over each variable for the portions of the expression dependent upon the variable.

Variable EliminationFirst, factor the equation.

Second, store the factor for E A 2x2 matrix fE(S,PR).

Third, store the factor for S. A 2x2 matrix.

F C S E

c s

PRSePCSPCPRPCPPRfPfePR ),|()|()|()()|(),|( P

PR

)|(

)|(

)|(

)|(),(

csP

csP

csP

csPCSSf

Variable EliminationFourth, Sum out S from the product of the first two factors.

This is called a pointwise product It creates a new factor whose variables are the

union of the two factors in the product.

Any factor that does not depend on the variable to be summed out can be moved outside the summation.

),( * ),( ),( * ),(

),( * ),(),(

PRsCsPRsCs

PRSCSPRCS

ES

ESEs

ES

ffff

fff

)...,...(*)...,...()...,...,...( 1111111 lkkjlkj ZZYYfYYXXfZZYYXXf

Variable Elimination

Fifth, store the factor for PR A 2x2 matrix.

Sixth, Store the factor for C.

),()|()()|(),|( PRCCPRPCPPRfPfePRES

C

fP

)|(

)|(

)|(

)|(),(

cprP

cprP

cprP

cprPCPRPRf

)(

)()(

cP

cPCCf


Seventh, sum out C from the product of the factors where

),(* ),( * )(

),(* ),( * )(

),(* ),( * )()(

PRccPRc

PRccPRc

PRCCPRCPR

S

S

Sc

ESPRC

EPRC

EPRC

EPRC

fff

fff

ffff

),()|()()|(),|( PRCCPRPCPPRfPfePRES

c

fP

Elimination SimplificationAny leaf node that is not a query variable or an evidence variable can be removed.Every variable that is not an ancestor of a query variable or an evidence variable is irrelevant to the query and can be eliminated.

Elimination SimplificationBook Example: What is the probability that John calls

if there is a burglary?

e a m

amPaJPebaPePbPbJP )|()|(),|()()()|(

Does this matter?Burglary

Alarm

Earthquake

MaryCallsJohnCalls

Complexity of Exact Inference

Variable elimination is more efficient than enumeration. Time and space requirements are

dominated by the size of the largest factor constructed which is determined by the order of variable elimination and the network structure.

PolytreesPolytrees are singly connected networks At most one directed path between any

two nodes. Time and space requirements are linear

in the size of the network. Size is the number of CPT entries.

Polytrees

Burglary

Alarm

Earthquake

MaryCallsJohnCalls

College

PartyStudy

FunExam

Are these trees polytrees?

Applying variable elimination to multiply connected networks has worst case exponential

time and space complexity.

baye’s rule. baye’s rule and reasoning allows use of uncertain causal knowledge knowledge: given...

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