1 Reasoning Under Uncertainty Artificial Intelligence Chapter 9

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Reasoning Under Uncertainty

Artificial Intelligence

Chapter 9

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ReasoningReasoning

Part 2

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Notation

Random variable (RV): a variable (uppercase)that takes on values (lowercase) from a domainof mutually exclusive and exhaustive values

A=a: a proposition, world state, event, effect, etc.– abbreviate: P(A=true) to P(a)– abbreviate: P(A=false) to P(a)– abbreviate: P(A=value) to P(value)– abbreviate: P(Avalue) to P(value)

Atomic event: a complete specification of the stateof the world about which the agent is uncertain

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Notation

P(a): a prior probability of RV A=awhich is the degree of belief proposition ain absence of any other relevant information

P(a|e): conditional probability of RV A=a given E=ewhich is the degree of belief in proposition awhen all that is known is evidence e

P(A): probability distribution, i.e. set of P(ai) for all i Joint probabilities are for conjunctions of propositions

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Reasoning under Uncertainty

Rather than reasoning about the truth or falsityof a proposition, instead reason about the belief that a proposition is true.

Use knowledge base of known probabilities to determine probabilities for query propositions.

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Reasoning under Uncertaintyusing Full Joint Distributions

Assume a simplified Clue game havingtwo characters, two weapons and two rooms:

each row is an atomic event- one of these must be true- list must be mutually exclusive- list must be exhaustive

Who What Whereplum rope hall

plum rope kitchen

plum pipe hall

plum pipe kitchen

green rope hall

green rope kitchen

green pipe hall

green pipe kitchen

Probability1/8

1/8

1/8

1/8

1/8

1/8

1/8

1/8

prior probability for each is 1/8- each equally likely- e.g. P(plum,rope,hall) = 1/8

hallpipeplum

kitchenpipeplum

hallropegreen

kitchenropegreen

hallpipegreen

hallropeplum

kitchenpipegreen

kitchenropeplum

P(atomic_eventi) = 1since each RV's domain isexhaustive & mutually exclusive

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Determining Marginal Probabilitiesusing Full Joint Distributions

The probability of any proposition is equal tothe sum of the probabilities of the atomic eventsin which it holds, which is called the set e(a).P(a) = P(ei)where ei is an element of e(a)– its the disjunction of atomic events in set e(a)– recall this property of atomic events:

any proposition is logically equivalent to the disjunctionof all atomic events that entail the truth of that proposition

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Determining Marginal Probabilitiesusing Full Joint Distributions


P(plum) = ?


plum rope kitchen

plum pipe hall

plum pipe kitchen

green rope hall

green rope kitchen

green pipe hall

green pipe kitchen

Probability1/8

1/8

1/8

1/8

1/8

1/8

1/8

1/8

when obtained in this manner it is called a marginal probability

can be just a prior probability (shown) or more complex (next)

this process is called marginalization or summing out

1/8+1/8+1/8+1/8 = 1/2

1/8

1/8

1/8

1/8

P(a) = P(ei)where ei is an element of e(a)

plum

plum

plum

plum

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P(green,pipe) =


plum rope kitchen

plum pipe hall

plum pipe kitchen

green rope hall

green rope kitchen

green pipe hall

green pipe kitchen

Probability1/8

1/8

1/8

1/8

1/8

1/8

1/8

1/8

1/8+1/8 = 1/4

1/8

1/8

1/8

P(rope, hall) =

1/8

1/8 1/8+1/8 = 1/4

P(rope hall) = 1/8+1/8+

1/8+1/8+1/8+1/8 = 3/4

1/8

1/8

1/8

1/8

1/8

1/8

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Independence

Using the game cluefor an example is uninteresting! Why?– Because the random variables

Who, What, Where are independent.– Does picking the murder from the deck of cards

affect which weapon is chosen? Location?

No! Each is randomly selected.

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Independence

Unconditional (absolute) Independence:RVs have no affect on each other's probabilities1. P(X|Y) = P(X)2. P(Y|X) = P(Y)3. P(X,Y) = P(X) P(Y)

Example (full clue):P(green | hall) = P(green, hall) / P(hall)

= 6/324 / 1/9= P(green) = 1/6

P(hall | green) = P(hall) = 1/9P(green, hall) = P(green) P(hall) = 1/54We need a more interesting example!

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Independence

Conditional Independence:RVs (X, Y) are dependent on another RV (Z)but are independent of each other1. P(X|Y,Z) = P(X|Z)2. P(Y|X,Z) = P(Y|Z) 3. P(X,Y|Z) = P(X|Z) P(Y|Z)

Idea:sneezing (x) and itchy eyes (y)are both directly caused by hayfever (z)

but neither sneezing nor itchy eyeshas a direct effect on each other

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HF SN IEfalse false false

false false true

false true false

false true true

true false false

true false true

true true false

true true true


Assume three boolean RVs: Hayfever HF, Sneeze SN, ItchyEyes IE

P(sn) = 0.1+ 0.1+ 0.04+ 0.1=0.34

Probability0.5

0.09

0.1

0.1

0.01

0.06

0.04

0.1

P(a) = P(ei)where ei is an element of e(a)

P(hf) = 0.01+ 0.06+ 0.04+ 0.1=0.21P(sn,ie) = 0.1+ 0.1=0.20

P(hf,sn) = 0.04+ 0.1=0.14

and fictional probabilities:

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false false true

false true false

false true true

true false false

true false true

true true false

true true true


Assume three boolean RVs: Hayfever HF, Sneeze SN, ItchyEyes IEand fictional probabilities:

Probability0.5

0.09

0.1

0.1

0.01

0.06

0.04

0.1

P(a|e) = P(a, e) / P(e)

P(hf | sn) = P(hf,sn) / P(sn) = 0.14 / 0.34 =

0.41P(hf | ie) = P(hf,ie) / P(ie) = 0.16 / 0.35 =

0.46

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false false true

false true false

false true true

true false false

true false true

true true false

true true true


Assume three boolean RVs: Hayfever HF, Sneeze SN, ItchyEyes IEand fictional probabilities:

P(hf | sn) = 0.14 / P(sn)

Probability0.5

0.09

0.1

0.1

0.01

0.06

0.04

0.1

Instead of computing P(e),could use normalization

also compute:P(hf | sn) = 0.20 / P(sn)since P(hf | sn) + P(hf | sn) = 1substituting and solving gives

P(sn) = 0.34 !

P(a|e) = P(a, e) / P(e)

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Combining Multiple Evidence

As evidence describing the state of the worldis accumulated, we'd like to be able to easily update the degree of belief in a conclusion.

Using the Full Joint Prob. Dist. Table:P(v1,...,vk|vk+1,...,vn) = P(V1=v1,...,Vn=vn) /

P(Vk+1=vk+1,...,Vn=vn)1. sum of all entries in the table, where V1=v1, ..., Vn=vn

2. divided by the sum of all entries in the tablecorresponding to the evidence, where Vk+1=vk+1, ..., Vn=vn

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false false true

false true false

false true true

true false false

true false true

true true false

true true true

Combining Multiple Evidenceusing Full Joint Distributions

Assume three boolean RVs and fictional probabilities Hayfever HF, Sneeze SN, ItchyEyes IE:

P(hf | sn, ie) = P(hf,sn,ie) / P(sn,ie)

= 0.10 / (0.1+0.1)

= 0.5

Probability0.5

0.09

0.1

0.1

0.01

0.06

0.04

0.1

P(a|b, c) = P(a,b,c) / P(b,c) as described in prior slide

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Combining Multiple Evidence (cont.)

FJDT techniques are intractable in general because the table size grows exponentially.

Independence assertions can help reducethe size of the domain and the complexityof the inference problem.

Independence assertions are usually basedon the knowledge of the domain enablingFJD table to be factored in to separate JD tables.– it's a good thing that problem domains are independent– but typically subsets of dependent RVs are quite large

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Probability Rulesfor Multi-valued Variables

Summing Out: P(Y) = P(Y, z)sum over all values z of RV Z

Conditioning: P(Y) = P(Y|z) P(z)sum over all values z of RV Z

Product Rule: P(X, Y) = P(X|Y) P(Y) = P(Y|X) P(X) Chain Rule: P(X, Y, Z) = P(X|Y, Z) P(Y|Z) P(Z)

– this is a generalization of product rule with Y= Y,Z– order of RVs doesn't matter, i.e. gives same result

Conditionalized Chain Rule:(let Y=A|B) P(X, A|B) = P(X|A, B) P(A|B)(order doesn't matter) = P(A|X, B) P(X|B)

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Bayes' Rule

Bayes' Rule:P(b|a) = (P(a|b) P(b)) / P(a)

– derived from P(a b) = P(b|a) P(a) = P(a|b) P(b)just divide both sides of equation by P(a)

– basis of AI systems using probabilistic reasoning

For Example:a=happy, b=sun a= sneeze, b= fallP(sun|happy) = ? P(fall|sneeze) = ?P(happy|sun) = 0.95 P(sneeze|fall) = 0.85P(sun) = 0.5 P(fall) = 0.25P(happy) = 0.75 P(sneeze) = 0.3(0.95 * 0.5)/0.75 = 0.63 (0.85 * 0.25)/0.3 = 0.71

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Bayes' Rule

P(b|a) = (P(a|b) P(b)) / P(a)What's the benefit of being able to calculateP(b|a) from the three probabilities on the right?

Usefulness of Bayes' Rule:– many problems have good estimates of probabilities on right– P(b|a) needed to identify cause, classification, diagnosis, etc– typical use is to calculate diagnostic knowledge

from causal knowledge

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Bayes' Rule

Causal knowledge: from causes to effects– e.g. P(sneeze|cold)

probability of effect sneeze given cause common cold– this probability the doctors obtains from experience

treating patients and understanding the disease process

Diagnostic knowledge: from effects to causes– e.g. P(cold|sneeze)

probability of cause common cold given effect sneeze– knowing this probability helps a doctor make a

disease diagnosis based on a patient's symptoms– diagnostic knowledge is more fragile that causal knowledge

since it can change significantly over time given variationsin rate of occurrence of its causes (due to epidemics, etc.)

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Bayes' Rule

Using Bayes' Rule with causal knowledge:– want to determine diagnostic knowledge (diagnostic reasoning)

that is difficult to obtain from a general population– e.g. symptom is s=stiffNeck, disease is m=meningitis

P(s|m) = 1/2 the casual knowledgeP(m) =1/50000, P(s) = 1/20 prior probabilitiesP(m|s) = ? desired diagnostic knowledge(1/2 * 1/50000)/ (1/20) = 1/5000

– doctor can now use P(m|s) to guide diagnosis

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Combining Multiple Evidenceusing Bayes' Rule

How do you update conditional probabilityof Y given two pieces of evidence A and B?

General Bayes' Rule for multi-valued RVs:P(Y|X) = (P(X|Y) * P(Y)) / P(X)let X=A,B:

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

= P(Y)*(P(A|Y)/P(A))*(P(B|A,Y)/P(B|A))conditionalized chain rule used, product rule used

Problems:– P(B|A,Y) generally hard to compute or obtain– doesn't scale well for n evidence RVs, table size grows O(2n)

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Problems can be circumvented: If A and B are conditionally independent given Y

then P(A,B|Y) = P(A|Y)P(B|Y) and for P(A,B) use product rule P(Y|A,B) = (P(Y) P(A,B|Y) ) / P(A,B)Bayes' Rule Multi-EP(Y|A,B) = P(Y) * (P(A|Y)/P(A)) * (P(B|Y)/P(B|A))

No joint probabilities, representation grows O(n) If A is unconditionally independent of B

then P(A,B|Y) = P(A|Y)P(B|Y) and P(A,B) = P(A)P(B)– P(Y|A,B) = (P(Y) P(A,B|Y) ) / P(A,B)Bayes' Rule Multi-E

P(Y|A,B) = P(Y) * (P(A|Y)/P(A)) * (P(B|Y)/P(B)) This equation used to define a naïve Bayes classifier.

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Example:– What is the likelihood that a patient has sclerosis colangitis?– doctor's initial belief: P(sc) = 1/1,000,000– examine reveals jaundice: P(j) = 1/10,000

P(j|sc) = 1/5– doctor's belief given test result: P(sc|j) = P(sc)P(j|sc)/P(j)

= 2/1000– tests reveal fibrosis of bile ducts: P(f|sc) = 4/5

P(f) = 1/100– doctor naïvely assumes jaundice and fibrosis are independent– doctor's belief now rises: P(sc|j,f) = 16/100

P(sc|j,f) = P(sc)*(P(j |sc)/P(j)) *(P(f |sc)/P(f )) P(Y|A,B) = P(Y) *(P(A|Y)/P(A))*(P(B|Y) /P(B))

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Naïve Bayes Classifier

Naïve Bayes Classifierused where single class is based on a number of featuresor where single cause influences a number of effects

based on P(Y|A,B) = P(Y) * (P(A|Y)/P(A)) * (P(B|Y)/P(B))– given RV C

domain is possible classifications say {c1,c2,c3} classifies input example of features F1, …, Fn

– compute: P(c1|F1, …, Fn), P(c2|F1, …, Fn), P(c3|F1, …, Fn) naïvely assume features are independent

– choose value for C that gives maximum probability– works surprising well in practice even

when independence assumption aren't true

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Bayesian Networks

AKA: Bayes Nets, Belief Nets, Causal Nets, etc.

Encodes the full joint probability distribution (FJPD) for the set of RVs defining a problem domain

Uses a space-efficient data structure by exploiting:– fact that dependencies between RVs are generally local– which results in lots of conditionally independent RVs

Captures both qualitative and quantitative relationships between RVs

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Bayesian Networks

Can be used to compute any value in FJPD Can be used to reason:

– predictive/causal reasoning:forward (top-down) from causes to effects

– diagnostic reasoning:backward (bottom-up) from effects to causes

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Bayesian Network Representation

Is an augmented DAG (i.e. directed, acyclic graph) Represented by V,E where

– V is a set of vertices– E is a set of directed edges joining vertices, no loops

Each vertex contains:– the RV's name– either a prior probability distribution or

a conditional probability distribution table (CDT)that quantifies the effects of the parents on this RV

Each directed arc:– is from cause (parent) to its immediate effects (children)– represents direct causal relationship between RVs

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Bayesian Network Representation

Example: in class– each row in conditional probability tables must sum to 1– columns don't need to sum to 1– values obtained from experts

Number of probabilities required is typicallyfar fewer than the number required for a FJDT

Quantitative information is usually givenby an expert or determined empirically from data

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Conditional Independence

Assume effects are conditionally independentof each other given their common cause

The net is constructed so that given its parents,a node is conditionally independent of its non-descendant RVs in the net:P(X1=x1, ..., Xn=xn) = P(xi | parents(Xi)) * ... * P(xn | parents(Xn))

Note the full joint probability distribution isn't needed, only need conditionals relative to their parent RVs

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Algorithm for ConstructingBayesian Networks

1. Choose a set of relevant random variables

2. Choose an ordering for them

3. Assume they're X1 .. Xm where X1 is first, X2 is second, etc.

4. For i = 1 to ma. add a new node for Xi to the networkb. set Parents(Xi) to be a minimal subset of {X1 .. Xi-1}

such that we have conditional independence of Xi

and all other members of {X1 ..Xi-1} given Parents(Xi)c. add directed arc from each node in Parents(X i) to Xi d. non-root nodes: define a conditional probability table

P(Xi =x | combinations of Parents(Xi)) root nodes: define prior probability distribution at Xi: P(Xi)

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Algorithm for ConstructingBayesian Networks

For a given set of random variables (RVs)there is not, in general, a unique Bayesian Netbut all of them represent the same information

For the best net, topologically sort RVs in step 2– each RV comes before all of its children– first nodes are roots, then nodes they directly influence

Best Bayesian Network for a problem has:– fewest number of probabilities and arcs– easy to determine probabilities for the CDT

Algorithm won't construct a net that violatesthe rules of probability

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Compute P(a,b,c,d) = P(d,c,b,a)order RVS in joint probability bottom up D,C,B,A= P(d|c,b,a) P(c,b,a) Product Rule P(d,c,b,a) = P(d|c) P(c,b,a) Conditional Independ. of D given C= P(d|c) P(c|b,a) P(b,a) Product Rule P(c,b,a) = P(d|c) P(c|b,a) P(b|a) P(a) Product Rule P(b,a) = P(d|c) P(c|b,a) P(b ) P(a) Independence of B and A

given no evidence

Computing Joint Probabilitiesusing a Bayesian Network

1. Use product rule

2. Simplify using independence

For Example:

A B

C

D

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Any entry in the full joint dist. table(i.e. atomic event) can be computed!

P(v1,...,vn) = P(vi|Parents(Vi)) over i from 1 to n

e.g. given boolean RVs what is P(a,..,h,k,..,p)?D E

G

A

K

B

HF

C

L

N O PM

= P(a)P(b)P(c)P(d|a,b)P(e|b,c)P(f)P(g|d,e)P(h) * P(k|f,g)P(l|gh)P(m|k)P(n|k)P(o|k,l)P(p|l)

Note this is fast, i.e. linearin the number of nodes in the net!

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How is any joint probability computed?

sum the relevant joint probabilities:

e.g. Compute: P(a,b)

= P(a,b,c,d) + P(a,b,c,d) + P(a,b,c,d) + P(a,b,c,d)e.g. Compute: P(c)= P(a,b,c,d) + P(a,b,c,d) + Pa,b,c,d) + Pa,b,c,d) + P(a,b,c,d) + P(a,b,c,d) + P(a,b,c,d) + P(a,b,c,d)

A BN can answer any query (i.e. probability) about the domain by summing the relevant joint probs.

Enumeration can require many computations!

A B

C

D

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Computing Conditional Probabilitiesusing a Bayesian Network

Basic task of probabilistic systemis to compute conditional probabilities.

Any conditional probability can be computed:P(v1,...,vk|vk+1,...,vn) = P(V1=v1,...,Vn=vn) /

P(Vk+1=vk+1,...,Vn=vn)

Key problem is that the technique of enumeratingjoint probabilities can make the computations intractable (exponential in the number of RVs).

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These computations generally relyon the simplifications resulting fromthe independence of the RVs.

Every variable that isn't an ancestorof a query variable or an evidence variableis irrelevant to the query.

What ancestors are irrelevant?

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D E

G

A

K

B

HF

C

L

N O PM

Independence in a Bayesian Network

Given a Bayesian Networkhow is independence established?

1. A node is conditionally independent (CI)of its non-descendants, given its parents.

e.g. Given D and E, G is CI of ?G

D E

A, B, C, F, H

A B

HF

C

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D E

G

A

K

B

HF

C

L

N O PM



1. A node is conditionally independent (CI)of its non-descendants, given its parents.

e.g. Given D and E, G is CI of ?

A, B, C, F, He.g. Given F and G, K is CI of ?

GF

KA, B, C, D, E ,

H, L, P

D E

A B

H

C

L

P

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D E

G

A

K

B

HF

C

L

N O PM



2. A node is conditionally independentof all other nodes in the network givenits parents, children, and children'sparents, which is called a Markov blanket

e.g. What is the Markov blanket for G? G

D E

K

HF

L D, E, F, H, K , LGiven this blanket G is CI of ?

A, B, C, M, N , O, PWhat about absolute independence?

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The general algorithm for computingconditional probabilities is complicated.

It is easy if the query involves nodesthat are directly connected to each other.examples assumed to use boolean RVs

Simple causal inference: P(E|C)– conditional prob. dist. of effect E given cause C as evidence– reasoning in same direction as arc, e.g. disease to symptom

Simple diagnostic inference: P(Q|E)– conditional prob. dist. of query Q given effect E as evidence– reasoning in direction opposite of arc, e.g. symptom to disease

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Computing Conditional Probabilities:Causal (Top-Down) Inference

Compute P(e|c)conditional probability of effect E=e given cause C=c as evidenceassume arc exists to E from C and C2 C C2

E1. Rewrite conditional probability of e in terms

of e and all of its parents (that aren't evidence)given evidence c

2. Re-express each joint probability backto the probability of e given all of its parents

3. Simplify using independence and Look Uprequired values in the Bayesian Network

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Computing Conditional Probabilities:Causal (Top-Down) Inference

Compute P(e|c)1. = P(e,c) / P(c) product rule

= (P(e,c,c2)+ P(e,c,c2)) / P(c) marginalizing

= P(e,c,c2) / P(c) + P(e,c,c2) / P(c) algebra

= P(e,c2|c) + P(e,c2|c) product rule, e.g. X=e,c2

2. = P(e|c2,c) P(c2|c) + P(e|c2,c) P(c2|c) cond. chain rule

3. Simplify given C and C2 are independent

P(c2|c) = P(c2)

P(c2|c) = P(c2)

= P(e|c2,c) P(c2) + P(e|c2,c) P(c2) algebra

now look up values to finish computation

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Computing Conditional Probabilities:Diagnostic (Bottom-Up) Inference

Compute P(c|e)conditional probability of cause C=c given effect E=e as

evidenceassume arc exists from C to Eidea: convert to casual inference using Bayes' rule

1. Use Bayes' rule P(c|e) = P(e|c) P(c) / P(e)2. Compute P(e|c) using causal inference method

3. Look up value of P(c) in Bayesian Net

4. Use normalization to avoid computing P(e)– requires computing P(c|e)– using steps as in 1 – 3 above

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Summary: the Good News

Bayesian Nets are the bread and butter ofAI-uncertainty community (like resolution to AI-logic)

Bayesian Nets are a compact representation– don't require exponential storage to hold

all of the info in the full joint probability distribution (FJPD) table– are a decomposed representation of the FJPD table– conditional probability distribution tables in non-root nodes are

only exponential in the max number of parents of any node Bayesian Nets are fast at computing joint probs:

P(V1, ..., Vk) i.e. prior probability of V1, ..., Vk

– computing the probability of an atomic event can be donein linear time with the number of nodes in the net

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Summary: the Bad News

Conditional probabilities can also be computed:P(Q|E1, ..., Ek)posterior probability of query Q given multiple evidence E1, ..., Ek

– requires enumerating all of the matching entries,which takes exponential time in the number of variables

– in special cases it can be done faster, <= polynomial timee.g. polytree: linear time for nets structured like trees

In general, inference in Bayesian Networks (BN)is NP-hard. but BNs are well studied so there exists many efficient exact solution methods as well as a variety of approximation techniques

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1 Reasoning Under Uncertainty Artificial Intelligence Chapter 9