oregon state university – cs430 intro to ai (c) 2003 thomas g. dietterich1 probabilistic...

(c) 2003 Thomas G. Dietterich 1

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AI Probabilistic Representation and

Reasoning

Given a set of facts/beliefs/rules/evidence Evaluate a given statement

Determine the probability of a statement Find a statement that optimizes a set of

constraintsMost probable explanation (MPE) (Setting of

hidden variables that best explains observations.)


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AI Probability Theory

Random Variables Boolean: W1,2 (just like propositional logic). Two

possible values {true, false} Discrete: Weather 2 {sunny, cloudy, rainy, snow} Continuous: Temperature 2 <

Propositions W1,2 = true, Weather = sunny, Temperature = 65 These can be combined as in propositional logic


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AI Example

Consider a car described by 3 random variables: Gas 2 {true, false}: There is gas in the tank Meter 2 {empty,full}: The gas gauge shows

the tank is empty or full Starts 2 {yes,no}: The car starts when you

turn the key in the ignition


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AI Joint Probability Distribution

Gas Meter Starts P(G,M,S)

false empty no 0.1386

false empty yes 0.0014

false full no 0.0594

false full yes 0.0006

true empty no 0.0240

true empty yes 0.0560

true full no 0.2160

true full yes 0.5040

Each row is called a “primitive event”

Rows are mutually exclusive and exhaustive

Corresponds to an “8-sided coin” with the indicated probabilities


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AI Any Query Can Be Answered from

the Joint Distribution








true full no 0.2160


P(Gas = false Æ Meter = full Æ Starts = yes) = 0.0006

P(Gas = false) = 0.2, this is the sum of all cells where Gas = false

In general: To compute P(Q), for any proposition Q, add up the probability in all cells where Q is true


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AI Notations

P(G,M,S) denotes the entire joint distribution (In the book, P is boldface). It is a table or function that maps from G, M, and S to a probability.

P(true,empty,no) denotes a single probability value:

P(Gas=true Æ Meter=empty Æ Starts=no)


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AI Operations on Probability Tables

(1)

Marginalization (“summing away”) M,S P(G,M,S) = P(G) P(G) is called a

“marginal probability” distribution. It consists of two probabilities:

Gas P(G)

false

P(false,empty,yes) +

P(false,empty,no) +

P(false,full,yes) +

P(false,full,no)

true

P(true,empty,yes) +

P(true,empty,no) +

P(true,full,yes) +

P(true,full,no)


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AI Conditional Probability

Suppose we observe that M=full. What is the probability that the car will start? P(S=yes | M=full)

Definition: P(A|B) = P(A Æ B) / P(B)


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AI Conditional Probability








true full no 0.2160


Select cells that match the condition (M=full)

Delete remaining cells and M column

Renormalize the table to obtain P(S,G|M=full)

Sum away Gas: G P(S,G | M=full) = P(S|M=full)

Read answer from P(S=yes | M=full) cell








true full no 0.2160


Gas Starts P(G,S|M=full)

false no 0.0594

false yes 0.0006

true no 0.2160

true yes 0.5040

Starts P(S|M=full)

no 0.3531

yes 0.6469

Gas Starts P(G,S|M=full)

false no 0.0762

false yes 0.0008

true no 0.2769

true yes 0.6462




true full no 0.2160





true full no 0.2160



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AI Operations on Probability Tables (2):

Conditionalizing

Construct P(G,S | M) by normalizing the subtable corresponding to M=full and normalizing the subtable corresponding to M=empty








true full no 0.2160


0.0762

0.0008

0.2769

0.6462

0.6300

0.0064

0.1091

0.2545


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AI Chain Rule of Probability

P(A,B,C) = P(A|B,C) ¢ P(B|C) ¢ P(C) Proof:


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AI Chain Rule (2)

Holds for distributions too:

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

This means that for each setting of A,B, and C, we can substitute into the equation, and it is true.


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AI Belief Networks (1):

Independence

Defn: Two random variables X and Y are independent iff

P(X,Y) = P(X) ¢ P(Y) Example:

X is a coin with P(X=heads) = 0.4 Y is a coin with P(Y=heads) = 0.8 Joint distribution:

P(X,Y) X=heads X=tails

Y=heads 0.32 0.48

Y=tails 0.08 0.12


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AI Belief Networks (2)

Conditional Independence

Defn: Two random variables X and Y are conditionally independent given Z iffP(X,Y | Z) = P(X|Z) ¢ P(Y|Z)

Example:P(S,M | G) = P(S | G) ¢ P(M | G)Intuition: G independently causes S and M

S M G=no G=yes

no empty 0.693 0.030

no full 0.297 0.270

yes empy 0.007 0.070

yes full 0.003 0.630

S G=no G=yes

no 0.99 0.30

yes 0.01 0.70

M G=no G=yes

empty 0.70 0.10

full 0.30 0.90


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AI Operations on Probability Tables (3):

Conformal Product

Allocate space for resulting table and then fill in each cell with the product of the corresponding cells:

P(S,M | G) = P(S | G) ¢ P(M | G)

S M G=no G=yes

no empty a00¢b00 a01¢b01

no full a00¢b10 a01¢b11

yes empy a10¢b00 a11¢b01

yes full a10¢b10 a11¢b11

S G=no G=yes

no a00 a01

yes a10 a11

M G=no G=yes

empty b00 b01

full b10 b11= ¢


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AI Properties of Conformal Products

Commutative Associative Work on normalized or unnormalized

tables Work on joint or conditional tables


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AI Conditional Independence Allows

Us to Simplify the Joint Distribution

P(G,M,S) = P(M,S | G) ¢ P(G) [chain rule]

= P(M | G) ¢ P(S | G) ¢ P(G) [CI]

Bayesian Networks

One node for each random variable Each node stores a probability

distribution P(node | parents(node)) Only direct dependencies are shown Joint distribution is conformal product of

node distributions:P(G,M,S) = P(G) ¢ P(M | G) ¢ P(S | G)

Gas

StartsMeter


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AI Inference in Bayesian Networks

Suppose we observe that M=full. What is the probability that the car will start? P(S=yes | M=full)

Before, we handled this by the following steps: Remove all rows corresponding to M=empty Normalize remaining rows to get P(S,G|M=full) Sum over G: G P(S,G|M=full) = P(S | M=full) Read answer from the S=yes entry in the table

We want to get the same result, but without constructing the joint distribution first.


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AI Inference in Bayesian Networks (2)

Remove all rows corresponding to M=empty from all nodes P(G) – unchanged P(M | G) becomes P[G] P(S | G) – unchanged

Sum over G: G P(G) ¢ P[G] ¢ P(S | G)

Normalize to get P(S|M=full) Read answer from the S=yes entry in the table


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AI Inference with Tables

P(M|G) G=no G=yes

empty 0.70 0.10

full 0.30 0.90

P(S|G) G=false G=true

no 0.99 0.30

yes 0.01 0.70

G P(G)

false 0.20

true 0.80

G


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no 0.99 0.30

yes 0.01 0.70

G P(G)

false 0.20

true 0.80

Step 1: Delete M=empty rows from all tables

G

G=false G=true

0.30 0.90


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no 0.99 0.30

yes 0.01 0.70

G P(G)

false 0.20

true 0.80


G

G=false G=true

0.30 0.90

Step 2: Perform algebra to push summation inwards (no-op in this case)


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G


Step 3: Form conformal product

S G=false G=true

no 0.0594 0.2160

yes 0.0006 0.5040


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Step 4: Sum away G

S

no 0.2754

yes 0.5046


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Step 4: Sum away G

Step 5: Normalize

S

no 0.3531

yes 0.6469


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Step 4: Sum away G

Step 5: Normalize

S

no 0.3531

yes 0.6469

Step 6: Read answer from table: 0.6469


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AI Notes

We never created the joint distribution Deleting the M=empty rows from the

individual table followed by conformal product has the same effect as performing the conformal product first and then deleting the M=empty rows

Normalization can be postponed to the end


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AI Another Example: “Asia”

(all variables Boolean)

Suppose we observe Sneeze What is P(Cold | Sneeze) = P(Co|S)?

Cat

ScratchSneeze

AllergyCold


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AI Answering the query

Joint distribution:A,Ca,Sc P(Co) ¢ P(A) ¢ P(Sn | Co,A) ¢ P(Ca) ¢ P(A | Ca) ¢ P(Sc | Ca)

Apply evidence: sn (Sneeze = true)A,Ca,Sc P(Co) ¢ P(A) ¢ P[Co,A]¢ P(Ca) ¢ P(A | Ca) ¢ P(Sc | Ca)

Push summations in as far as possible: P(Co) ¢ A P(A) ¢ P[Co,A] ¢ Ca P(A | Ca) ¢ P(Ca) ¢ Sc P(Sc | Ca)

Evaluate: P(Co) ¢ A P(A) ¢ P[Co,A] ¢ Ca P(A | Ca) ¢ P(Ca) ¢ P[Ca]

P(Co) ¢ A P(A) ¢ P[Co,A] ¢ P[A]P(Co) ¢ P[Co]P[Co]

Normalize and extract answer


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AI Pruning Leaves

Leaf nodes not involved in the evidence or the query can be pruned. Example: Scratch

Cat

ScratchSneeze

AllergyCold

Evidence

Query


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AI Greedy algorithm for choosing the

elimination order

nodes = set of tables (after evidence)V = variables to sum overwhile |nodes| > 1 do

Generate all pairs of tables in nodes that share at least one variableCompute size of table that would result from conformal product of each pair (summing over as many variables in V as possible)Let (T1,T2) be the pair with smallest resulting sizeDelete T1 and T2 from nodesAdd conformal product V T1¢T2 to nodes

end


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AI Example of Greedy Algorithm

Given tables P(Co), P[Co,A], P(A|Ca), P(Ca) Variables to sum: A, Ca

Choose: P[A] = Ca P(A|Ca) ¢ P(Ca)

candidate pair size of product sum over size of result

P(Co) ¢ P[Co,A] 2 nil 2

P[Co,A] ¢ P(A|Ca) 3 A 2

P(A|Ca) ¢ P(Ca) 2 Ca 1


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AI Example of Greedy Algorithm (2)

Given tables P(Co), P[Co,A], P[A] Variables to sum: A

Choose: P[Co] = A P[Co,A] ¢ P[A]


P(Co) ¢ P[Co,A] 2 nil 2

P[Co,A] ¢ P[A] 2 A 2


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AI Example of Greedy Algorithm (3)

Given tables P(Co), P[Co] Variables to sum: none

Choose: P2[Co] = P(Co) ¢ P[Co]

Normalize and extract answer


P(Co) ¢ P[Co] 1 nil 1


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AI Bayesian Network For WUMPUS

P(P1,1,P1,2, …, P4,4, B1,1, B1,2, …, B4,4) 1,4 2,4 3,4 4,4

1,3 2,3 3,3 4,3

1,2 2,2 3,2 4,2

1,1 2,1 3,1 4,1

P2,2P1,1 P1,2 P2,1

…

B1,1 B1,2 B2,1 B1,3 B2,2 B3,1 B1,4 B2,3 B4,1

P1,3 P3,1 P1,4

B2,4…


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AI Probabilistic Inference in

WUMPUS

Suppose we have observed No breeze in 1,1 Breeze in 1,2 and 2,1 No pit in 1,1, 1,2, and 1,3

What is the probability of a pit in 1,3?

P(P1,3|:B1,1,B1,2,B2,1, :P1,1,:P1,2,:P2,1)

1,3

????

1,2

B

OK

2,2

1,1

OK

2,1

B

OK

3,1


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AI What is

P(P1,3|:B1,1,B1,2,B2,1, :P1,1,:P1,2,:P2,1)?

P2,2P1,1 P1,2 P2,1

…

B1,1 B1,2 B2,1 B1,3 B2,2 B3,1 B1,4 B2,3 B4,1

P1,3 P3,1 P1,4

B2,4…

false true true

false false false query


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AI Prune Leaves Not Involved in Query or

Evidence

P2,2P1,1 P1,2 P2,1

…

B1,1 B1,2 B2,1 B1,3 B2,2 B3,1 B1,4 B2,3 B4,1

P1,3 P3,1 P1,4

B2,4…

false true true



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AI Prune Independent Nodes

P2,2P1,1 P1,2 P2,1

…

B1,1 B1,2 B2,1

P1,3 P3,1 P1,4

false true true



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AI Solve Remaining Network

P2,2P1,1 P1,2 P2,1

B1,1 B1,2 B2,1

P1,3 P3,1

false true true


P2,2,P3,1 P(B1,1|P1,1,P1,2,P2,1) ¢ P(B1,2|P1,1,P1,2,P1,3) ¢ P(B2,1|P1,1,P2,1,P2,2,P3,1) ¢ P(P1,1) ¢ P(P1,2) ¢ P(P2,1) ¢ P(P2,2) ¢ P(P1,3) ¢ P(P3,1)


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AI Performing the Inference

NORM{ P2,2,P3,1 P(B1,1|P1,1,P1,2,P2,1) ¢ P(B1,2|P1,1,P1,2,P1,3) ¢ P(B2,1|P1,1,P2,1,P2,2,P3,1) ¢ P(P1,1) ¢ P(P1,2) ¢ P(P2,1) ¢ P(P2,2) ¢ P(P1,3) ¢ P(P3,1) }

NORM{ P2,2,P3,1 P[P1,3] ¢ P[P2,2,P3,1] ¢ P(P2,2) ¢ P(P1,3) ¢ P(P3,1) }

NORM{ P[P1,3] ¢ P(P1,3) ¢ P2,2 P(P2,2) ¢ P3,1 P[P2,2,P3,1] ¢ P(P3,1) }

We have reduced the inference to a simple computation over 2x2 tables.

P(P1,3) = h0.69, 0.31i 31% chance of WUMPUS!


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AI Summary

The Joint Distribution is analogous to the truth table for propositional logic. It exponentially large, but any query can be answered using it

Conditional independence allows us to factor the joint distribution using conformal products

Conditional independence relationships are conveniently visualized and encoded in a belief network DAG

Given evidence, we can reason efficiently by algebraic manipulation of the factored representation

oregon state university – cs430 intro to ai (c) 2003 thomas g. dietterich1 probabilistic...

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