Computer Science CPSC 322 Lectur ture 2 e 23 Planni anning U ng Under er U Uncer ertai tainty nty a and Decision N on Networ works 1
Announ nouncem emen ents • Final exam • Mon, Dec. 18, 12noon • Same general format as midterm Part short questions, part longer problems List from which I will draw the short questions is posted on Connect (“Final” folder) I will also post there some Practice problems Covers material from the beginning of the course See list of posted learning goals for what you should know Office hours will continue as usual after the end of • classes
Lect cture re Overvi rview • Recap • Intro to Decision theory • Utility and expected utility • Decision Networks for Single-stage decision problems 3
Inf nference in n Gen eneral • Y: subset of variables that is queried (e.g. Temperature in previous example) • E: subset of variables that are observed . E = e ( W = yes in previous example) • Z 1 , …,Z k remaining variables in the JPD ( Cloudy in previous example) We need to compute this numerator for each value of Y, Y, y i We need to marginalize over all the variables Z 1 ,…Z k not involved in the query ∑ ∑ = = = … = = P ( Y y , E e ) ... P(Z , ,Z ,Y y ,E e) i 1 k i Z Z 1 k = Def of conditional P ( Y , E e ) = = P ( Y | E e ) probability = P ( E e ) = To compute the denominator, marginalize over Y P ( Y , E e ) - Same value for every P(Y=y i ). Normalization ∑ = P ( Y , E e ) constant ensuring that ∑ = = P(Y y | E ) 1 i Y Y • All we need to compute is the numerator: joint probability of the query variable(s) and the evidence! • Variable Elimination is an algorithm that efficiently performs this operation by casting it as operations between factors - introduced next
Factors Fa • A factor is a function from a tuple of random variables to the real numbers R • We write a factor on variables X 1 ,… ,X j as f( X 1 ,… ,X j ) • A factor denotes one or more (possibly partial) distributions over the given tuple of variables, e.g., • P(X 1 , X 2 ) is a factor f(X 1 , X 2 ) Distribution X Y Z val Set of Distributions t t t 0.1 • P(Z | X,Y) is a factor One for each combination t t f 0.9 of values for X and Y f(Z,X,Y) t f t 0.2 t f f 0.8 • P(Z=f|X,Y) is a factor f(X,Y) f(X, Y ) Z = f f t t 0.4 f t f 0.6 Set of partial Distributions f f t 0.3 • Note: Factors do not have to sum to one f f f 0.7
Rec ecap: F Fac actors and and Ope perations on T on Them hem If we assign variable A=a in factor f 7 (A,B), what is the correct form for the resulting factor? • f(B). When we assign variable A we remove it from the factor’s domain If we marginalize variable A out from factor f 7 (A,B), what is the correct form for the resulting factor? • f(B). When we marginalize out variable A we remove it from the factor’s domain If we multiply factors f 4 (X,Y) and f 6 (Z,Y), what is the correct form for the resulting factor? • f(X,Y,Z) • When multiplying factors, the resulting factor’s domain is the union of the multiplicands’ domains What is the correct form for ∑ B f 5 (A,B) × f 6 (B,C) • • As usual, product before sum: ∑ B ( f 5 (A,B) × f 6 (B,C) ) • Result of multiplication: f 7 (A,B,C). Then marginalize out B: f 8 (A,C)
Th The e variable elimi mination algo gorithm, n ∏ … = The JPD of a Bayesian network is P( X , , X ) P ( X | pa ( X ) ) 1 n i i = i 1 observed Other variables not involved in the query Given: P(Y, E 1 …, E j , Z 1 …,Z k ) = = … = P ( Y y , E e , , E e ) i 1 1 j j To compute P(Y=y i | E 1 =e 1 , …, E j =e j ) = ∑ = = … = P ( Y y , E e , , E e ) 1 1 j j = Y y = 1. Construct a factor for each conditional probability. P ( X | pa ( X )) f ( X , pa ( X )) i i i i i n ∑ ∑∏ = = = P ( Y , E e , , E e ) ( f ) = = 1 1 j j i E e , , E e 1 1 j j = Z Z i 1 k 1 2. For each factor, assign the observed variables E to their observed values. 3. Given an elimination ordering, decompose sum of products 4. 4. Sum um out out all variables Z i not involved in the query (one a time) • Multiply factors containing Z i • Then marginalize out Z i from the product 5. Multiply the remaining factors (which only involve Y ) ∑ f ( Y ) y
The variable elimination algorithm, n ∏ … = The JPD of a Bayesian network is P( X , , X ) P ( X | pa ( X ) ) 1 n i i = i 1 observed Other variables not involved in the query Given: P(Y, E 1 …, E j , Z 1 …,Z k ) = = … = P ( Y y , E e , , E e ) i 1 1 j j To compute P(Y=y i | E 1 =e 1 , …, E j =e j ) = ∑ = = … = P ( Y y , E e , , E e ) 1 1 j j = Y y = 1. Construct a factor for each conditional probability. P ( X | pa ( X )) f ( X , pa ( X )) i i i i i n ∑ ∑∏ = = = P ( Y , E e , , E e ) ( f ) = = 1 1 j j i E e , , E e 1 1 j j = Z Z i 1 k 1 2. For each factor, assign the observed variables E to their observed values. 3. Given an elimination ordering, decompose sum of products Sum out all variables Z i not involved in the query (one a time) 4. • Multiply factors containing Z i Then marginalize out Z i from the product • 5. Multiply the remaining factors (which only involve Y ) 6. Normalize by dividing the resulting factor f(Y) by ∑ f ( Y ) y
Step 1: Construct a factor for each cond. probability Compute P(G | H=h 1 ) . P(G,H) = ∑ A,B,C,D,E,F,I P(A)P(B|A)P(C)P(D|B,C)P(E|C)P(F|D)P(G|F,E)P(H|G)P(I|G) P(G,H) = ∑ A,B,C,D,E,F,I f 0 (A) f 1 (B,A) f 2 (C) f 3 (D,B,C) f 4 (E,C) f 5 (F, D) f 6 (G,F,E) f 7 (H,G) f 8 (I,G) 9
Step 2: assign to observed variables their observed values. Compute P(G | H=h 1 ) . Previous state: P(G,H) = ∑ A,B,C,D,E,F,I f 0 (A) f 1 (B,A) f 2 (C) f 3 (D,B,C) f 4 (E,C) f 5 (F, D) f 6 (G,F,E) f 7 (H,G) f 8 (I,G) Observe H : P(G,H=h 1 )= ∑ A,B,C,D,E,F,I f 0 (A) f 1 (B,A) f 2 (C) f 3 (D,B,C) f 4 (E,C) f 5 (F, D) f 6 (G,F,E) f 9 (G) f 8 (I,G) 10
Step 3: Decompose sum of products Compute P(G | H=h 1 ) . Previous state: P(G,H=h 1 ) = ∑ A,B,C,D,E,F,I f 0 (A) f 1 (B,A) f 2 (C) f 3 (D,B,C) f 4 (E,C) f 5 (F, D) f 6 (G,F,E) f 9 (G) f 8 (I,G) Elimination ordering A, C, E, I, B, D, F : P(G,H=h 1 ) = f 9 (G) ∑ F ∑ D f 5 (F, D) ∑ B ∑ I f 8 (I,G) ∑ E f 6 (G,F,E) ∑ C f 2 (C) f 3 (D,B,C) f 4 (E,C) ∑ A f 0 (A) f 1 (B,A) 11
Step 4: sum out non query variables (one at a time) Compute P(G | H=h 1 ) . Elimination order: A ,C,E,I,B,D,F Previous state: P(G,H=h 1 ) = f 9 (G) ∑ F ∑ D f 5 (F, D) ∑ B ∑ I f 8 (I,G) ∑ E f 6 (G,F,E) ∑ C f 2 (C) f 3 (D,B,C) f 4 (E,C) ∑ A f 0 (A) f 1 (B,A) Eliminate A: perform product and sum out A in P(G,H=h 1 ) = f 9 (G) ∑ F ∑ D f 5 (F, D) ∑ B f 10 (B) ∑ I f 8 (I,G) ∑ E f 6 (G,F,E) ∑ C f 2 (C) f 3 (D,B,C) f 4 (E,C) f 10 (B) does not depend on C, E, or I, so we can push it outside of those sums. 12
Step 4: sum out non query variables (one at a time) Compute P(G | H=h 1 ) . Elimination order: A, C ,E,I,B,D,F Previous state: P(G,H=h 1 ) = f 9 (G) ∑ F ∑ D f 5 (F, D) ∑ B f 10 (B) ∑ I f 8 (I,G) ∑ E f 6 (G,F,E) ∑ C f 2 (C) f 3 (D,B,C) f 4 (E,C) Eliminate C: perform product and sum out C in P(G,H=h 1 ) = f 9 (G) ∑ F ∑ D f 5 (F, D) ∑ B f 10 (B) ∑ I f 8 (I,G) ∑ E f 6 (G,F,E) f 11 (B,D,E) 13
Step 4: sum out non query variables (one at a time) Compute P(G | H=h 1 ) . Elimination order: A,C, E ,I,B,D,F Previous state: P(G,H=h 1 ) = P(G,H=h 1 ) = f 9 (G) ∑ F ∑ D f 5 (F, D) ∑ B f 10 (B) ∑ I f 8 (I,G) ∑ E f 6 (G,F,E) f 11 (B,D,E) Eliminate E: perform product and sum out E in P(G,H=h 1 ) = P(G,H=h 1 ) = f 9 (G) ∑ F ∑ D f 5 (F, D) ∑ B f 10 (B) f 12 (B,D,F,G) ∑ I f 8 (I,G) 14
Step 4: sum out non query variables (one at a time) Compute P(G | H=h 1 ) . Elimination order: A,C,E, I ,B,D,F Previous state: P(G,H=h 1 ) = P(G,H=h 1 ) = f 9 (G) ∑ F ∑ D f 5 (F, D) ∑ B f 10 (B) f 12 (B,D,F,G) ∑ I f 8 (I,G) Eliminate I: perform product and sum out I in P(G,H=h 1 ) = P(G,H=h 1 ) = f 9 (G) f 13 (G) ∑ F ∑ D f 5 (F, D) ∑ B f 10 (B) f 12 (B,D,F,G) 15
Step 4: sum out non query variables (one at a time) Compute P(G | H=h 1 ) . Elimination order: A,C,E,I, B ,D,F Previous state: P(G,H=h 1 ) = P(G,H=h 1 ) = f 9 (G) f 13 (G) ∑ F ∑ D f 5 (F, D) ∑ B f 10 (B) f 12 (B,D,F,G) Eliminate B: perform product and sum out B in P(G,H=h 1 ) = P(G,H=h 1 ) = f 9 (G) f 13 (G) ∑ F ∑ D f 5 (F, D) f 14 (D,F,G) 16
Step 4: sum out non query variables (one at a time) Compute P(G | H=h 1 ) . Elimination order: A,C,E,I,B, D ,F Previous state: P(G,H=h 1 ) = P(G,H=h 1 ) = f 9 (G) f 13 (G) ∑ F ∑ D f 5 (F, D) f 14 (D,F,G) Eliminate D: perform product and sum out D in P(G,H=h 1 ) = P(G,H=h 1 ) = f 9 (G) f 13 (G) ∑ F f 15 (F,G) Multiply remaining factors (all in G): P(G,H=h 1 ) = f 17 (G) 17
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