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Basic Properties of Belief Networks

Basic Properties of Belief Networks. Simplifying Assumption : Let X 1 ,…,X n be the variables of a belief network and all variables have binary states: P(X 1 ,…,X n )= P P(X i |Parents(X i )) P(X 1 ,…,X p-1 )= P(X 1 ,…,X p-1 ,X p ) + P(X 1 ,…,X p-1 ,~X p )

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Basic Properties of Belief Networks

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  1. Basic Properties of Belief Networks Simplifying Assumption: Let X1,…,Xn be the variables of a belief network and all variables have binary states: • P(X1,…,Xn)= P P(Xi|Parents(Xi)) • P(X1,…,Xp-1)= P(X1,…,Xp-1,Xp) + P(X1,…,Xp-1,~Xp) • P(X|Y) = a* P(X,Y) where a =1/P(Y) Remark: These 3 equations are sufficient to compute any probability in a belief network; however, using this approach is highly inefficient; e.g. with n=20 computing P(X1|X2) would require the addition of 218+219 probabilities. Therefore, more efficient ways to compute probabilities are needed; e.g. if X1 and X2 are independent, only P(X1) needs to be computed. Another way to speedup computations is using probabilities that are already known and no not need to be computed and taking advantage of the fact that probabilities add up to 1 n i=1

  2. Hurricane Point (580 feet) Big Sur 2006 Plan • Speed Work1 • March 8: t: 3x1M (9) • March 16: t: 3x1M (10) • March 22: t: 2.5M • March 25: mp: 7M (12) • March 29: 8k: 8x800 • April 2: ct: 4x1M • April 7: 8k: 10x800 • April 12: ct: 3x1.5M • April 18: t: 2x2.5M (11) • April 22: ct: 2x2800 • April 26: 30k: 3x1M • Long Runs and Races • Nov. 19: 11.5M • Nov. 26: 14.3M • Dec. 4: 12M • Dec. 11: 16.5M • Dec. 18: 18M • Dec. 28: 20M • Jan. 6: 13Mhi • Jan. 14: 20M • Jan. 21: 22M • Jan. 30: 20M • Feb. 2005: kind of sick • March 5: 16 • March 12: 19 • March 18: 21 • March 30: 21 • April 9: 19 • April 15: 17 • April 30: Big Sur Marathon Bixby Creek Bridge (1932) Bixby Creek Bridge (110 feet)

  3. D-Separation • Belief Networks abandon the simple independence assumptions of naïve Bayesian systems and replace them by a more complicated notion of independence called d-separation. • Problem: Given evidence involving a set of variables E; when are two sets of variables X and Y of a belief network independent (d-separated)? • Why is this question important? If X and Y are d-separated (given E) P(X&Y|E)=P(X|E)*P(Y|E) and P(X|E&Y)=P(X|E) easy to show; how? • D-separation is used a lot in belief network computations (see P(D|S1,S2) example to be discussed later); particularly to speed up belief network computations.

  4. D-Separation :=All paths between members of X and Y must match one of the following 4 patters: Y X E(in E, not in E) (1a) (1b) (2) (3)

  5. D-Separation A D a)  Which of the following statements are implied by the indicated network structure; answer yes and no; and give a brief reason for your answer! [6] i) P(A,B|C) = P(A|C)*P(B|C)  yes, because… ii) P(C,E|D) = P(C|D)*P(E|D)  no, because iii) P(C|A)=P(C)  no, because C B E

  6. Compute P(D|S1,S2)!! S1 • All 3 variables of B have binary states: {T,F} • P(D) is a short notation for P(D=T) and P(S2|~D) is a short notation for P(S2=T|D=F). • B’s probability tables contain: P(D)=0.1, P(S1|D)=0.95, P(S2|D)=0.8, P(S1|~D)=0.2, P(S2|~D)=0.2 Task: Compute P(D|S1,S2) D B S2

  7. Computing P(D|S1,S2) • P(D|S1,S2)=P(D)*P(S1|D)*P(S2|D)/P(S1,S2) because S1|D indep S2|D • P(~D|S1,S2)=P(~D)*P(S1|~D)*P(S2|~D)/P(S1,S2) S1|D indep S2|D • (1+2) 1=(P(D)*P(S1|D)*P(S2|D) + P(~D)*P(S1|~D)*P(S2|~D))/P(S1,S2) • P(S1,S2)= P(D)*P(S1|D)*P(S2|D) + P(~D)*P(S1|~D)*P(S2|~D)=g • P(D|S1,S2)= a / a + b with • a=P(D)*P(S1|D)*P(S2|D) and b =P(~D)*P(S1|~D)*P(S2|~D) • For the example a=0.1*0.95*0.8=0.076 and b =0.9*0.2*0.2=0.036 • P(D|S1,S2)=0.076/0.112=0.678 S1 D S2

  8. John/Fred Late Problem Assume that John and Fred do not know each other, but take the same bus to come to UH to attend the same class; moreover, both use an alarm clock, and will be late to school if their alarm clock does not work; moreover, if the bus is late, both will come late to school. Assume you have to model this information using a belief network that consists of the following variables that have boolean ({true,false}) states: • John-no-alarm: John’s alarm clock does not work (short JNOA) • Fred-no-alarm: Fred’s alarm clock does not work (short FNOA) • Bus-late: The bus that John and Fred take is late (short BL) • John-late: John is late at school (short JL) • Fred-late: Fred is late at school (short FL) Moreover, assume that probability that John’s alarm clock does not work is 0.02, the probability that Fred’s alarm clock malfunctions is 0.02, and the probability that the bus is late is 0.1. You can also assume that Fred and John are on time if their alarm clock works and the bus they take isn’t late. a)      Design the structure of a belief network that involves the above variables! b) Using your results from the previous step, compute P(Bus-late=true|Fred-late=true) !

  9. Solution Student Late Problem Ass4 • P(FBL|FL)= P(FL|FBL)*P(FBL)/P(FL)=1x0.1/P(FL)=0.1/0.11791=0.848 • P(FL)=P(FL,WSF, FBL) + P(FL,WSF, ~FBL) + P(FL,~WSF, FBL) + P(FL,~WSF, ~FBL) • P(FL,WSF, FBL) =P(FL | WSF,FBL)*P(FBL,WSF)= (because WSF and FBL are d-separable giving no evidence) P(FL | WSF,FBL)*P(WSF)* P(FBL)= 1*0.01*0.1 =0.001 • Same for the other 3 formulas in (2) • P(FL)= 1*0.01*0.1 + 1*0.01*0.9 + 1*0.99*0.1 + 0.01*0.99*0.9=0.001 + 0.009 + 0.099 + 0.00891=0.11791 Problem Specification FBL FL MSF JL JBL

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