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Fred/John Complains Problem Problem 12 Assignment3 Fall 2002

Fred/John Complains Problem Problem 12 Assignment3 Fall 2002. P(FC)=P(FC|GF=A)*P(GF=A) + P(FC|GF=C)*P(GF=C) + P(FC|GF=C)*P(GF=C) = 0*0.2 + 0.5x0.6 + 1x0.2 = 0.5 P(JC)= … (problem description) = P(FC,GJ=B) + (FC,GJ=C) = (d-separation of FC and GJ) = P(FC)*0.8 + P(FC)*0.1=P(FC)*0.9=0.45

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Fred/John Complains Problem Problem 12 Assignment3 Fall 2002

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  1. Fred/John Complains Problem Problem 12 Assignment3 Fall 2002 • P(FC)=P(FC|GF=A)*P(GF=A) + P(FC|GF=C)*P(GF=C) + P(FC|GF=C)*P(GF=C) = 0*0.2 + 0.5x0.6 + 1x0.2 = 0.5 • P(JC)= … (problem description) = P(FC,GJ=B) + (FC,GJ=C) = (d-separation of FC and GJ) = P(FC)*0.8 + P(FC)*0.1=P(FC)*0.9=0.45 • P(JC|FC)= P(JC,GJ=A|FC) + P(JC,GJ=B|FC) + P(JC,GJ=A|FC) = P(GJ=A|FC)*P(JC|GJ=A,FC) + … + … = (GJ and FC are d-separated) = P(GJ=A)*P(JC|GJ=A,FC) + P(GJ=B)*P(JC|GJ=B,FC) + P(GJ=A)* P(JC|GJ=A,FC) = 0.1*0 + 0.8x1 + 0.1x1 = 0.9 • P(JC|GF=C)= P(JC,FC|GF=C) + P(JC,~FC|GF=C) = P(FC|GF=C)*P(JC|FC,GF=C) + P(~FC|GF=C)*P(JC|~FC,GF=C) = (given FC: JC and GF are d-separated) = P(FC|GF=C)*P(JC|FC) + P(~FC)GF=C)*P(JC|~FL) = 1*(JC|FC) + 0= 0.9 • P(GF=C|JC)= (Bayes’ Theorem) = P(JC|GF=C)* P(GF=C) / P(JC) = 0.9*0.2/0.45=0.4

  2. Solution Sketch Problem 13.a • The belief network (iii) is not correct, because it specifies dependencies that are not described in the problem specification (e.g. it suggests that there is a dependency between the numbers of stars in the sky and telescopes being out of focus which violates common sense; the specification also does not suggest that astronomers communicate with each other about their measurements (but (iii) suggests a dependency between M1 and M2)). More importantly, using belief network (iii) is not feasible because it would require the knowledge of probabilities that are not given in the problem specification. • Both, belief network (i) and (ii), are correct; as a matter of fact one belief network can be obtained from the other belief network by reversing the direction of the arrows. This implies that given one of the two belief networks the probability tables for the other belief network can be easily be computed by hand using Bayes’ theorem. Network (ii) is more “convenient”, because the given probabilities can be directly used in the probability tables, whereas using (i) would require computing some probabilities by hand (using Bayes’ theorem).

  3. Fred/John Complains ProblemFinal Exam Fall 2001 • P(FC)=P(FC|GF=A)*P(GF=A) + P(FC|GF=C)*P(GF=C) + P(FC|GF=C)*P(GF=C) = 0*0.2 + 0.5x0.6 + 1x0.2 = 0.5 • P(JC)= … (problem description) = P(FC,GJ=B) + (FC,GJ=C) = (d-separation of FL and GJ) = P(FC)*0.8 + P(FC)*0.1=P(FC)*0.9=0.45 • P(JC|FC)= P(JC,GJ=A|FC) + P(JC,GJ=B|FC) + P(JC,GJ=A|FC) = P(GJ=A|FC)*P(JC|GJ=A,FC) + … + … = (GJ and FC are d-separated) = P(GJ=A)*P(JC|GJ=A,FC) + P(GJ=B)*P(JC|GJ=B,FC) + P(GJ=A)* P(JC|GJ=A,FC) = 0.1*0 + 0.8x1 + 0.1x1 = 0.9 • P(JC|GF=B)= P(JC,FC|GF=B) + P(JC,~FC|GF=B) = P(FC|GF=B)*P(JC|FC,GF=B) + P(~FL|GF=B)*P(JC|~FL,GF=B) = (given FC: JC and GF are d-separated) = P(FC|GF=B)*P(JC|FC) + P(~FC|GF=B)*P(JC|~FC) =0.5xP(JC|FC) + 0= 0.5xP(JC|FC) = 0.45 Remark: JC and JC|GF=B just have by coincidence the same numbers; for the example, JC and JC|GF=C have different probabilities. • P(GF=B|JC)= (Bayes’ Theorem) = P(JC|GF=B)* P(GF=B) / P(JC) = 0.45*0.6/0.45=0.6 Remark: In the example P(GF=B) and P(GF=B|JC) are both 0.6, but P(GF=C) is 0.2 whereas P(GF=C|JC)=0.4

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