Conditional Independence"A and P are independent given C"Pr(A I P,C) = Pr(A I C)PcAProbabilitFFF0.534AcheFFT0.356FFT0.006FTT0.004CavityTFF0.048TFT0.012ProbeFTT0.032CatchesTTT0.008
Conditional Independence n “A and P are independent given C” n Pr(A | P,C) = Pr(A | C) Cavity Probe Catches Ache C A P Probability F F F 0.534 F F T 0.356 F T F 0.006 F T T 0.004 T F F 0.048 T F T 0.012 T T F 0.032 T T T 0.008
Conditional Independence"A and P are independent given C"Pr(A I P,C) = Pr(A I C)and also Pr(P / A,C) = Pr(P / C)cPAProbSuppose C=TrueFFF0.534Pr(A/P,C) = 0.032/(0.032+0.048)FFT0.350=0.032/0.080FFT0.000= 0.4FTT0.00FTF0.01Pr(A/C) = 0.032+0.008/TFT0.04(0.048+0.012+0.032+0.008)FTT0.000TT0.T03=0.04/0.1=0.4
Pr(A|C) = 0.032+0.008/ (0.048+0.012+0.032+0.008) = 0.04 / 0.1 = 0.4 Suppose C=True Pr(A|P,C) = 0.032/(0.032+0.048) = 0.032/0.080 = 0.4 Conditional Independence n “A and P are independent given C” n Pr(A | P,C) = Pr(A | C) and also Pr(P | A,C) = Pr(P | C) C A P Probability F F F 0.534 F F T 0.356 F T F 0.006 F T T 0.004 T F F 0.012 T F T 0.048 T T F 0.008 T T T 0.032
Conditional IndependenceCan encode joint probability distribution in compactformConditional probability table (CPT)cPAProboCP(A)FFF0.534AcheT0.4FFT0.35FFFT0.020.000FTT0.00CavityTFF0.01:TFT0.04ProbeFTT0.00P(C)CCatchesP(P)TTT0.03.01T0.8F0.4
Conditional Independence n Can encode joint probability distribution in compact form C A P Probability F F F 0.534 F F T 0.356 F T F 0.006 F T T 0.004 T F F 0.012 T F T 0.048 T T F 0.008 T T T 0.032 Cavity Probe Catches Ache P(C) .01 C P(P) T 0.8 F 0.4 C P(A) T 0.4 F 0.02 Conditional probability table (CPT)