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Ch. 14 – Probabilistic Reasoning

Ch. 14 – Probabilistic Reasoning. Supplemental slides for CSE 327 Prof. Jeff Heflin. Conditional Independence. if effects E 1 ,E 2 ,…,E n are conditionally independent given cause C. can be used to factor joint distributions P ( Weather,Cavity,Toothache,Catch ) =

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Ch. 14 – Probabilistic Reasoning

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  1. Ch. 14 – Probabilistic Reasoning Supplemental slides for CSE 327 Prof. Jeff Heflin

  2. Conditional Independence • if effects E1,E2,…,En are conditionally independent given cause C • can be used to factor joint distributions • P(Weather,Cavity,Toothache,Catch) = • P(Weather)P(Cavity,Toothache,Catch) = • P(Weather)P(Cavity)P(Toothache|Cavity)P(Catch|Cavity)

  3. P(B) P(E) 0.001 0.002 B E P(A|B,E) T T 0.95 T F 0.94 F T 0.29 F F 0.001 A P(M|A) A P(J|A) T 0.70 T 0.90 F 0.01 F 0.05 Bayes Net Example Burglary Earthquake Alarm MaryCalls JohnCalls From Fig. 14.2, p. 512

  4. Global Semantics • atomic event using a Bayesian Network P(b,e,a, j,m) = P(b)P(e)P(a|b,e)P(j|a)P(m|a) • atomic event using the chain rule P(b,e,a, j,m) = P(b)P(e|b)P(a|b,e)P(j| b,e,a)P(m| b,e,a,j)

  5. Bayes Net Inference Formula: Example: P(b|j,m)=αP(b)[P(e)[P(a|b,e)P(j|a)P(m|a) +P(a|b,e)P(j|a)P(m|a)] +P(e)[P(a|b,e)P(j|a)P(m|a) + P(a|b,e)P(j|a)P(m|a)]

  6. Tree of Inference Calculations P(b)=.001 + P(e)=.998 P(e)=.002 + + P(a|b,e)=.95 P(a|b,e)=.05 P(a|b,e)=.94 P(a|b,e)=.06 P(j|a)=.05 P(j|a)=.90 P(j|a)=.90 P(j|a)=.05 P(m|a)=.99 P(m|a)=.30 P(m|a)=.30 P(m|a)=.99

  7. Calculating P(b|j,m)and P(b|j,m) P(b|j,m)=αP(b)[P(e)[P(a|b,e)P(j|a)P(m|a) + P(a|b,e)P(j|a)P(m|a)] +P(e)[P(a|b,e)P(j|a)P(m|a) + P(a|b,e)P(j|a)P(m|a)]] = α(0.001)[(0.002)[(0.95)(0.9)(0.3) + (0.05)(0.05)(0.99)] + (0.998)[(0.94)(0.9)(0.3) + (0.06)(0.05)(0.99)]] = α(0.001)[(0.002)[0.2565 + 0.002475] + (0.998)[0.2538 + 0.00297]] = α(0.001)[(0.002)(0.258975) + (0.998)(0.25677)] = α(0.001)[0.00051795 + 0.25625646] = α(0.001)(0.25677441) = α(0.00025677441) P(b|j,m)=αP(b)[P(e)[P(a| b,e)P(j|a)P(m|a) + P(a| b,e)P(j|a)P(m|a)] +P(e)[P(a| b,e)P(j|a)P(m|a) + P(a| b,e)P(j|a)P(m|a)]] = α(0.999)[(0.002)[(0.29)(0.9)(0.3) + (0.71)(0.05)(0.99)] + (0.998)[(0.001)(0.9)(0.3) + (0.999)(0.05)(0.99)]] = α(0.999)[(0.002)[0.0783 + 0.035145] + (0.998)[0.00027 + 0.0494505]] = α(0.999)[(0.002)(0.113445) + (0.998)(0.497205)] = α(0.999)[0.00022689 + 0.049621059] = α(0.999)(0.049847949) = α(0.049798101051)

  8. Normalizing the Answer P(b|j,m) =α(0.00025677441) P(b|j,m) = α(0.04979801051) α = 1 / (0.00025677441 + 0.04979801051) α = 1 / 0.050054875461 α19.97807 P(b|j,m) (19.97807)(0.00025677441)  0.0051 P(b|j,m) (19.97807) (0.04979801051)  0.9949 P(B|j,m) = <0.0051, 0.9949>

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