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

2. 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) = P(Weather)P(Cavity,Toothache,Catch) = P(Weather)P(Cavity)P(Toothache|Cavity)P(Catch|Cavity)

3. Bayes Net Example P(M|A) 0.70 A T F0.01 P(J|A) 0.90 A T F0.05 P(B) 0.001 Burglary Earthquake Alarm JohnCalls MaryCalls P(E) 0.002 0.94 0.29 P(A|B,E) 0.95 E T F B T T T F0.001 F F From Fig. 14.2, p. 512

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

5. Local Semantics Each node is conditionally independent of its non-descendants, given its parents. U1U1 U2U2 XZ1Z1 Y1Y1 Z2Z2 Y2Y2 P(X|U 1,U 2,Z 1,Z 2 ) = P(X| U 1,U 2 )

6. Bayes Net Inference 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)] Formula: Example:

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

8. 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)

9. 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) =