Probabilistic Reasoning in Bayesian Networks: Insights & Calculations
Learn about conditional independence, Bayesian networks, global and local semantics, tree of inference calculations, and how to calculate probabilities in a Bayesian network example. Understand the fundamentals and intricacies of probabilistic reasoning. Dive into the world of Bayesian networks and explore its practical implications.
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Ch. 14 Probabilistic Reasoning Supplemental slides for CSE 327 Prof. Jeff Heflin
Conditional Independence if effects E1,E2, ,Enare conditionally independent given cause C E E E C 2 1 ,..., , , ( n = i = ) C ( ) ( | ) E C n i 1 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)
Bayes Net Example P(B) Burglary P(E) Earthquake 0.001 0.002 B E P(A|B,E) T T 0.95 Alarm T F 0.94 F T 0.29 F F 0.001 A P(M|A) A P(J|A) MaryCalls T 0.70 JohnCalls T 0.90 F 0.01 F 0.05 From Fig. 14.2, p. 512
Global Semantics atomic event using a Bayesian Network = i 1 P(b, e,a, j, m) = P(b)P( e)P(a|b, e)P(j|a)P( m|a) n = ( , ,..., ) ( | ( )) P x x x P x parvals X 1 2 n i i atomic event using the chain rule = n x x x P 2 1 ) ,..., , ( n = i ( | ,..., ) P x x x 1 1 i i 1 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)
Local Semantics Each node is conditionally independent of its non-descendants, given its parents. U1 U2 Z1 X Z2 Y1 Y2 P(X|U1,U2,Z1,Z2) = P(X| U1,U2)
Bayes Net Inference y = P e P e y ( | ) ( , , ) X X Formula: Example: e a = ) ( | ( ) , ) | ) | ( | , ) ( ) ( ( P b j m P b P e P a b e P j a P m a P(b|j, m)= P(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)]
Tree of Inference Calculations P(b)=.001 + P( e)=.998 P(e)=.002 + + P( a|b,e)=.05 P(a|b, e)=.94 P( a|b, e)=.06 P(a|b,e)=.95 P(j| a)=.05 P(j| a)=.05 P(j|a)=.90 P(j|a)=.90 P( m| a)=.99 P( m|a)=.30 P( m|a)=.30 P( m| a)=.99
Calculating P(b|j,m)and P(b|j,m) P(b|j, m)= P(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)]] = (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)]] = (0.001)[(0.002)[0.2565 + 0.002475] + (0.998)[0.2538 + 0.00297]] = (0.001)[(0.002)(0.258975) + (0.998)(0.25677)] = (0.001)[0.00051795 + 0.25625646] = (0.001)(0.25677441) = (0.00025677441) P( b|j, m)= P( 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)]] = (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)]] = (0.999)[(0.002)[0.0783 + 0.035145] + (0.998)[0.00027 + 0.0494505]] = (0.999)[(0.002)(0.113445) + (0.998)(0.497205)] = (0.999)[0.00022689 + 0.049621059] = (0.999)(0.049847949) = (0.049798101051)
Normalizing the Answer P(b|j, m) = (0.00025677441) P( b|j, m) = (0.04979801051) = 1 / (0.00025677441 + 0.04979801051) = 1 / 0.050054875461 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>
