Bayesia n net wor k s Chapter 14.1 – 3 Extracted from : http://aima.eecs.berkeley.edu/slides-pdf/chapter14a.pdf Chapter 14.1 – 1 3
Outline ♦ Syntax ♦ Semantics ♦ Parameterized distributions Chapter 14.1 – 2 3
B ayesian networks A simple, graphical notation for conditional independence assertions and hence for compact specification of full joint distributions Syntax: a set of nodes, one per variable a directed, acyclic graph (link ≈ “directly influences”) a conditional distribution for each node given its parents: P ( X i | Parents ( X i )) In the simplest case, conditional distribution represented as a conditional probability table (CPT) giving the distribution over X i for each combination of parent values Chapter 14.1 – 3 3
Exam ple Topology of netw ork encodes conditional independence assertions: Cavity Weather Toothache Catch Weather is independent of the other variables Toothache and Catch are conditionally independent given Cavity Chapter 14.1 – 4 3
Exam ple I’m at work, neighbor John calls to say my alarm is ringing, but neighbor Mary doesn’t call. Sometimes it’s set off by minor earthquakes. Is there a burglar? Variables: Burglar , Earthquake , Alarm , JohnCalls , MaryCalls Network topology reflects “causal” knowledge: – A burglar can set the alarm off – An earthquake can set the alarm off – The alarm can cause Mary to call – The alarm can cause John to call Chapter 14.1 – 5 3
Exam ple contd. P(E) P(B) Burglary Earthquake .002 .001 B E P(A|B,E) T T .95 Alarm T F .94 F T .29 F F .001 A P(J|A) A P(M|A) T .90 T .70 JohnCalls MaryCalls F F .05 .01 Chapter 14.1 – 6 3
C om pactness A CPT for Boolean X i with k Boolean parents has 2 k rows for the combinations of parent values B E A Each row requires one number p for X i = true (the number for X i = false is just 1 − p ) J M If each variable has no more than k parents, the complete network requires O ( n · 2 k ) numbers I.e., grows linearly with n , vs. O (2 n ) for the full joint distribution For burglary net, 1 + 1 + 4 + 2 + 2 = 10 numbers (vs. 2 5 − 1 = 31) Chapter 14.1 – 7 3
G lobal sem antics “Global” semantics defines the full joint distribution as the product of the local B E conditional distributions: A n P ( x , . . . , x ) = Π P ( x | parents ( X )) J M 1 n i i i = 1 e.g., P ( j ∧ m ∧ a ∧ ¬ b ∧ ¬ e ) P ( j | a ) P ( m | a ) P ( a |¬ b, ¬ e ) P ( ¬ b ) P ( ¬ e ) = 0 . 9 × 0 . 7 × 0 . 001 × 0 . 999 × 0 . 998 = ≈ 0 . 00063 Chapter 14.1 – 8 3
C onstru cting B ayesian networks Need a method such that a series of locally testable assertions of conditional independence guarantees the required global semantics 1.Choose an ordering of variables X 1 , . . . , X n 2.For i = 1 to n add X i to the network select parents from X 1 , . . . , X i − 1 such that P ( X i | Parents ( X i )) = P ( X i | X 1 , . . . , X i − 1 ) This choice of parents guarantees the global semantics: n P P ( X i | X 1 , . . . , X i-1 ) (chain rule) P ( X , . . . , X ) = 1 n i=1 n = P P(X i | Parents(X i )) ( by construction) i=n Chapter 14.1 – 9 3
Exam ple Suppose we choose the ordering M , J , A , B , E MaryCalls JohnCalls | M ) = P ( J P ( J )? Chapter 14.1 – 10 3
Exam ple Suppose we choose the ordering M , J , A , B , E MaryCalls JohnCalls A larm | M ) = P ( J P ( J )? No P ( A | J, M ) = P ( A | J )? P ( A | J, M ) = P ( A )? Chapter 14.1 – 11 3
Exam ple Suppose we choose the ordering M , J , A , B , E MaryCalls JohnCalls Alarm Burglary P ( J | M ) = P ( J )? No P ( A | J, M ) = P ( A | J )? P ( A | J, M ) = P ( A )? No P ( B | A, J, M ) = P ( B | A )? P ( B | A, J, M ) = P ( B )? Chapter 14.1 – 12 3
Exam ple Suppose we choose the ordering M , J , A , B , E MaryCalls JohnCalls Alarm Burglary Earthquake P ( J | M ) = P ( J )? No P ( A | J, M ) = P ( A | J )? P ( A | J, M ) = P ( A )? No P ( B | A, J, M ) = P ( B | A )? Yes P ( B | A, J, M ) = P ( B )? No P ( E | B, A, J, M ) = P ( E | A )? P ( E | B, A, J, M ) = P ( E | A, B )? Chapter 14.1 – 13 3
Exam ple Suppose we choose the ordering M , J , A , B , E MaryCalls JohnCalls Alarm Burglary Earthquake P ( J | M ) = P ( J )? No P ( A | J, M ) = P ( A | J )? P ( A | J, M ) = P ( A )? No P ( B | A, J, M ) = P ( B | A )? Yes P ( B | A, J, M ) = P ( B )? No P ( E | B, A, J, M ) = P ( E | A )? No P ( E | B, A, J, ( E | A, B )? ) = P M Yes Chapter 14.1 – 14 3
Exam ple contd. MaryCalls JohnCalls Alarm Burglary Earthquake Deciding conditional independence is hard in noncausal directions (Causal models and conditional independence seem hardwired for humans!) Assessing conditional probabilities is hard in noncausal directions Network is less compact: 1 + 2 + 4 + 2 + 4 = 13 numbers needed Chapter 14.1 – 3 15
Exam ple: C ar diagnosis Initial evidence: car won’t start Testable variables (green), “broken, so fix it” variables (orange) Hidden variables (gray) ensure sparse structure, reduce parameters fanbelt alternator battery age broken broken battery no charging dead battery fuel line starter battery no oil no gas flat blocked broken meter car won’t gas gauge oil light lights dipstick start Chapter 14.1 – 16 3
Exam ple: C ar insurance SocioEcon Age GoodStudent ExtraCar Mileage RiskAversion VehicleYear SeniorTrain MakeModel DrivingSkill DrivingHist Antilock DrivQuality CarValue HomeBase AntiTheft Airbag Accident Ruggedness Theft OwnDamage Cushioning OwnCost OtherCost MedicalCost LiabilityCost PropertyCost Chapter 14.1 – 17 3
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