1. 1 22c:145 Artificial Intelligence Bayesian Networks Reading: Ch 14. Russell & Norvig

2. 2 Review of Probability Theory Random Variables The probability that a random variable X has value val is written as P(X=val) P: domain ! [0, 1] –Sums to 1 over the domain: »P(Raining = true) = P(Raining) = 0.2 »P(Raining = false) = P( : Raining) = 0.8 Joint distribution: P(X1, X2, …, Xn) Probability assignment to all combinations of values of random variables and provide complete information about the probabilities of its random variables. A JPD table for n random variables, each ranging over k distinct values, has k n entries! Toothache : Toothache Cavity 0.040.06 : Cavity 0.010.89

3. 3 Review of Probability Theory Conditioning P(A) = P(A | B) P(B) + P(A | : B) P( : B) = P(A Æ B) + P(A Æ : B) A and B are independent iff P(A Æ B) = P(A) ¢ P(B) P(A | B) = P(A) P(B | A) = P(B) A and B are conditionally independent given C iff P(A | B, C) = P(A | C) P(B | A, C) = P(B | C) P(A Æ B | C) = P(A | C) ¢ P(B | C) Bayes’ Rule P(A | B) = P(B | A) P(A) / P(B) P(A | B, C) = P(B | A, C) P(A | C) / P(B | C)

4. 4 Bayesian Networks To do probabilistic reasoning, you need to know the joint probability distribution But, in a domain with N propositional variables, one needs 2 N numbers to specify the joint probability distribution We want to exploit independences in the domain Two components: structure and numerical parameters

5. Lecture 14 5 Bayesian 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

6. 6 Bayesian (Belief) Networks Set of random variables, each has a finite set of values Set of directed arcs between them forming acyclic graph, representing causal relation Every node A, with parents B 1, …, B n, has P(A | B 1,…,B n ) specified

7. 7 Key Advantage The conditional independencies (missing arrows) mean that we can store and compute the joint probability distribution more efficiently How to design a Belief Network? Explore the causal relations

8. 8 Icy Roads “Causal” Component Holmes Crash Icy Watson Crash Inspector Smith is waiting for Holmes and Watson, who are driving (separately) to meet him. It is winter. His secretary tells him that Watson has had an accident. He says, “It must be that the roads are icy. I bet that Holmes will have an accident too. I should go to lunch.” But, his secretary says, “No, the roads are not icy, look at the window.” So, he says, “I guess I better wait for Holmes.”

9. 9 Icy Roads “Causal” Component Holmes Crash Icy Watson Crash Inspector Smith is waiting for Holmes and Watson, who are driving (separately) to meet him. It is winter. His secretary tells him that Watson has had an accident. He says, “It must be that the roads are icy. I bet that Holmes will have an accident too. I should go to lunch.” But, his secretary says, “No, the roads are not icy, look at the window.” So, he says, “I guess I better wait for Holmes.”

10. 10 Icy Icy Roads “Causal” Component Holmes Crash Watson Crash Inspector Smith is waiting for Holmes and Watson, who are driving (separately) to meet him. It is winter. His secretary tells him that Watson has had an accident. He says, “It must be that the roads are icy. I bet that Holmes will have an accident too. I should go to lunch.” But, his secretary says, “No, the roads are not icy, look at the window.” So, he says, “I guess I better wait for Holmes.”

11. 11 Icy Icy Roads “Causal” Component Holmes Crash Watson Crash Inspector Smith is waiting for Holmes and Watson, who are driving (separately) to meet him. It is winter. His secretary tells him that Watson has had an accident. He says, “It must be that the roads are icy. I bet that Holmes will have an accident too. I should go to lunch.” But, his secretary says, “No, the roads are not icy, look at the window.” So, he says, “I guess I better wait for Holmes.” H and W are dependent,

12. 12 Icy Roads “Causal” Component Holmes Crash Icy Watson Crash Inspector Smith is waiting for Holmes and Watson, who are driving (separately) to meet him. It is winter. His secretary tells him that Watson has had an accident. He says, “It must be that the roads are icy. I bet that Holmes will have an accident too. I should go to lunch.” But, his secretary says, “No, the roads are not icy, look at the window.” So, he says, “I guess I better wait for Holmes.” H and W are dependent, but conditionally independent given I

13. 13 Holmes and Watson in IA Holmes and Watson have moved to IA. He wakes up to find his lawn wet. He wonders if it has rained or if he left his sprinkler on. He looks at his neighbor Watson’s lawn and he sees it is wet too. So, he concludes it must have rained. Holmes Lawn Wet Sprinkler Watson Lawn Wet Rain

14. 14 Holmes and Watson in IA Holmes and Watson have moved to IA. He wakes up to find his lawn wet. He wonders if it has rained or if he left his sprinkler on. He looks at his neighbor Watson’s lawn and he sees it is wet too. So, he concludes it must have rained. Holmes Lawn Wet Sprinkler Watson Lawn Wet Rain

15. 15 Holmes and Watson in IA Holmes and Watson have moved to IA. He wakes up to find his lawn wet. He wonders if it has rained or if he left his sprinkler on. He looks at his neighbor Watson’s lawn and he sees it is wet too. So, he concludes it must have rained. Holmes Lawn Wet Sprinkler Watson Lawn Wet Rain

16. 16 Holmes and Watson in IA Holmes and Watson have moved to IA. He wakes up to find his lawn wet. He wonders if it has rained or if he left his sprinkler on. He looks at his neighbor Watson’s lawn and he sees it is wet too. So, he concludes it must have rained. Holmes Lawn Wet Sprinkler Watson Lawn Wet Rain

17. 17 Rain Holmes Lawn Wet Holmes and Watson in IA Holmes and Watson have moved to IA. He wakes up to find his lawn wet. He wonders if it has rained or if he left his sprinkler on. He looks at his neighbor Watson’s lawn and he sees it is wet too. So, he concludes it must have rained. Sprinkler Watson Lawn Wet Given W, P(R) goes up

18. 18 Rain Holmes Lawn Wet Holmes and Watson in IA Holmes and Watson have moved to IA. He wakes up to find his lawn wet. He wonders if it has rained or if he left his sprinkler on. He looks at his neighbor Watson’s lawn and he sees it is wet too. So, he concludes it must have rained. Sprinkler Watson Lawn Wet Given W, P(R) goes up and P(S) goes down – “explaining away”

19. 19 Inference in Bayesian Networks Query Types Given a Bayesian network, what questions might we want to ask? Conditional probability query: P(x | e) Maximum a posteriori probability: What value of x maximizes P(x|e) ? General question: What’s the whole probability distribution over variable X given evidence e, P(X | e)?

20. 20 Using the joint distribution To answer any query involving a conjunction of variables, sum over the variables not involved in the query.

21. 21 Using the joint distribution To answer any query involving a conjunction of variables, sum over the variables not involved in the query.

22. 22 Chain Rule Variables: V 1, …, V n Values: v 1, …, v n P(V 1 =v 1, V 2 =v 2, …, V n =v n ) =  i P(V i =v i | parents(V i )) AB C D P(A)P(B) P(C|A,B) P(D|C)

23. 23 Chain Rule Variables: V 1, …, V n Values: v 1, …, v n P(V 1 =v 1, V 2 =v 2, …, V n =v n ) =  i P(V i =v i | parents(V i )) AB C D P(A)P(B) P(C|A,B) P(D|C) P(ABCD) = P(A=true, B=true, C=true, D=true)

24. 24 Chain Rule Variables: V 1, …, V n Values: v 1, …, v n P(V 1 =v 1, V 2 =v 2, …, V n =v n ) =  i P(V i =v i | parents(V i )) AB C D P(A)P(B) P(C|A,B) P(D|C) P(ABCD) = P(A=true, B=true, C=true, D=true) P(ABCD) = P(D|ABC)P(ABC)

25. 25 Chain Rule Variables: V 1, …, V n Values: v 1, …, v n P(V 1 =v 1, V 2 =v 2, …, V n =v n ) =  i P(V i =v i | parents(V i )) AB C D P(A)P(B) P(C|A,B) P(D|C) P(ABCD) = P(A=true, B=true, C=true, D=true) P(ABCD) = P(D|ABC)P(ABC) = P(D|C) P(ABC) = A independent from D given C B independent from D given C

26. 26 Chain Rule Variables: V 1, …, V n Values: v 1, …, v n P(V 1 =v 1, V 2 =v 2, …, V n =v n ) =  i P(V i =v i | parents(V i )) AB C D P(A)P(B) P(C|A,B) P(D|C) P(ABCD) = P(A=true, B=true, C=true, D=true) P(ABCD) = P(D|ABC)P(ABC) = P(D|C) P(ABC) = P(D|C) P(C|AB) P(AB) = A independent from D given C B independent from D given C

27. 27 Chain Rule Variables: V 1, …, V n Values: v 1, …, v n P(V 1 =v 1, V 2 =v 2, …, V n =v n ) =  i P(V i =v i | parents(V i )) AB C D P(A)P(B) P(C|A,B) P(D|C) P(ABCD) = P(A=true, B=true, C=true, D=true) P(ABCD) = P(D|ABC)P(ABC) = P(D|C) P(ABC) = P(D|C) P(C|AB) P(AB) = P(D|C) P(C|AB) P(A)P(B) A independent from D given C B independent from D given C A independent from B

28. 28 Chain Rule Variables: V 1, …, V n Values: v 1, …, v n P(V 1 =v 1, V 2 =v 2, …, V n =v n ) =  i P(V i =v i | parents(V i )) AB C D P(A)P(B) P(C|A,B) P(D|C) P(ABCD) = P(A=true, B=true, C=true, D=true) P(ABCD) = P(D|ABC)P(ABC) = P(D|C) P(ABC) = P(D|C) P(C|AB) P(AB) = P(D|C) P(C|AB) P(A)P(B) A independent from D given C B independent from D given C A independent from B

29. 29 Icy Roads with Numbers Holmes Crash Icy Watson Crash P(I=t)P(I=f) 0.70.3 t = true f = false The right-hand column in these tables is redundant, since we know the entries in each row must add to 1. NB: the columns need NOT add to 1.

30. 30 Icy Roads with Numbers P(W=t | I)P(W=f | I) I=t0.80.2 I=f0.10.9 Holmes Crash Icy Watson Crash P(I=t)P(I=f) 0.70.3 t = true f = false The right-hand column in these tables is redundant, since we know the entries in each row must add to 1. Note: the columns need NOT add to 1.

31. 31 Icy Roads with Numbers P(W=t | I)P(W=f | I) I=t0.80.2 I=f0.10.9 Holmes Crash Icy Watson Crash P(I=t)P(I=f) 0.70.3 P(H=t | I)P(H=f | I) I=t0.80.2 I=f0.10.9 t = true f = false The right-hand column in these tables is redundant, since we know the entries in each row must add to 1. Note: the columns need NOT add to 1.

32. 32 Probability that Watson Crashes Holmes Crash Icy Watson Crash P(I)=0.7 P(H| I) I0.8 -I-I0.1 P(W| I) I0.8 -I-I0.1 P(W) =

33. 33 Probability that Watson Crashes Holmes Crash Icy Watson Crash P(I)=0.7 P(H| I) I0.8 -I-I0.1 P(W| I) I0.8 -I-I0.1 P(W) = P(W| I) P(I) + P(W| - I) P( - I) = 0.8*0.7 + 0.1*0.3 = 0.56 + 0.03 = 0.59

34. 34 Probability of Icy given Watson Holmes Crash Icy Watson Crash P(I)=0.7 P(H| I) I0.8 -I-I0.1 P(W| I) I0.8 -I-I0.1 P(I | W) =

35. 35 Probability of Icy given Watson Holmes Crash Icy Watson Crash P(I)=0.7 P(H| I) I0.8 -I-I0.1 P(W| I) I0.8 -I-I0.1 P(I | W) = P(W | I) P(I) / P(W) = 0.8*0.7 / 0.59 = 0.95 We started with P(I) = 0.7; knowing that Watson crashed raised the probability to 0.95

36. 36 Probability of Holmes given Watson Holmes Crash Icy Watson Crash P(I)=0.7 P(H| I) I0.8 -I-I0.1 P(W| I) I0.8 -I-I0.1 P(H|W) =

37. 37 Probability of Holmes given Watson Holmes Crash Icy Watson Crash P(I)=0.7 P(H| I) I0.8 -I-I0.1 P(W| I) I0.8 -I-I0.1 P(H|W) = P(H, I | W) + P(H, -I | W) = P(H|W,I)P(I|W) + P(H|W, - I) P( - I| W) = P(H|I)P(I|W) + P(H| - I) P( - I| W) = 0.8*0.95 + 0.1*0.05 = 0.765 We started with P(H) = 0.59; knowing that Watson crashed raised the probability to 0.765

38. 38 Prob of Holmes given Icy and Watson Holmes Crash Icy Watson Crash P(I)=0.7 P(H| I) I0.8 ~I0.1 P(W| I) I0.8 ~I0.1 P(H|W, ~I I) = P(H ~I ) = 0.1 H and W are independent given I, so H and W are conditionally independent given I

39. Lecture 14 39 Example Topology of network encodes conditional independence assertions: Weather is independent of the other variables Toothache and Catch are conditionally independent given Cavity

40. Lecture 14 40 Example 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: Burglary, 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

41. Lecture 14 41 Example contd.

42. Lecture 14 42 Compactness A CPT for Boolean X i with k Boolean parents has 2 k rows for the combinations of parent values Each row requires one number p for X i = true (the number for X i = false is just 1-p) 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)

43. Lecture 14 43 Semantics The full joint distribution is defined as the product of the local conditional distributions: P (X 1, …,X n ) = π i = 1 P (X i | Parents(X i )) e.g., P(j  m  a  b  e) = P (j | a) P (m | a) P (a | b, e) P (b) P (e) n

44. Lecture 14 44 Constructing Bayesian networks 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: P (X 1, …,X n ) = π i =1 P (X i | X 1, …, X i-1 ) (chain rule) = π i =1 P (X i | Parents(X i )) (by construction) n n

45. Lecture 14 45 Suppose we choose the ordering M, J, A, B, E P(J | M) = P(J)? Example

46. Lecture 14 46 Suppose we choose the ordering M, J, A, B, E P(J | M) = P(J)? No P(A | J, M) = P(A | J)? P(A | J, M) = P(A)? Example

47. Lecture 14 47 Suppose we choose the ordering M, J, A, B, E 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)? Example

48. Lecture 14 48 Suppose we choose the ordering M, J, A, B, E 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)? Example

49. Lecture 14 49 Suppose we choose the ordering M, J, A, B, E 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, M) = P(E | A, B)? Yes Example

50. Lecture 14 50 Example contd. Deciding conditional independence is hard in noncausal directions (Causal models and conditional independence seem hardwired for humans!) Network is less compact: 1 + 2 + 4 + 2 + 4 = 13 numbers needed

51. 51 Excercises P(J, M, A, B, E ) = ? P(M, A, B) = ? P(-M, A, B) = ? P(A, B) = ? P(M, B) = ? P(A | J) = ?

52. Lecture 14 52 Summary Bayesian networks provide a natural representation for (causally induced) conditional independence Topology + CPTs = compact representation of joint distribution Generally easy for domain experts to construct