1. Bayes Nets and Probabilities
Oliver Schulte
Machine Learning 726
If you use “insert slide number” under “Footer”, that text box only displays the slide number, not the total number of slides. So I use a new textbox for the slide number in the master.

2. Bayes Nets: General Points
Represent domain knowledge.
Allow for uncertainty.
Complete representation of probabilistic knowledge.
Represent causal relations.
Fast answers to types of queries:
Probabilistic: What is the probability that a patient has strep throat given that they have fever?
Relevance: Is fever relevant to having strep throat?

3. Bayes Net Links
Judea Pearl's Turing Award
See UBC’s AISpace

4. Probability Reasoning (With Bayes Nets)

5. Random Variables
A random variable has a probability associated with each of its values.
A basic statement assigns a value to a random variable.
Variable
Value
Probability
Weather
Sunny
0.7
Rainy
0.2
Cloudy
0.08
Snow
0.02
Cavity
True
False
0.8
Fill in Cavity value later

6. Probability for Sentences
A sentence or query is formed by using “and”, “or”, “not” recursively with basic statements.
Sentences also have probabilities assigned to them.
Sentence
Probability
P(Cavity = false AND Toothache = false)
0.72
P(Cavity = true OR Toothache = false)
0.08
Use sentences that are used later.
2nd sentence should have higher probability.
Exercise: Prove that if A entails B, then P(A) >= P(B).

7. Probability Notation
Often probability theorists write A,B instead of A  B (like Prolog).
If the intended random variables are known, they are often not mentioned.
Shorthand
Full Notation
P(Cavity = false,Toothache = false)
P(Cavity = false Toothache = false)
P(false, false)

8. Axioms of probability 0 ≤ P(A) ≤ 1 P(true) = 1 and P(false) = 0
P(A  B) = P(A) + P(B) - P(A  B)
P(A) = P(B) if A and B are logically equivalent.
Logical equivalence connects probability and logic.
“True” is a constant sentence that is true in all possible worlds.
Formulas considered as sets of complet

9. Rule 1: Logical Equivalence
P(NOT (NOT Cavity))
P(Cavity)
0.2
P(NOT (Cavity AND Toothache))
P(Cavity = F OR Toothache = F)
0.88
Spot the pattern
P(NOT (Cavity OR Toothache)
P(Cavity = F AND Toothache = F)
0.72

10. The Logical Equivalence Pattern
P(NOT (NOT Cavity)) =
P(Cavity)
0.2
Rule 1: Logically equivalent expressions have the same probability.
P(NOT (Cavity AND Toothache)) =
P(Cavity = F OR Toothache = F)
0.88
This shows how logical reasoning is an important part of probabilistic reasoning.
Often easier to determine probability after transforming expression into a logically equivalent form.
P(NOT (Cavity OR Toothache) =
P(Cavity = F AND Toothache = F)
0.72

11. Rule 2: Marginalization
P(Cavity, Toothache)
P(Cavity, Toothache = F)
P(Cavity)
0.12
0.08
0.2
P(Cavity = F, Toothache)
P(Cavity = F, Toothache = F)
P(Cavity = F)
0.08
0.72
0.8
Assignment: fill in marginal over 2 variables.
P(Cavity, Toothache)
P(Cavity = F, Toothache)
P(Toothache)
0.12
0.08
0.2

12. The Marginalization Pattern
P(Cavity, Toothache) +
P(Cavity, Toothache = F) =
P(Cavity)
0.12
0.08
0.2
P(Cavity = F, Toothache) +
P(Cavity = F, Toothache = F) =
P(Cavity = F)
0.08
0.72
0.8
Assignment: fill in marginal over 2 variables.
P(Cavity, Toothache) +
P(Cavity = F, Toothache) =
P(Toothache)
0.12
0.08
0.2

13. Prove the Pattern: Marginalization
Theorem. P(A) = P(A,B) + P(A, not B)
Proof.
A is logically equivalent to [A and B)  (A and not B)].
P(A) = P([A and B)  (A and not B)]) = P(A and B) + P(A and not B) – P([A and B)  (A and not B)]). Disjunction Rule.
[A and B)  (A and not B)] is logically equivalent to false, so P([A and B)  (A and not B)]) =0.
So 2. implies P(A) = P(A and B) + P(A and not B).

14. Completeness of Bayes Nets
A probabilistic query system is complete if it can compute a probability for every sentence.
Proposition: A Bayes net is complete. Proof has two steps.
Any system that encodes the joint distribution is complete.
A Bayes net encodes the joint distribution.

15. The Joint Distribution

16. Assigning Probabilities to Sentences
A complete assignment is a conjunctive sentence that assigns a value to each random variable.
The joint probability distribution specifies a probability for each complete assignment.
A joint distribution determines an probability for every sentence.
How? Spot the pattern.
Give examples: P(toothache, not cavity, toothache or cavity, as needed before).

17. Probabilities for Sentences: Spot the Pattern
Probability
P(Cavity = false AND Toothache = false)
0.72
P(Cavity = true OR Toothache = false)
0.08
P(Toothache = false)
0.8

18. Inference by enumeration

19. Inference by enumeration
Marginalization: For any sentence A, sum the joint probabilities for the complete assignments where A is true.
P(toothache) = = 0.2.

20. Completeness Proof for Joint Distribution
Theorem [from propositional logic] Every sentence is logically equivalent to a disjunction of the form A1 or A2 or ... or Ak where the Ai are complete assignments.
All of the Ai are mutually exclusive (joint probability 0). Why?
So if S is equivalent to A1 or A2 or ... or Ak, then P(S) = Σi P(Ai) where each Ai is given by the joint distribution.

21. Bayes Nets and The Joint Distribution

22. Example: Complete Bayesian Network
Example Horn Clauses: Alarm -> JohnClass p:0.9

23. The Story You have a new burglar alarm installed at home.
It’s reliable at detecting burglary but also responds to earthquakes.
You have two neighbors that promise to call you at work when they hear the alarm.
John always calls when he hears the alarm, but sometimes confuses alarm with telephone ringing.
Mary listens to loud music and sometimes misses the alarm.

24. Computing The Joint Distribution
A Bayes net provides a compact factored representation of a joint distribution.
In words, the joint probability is computed as follows.
For each node Xi:
Find the assigned value xi.
Find the values y1,..,yk assigned to the parents of Xi.
Look up the conditional probability P(xi|y1,..,yk) in the Bayes net.
Multiply together these conditional probabilities.

25. Product Formula Example: Burglary
Query: What is the joint probability that all variables are true?
P(M, J, A, E, B) = P(M|A) p(J|A) p(A|E,B)P(E)P(B) = .7 x .9 x .95 x .002 x .001

26. Compactness of Bayesian Networks
Consider n binary variables
Unconstrained joint distribution requires O(2n) probabilities
If we have a Bayesian network, with a maximum of k parents for any node, then we need O(n 2k) probabilities
Example
Full unconstrained joint distribution
n = 30: need 230 probabilities for full joint distribution
Bayesian network
n = 30, k = 4: need 480 probabilities
Recall toothache example where we directly represented the joint distribution.

27. Summary: Why are Bayes nets useful?
- Graph structure supports
- Modular representation of knowledge
- Local, distributed algorithms for inference and learning
- Intuitive (possibly causal) interpretation
- Factored representation may have exponentially fewer parameters than full joint P(X1,…,Xn) =>
lower sample complexity (less data for learning)
lower time complexity (less time for inference)

28. Is it Magic?
How can the Bayes net reduce parameters? By exploiting conditional independencies.
Why does the product formula work?
The Bayes net topological or graphical semantics.
The graph by itself entails conditional independencies.
The Chain Rule.

29. Conditional Probabilities and Independence

30. Conditional Probabilities: Intro
Given (A) that a die comes up with an odd number, what is the probability that (B) the number is
a 2
a 3
Answer: the number of cases that satisfy both A and B, out of the number of cases that satisfy A.
Examples:
#faces with (odd and 2)/#faces with odd = 0 / 3 = 0.
#faces with (odd and 3)/#faces with odd = 1 / 3.

31. Conditional Probs ctd.
Suppose that 50 students are taking 310 and 30 are women. Given (A) that a student is taking 310, what is the probability that (B) they are a woman?
Answer: #students who take 310 and are a woman/ #students in = 30/50 = 3/5.
Notation: P(A|B)

32. Conditional Ratios: Spot the Pattern
P(die comes up with odd number)
P(die comes up with 3)
P(3|odd number)
1/2
1/6
1/3
P(Student takes 310)
P(Student takes 310 and is woman)
P(Student is woman|Student takes 310)
=50/15,000
30/15,000
3/5

33. Conditional Probs: The Ratio Pattern
Spot the Pattern
P(die comes up with odd number) /
P(die comes up with 3) =
P(3|odd number)
1/2
1/6
1/3
P(Student takes 310) /
P(Student takes 310 and is woman) =
P(Student is woman|Student takes 310)
=50/15,000
30/15,000
3/5
Exercise: prove that conditioning leads to a well-defined normalized probability measure.
P(A|B) = P(A and B)/ P(B) Important!

34. Conditional Probabilities: Motivation
Much knowledge can be represented as implications B1,..,Bk =>A.
Conditional probabilities are a probabilistic version of reasoning about what follows from conditions.
Cognitive Science: Our minds store implicational knowledge.

35. The Product Rule: Spot the Pattern
P(Cavity)
P(Toothache|Cavity)
P(Cavity,Toothache)
0.12
0.08
0.2
P(Toothache)
P(Cavity| Toothache)
P(Cavity,Toothache)
0.08
0.72
0.8
P(Cavity =F)
P(Toothache|Cavity = F)
P(Toothache,Cavity =F)
0.12
0.08
0.2

36. The Product Rule Pattern
P(Cavity) x
P(Toothache|Cavity) =
P(Cavity,Toothache)
0.12
0.08
0.2
P(Toothache) x
P(Cavity| Toothache) =
P(Cavity,Toothache)
0.08
0.72
0.8
Moral: Joint proabilities can be inferred from conditional probabilities.
P(Cavity =F) x
P(Toothache|Cavity = F) =
P(Toothache,Cavity =F)
0.12
0.08
0.2

37. Independence A and B are independent iff
P(A|B) = P(A) or P(B|A) = P(B) or P(A, B) = P(A) P(B)
Suppose that Weather is independent of the Cavity Scenario. Then the joint distribution decomposes:
P(Toothache, Catch, Cavity, Weather)
= P(Toothache, Catch, Cavity) P(Weather)
Absolute independence powerful but rare
Dentistry is a large field with hundreds of variables, none of which are independent. What to do?

38. Exercise
Prove that the three definitions of independence are equivalent (assuming all positive probabilities).
A and B are independent iff
P(A|B) = P(A)
or P(B|A) = P(B)
or P(A, B) = P(A) P(B)

39. Conditional independence
If I have a cavity, the probability that the probe catches in it doesn't depend on whether I have a toothache:
(1) P(catch | toothache, cavity) = P(catch | cavity)
The same independence holds if I haven't got a cavity: (2) P(catch | toothache,cavity) = P(catch | cavity)
Catch is conditionally independent of Toothache given Cavity: P(Catch | Toothache,Cavity) = P(Catch | Cavity)
The equivalences for independence also holds for conditional independence, e.g.:
P(Toothache | Catch, Cavity) = P(Toothache | Cavity)
P(Toothache, Catch | Cavity) = P(Toothache | Cavity) P(Catch | Cavity)
Conditional independence is our most basic and robust form of knowledge about uncertain environments.

40. Bayes Nets Graphical Semantics

41. Common Causes: Spot the Pattern
Cavity
Catch
toothache
Catch is independent of toothache given Cavity.
In UBC “Simple Diagnostic Example:
Fever is independent of Coughing Given Influenza.
Coughing is independent of Fever given Bronchitis.

42. Burglary Example
JohnCalls, MaryCalls are conditionally independent given Alarm.

43. Spot the Pattern: Chain Scenario
MaryCalls is independent of Burglary given Alarm.
JohnCalls is independent of Earthquake given Alarm.
This is typical for sequential data.
In UBC “Simple Diagnostic Example:
Wheezing is independent of Influenza Given Bronchitis.
Coughing is independent of Smoke given Bronchitis.
Influenze is independent of Smokes. [check assignment]

44. The Markov Condition
A Bayes net is constructed so that: each variable is conditionally independent of its nondescendants given its parents.
The graph alone (without specified probabilities) entails conditional independencies.
Causal Interpretation: Each parent is a direct cause.
This can always be achieved by letting a node have enough parents. See text for details on how to construct a Bayesian network.
2. Helps with the retrieve relevant problem: from the graph we can tell that certain information is ignorable.
Can work on this using “spot the pattern”.

45. Derivation of the Product Formula

46. The Chain Rule We can always write
P(a, b, c, … z) = P(a | b, c, …. z) P(b, c, … z) (Product Rule)
Repeatedly applying this idea, we obtain
P(a, b, c, … z) = P(a | b, c, …. z) P(b | c,.. z) P(c| .. z)..P(z)
Order the variables such that children come before parents.
Then given its parents, each node is independent of its other ancestors by the topological independence.
P(a,b,c, … z) = Πx. P(x|parents)

47. Example in Burglary Network
P(M, J,A,E,B) = P(M| J,A,E,B) p(J,A,E,B)= P(M|A) p(J,A,E,B)
= P(M|A) p(J|A,E,B) p(A,E,B) = P(M|A) p(J|A) p(A,E,B)
= P(M|A) p(J|A) p(A|E,B) P(E,B)
= P(M|A) p(J|A) p(A|E,B) P(E)P(B)
Colours show applications of the Bayes net topological independence.

48. Explaining Away

49. Common Effects: Spot the Pattern
Influenza and Smokes are independent.
Given Bronchitis, they become dependent.
Influenza
Smokes
Bronchitis
Battery Age
Charging System OK
Battery Age and Charging System are independent.
Given Battery Voltage, they become dependent.
From UBC “Simple diagnostic problem”
Does this make sense?
Battery Voltage

50. Conditioning on Children
Independent Causes:
A and B are independent.
“Explaining away” effect:
Given C, observing A makes B less likely.
E.g. Bronchitis in UBC “Simple Diagnostic Problem”.
A and B are (marginally) independent, become dependent once C is known. A B C
Another example: Wumpus on one square explains stench.
Characterizes Bayes nets.

51. D-separation
A, B, and C are non-intersecting subsets of nodes in a directed graph.
A path from A to B is blocked if it contains a node such that either
the arrows on the path meet either head-to-tail or tail-to-tail at the node, and the node is in the set C, or
the arrows meet head-to-head at the node, and neither the node, nor any of its descendants, are in the set C.
If all paths from A to B are blocked, A is said to be d-separated from B by C.
If A is d-separated from B by C, the joint distribution over all variables in the graph satisfies

52. D-separation: Example

53. Mathematical Analysis
Theorem: If A, B have no common ancestors and neither is a descendant of the other, then they are independent of each other.
Proof for our example:
P(a,b) = Σc P(a,b,c) = Σc P(a) P(b) P(c|a,b)
Σc P(a) P(b) P(c|a,b) = P(a) P(b) Σc P(c|a,b) = P(a) P(b) A B C
Not quite a proof because we A,B may have parents. But illustrates general idea.
First step follows from marginilization. Second step follows from product formula.
3rd step follows since P(a), P(b) do not depend on c. Last step follows P(c|a,b) adds up to 1 over possible c values.

54. Bayes’ Theorem

55. Abductive Reasoning Burglary Alarm Cavity Toothache
Implications are often causal, from cause to effect.
Many important queries are diagnostic, from effect to cause.
Burglary
Alarm
Cavity
But not impossible in logic: see nonmonotonic reasoning.
Black: causal direction.
Red: direction of inference.
I mean that wumpus and stench are on adjacent squares.
Toothache

56. Bayes’ Theorem: Another Example
A doctor knows the following.
The disease meningitis causes the patient to have a stiff neck 50% of the time.
The prior probability that someone has meningitis is 1/50,000.
The prior that someone has a stiff neck is 1/20.
Question: knowing that a person has a stiff neck what is the probability that they have meningitis?

57. Spot the Pattern: Diagnosis
P(Cavity)
P(Toothache|Cavity)
P(Toothache)
P(Cavity|Toothache)
0.2
0.6
P(Wumpus)
P(Stench|Wumpus)
P(Stench)
P(Wumpus|Stench)
0.2
0.6
How does the last number depend on the first three?
P(Meningitis)
P(Stiff Neck| Meningitis)
P(Stiff Neck)
P(Meningitis|Stiff Neck)
1/50,000
1/2
1/20
0.6

58. Spot the Pattern: Diagnosis
P(Cavity) x
P(Toothache|Cavity) /
P(Toothache) =
P(Cavity|Toothache)
0.2
0.6
P(Wumpus) x
P(Stench|Wumpus) /
P(Stench) =
P(Wumpus|Stench)
0.2
0.6
P(Meningitis) x
P(Stiff Neck| Meningitis) /
P(Stiff Neck) =
P(Meningitis|Stiff Neck)
1/50,000
1/2
1/20
1/5,000

59. Explain the Pattern: Bayes’ Theorem
Exercise: Prove Bayes’ Theorem
P(A | B) = P(B | A) P(A) / P(B).

60. On Bayes’ Theorem
P(a | b) = P(b | a) P(a) / P(b).
P(Cause|Effect) = P(Effect|Cause) P(Cause) / P(Effect).
The better the cause explains the effect, the more likely it is.
The more plausible the cause is, the more likely it is.
The more surprising the evidence (the lower its prior probability), the greater its impact.
Likelihood: how well does the cause explain the effect?
Prior: how plausible is the explanation before any evidence?
Evidence Term/Normalization Constant: how surprising is the evidence?