1. Representations for KBS: Uncertainty & Decision Support
Knowledge Based Systems
Tuesday March 20, 2006
Howard Shrobe
Randall Davis

2. Outline The Other Problem with Mycin
Brief review of history of uncertainty in AI
Bayes Theorem
Some tractable Bayesian situations
Bayes Nets
Decision Theory and Rational Choice

3. The Other Problem with Mycin
In an earlier class we argued that Mycin used an extremely impoverished language for stating facts and rules (A-O-V triples)
Here we argue that it’s notion of uncertainty was broken:
In mycin the certainty factor for OR is Max
CF (OR A B) = (Max (Cf A) (Cf B))
Consider
Rule-1 IF A then C, certainty factor 1
Rule-2 If B then C, certainty factor 1
This is logically the same as:
If (Or A B) then C, certainty factor 1 I

4. More Problems If CF(A) = .8 and CF(B) = .3
Then CF (C ) = * (1 - .8) = = .86
CF (OR A B) = (Max .8 .3) = .8 and CF(C ) = .8
IF A -> B, A -> C, B ->D, C->D there will also be a mistake: (why?) B A D C

5. History of Uncertainty Representations
Probability tried and Rejected
too many numbers
Focus on Logical, qualitative
reasoning by cases
non-monotonic reasoning
Numerical Approaches retried
Certainty factors
Dempster-Schafer
Fuzzy
Bayes Networks

6. Understanding Bayes Theorem
Has it and
Tests for it
10 • .95 = 9.5
Positive: .95
Test?
Yes: 010
Has it and
Doesn’t Test for it
Has Cancer?
Doesn’t Have it
But Tests for it
990 • .05 = 49.5
No: 990
Test?
Doesn’t Have it and doesn’t Test for it
Number that test positive = = 59
If you test positive your probability of having
Cancer is 9.5 / 59 = 16.1 %

7. Reviewing Bayes Theorem
Symptom S U
Conditional Probability of S given D D S

8. Independence & Conditional Independence
P(A) • P(B) = P(A&B)
A varies the same within B as it does in the universe
Conditional Independence within C:
P(A|C) • P(B|C) = P(A&B|C)
When we restrict attention to C, A and B are independent

9. Examples D A B A’ C B A and B are Independent A’ and B are Dependent
A and B are conditionally
Dependent, given C
A’ and B are Conditionally
Independent, given C.

10. “IDIOT” BAYES Model Single Disease Conditionally Independent Symptoms:SK
Single Disease
Conditionally Independent Symptoms:
P(S1,S2|D) = P(S1|D) * P(S2|D)
N Symptoms means N probabilities
Without conditional independence need joint probabilities 2^N

11. Using Multiple Pieces of Evidence
P(H|E1,E2) =
P(E1,E2|H) · P(H)
P(E1,E2) H E2
If you assume "conditional independence" between the evidence then this takes on a nice multiplicative form.
Conditional Independence is the notion that the various pieces of evidence are statistically independent of one another, given that the hypothesis obtains, i.e. the hypothesis "separates" the different pieces of evidence:
P(E1,E2|H) = P(E1|H) · P(E2|H)
P(~E1,~E2|H) = P(~E1|H) · P(~E2|H)
Without conditional independence you need to build up a very large database of joint probabilities and joint conditional probabilities.

12. Sequential Bayesian Inference
Consider symptoms one by one
Prior Probabilities P(Di)
Observe Symptom Sj
Updates Priors using Bayes Rule:
Repeat for Other Symptoms using the resulting Posterior as the new Prior
If symptoms are conditionally independent, same as doing it all at once
Allows choice of what symptom to observe (test to perform) next in terms of cost/benefit.

13. Bipartite Graphs Multiple Symptoms, multiple diseases
Diseases are probabilistically independent
Symptoms are conditionally independent
Symptoms probabilities depend only the diseases causing them
Symptoms with multiple causes require joint probabilities P(S2|D1,D2,D3)
Information explosion D1 D2 D3 S1 S2 S3 S4

14. Noisy OR A useful element in the modeling vocabulary
Make the simplifying assumption that only 1 disease is present at a time:
Probability that all diseases cause the symptom is just the probability that at least 1 does
Therefore: Symptom is absent only if no disease caused it.
1 - P(S2|D1,D2,D3) = (1 - P(S2|D1))
* (1 - P(S2|D2))
* (1 - P(S2|D3))
P(S2|D1) = P(S2|D1), all other Di absent c
Use Causal Probabilities for the basic data
Reduces probability table size: if n diseases and k symptoms, from k2^n to nk

15. Polytrees What if diseases cause or influence each other?
Are there still well behaved versions?
Yes, Polytrees: At most one path between any two nodes
Don’t have to worry about “double-counting”
Efficient Sequential updating is still possible D1 D2 D3 S1 S5 S2 S4 S3

16. Bayes Nets B A E D C Directed Acyclic Graphs
Absence of link --> conditional independence
P(X1,...,Xn) = Product P(Xi|{parents (Xi)})
Specify joint probability tables over parents for each node
Probability A,B,C,D,E all present:
P(A,B,C,D,E) =
P(A) * P(B|A) * P(C|A) * P(D|B,C) * P(E|C)
Probability A,C,D present B,E absent:

17. Example Burglary Earthquake Alarm Radio Report Phone Call
P(Call|Alarm) t f
P(RadioReport|Earthquake) t f t .9
.01 t 1 f .1
.99 f 1
P(Alarm|B,E)
t,t
t,f
f,t
f,f t .8
.99 .6
.01 f .2
.01 .4
.99
16 vs. 32 probabilites

18. Computing with Partial InformationB A E D C
Probability that A present and E absent:
Graph separators (e.g. C) correspond to factorizations
General problem of finding separators is NP-hard

19. Odds Likelihood Formulation
Define Odds as
Define Likelihood as:
Divide complementary instances of Bayes Rule:
Bayes Rule is Then:
In Logarithmic Form: Log Odds = Log Odd + Log Likelihood

20. Certainty Factors Rules A C D B Parallel Combination If A, then C (x)z C D B y
Parallel
Combination
Rules
If A, then C (x)
If B, then C (x)
If C, then D(x)
x+y -xy x,y>0
x+y +xy x,y<0
otherwise
CF(C) =
Series
Combination
CF(C) = z • max(0, CF(C))

21. Issues with Certainty Factors
Results obtained depend on order in which evidence is considered in some cases
Reasoning is often fairly insensitive to them. 20% variations yield no change in MYCIN
What do they mean? (in some cases the answer is:)
Conditional Probability
Likelihood
Certainty Factor

22. Decision Making
So far, what we’ve considered is how to use evidence to evaluate a situation.
In many cases, this is only the first part of the problem
What we want to do is to take actions to improve the situation
But which action should we take?
The one which is most likely to leave us in the best condition
Decision Analysis helps us calculate which action that is

23. A Decision Making Problem
There are two types of Urn U1 and U2 (80% are U1)
U1 contains 4 Red balls and 6 Black balls
U2 contains nine Red balls and one Black ball
An urn is selected at random and you are to guess which type it is.
You have several courses of action:
Refuse to play No Payoff no cost
Guess it is of type 1 $40 Payoff if right, $20 penalty if wrong
Guess it is of type 2 $100 Payoff if right, $5 penalty if wrong
Sample a ball $8 payment for the right to sample

24. Decision Flow Diagrams
Decision Fork
Chance Fork
(e1,R, a1)
$40.00 a1
(e1,R)
-$20.00
$0.00
Refuse to Play
-$5.00 a2
-$8.00 R
$100.00
Make an
Observation
$40.00 B a1
-$20.00 No
Observation
-$5.00
(e1,B) a2
(e1,R, a1)
$100.00
$40.00 a1
-$20.00
-$5.00 a2
$100.00

25. Expected Monetary Value
Suppose there are several possible outcomes
Each has a monetary payoff or penalty
Each has a probability
The Expected Monetary Value is the sum of the products of the monetary payoffs times their corresponding probabilities. .8
$40 .2
-$20
EMV = .8 · $ · -$20 = $32 + (-$4) = $28
EMV is a normative notion of what a person who has no other biases (risk aversion, e.g.) should be willing to accept in exchange for the situation facing him. In the picture above, you should be indifferent to the choice of taking $28 or playing the game.
Most people have some extra biases and these can be incorporated in the form of a utility function applied to the calculated value.
A rational person should choose that course of action which has the highest EMV.

26. Averaging Out and Folding Back
EMV of Decision Node is Max over all branches
EMV of Chance Node is Probability Weighted Sum over all branches .8 .2
$40.00
-$20.00
-$5.00
$100.00
Action
State A1 A A3 Probability U U
EMV
$32.00
$28.00
-$4.00
$28.00
-$4.00
$16.00
$20.00

27. The Effect of Observation
Bayes theorem is used to calculate probabilities at chance nodes following decision nodes that provide relevant evidence. R B a1 a2
$40.00
-$20.00
-$5.00
$100.00 U1 U2
Action
State A1 A A3 Probability U U
EMV
P(R) = P(R|U1) • P(U1) + P(R|U2) • P(U2)
P(U1|R) = P(R|U1) • P(U1) / P(R)

28. Calculating the Updated Probabilities
Initial Probabilities
P(Outcome|State) State
Outcome U1 U2
Red .4 .9
Black .6 .1
.8 .2
Joint & Marginal Probabilities
P(Outcome & State) State Marginal Probability
Outcome U1 U2 of Outcome
Red .8 • .4 = • .9 =
Black .8 • .6 = • .1 =
Updated Probabilities
P(State |Outcome) State
Outcome U1 U2
Red
Black

29. Illustrating Evaluation.5 R B a1 a2
$100.00
$40.00
-$20.00
-$5.00
.64
.36
.96
.04
-$8.00
+25.60
+18.40
-7.20
-3.20
+32.80
16.40
+36.00
27.20
35.20
38.40
18.80
37.60
-.80
U1 U2 R B
-4.04
-0.04
4.00

30. Final Value of Decision Flow Diagram
(e1,R, a1)
$40.00 a1
(e1,R)
-$20.00
$0.00
Refuse to Play
-$5.00 a2
$28.00
27.20
-$8.00 R
$100.00
Make an
Observation
$40.00 B a1
-$20.00 No
Observation
$28.00
-$5.00
(e1,B) a2
(e1,R, a1)
$100.00
$40.00 a1
-$20.00
-$5.00 a2
$100.00

31. Maximum Entropy Suppose there are several tests you can make..1 .2 .5 .2
Several Competing Hypotheses
Each with a Probability rating.
Suppose there are several tests you can make.
Each test can change the probability of some (or all) of the hypotheses (using Bayes Theorem).
Each outcome of the test has a probability.
We’re only interested in gathering information at this point
Which test should you make?
Entropy = Sum -2 · P(i) · Log P(i) is a standard measure of Information.
For each outcome of a test calculate the change in entropy.
Weigh this by the probability of that outcome.
Sum these to get an expected change of entropy for the test.

32. Maximum Entropy (2)
Chose that test which has the greatest expected change in entropy.
This is equivalent to choosing the test which is most likely to provide the most information.
Tests have different costs (sometimes quite drastic ones like life and death).
Normalize the benefits by the costs and then make choice.