1. Discrete Random Events n Thurs. March 11, 1999 – Review with Examples – Discrete Outcome Sample Spaces – Influence Diagrams & Probability Trees – Moments (e.g., expectation, variance, kth central) – Probability Functions (Marginal, Joint, Conditional) n Next week: Influence Diagrams (Tues.) and Dr. Anil Varma, Guest Speaker from GE’s AI/Expert Systems Group (Thurs.)

2. Example: Venn Diagram and Independence n We know from the Venn Diagram – Pr(A|B) = 1 – Pr(A) – Pr(B) n Are A and B independent? Is Pr(AB)=Pr(A)Pr(B)? B A AB

3. Disk Drive Failure Example c (C, C’)x (X, X’) Pr(C) =Pr(CI) = Pr(CX) + Pr(CX’) = 0.6 + 0.1 = 0.7 Pr(X) = Pr (XI) =Pr(XC) + Pr(XC’) 0.6 + 0.2 = 0.8 Pr(C|X) =Pr(CX)/Pr(X) = 0.6/0.8 = 0.75

4. Outcome Sample Spaces n It is useful in performing probability manipulations to define the random events "E i " as unions of elementary mutually exclusive and collectively exhaustive events. n Outcome sample spaces are analogous to Venn diagrams where the universe I, is made up of these elementary mutually exclusive and collectively exhaustive events. n The probability of any event E i is then the sum of the probabilities (due to the mutual exclusivity) of the elementary events which comprise E i.

5. Talus Bone Example n Assume that we are to roll two "fair" talus bones; one is the heel bone of a sheep, the other is a heelbone of a deer. Each one has four sides, and if it is "fair" the likelihood of any one side coming up on a roll is equally likely. n Let us further assume that each side of the talus bone has dots on it one dot, two dots, three dots and four dots (similar to a conventional die but with four sides instead of six).

6. Talus Bone Events n Assume both the sheep and deer talus bones are thrown randomly on the ground. Define the events: – Si = the number of dots on the up side of the sheep talus bone. – Dj = the number of dots on the up side of the deer talus bone. n Assume the throw of the sheep bone is independent of the throw of the deer bone, – Pr(SiDj ) = Pr(Si )Pr(Dj )

7. Sample Space for Joint Events n The joint events D i S j (i,j = 1, 2, 3, 4) represent the 16 possible outcomes independent rolls for each talus bone. n The probability of rolling any side is equal to 1/4 for each bone. n Thus Pr(D i S j ) = Pr(D i ) Pr(S j ) = (1/4)(1/4) = 1/16 Deer Talus Sheep Talus

8. Example: Find probability of E4 = the event that the sum of the two numbers is four. n Because the events are mutually exclusive, the probability of event E 4 is 1/16 times the number of ways a sum of four can occur. n There are three events that sum to four and thus: Pr(E 4 ) = 3(1/16) = 3/16 Deer Talus Sheep Talus

9. Example: F<10 = the event that the product of the two numbers is less than ten n Because the events are mutually exclusive, the probability of event F<10 is 1/16 times the number of ways a product less than ten can occur. n There are thirteen events with a product less than ten and thus: n Pr( F<10 ) = 13(1/16) = 13/16 Deer Talus Sheep Talus

10. What if the Talus Bones are Biased? n Suppose the probability of getting four dots is higher than getting three and so on as defined by the probability distribution: 2i-1 Pr(S i ) = Pr(D i ) =  16 Probability Mass Function on S i and D i

11. Biased Talus Bones - Probability Mass Function? n Because these events are mutually exclusive and collectively exhaustive, the sum of the probabilities of all events S i or D i equal one. Consider the sheep talus: n ∑ i=1,4 Pr(S i ) = 1 = 1 + 3 + 5 + 7 16 Probability Mass Function on S i and D i

12. Biased Talus Bones - Joint Probabilities Probability Mass Function on S i and D i Deer Talus Sheep Talus Joint Probabilities, Pr(S i, D i ) x 256

13. Conditional Probability Expansions of Talus Bone Example n How would one represent events E 4 and F <10 as a conditional probability expansion? There are many possible and consistent representations. n If you find it natural to think of the sum of the outcomes of the tosses to be influenced by the outcome of each individual toss, state E would be expressed mathematically by the following expansion: Pr(E ) = Pr(Sum(D i and S j ) ) =Pr( Sum(D i and S j ) | D i,S j ) Pr( D i | S j ) Pr(S j ) n Because D i and S j are assumed to be independent, this reduces to: Pr(E) =Pr(Sum(D i and S j )) =Pr( Sum(D i and S j ) | D i,S j ) Pr( D i ) Pr(S j )

14. Influence Diagrams & Talus Bone Example: E 4 n Because D i and S j are assumed to be independent, this reduces to: n Pr(E) =Pr(Sum(D i and S j )) =Pr( Sum(D i and S j ) | D i,S j ) Pr( D i ) Pr(S j )

15. Influence Diagrams & F <10 Example: n Pr(F) = Pr(Product(D i and S j ) ) =Pr( Product(D i and S j ) | D i,S j ) Pr( D i | S j ) Pr(S i j) n Because D i and S j are assumed to be independent, this reduces to: n Pr( Product(D i and S j ) | D i,S j ) Pr( D i ) Pr(S j )

16. How do we represent both events E and F in the same influence diagram? n Both diagrams show D i and S j as being independent, which would remain unchanged for the combined diagram. But should there be an arc between E and F? n The question to be answered is: given we know the outcome of D i and S j, would knowing the product give us any new information about the sum and vice versa. The answer is NO, and thus there is no arc between E and F. E and F are said to be conditionally independent of each other given D i and S j.

17. E and F are said to be conditionally independent of each other given Di and Sj n Given we know the outcome of D i and S j, knowing the product give us any new information about the sum and vice versa. n The lack of an arc reveals this conditional independence graphically.

18. Are E and F independent if we do not conditional information on D and S? n Given we know the outcome of D i and S j, knowing the product give us any new information about the sum and vice versa. n The lack of an arc reveals this conditional independence graphically. E Sum F Product

19. Probability Tree Representation

20. Expectation of a Random Variable n E(x|H) = ∑ x i Pr(x i )  x i orE(x|H) =

21. Expectation of a Function of a Random Variable n Let g(x) be any single-valued function of the random variable x. n The expected value of g(x) is designated as E[g(x)] and defined as: E[g(x)] = ∑ g(x i )Pr(x=x i ) = n Is E[g(x)] = g[E(x)]?

22. Probability Mass Function, Pr(x) n The probability that the random variable x takes on the discrete value x o is defined by means of the probability mass function Pr(x) with the notation below: Pr(x) = Pr(x=x i ) for all x i Biased Talus Bone Example

23. Cumulative Probability Distribution, Pr(x≤y|y) Pr(x≤y|y) = ∑ x≤y Pr(x) = Pr(x≤y|y=y i ) for all y i Biased Talus Bone Example

24. Cumulative Probability is Summation (Integral) of the Mass Function Probability Mass Function Cumulative Probability Distribution

25. Complementary Probability Distribution, Pr(x>y|y) Pr(x>y|y) = 1 - Pr(x≤y|y) Biased Talus Bone Example

26. Joint Discrete Mass Functions Two Representations for Joint Probability Mass Function for Two Biased Talus Tosses

27. Bayesian Pair of Bolts n Suppose that a "pick and place" robot with two manipulators has been designed to pick two bolts of the same size from a bin and place each pair in separate containers on a belt for assembly. Suppose also that there are three different kinds of pairs of bolts that are used in the assembly of this product: Pair 1two bolts both made of alloy A Pair 2two bolts both made of alloy B Pair 3one of alloy A and the other of alloy B

28. Bayesian Pair of Bolts: Bayes Theorem n The alloys look identical and the only way to determine whether one is A or B requires performing a test. n Suppose that one of the pairs of bolts falls on the ground and the operator picks it up. The probability of getting any specific pair is 1/3. n Now suppose she randomly picks one of the bolts from the pair, performs a test on it and determines that it is made of alloy A. n What is the probability that the other bolt is of alloy A?

29. Bayesian Pair of Bolts: Define Events n A common erroneous answer to this question is 1/2. But this is incorrect! Define the following events: – P1: The event that pair 1 was picked. – P2: The event that pair 2 was picked. – P3: The event that pair 3 was picked. – A1: The event that the first member of the pair is of alloy A. – A2: The event that the second member of the pair is of alloy A. – B1: The event that the first member of the pair is of alloy B. – B2: The event that the second member of the pair is of alloy B. n What we want to know is Pr(A2|A1,S). We know that Pr(P1|S) = Pr(P2|S) = Pr(P3|S) = 1/3. Using Bayes' theorem and conditional probability find Pr(A2|A1,S).