Introduction to Stochastic Models GSLM 54100

1. Introduction to Stochastic ModelsGSLM 54100 1

2. Outline • conditional probability • conditional distribution • recursive relationships 2

3. Conditional Probability 3

4. Example 3.1.3 • a car behind one of the 3 doors, I, II, and III • door I chosen by a guest, getting whatever behind the door • door III opened by the host, empty behind   • the guest given an option to switch his choice • good for the guest to switch? 4

5. Example 3.1.3 • the correct answer dependent on the assumption of the problem • assumption 1: an empty door deliberately opened by the host • defining events • G = door IIIopened (given that door I is chosen) • Gc = door IIopened (given that door I is chosen) • A (Band C) = the car behind door I (IIand III) 5

6. Example 3.1.3 • P(G) • = P(G|A)P(A) + P(G|B)P(B) + P(G|C)P(C) • = 1/2 • . • . 6

7. Example 3.1.3 • assumption 2: one of the three doors randomly opened by the host, which incidentally was an empty door III • P(G|A) = P(G|B) = P(G|C) = 1/3 • P(G) = P(G|A)P(A) + P(G|B)P(B) + P(G|C)P(C) • G being independent of A, B, and C • . • . 7

8. Conditional Distributions 8

9. Conditional Distribution • X ~ {pn} and A is an event • 0  P(X = n|A)  1 • nP(X = n|A) = •  {P(X = n|A)} is a probability mass function, called the conditional distribution of X given A 9

10. Conditional Distribution • event A, discrete random variables X & Y • P(X = x|A) = the conditional p.m.f. of X given A • P(X = x|Y = y) = the conditional p.m.f. of X given Y = y 10

11. Conditional Distribution • event A, continuous random variables X & Y • fX|A(x|A) = the conditional density function of X given A • fX|Y(x|y) = the conditional density function of X given Y = y 11

12. Modified from Example 3.2 of Ross • X1 ~ Bin(5, 0.3), X2 ~ Bin(10, 0.3), independent • P(X1 = 2|X1+X2 = 8) = (X1| X1+X2=8) ~ hypergeometric 12

13. Preparation for Example 3.3 of Ross • X ~ Poisson(λ); Y ~ Poisson(); independent • distribution of X+Y i.e., X+Y ~ Poisson(+) 13

14. Example 3.3 of Ross • X ~ Poisson(λ1); Y ~ Poisson(λ2); independent • find (X|X + Y = n) (X|X + Y = n) ~ Bin(n, 1/(1+2)) 14

15. Conditional Distribution • X ~ {pn} and A is an event • 0  P(X = n|A)  1 • nP(X = n|A) = •  {P(X = n|A)} is a probability mass function, called the conditional distribution of X given A 15

16. Conditional Distribution • define Z = (X|A) • Z is a random variable • E(Z) and Var(Z) being well-defined • E(X|A), the conditional mean of X given A • Var(X|A), the conditional variance of X given A • event A can defined by a random variable, e.g., A = {Y = 3} 16

17. Ex #1 of WS #5 • Exercise 1. (Joint and conditional distributions) The joint distribution of X and Y is shown below, where pm,n = P(X = m, Y = n). • p1,1 = 0; p1,2 = 1/8; p1,3 = 1/8; • p2,1 = 1/4; p2,2 = 1/4; p2,3 = 0; • p3,1 = 1/8; p3,2 = 0; p3,3 = 1/8. • Find the (marginal) distribution of X. • Find the (marginal) distribution of Y. • Find the conditional distribution of (X|Y = 1), (X|Y = 2), and (X|Y = 3). • Find the conditional means E(X|Y = 1), E(X|Y = 2), and E(X|Y = 3). • Find the conditional variances V(X|Y = 1), V(X|Y = 2), and V(X|Y = 3). 17

18. Ex #1 of WS #5 • Exercise 1. (Joint and conditional distributions) The joint distribution of X and Y is shown below, where pm,n = P(X = m, Y = n). • p1,1 = 0; p1,2 = 1/8; p1,3 = 1/8; • p2,1 = 1/4; p2,2 = 1/4; p2,3 = 0; • p3,1 = 1/8; p3,2 = 0; p3,3 = 1/8. • distribution of X: p1 = 1/4, p2 = 1/2, p3 = 1/4 • distribution of Y: p1 = 3/8, p2 = 3/8, p3 = 1/4 18

19. Ex #1 of WS #5 • Exercise 1. (Joint and conditional distributions) The joint distribution of X and Y is shown below, where pm,n = P(X = m, Y = n). • p1,1 = 0; p1,2 = 1/8; p1,3 = 1/8; • p2,1 = 1/4; p2,2 = 1/4; p2,3 = 0; • p3,1 = 1/8; p3,2 = 0; p3,3 = 1/8. • conditional distribution of • (X|Y = 1): p(X=1|Y=1) = 0; p(X=2|Y=1) = 2/3; p(X=3|Y=1) = 1/3 • (X|Y = 2): p(X=1|Y=2) = 1/3; p(X=2|Y=2) = 2/3; p(X=3|Y=2) = 0 • (X|Y = 3): p(X=1|Y=3) = 1/2; p(X=2|Y=3) = 0; p(X=3|Y=3) = 1/2 19

20. Ex #1 of WS #5 • Exercise 1. (Joint and conditional distributions) The joint distribution of X and Y is shown below, where pm,n = P(X = m, Y = n). • p1,1 = 0; p1,2 = 1/8; p1,3 = 1/8; • p2,1 = 1/4; p2,2 = 1/4; p2,3 = 0; • p3,1 = 1/8; p3,2 = 0; p3,3 = 1/8. • (X|Y = 1) being a random variable with well-defined distribution • the conditional means being well-defined • E[(X|Y = 1)] = (2)(2/3)+(3)(1/3) = 7/3 • E[(X|Y = 2)] = 5/3 • E[(X|Y = 3)] = 2 20

21. Ex #1 of WS #5 • Exercise 1. (Joint and conditional distributions) The joint distribution of X and Y is shown below, where pm,n = P(X = m, Y = n). • p1,1 = 0; p1,2 = 1/8; p1,3 = 1/8; • p2,1 = 1/4; p2,2 = 1/4; p2,3 = 0; • p3,1 = 1/8; p3,2 = 0; p3,3 = 1/8. • (X|Y = 1) being a random variable with well-defined distribution • the conditional variances being well-defined • V(X|Y = 1) = E(X2|Y = 1) E2(X|Y = 1) = 2/9 • V(X|Y = 2) = 2/9 • V(X|Y = 3) =1 21

22. Ex #1 of WS #5 • note the mapping defined by the conditional means E[(X|Y = 1)] = 7/3, E[(X|Y = 2)] = 5/3, E[(X|Y = 3)] = 2 • at {1|Y(1) = 1}, the mapping gives 7/3 • at {2|Y(2) = 2}, the mapping gives 5/3 • at {3|Y(3) = 3}, the mapping gives 2 • the mapping E(X|Y), i.e., the conditional mean, defines a random variable • E[E(X|Y)] = (3/8)(7/3)+(3/8)(5/3)+(1/4)(2) = 2 • incidentally E(X) = 2 22

23. Ex #1 of WS #5 • note the mapping defined by the conditional means V[(X|Y = 1)] = 2/9, V[(X|Y = 2)] = 2/9, V[(X|Y = 3)] = 1 • at {1|Y(1) = 1}, the mapping gives 2/9 • at {2|Y(2) = 2}, the mapping gives 2/9 • at {3|Y(3) = 3}, the mapping gives 1 • the mapping V(X|Y), i.e., the conditional variance, defines a random variable • E[V(X|Y)] = (3/8)(2/9)+(3/8)(2/9)+(1/4)(1) = 5/12 23

24. A former Mid-Term Question (Compared to Example 3.4 in Ross ) • three types of cakes, chocolate, mango, and strawberry in a bakery • each customer choosing chocolate, mango, and strawberry w.p. 1/2, 1/3, and 1/6, respectively, independent of everything else • profit from each piece of chocolate, mango, and strawberry cake ~ $3, $2, and $1, respectively • 4 cream cakes sold on a particular day • (a). Let Xc be the number of chocolate cream cakes sold on that day. Find the distribution of Xc. • (b). Find the expected total profit of the day from the 4 cream cakes. • (c). Given that no chocolate cream cake is sold on that day, find the variance of the total profit of the day. 24

25. A former Mid-Term Question (Compared to Example 3.4 in Ross ) • (a). Xc = the number of chocolate cream cakes sold on that day • Xc ~ Bin(4, 1/2) • (b). E(total profit of the day) = E(3Xc+ 2Xm + Xs) = 3E(Xc) + 2E(Xm) + E(Xs) = 6+(8/3)+(2/3) = 28/3. 25

26. A former Mid-Term Question (Compared to Example 3.4 in Ross ) • (c). Given that no chocolate cream cake is sold on that day, find the variance of the total profit of the day. • given Xc = 0, each cake is of mango of probability 2/3 and of strawberry of probability 1/3. • (Xm|Xc = 0) ~ Bin(4, 2/3) and (Xs|Xc = 0) ~ Bin(4, 1/3). • V(Xm|Xc = 0) = V(Xs|Xc = 0) = 8/9 • the total profit = (Y|Xc = 0) = 2(Xm|Xc = 0) + (Xs|Xc = 0) (Xm|Xc = 0) + (Xs|Xc = 0) = 4 • (Y|Xc = 0) = 4 + (Xm|Xc = 0)  • V(Y) = V(4+Xm|Xc = 0) = V(Xm|Xc = 0) = 8/9 26

27. Recursive Relationships 27

28. Two Innocent Equations • A and B : two events • P(A) = P(A|B)P(B) + P(A|Bc)P(Bc) •  generalization: jBj =  and BiBj =  for ij • P(A) = • recursive equations induced by these equations 28

29. Recursive Relationship • a special property in some random phenomena: changing back to oneself, or to something related • flipping a coin until getting the first head first flip = H THE END flipping a coin until getting the first head flipping a coin until getting the first head + first flip = T one flip 29

30. Recursive Relationship type 1 outcome simple type 1 problem . . . . random phenomenon type k+1 problem being related to the original random phenomenon type k outcome simple type k problem type k+1 outcome type 1 outcome simple type 1 analysis difficult type k+1 problem . . . . the problem may become easy if the type k+1 problem is related to the original random phenomenon random phenomenon type k outcome simple type k analysis type k+1 outcome 30

31. More Recursive Relationships type 1 outcome simple type 1 problem The two problems for random phenomena A and B may be solved easily if they evolve into each other. random phenomenon B random phenomenon A type 2 outcome difficult problem related to random phenomenon B difficult problem related to random phenomenon A type 1’ outcome simple type 1’ problem type 2’ outcome 31

32. About Recursive Relationships • more forms, possibly involving more than 2 random phenomena • identifying the relationships among random phenomena being an art, not necessarily science 32

33. Exercise 3.1.2 of Notes • n contractors bidding for m projects (n  m) • one project for each contractor • all projects being equally profitable • random independent bids by contractors • Ai = the project i, im is bid (by at least one contractor) • (a). Find   • (b). Find P( A1) • (c). Find • (d). Find P(A2| A1)  33

34. Exercise 3.1.2 of Notes random bids by n contractors on m projects (n  m), one project for each contractor • Ai = the project i, im is bid (by at least one contractor) • (a). • (b). P( A1) = • (c). • (d). to find P(A2| A1), note that   34

35. Examples of Ross in Chapter 3 • Examples 3.2, 3.3, 3.4, 3.5, 3.6, 3.7 35