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# Introduction to Stochastic Models GSLM 54100

Introduction to Stochastic Models GSLM 54100. Outline. independence of random variables variance and covariance two useful ideas examples conditional distribution. Independent Random Variables.

## Introduction to Stochastic Models GSLM 54100

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1. Outline • independence of random variables • variance and covariance • two useful ideas • examples • conditional distribution 2

2. Independent Random Variables • two random variables X and Y being independent  all events generated by X and Y being independent • discrete X and Y P(X = x, Y = y) = P(X = x) P(Y = y) for all x, y • continuous X and Y fX ,Y(x, y) = fX(x) fY(y) for all x, y • any X and Y FX ,Y(x, y) = FX(x) FY(y) for all x, y 3

3. Proposition 2.3 • E[g(X)h(Y)] = E[g(X)]E[h(Y)] for independent X, Y • different meanings of E() • Ex #7 of WS #5 (Functions of independent random variables) • X and Y be independent and identically distributed (i.i.d.) random variables equally likely to be 1, 2, and 3 • Z = XY • E(X) = ? E(Y) = ? distribution of Z? E(Z) = E(X)E(Y)?  E(Z) as the mean of a function of X and Y, or as the mean of a random variable Z 4

4. Proposition 2.3 • E[g(X)h(Y)] = E[g(X)]E[h(Y)] for independent X, Y • different meanings of E() • E[g(X)] = • E[h(Y)] = • E[g(X)h(Y)] = x and y are dummy variables 5

5. Variance and Covariance (Ross, pp 52-53) • Cov(X, Y) = E(XY)  E(X)E(Y) • Cov(X, X)= Var(X) • Cov(X, Y)= Cov(Y, X) • Cov(cX, Y)= cCov(X, Y) • Cov(X, Y + Z)= Cov(X, Y)+ Cov(X, Z) • Cov(iXi, jYj) = i j Cov(Xi, Yj) • . 6

6. Two Useful Ideas • for X = X1 + … + Xn, E(X) = E(X1) + … + E(Xn), no matter whether Xi are independent or not • for a prize randomly assigned to one of the n lottery tickets, the probability of winning the price = 1/n for all tickets • the order of buying a ticket does not change the probability of winning 8

7. Applications of the Two Ideas • the following are interesting applications • mean of Bin(n, p) (Ex #7(b) of WS #8) • variance of Bin(n, p) (Ex #8(b) of WS #8) • the probability of winning a lottery (Ex #3(b) of WS #9) • mean of hypergeometric random variable (Ex #4 of WS #9) • mean and variance of random number of matches (Ex #5 of WS #9) 9

8. Mean of Bin(n, p) Ex #7(b) of WS #8 • X ~ Bin(n, p) • find E(X) from E(I1+…+In) • E(X) = E(I1+…+In) = np 10

9. Variance of Bin(n, p) Ex #8(b) of WS #8 • X ~ Bin(n, p) • find V(X) from V(I1+…+In) • V(X) = V(I1+…+In) = nV(I1) = np(1p) 11

10. Probability of Winning a Lottery Ex #3(b) & (c) of WS #9 • a grand prize among n lotteries • (b) Let n 3. Find the probability that the third person who buys a lottery wins the grand prize • (c). Let Ii = 1 if the ith person buys the lottery wins the grand prize, and Ii = 0 otherwise, 1 in • (i). Show that all Ii have the same (marginal) distribution • Find cov(Ii, Ij) for ij • Verify 12

11. Probability of Winning a Lottery Ex #3(b) & (c) of WS #9 • (b) A = the third person buying a lottery wins the grand prize • find P(A) when there are 3 persons • Sol. P(A) = • actually the order does not matter • thinking about randomly throwing a ball into one of three boxes 13

12. Probability of Winning a Lottery Ex #3(b) & (c) of WS #9 • (c)(i). P(Ij = 1) = 1/n for any j • . • for ij, cov(Ii, Ij) = E(IiIj) E(Ii)E(Ij) • E(IiIj) = 0  cov(Ii, Ij) = -1/n2 • checking: 14

13. Hypergeometric in the Context of Ex #4 of WS #9 • 3 balls are randomly picked from 2 white & 3 black balls • X = the total number of white balls picked E(X) = 6/5 15

14. Hypergeometric in the Context of Ex #4 of WS #9 • Ex #4(c). Assume that the three picked balls are put in bins 1, 2, and 3 in the order of being picked • (i). Find P(bin i contains a white ball), i = 1, 2, & 3 • (ii). Define Bi = 1 if the ball in bin i is white in color, i = 1, 2, and 3. Find E(X) by relating X to B1, B2, and B3 16

15. Hypergeometric in the Context of Ex #4 of WS #9 • (i). P(bin i contains a white ball) = 2/5 • each ball being equally likely to be in bin i • (ii). Bi = 1 if the ball in bin i is white in color, and = 0 otherwise • X = B1 + B2 + B3 • E(Bi) = P(bin i contains a white ball) = 2/5 • E(X) = E(B1) + E(B2) + E(B3) = 6/5 17

16. Hypergeometric in the Context of Ex #4 of WS #9 • Ex #4(d). Arbitrarily label the white balls as 1 and 2. •  (i). Find P(white ball 1 is put in a bin); find P(white ball 2 is put in a bin) • (ii). let Wi = 1 if the white ball i is put in a bin, and Wi = 0 otherwise, i = 1, 2; find E(X) from Wi 18

17. Hypergeometric in the Context of Ex #4 of WS #9 • (i) P(white ball 1 is put in a bin) = 3/5 • each ball being equally likely to be in a bin • (ii) Wi = 1 if the white ball i is put in a bin, and Wi = 0 otherwise, i = 1, 2. Find E(X) by relating X to W1 and W2 • X = W1 + W2 • E(Wi) = P(white ball 1 is put in a bin) = 3/5 • E(X) = E(W1) + E(W2) = 6/5 19

18. Mean and Variance of Random Number of Matches Ex #5 of WS #9 • gift exchange among n participants • X = total # of participants who get back their own gifts • (a). Find P(the ith participant gets back his own gift) • (b). Let Ii = 1 if the ith participant get back his own gift, and Ii = 0 otherwise, 1 in. Relate X to I1, …, In • (c). Find E(X) from (b) • (d). Find cov(Ii, Ij) for ij • (e). Find V(X) 20

19. Mean and Variance of Random Number of Matches Ex #5 of WS #9 • (a). P(the ith participant gets back his own gift) = 1/n • each hat being equally likely be picked by the person • (b). Ii = 1 if the ith participant get back his own gift, and Ii = 0 otherwise, 1 in;  X = I1 + …+ In • (c). E(X) = E(I1+ …+In) = 1 • (d). for ij, cov(Ii, Ij) = E(IiIj)  E(Ii)E(Ij) • E(IiIj) = P(Ii = 1, Ij = 1) = P(Ii = 1|Ij = 1)P(Ij = 1) = 1/[n(n-1)] • cov(Ii, Ij) = 1/[ n2(n-1)] • (e). V(X) = 21

20. Example 1.11 of Ross It is still too complicated to discuss. Let us postpone its discussion until covering the condition probability and the condition probability 22

21. Chapter 2 • material to read: from page 21 to page 59 (section 2.5.3) • Examples highlighted: Examples 2.3, 2.5, 2.17, 2.18, 2.19, 2.20, 2.21, 2.30, 2.31, 2.32, 2.34, 2.35, 2.36, 2.37 • Sections and material highlighted: 2.2.1, 2.2.2, 2.2.3, 2.2.4, 2.3.1, 2.3.2, 2.3.3, 2.4.3, Proposition 2.1, Corollary 2.2, 2.5.1, 2.5.2, Proposition 2.3, 2.5.3, Properties of Covariance 23

22. Chapter 2 • Exercises #5, #11, #20, #23, #29, #37, #42, #43, #44, #45, #46, #51, #57, #71, #72 24

23. 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 26

24. 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} 27

25. 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). 28

26. 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 29

27. 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 30

28. 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 31

29. 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 32

30. 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 33

31. 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 34

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