1 introduction to stochastic models gslm 54100. 2 outline independence of random variables ...

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Introduction to Stochastic ModelsIntroduction to Stochastic ModelsGSLM 54100GSLM 54100

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OutlineOutline

independence of random variables

variance and covariance

two useful ideas examples

conditional distribution

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Independent Random VariablesIndependent 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

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Proposition 2.3Proposition 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

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Proposition 2.3Proposition 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)] =

( ) ( )Xg x f x dx

( ) ( )Yh y f y dy

,( ) ( ) ( , )X Yg x h y f x y dxdy

x and y are dummy variables

( ) ( ) ( ) ( )X Yg x h y f x f y dxdy

( ) ( ) ( ) ( )X Yg x f x dx h y f y dy

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Variance and Covariance Variance and Covariance (Ross, pp 52-53)(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)

. 1 1 1

( ) ( ) 2 ( , )n n

i i i ji i i j n

Var X Var X Cov X X

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Two Useful IdeasTwo Useful Ideas

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Two Useful IdeasTwo 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

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Applications of the Two IdeasApplications 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)

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Mean of Bin(Mean of Bin(nn, , pp) ) Ex #7(b) of WS #8Ex #7(b) of WS #8

X ~ Bin(n, p)

find E(X) from E(I1+…+In)

E(X) = E(I1+…+In) = np

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Variance of Bin(Variance of Bin(nn, , pp) ) Ex #8(b) of WS #8Ex #8(b) of WS #8

X ~ Bin(n, p)

find V(X) from V(I1+…+In)

V(X) = V(I1+…+In) = nV(I1) = np(1p)

1 1 1( ) ( ) 2 ( , )n n

i i i ji i i j n

Var X Var X Cov X X

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Probability of Winning a Lottery Probability of Winning a Lottery Ex #3(b) & (c) Ex #3(b) & (c) of WS #9of 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 1 1 1( ) ( ) 2 ( , )n n

i i i ji i i j n

Var X Var X Cov X X

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(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

2 1 13 2 3

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(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: 1 1 1( ) ( ) 2 ( , )n n

i i i ji i i j n

Var X Var X Cov X X

1

n

jiI

1

1

n

ji

Var I

0

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Hypergeometric Hypergeometric in the Context of in the Context of Ex #4 of WS #9Ex #4 of WS #9

3 balls are randomly picked from 2 white & 3 black balls

X = the total number of white balls picked 2 30 3

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1( 0)

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C CP X

C

2 31 2

53

3( 1)

5

C CP X

C

2 32 1

53

3( 2)

10

C CP X

C E(X) = 6/5

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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

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(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

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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

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(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

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Mean and Variance Mean and Variance of Random Number of Matches of Random Number of Matches

Ex #5 Ex #5 of WS #9of 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)

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Mean and Variance Mean and Variance of Random Number of Matches of Random Number of Matches

Ex #5 Ex #5 of WS #9of 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) = 2

( 1)1 1( 1)

1 1n nn n n n

n

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Example 1.11 of RossExample 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

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Chapter 2 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

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Chapter 2 Chapter 2

Exercises #5, #11, #20, #23, #29, #37, #42, #43, #44, #45, #46, #51, #57, #71, #72

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Conditional DistributionsConditional Distributions

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Conditional DistributionConditional Distribution

X ~ {pn} and A is an event

0 P(X = n|A) 1

n P(X = n|A) =

{P(X = n|A)} is a probability mass

function, called the conditional distribution

of X given A

( , )1

( )n

P X n A

P A

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Conditional DistributionConditional 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}

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Ex #1 of WS #5Ex #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).

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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

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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

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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

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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

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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

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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

1 introduction to stochastic models gslm 54100. 2 outline independence of random variables ...

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