1 Introduction to Introduction to Stochastic Models Stochastic Models GSLM 54100 GSLM 54100
Dec 23, 2015
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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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Probability of Winning a Lottery Probability of Winning a Lottery Ex #3(b) & (c) Ex #3(b) & (c) of WS #9of 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
2 1 13 2 3
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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
(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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Hypergeometric Hypergeometric in the Context of in the Context of Ex #4 of WS #9Ex #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
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Hypergeometric Hypergeometric in the Context of in the Context of Ex #4 of WS #9Ex #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
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Hypergeometric Hypergeometric in the Context of in the Context of Ex #4 of WS #9Ex #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
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Hypergeometric Hypergeometric in the Context of in the Context of Ex #4 of WS #9Ex #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
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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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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.
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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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. 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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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. (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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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. (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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Ex #1 of WS #5Ex #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
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Ex #1 of WS #5Ex #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