Transcript 投影片 1

Introduction to Stochastic Models
GSLM 54100
1
Outline
 independence
 variance
 two
of random variables
and covariance
useful ideas
 examples
 conditional
distribution
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
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
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Proposition 2.3
 E[g(X)h(Y)]
= E[g(X)]E[h(Y)] for
independent X, Y
 different
 E[g(X)]
 E[h(Y)]

meanings of E()
=

 g ( x) f X
=
 h( y ) fY ( y )dy

( x)dx
x and y are
dummy variables
 
E[g(X)h(Y)] =   g ( x)h( y ) f X ,Y ( x, y )dxdy
 
 
 g ( x)h( y ) f X ( x) fY ( y )dxdy


 
g ( x) f X ( x)dx 
h( y ) fY ( y )dy
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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,
 Cov(X,
Y) = cCov(X, Y)
Y + Z) = Cov(X, Y) + Cov(X, Z)
 Cov(iXi, jYj) = i j Cov(Xi, Yj)
n
n
 Var
. (  X i )   Var ( X i )  2  Cov( X i , X j )
i 1
i 1
1i j n
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Two Useful Ideas
7
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
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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)
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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
Variance of Bin(n, p)
Ex #8(b) of WS #8
X
~ Bin(n, p)
 find
n
n
i 1
i 1
Var (  X i )   Var ( X i )  2  Cov( X i , X j )
1i j n
V(X) from V(I1+…+In)
 V(X)
= V(I1+…+In) = nV(I1) = np(1p)
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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

n
n
i 1
i 1
Verify Var (  X i )   Var ( X i )  2  Cov( X i , X j )
1i j n
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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.
2 1 1

P(A) = 3   2  3
 actually
the order does not matter
 thinking
about randomly throwing a ball into
one of three boxes
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Probability of Winning a Lottery
Ex #3(b) & (c) of WS #9
 (c)(i).
P(Ij = 1) = 1/n for any j
n
 . I j  1
i 1
 for
n 
 Var   I j  
 i 1 
0
i  j, cov(Ii, Ij) = E(IiIj) E(Ii)E(Ij)
 E(IiIj)
= 0  cov(Ii, Ij) = -1/n2
n
n
 checking: Var (  X i )   Var ( X i )  2 
i 1
i 1
1i j n
Cov( X i , X j )
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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
P( X  0) 
P( X  2) 
C02C33
C35
C22C13
C35
1

10
3

10
P ( X  1) 
C12C23
C35
3

5
E(X) = 6/5
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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
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Hypergeometric
in the Context of Ex #4 of WS #9
 (i). P(bin
 each
i contains a white ball) = 2/5
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
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
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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 + W 2
 E(Wi)
 E(X)
= P(white ball 1 is put in a bin) = 3/5
= E(W1) + E(W2) = 6/5
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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)
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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) = n
 
1
n

1  1n 
n( n1)
n2 ( n1)
1
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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
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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
 Exercises
#5, #11, #20, #23, #29, #37, #42,
#43, #44, #45, #46, #51, #57, #71, #72
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Conditional Distributions
25
Conditional Distribution
X
~ {pn} and A is an event
0
 P(X = n|A)  1
 n
P(X = n|A) =
P( X  n, A)
1

P( A)
n

{P(X = n|A)} is a probability mass
function, called the conditional distribution
of X given A
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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}
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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
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
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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
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
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
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
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