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
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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
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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
1i j n
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Two Useful Ideas
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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
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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 )
1i j n
V(X) from V(I1+…+In)
V(X)
= V(I1+…+In) = nV(I1) = np(1p)
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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 )
1i 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
1i 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( n1)
n2 ( n1)
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
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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).
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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.
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
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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.
(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 #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 #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 #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
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