ECE 313 Problem Set # 14 Solutions Fall 2002
1. Correlation of histogram values
(a) X1is Bernoulli with parameter p=1
6, so E[X1] = 1
6and V ar(X1) = 1
6(1 −1
6) = 5
36 .
(b) E[X] = nE[X1] = n
6and V ar(X) = nV ar(X1) = 5n
36 .
(c) We begin by computing Cov(X1, Y1). Since X1Y1= 0 with probability one, E[X1Y1] = 0.
Therefore Cov(X1, Y1) = E[X1Y1]−E[X1]E[Y1] = 0 −1
6
1
6=−1
36 . So Cov(Xi, Yi) = −1
36 for any i.
On the other hand, if i6=jthen Xiis independent of Xj. So
Cov(Xi, Yj) = (−1
36 if i=j
0 if i6=j
(d)
Cov(X, Y ) = X
iX
j
Cov(Xi, Yj) = X
i
Cov(Xi, Yi) = nCov(X1, Y1) = −n
36
and
ρ(X, Y ) = Cov(X, Y )
pV ar(X)V ar(Y)=−1
5
2. Conditional expectation of histogram values
(a) Given that xof the dice show a 1, each of the remaining dice is equally likely to show 2,3,4,5,
or 6. Thus, each of the remaining n−xdice shows a 2 with conditional probability 1
5. Therefore
E[Y|X=x] = n−x
5.
(b) This is the same function found in part (a) by general theory. That is, g(x) = E[Y|X=x] =
n−x
5.
3. Covariance for a Poisson process
(a) As found in Problem 3(d) of Problem Set 10, the conditional distribution of N(τ) given N(T) =
nis binomial with parameters nand p=τ
T. Therefore E[N(τ)|N(T) = n] = nτ
T.
(b) As found in Problem 3(e) of Problem Set 10, the conditional distribution of N(T) given N(τ) = i
is the same as the distribution of iplus a Poisson random variable with mean (T−τ)λ. Therefore
E[N(T)|N(τ) = i] = i+ (T−τ)λ.
(c) Since N(τ) is a Poisson random variable with mean λτ, and since the variance of a Poisson
random variable is the same as the mean, Cov(N(τ), N (τ)) = V ar(N(τ)) = λτ .
(d) Since N(τ) is independent of N(T)−N(τ), Cov(N(τ), N (T)−N(τ)) = 0.
(e) Use N(T) = N(τ) + (N(T)−N(τ)) to get Cov(N(τ), N (T)) = Cov(N τ ), N (τ) + (N(T)−
N(τ))) = Cov(N(τ), N (τ)) + Cov(N(τ),(N(T)−N(τ))) = λτ + 0 = λτ.
4. Conditional expectation for uniform density over a triangle
(a) The triangle has base and height one, so the area of the triangle is 0.5. Thus the joint pdf is 2
inside the triangle.
(b)
fX(x) = Z∞
−∞
fXY (x, y )dy =
Rx/2
02dy =xif 0 <x<1
Rx/2
x−12dy = 2 −xif 1 <x<2
0 else
(c) In view of part (c), the conditional density fY|X(y|x) is not well defined unless 0 < x < 2. In
