ECE 534: Random Pro cesses Spring 2010
Solutions to Problem Set 2
1.15 Transformation of a random variable
(a) Observe that Ytakes values in the interval [1,+∞).
FY(c) = P[exp(X)≤c] = P[X≤ln c]=1−exp(−λln c)=1−c−λc≥1
0c < 1
Differentiate to obtain
fY(c) = λc−(1+λ)c≥1
0c < 1
(b) Observe that Ztakes values in the interval [0,3].
FZ(c) = P[min{X, 3} ≤ c] =
0c < 0
P[X≤c] = 1 −exp(−λc) 0 ≤c < 3
1c≥3
The random variable Zis neither discrete nor continuous type. Rather it is a mixture,
having a density over the interval [0,3) and a discrete mass at the point 3.
1.19 Using the Gaussian Qfunction
(a) P[X≥16] = P[X−10
3>16−10
3] = Q(16−10
3) = Q(2).
(b) P[X2≥16] = P[X≥4]+P[X≤ −4] = Q(4−10
3)+1−Q(−4−10
3) = Q(−2)+1−Q(−14
3) =
1−Q(2) + Q(14
3).
(c) Zis N(0,5) so P[|Z|>1] = P[Z > 1] + P[Z < −1] = Q(1
√5) + 1 −Q(−1
√5)=2Q(1
√5).
1.21 Correlation of histogram values
(a) X1is Bernoulli with parameter p=1
6, so EX1=1
6and Var(X1) = 1
6(1 −1
6) = 5
36 .
(b) EX =nE X1=n
6and Var(X) = nVar(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]−EX1E 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 Yj. So
Cov(Xi, Yj) = −1
36 if i=j
0 if i6=j
1
