University of Illinois at Urbana-Champaign
Department of Electrical and Computer Engineering
ECE 434: Random Processes
Spring 2005
Solutions to Probability Review
Problem 1.1
(a) True. If X, Y are conditionally independent on Z, then fX Y |Z(x, y|z) = fX|Z(x|z)·
fY|Z(y|z). Here fXY |Z(x, y|z) = 4
fZ(z)e−zx ·e−zy ,Xand Ycomponents are uncoupled,
thus they are conditionally independent.
(b) True.
V ar(X+Y) = E[(x+y−mx−my)2] = E[((x−mx)+(y−my))2]
=E[(x−mx)2] + E[(y−my)2] = V ar(X) + V ar(Y).
(c) False.P(A∪B) = P(A)+ P(B)−P(A∩B) = P(A)+P(B)−P(A)P(B)6=P(A) +P(B).
(d) True.P(A∪B) = P(A) + P(B)−P(A)P(B) = P(A) + P(A)−P2(A) = 3
2P(A).
P(A) = 1
2.
(e) False.P(Y= 0) = P[X≤0] = 1
2, so this is not a continuous random variable.
Problem 1.2
(a)
P(Y= 1) = P(Y= 1, X = 0) + P(Y= 1, X = 1)
=P(Y= 1|X= 0)P(X= 0) + P(Y= 1|X= 1)P(X= 1) = p2·1
2+ (1 −p1)·1
2
=1 + p2−p1
2
(b)
P(X= 0|Y= 1) = P(X= 0, Y = 1)
P(Y= 1) =p2·1
2
1+p2−p1
2
=p2
1−p1+p2
.
(c) We are asked to find the Bayes rule. Math aside, we can reason this out. The probability
of making a mistake during the transmission, p1and p2, are both less than 1/2. That
means more than half of the time, what we receive at Yis what was sent at source X.
