University Second Hour Exam: Solutions ECE 313
of Illinois Page 1 of 1 Summer 2003
1. TRUE FX(b) is right continuous.
FALSE It is possible that FX(a) = FX(b).
TRUE This is the certain event.
FALSE This need not hold.
TRUE This is true, since X is a continuous random variable.
TRUE This has to hold for all density functions.
2. P{|X – 4| > 3} = P{X > 7} + P{X < 1} = 1 – Φ
−
4
27 + Φ
−
4
21 = 1 – Φ(1.25) + Φ(–0.25)
= 1 – Φ(1.25) + 1 – Φ(0.25) = 2 – 0.8944 – 0.5987 = 0.5069.
P{X < 3 | X > 2} = P{2 < X < 3}/P{X > 2} = 2(Φ(0.25) – Φ(0)) = 2 Φ(0.25) – 1 = 1.1974– 1 = 0.1974
3.(a) The PDF of X has value 0.2 for –1 ≤ u ≤ 4. Using LOTUS we have
E[Y] = E[|X–1|] = .
∫∫
−
−=−+−=−+−
1
1
4
1
4
1
21
1
23.1|)2/(2.0|)2/(2.0)1(2.0)1(2.0 uuuuduuduu
(b) Y takes on values in the range [0,3]. Thus, for v < 0 or v > 3, FY(v) = 0. For any v, 0 ≤ v ≤ 2,
FY(v) = P[Y ≤ v] = P[1 – v ≤ X ≤ v + 1] = 0.2(v + 1 – (1 – v)) = 0.4v, while for any v, 2 ≤ v ≤ 3,
FY(v) = P[Y ≤ v] = P[– 1 ≤ X ≤ v + 1] = 0.2(v + 1 – (– 1)) = 0.2(v+2). Differentiating, we obtain
f
Y(v) =
≤≤
≤≤
elsewhere
v
v
,0
32,2.0
20,4.0
4. Let µ = 2 denote the arrival rate of the process. Then, both N(0,T] and N(0.5T, 1.5T] are Poisson random
variables with parameter µT = 2T.
(a) Hence, P(A) = P{N(0,T] = 0} = exp(–2T) and P(B) = P{N(0.5T, 1.5T] = 1} = 2T•exp(–2T).
(b) P(AB) = P{N(0,T] = 0, N(0.5T, 1.5T] = 1} = P{N(0,0.5T] = 0, N(0.5T, 1.5T] = 1} since the single arrival
must have occurred during (0.5T, 1.5T]. But, N(0,0.5T] and N(0.5T, 1.5T] are independent random
variables because the time intervals are disjoint. Hence,
P(AB) = P{N(0,0.5T] = 0, N(0.5T, 1.5T] = 1} = P{N(0,0.5T] = 0}•PN(0.5T, 1.5T] = 1}
= (2T/2)•exp(–2T/2)•exp(–2T) = T•exp(–3T), and P(B|A) = P(AB)/P(A) = T•exp(–T).
5. Since X is uniform its density function is
fX(p) =
≤≤
elsewhere
p
,0
6.04.0,5
Let A be the event A={1 head in 1 toss}. We know P[A | X = p] = p. The posterior density is
fX|A(p) =
∫=
=
6.0
4.0
)(]|[
)(]|[
pfpXAP
pfpXAP
X
X. Note the limits.
Carrying out the integral we obtain
f
X|A(p) =
≤≤
elsewhere
pp
,0
6.04.0,10
The Bayesian estimate of heads at the next tossing is just
P{Heads} = ∫∫ ==−===
6.0
4.0
6.0
4.0
6.0
4.0
3
|5067.0
3
52.1
)064.0216.0(
3
10
|
3
10
10)( ppdppdpppf AX > 0.5
Note that, based on the observation our estimate of the probability of observing heads at the next tossing
has increased.