University of Illinois Spring 2008
ECE 313: Solutions to Problem Set 9
1. (a) E[Y] = E[X2] = Z+1
−1
u21
2du =1
3.E[Y2] = E[X4] = Z+1
−1
u41
2du =1
5.
Hence, var(Y) = E[Y2]−(E[Y])2=1
5−1
9=4
45.
(b) E[Z] = E[g(X)] = Z0
−1−u21
2du +Z+1
0
u21
2du =−1
6+1
6= 0.
(c) Utakes on integer values ≥1. V= sin(πU/2) = 0 if Uis even. On the other hand, if
U= 4k+ 1, then V= +1 while if U= 4k−1, then V=−1. We get
E[V] =
∞
X
k=1
sin(πk/2) 1
2k
=1
2−1
8+1
32 − · ·· =1
21−x+x2− · ·· =1
2(1 + x)
where x=1
4⇒E[V] = 2
5.
2. (a) E[R] = Z1
0
u·2u du =2
3.E[V] = Z1
0
4π
3u3·2u du =8π
15 .E[A] = Z1
0
4πu2·2u du = 2π.
The average sphere does not have average volume or average surface area. More generally,
E[g(X)] 6=g(E[X]).
(b) For 0 <u<1, un> un+1.
Hence, Z1
0
un·fR(u)du =E[Rn]>Z1
0
un+1 ·fR(u)du =E[Rn+1].
3. I’m considering, I’m considering!
(a) The number of arrivals in (0,4] is N(0,4], a Poisson random variable with parameter
4λ. Hence, E[N(0,4]] = 4λ.
(b) The event {N(0,3] = 3}∩{N(2,6] = 0}is the same as {N(0,2] = 3}∩{N(2,6] = 0}
and thus P[{N(0,3] = 3}∩{N(2,6] = 0}] = P[{N(0,2] = 3}∩{N(2,6] = 0}]. We get
P[{N(0,3] = 3}∩{N(2,6] = 0}] = (2λ)3
3! exp(−2λ)×exp(−4λ) = 4λ3
3exp(−6λ)
by independence of Poisson variables on disjoint intervals of time.
(c) The number of arrivals in (0,6] is N(0,6], a Poisson random variable with parameter 6λ.
If it is observed that N(0,6] = k, then the maximum-likelihood estimate of the Poisson
parameter 6λis k(cf. Problem 2(c) of Problem Set 5), and hence the maximum-
likelihood estimate of the arrival rate is k/6 = # of arrivals/length of interval which is
exactly what one would expect.
(d) P{N(0, t)≥1}= 1 −P{N(0, t) = 0}= 1 −exp(−λt)=1−2−twhen λ= ln 2.
4. (a) The inter-arrival time in a Poisson process with arrival rate λ(time between two suc-
cessive chalk breaks on this instance) is an exponential random variable with parameter
λ. Hence, the average length of time between successive chalk-breaks is the mean of this
exponential random variable, which is λ−1= 10 minutes.
(b) The number of times that Todd breaks the chalk during a 50 minute lecture is a Poisson
random variable N(0,50] with parameter λ×50 = 5 and mean value E[N(0,50]] = 5.
