University of Illinois Fall 2006
ECE 413: Solutions to Problem Set 9
1. (a) Given X=n, it must be that Y+W=n. Therefore, whenever k+l6=n,
P[{Y =k} ∩ {W =l} ∩ {X =n}] = 0 ⇒P[{Y =k} ∩ {W =l}|{X =n}] = 0. When n=k+l,
P[{Y =k} ∩ {W =l}|{X =k+l}] = P[{Y =k}|{X =k+l}] = k+l
kpk(1 −p)l.
(b) Yand Ware not conditionally independent given the event {X =n}. In fact, they are very much
dependent since their sum is n.
(c) P[{Y =k} ∩ {W =l}] =
∞
X
n=0
P[{Y =k} ∩ {W =l}|{X =n}]P{X =n}
=k+l
kpk(1 −p)lexp(−λ)λk+l
(k+l)! = exp(−λp)(λp)k
k!·exp(−λ(1 −p))(λ(1 −p))l
l!
=P{Y =k} · P{W =l}.
(d) Yes, Yand Ware unconditionally independent, even though they are conditionally very much
dependent.
This particular result (called Poisson splitting) has many applications. For example, suppose
that packets arriving at a server are placed into one of two queues with probabilities pand
(1 −p) respectively. If the number of packets arriving per unit time is modeled as a Poisson
random variable with parameter λ, then the numbers of packets in the queues can be modeled
as independent Poisson random variables with parameters λp and λ(1 −p). Conversely, if the
numbers of packets arriving per unit time on different ports are modeled as independent Poisson
random variables with parameters λi, then the total number of packets is a Poisson random
variable with parameter Pλi.
2. (a) F(u) =
0u < 0,
u2,0≤u < 1,
1, u ≥1.
is a valid CDF. P{|X| >0.5}=P{X >0.5}= 1 −F(0.5) = 3
4.
(b) F(u) =
0u < 1,
2u−u2,1≤u≤2,
1, u > 2.
is not a valid CDF since F(1) = 1 > F (2) = 0.
(c) F(u) = 1
2exp(2u)u≤0,
1−1
4exp(−3u), u > 0,is not a valid CDF since it is not right-continuous at 0.
(d) F(u) = 1
2exp(2u)u < 0,
1−1
4exp(−3u), u ≥0,is a valid CDF.
P{|X | >0.5}= 1 −P{|X | ≤ 0.5}= 1 −(F(0.5) −F(−0.5)) = 1
2exp(−1) −1
4exp(−1.5).
3. (a) P{X = 2}=FX(2+)−FX(2−) = 1
4,P{X <2}=FX(2−) = 1
2,P{X >2}= 1 −FX(2) = 1
4,
P{1≤ X ≤ 3}=FX(3) −FX(1−) = 7
8−1
4=5
8, and P{X >2| X >0}=P{X >2}
P{X >0}=2
7.
(b) E[X] is the area between the CDF and the line at height 1 as shown in the figure below. Elementary
geometry gives E[X] = 1
2×1×7
8+6
8+1
2+1
2×2×1
4=22
16 = 1.375.
University Problem Set #8: Solutions ECE 313
of Illinois Page 1 of 2 Spring 2003
1. (a) FX(1) = 1, FX(3/2) = 3/4. Thus, FX(u) is not a nondecreasing function and thus cannot be a valid CDF.
(b) Yes. FX(u) is a valid CDF, and is continuous except at u = 0.
P{|X| > 0.5} = 1 -–P{|X| ≤ 0.5} = 1 – (FX(0.5) – FX(–0.5–)) = 1 – (FX(0.5) – FX(–0.5))
= 1 -–(1 – (1/4)exp(–3/2) – (1/2)exp(–1)) = (1/4)exp(–3/2) + (1/2) exp(–1).
(c) FX(u) is not right continuous at u = 0 and thus cannot be a valid CDF.
2.(a) P(works for exactly 2 hours) = P{X = 2} = FX(2+) – FX(2–) = 1/4.
(b) P(works for more than 2 hours) = P{X > 2} = 1 – FX(2+) = 1/4.
(c) P(works for less than 2 hours) = P{X < 2} = FX(2–) = 1/2.
(d) P(works for exactly 3 hours) = P{X = 3} = FX(3+) – FX(3–) = 0.
(e) P(works for more than 1/2 but less than 3 hours) = P{1/2 < X < 3} = FX(3–) – FX((1/2)+) = 7/8 – 3/16 =
11/16.
(f) P(works for more than 2 hours given that the student works at all) = P{X > 2|X > 0}
= P({X > 2}∩{X > 0})
P{X > 0} = P{X > 2}
P{X > 0} = 1 – FX(2+)
1 – FX(0+) = 1
4×8
7 = 2/7.
1
1234
F(u)
u
0.5
0.75
0.25
(g) The CDF is as shown, and as discussed in class, E[X] = shaded area between CDF and the line at height 1.
By elementary geometry, the shaded area is 1×(7/8 + 6/8)/2 + 1/2 + (1/2)×(2)×(1/4) = 22/16 = 1.375 hours.
Noncredit Exercise: How does this compare with the number reported by Eta Kappa Nu in their survey?
3.(a) E[X] = ∫
0
∞
P{X>u}du = ∫
0
1
P{X>u}du + ∫
1
2
P{X>u}du + ∫
2
3
P{X>u}du + … But, X takes on only integer values.
Thus, for any given nonnegative integer k, it must be that for all u ∈ [k,k+1), P{X > u} = P{X > k}.
Hence, ∫
k
k+1
P{X > u} du = P{X > k} and thus E[X] = P{X >0} + P{X >1} + … = ∑
k=0
∞
P{X > k}. This
is the CompE version: some spoilsport EEs insist on writing this result as E[X] = ∑
k=1
∞
P{X ≥ k} .
(b) If X is a geometric random variable with parameter p, then P{X > k} = P{first k trials ended in failure}
= (1–p)k. This formula holds even when k = 0, since obviously P{X > 0} = 1.
Hence, E[X] = ∑
k=1
∞
P{X ≥ k} = ∑
k=0
∞
P{X > k} = ∑
k=0
∞
(1–p)k = 1 + (1–p) + (1–p)2 + … = 1
1 – (1–p) = 1
p
4.(a) This is a valid pdf. (b) This is a valid pdf.
(c) This is not a valid pdf because the function is negative for 0 < u < 1. However, ∫
0
1
ln u du = u ln u – u
1
0
= –1, so –ln u, u ∈ (0,1) is a valid pdf.
(d) This is not a valid pdf because the function is negative for u ∈ (0,1). Also, since the function is positive
for u ∈ (1,2), Cf(u) cannot be a valid pdf for any C.
(e) This is a valid pdf.
