University Solutions to Second Hour Exam ECE 313
of Illinois Fall 2001
1. (a) ① P{X > b} = 1 – FX(b). ➁If FX(a) < FX(b), then a < b.
➂If a < b, then FX(a) < FX(b). ➃FX(u) = 1/2 for some u, –∞ < u < ∞.
The first two properties hold for all CDFs since FX(u) is a right-continuous non-decreasing function. But,
if a < b, it is possible that FX(a) = FX(b) because the CDF did not increase between those points (it cannot
decrease, of course.) As a counterexample for ➃, note that for a Bernoulli random variable with parameter
p ≠ 1/2, FX(u) takes on values 0, p, and 1 only. Thus, ➂ and ➃ are not properties of all CDFS.
(b) ① fX(u) ≤ 1 for all u, –∞ < u < ∞.➁limu→ –∞ fX(u) = 0
➂ limu→ ∞ fX(u) = 1 ➃P{a < X < b} = P{a ≤ X ≤ b}
The second and fourth properties are satisfied by all pdfs. The first need not hold: the value of the function
is a probability density, not a probability mass, and can exceed 1 for some pdfs. On the other hand, no pdf
can possibly satisfy the third property.
2. P{{|X – 4| > 3} = P{X > 7} + P{X < 1} = 1 – Φ( )
7–2
5 + Φ( )
1–2
5 = 1 – Φ(1) + Φ(–0.2)
= 1 – Φ(1) + 1 – Φ(0.2) = 2 – 0.8413 – 0.5793 = 0.5794.
P{X < 3 | X > 2} = P{2 < X < 3}/P{X > 2} = 2[ ]
Φ( )
3–2
5 – Φ( )
2–2
5 = 2(Φ(0.2) – 0.5)
= 1.1586 – 1 = 0.1586
3.(a) The pdf of X has value 0.2 for –1 ≤ u ≤ 4, is as shown below in the left-hand figure below. We have that
E[Y] = E[|X–1|] = ∫
–1
1 0.2(1–u) du + ∫
1
4
0.2(u–1) du = 0.2(u – u2/2)
1
–1 + 0.2(u2/2–u)
4
1 = 1.3.
0.2
–1 4 –1
v
–1
v1
1
2
v+1
1–v v+1
(b) Y takes on values in the range [0, 3]. From the middle and right-hand figures above, we see that
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+1} = P{X ≤ v} = 0.2(v+1–(–1)) = 0.2(v+2).
Hence, fY(v) =
0.4, 0 ≤ v ≤ 2,
0.2, 2 ≤ v ≤ 3
0, elsewhere which is easily verified to be a valid pdf.
4.(a) N(3, 6] is a Poisson random variable with parameter 3µ. Hence, P{N(3, 6] = 2} = exp(–3µ)•(3µ)2/2!.
(b) If there are 2 arrivals in (0, 6], and also 2 arrivals in (3, 9], then it must be that in the disjoint intervals
(0, 3], (3, 6], and (6, 9], there are (2, 0 ,2) arrivals respectively, or (1, 1, 1) arrivals respectively, or
(0, 2, 0) arrivals respectively, which correspond to values of N(3, 6] of 0, 1, and 2 respectively.
Hence, by the independence of arrivals in disjoint intervals, we get that
P(AB) = P{N(3,6] = 0}•P{two arrivals in (0, 3] and two arrivals in (6, 9]}
+ P{N(3,6] = 1}•P{one arrival in (0, 3] and one arrival in (6, 9]}
+ P{N(3,6] = 2}•P{no arrivals in (0, 3] and no arrivals in (6, 9]}
= exp(–3µ)•[exp(–3µ)•(3µ)2/2!]2 + [exp(–3µ)•(3µ)]•[exp(–3µ)•(3µ)]2 + [exp(–3µ)•(3µ)2/2!]•[exp(–3µ)]2
= exp(–9µ)•[81µ4/4 + 27µ3 + 9µ2/2]. Hence, we have
P{N(3, 6] = 1 | AB} = P{{N(3, 6] = 1} ∩ AB}/P(AB) = 27µ3
81µ4/4 + 27µ3 + 9µ2/2 = 12µ
9µ2 + 12µ + 2
5.(a) X is an exponential random variable with parameter λ0 = 1 or λ1 = 2 according as hypothesis H0 or H1 is
true. Thus, the pdfs f0(u) and f1(u) are as shown below where the crossing-point is easily found by equating
2•exp(–2t) = exp(–t) and solving to get t = ln 2. Thus, we see that the maximum-likelihood decision rule
chooses H1 if T < θ = ln 2, and H0 if T > ln 2. Hence, i = 0, θ = ln 2 in the decision rule exhibited on the
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