Comprehensive Solutions Manual for "Random Signals and Noise: A Mathematical Introduction" 1st Edition by Shlomo Engelberg

SOLUTIONS MANUAL for Random Signals and Noise A
Mathematical Introduction 1st edition by Shlomo Engelberg|Latest
2026/2027 Update

,Solutions Manual


SUMMARY: In this chapter we present complete solution to the
exercises set in the text.




Chapter 1
1. Problem 1. As defined in the problem, A—B is composed of the elements
in A that are not in B. Thus, the items to be noted are true. Making
use of the properties of the probability function, we find that:

P (A ∪ B) = P (A) + P (B — A)
and that:
P (B) = P (B — A) + P (A ∩ B).
Combining the two results, we find that:
P (A ∪ B) = P (A) + P (B) — P (A ∩ B).

2. Problem 2.
(a) It is clear that fX (α) ≥ 0. Thus, we need only check that the
integral of the PDF is equal to 1. We find that:
∫∞
∫ ∞
(α) dα = 0.5 e−|α| dα
fX
−∞ −∞
∫ 0 ∫ ∞
= 0.5 α
e dα + e−α dα
−∞ 0
= 0.5(1 + 1)
= 1.
Thus fX (α) is indeed a PDF.
(b) Because fX (α) is even, its expected value must be zero. Addition-
ally, because α2fX (α) is an even function of α, we find that:
∫ ∞ ∫ ∞
α2f X (α) dα = 2 α2f X (α) dα
−∞ 0


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,2 Random Signals and Noise: A Mathematical Introduction
∫∞
= α2e−α dα
0
∫ ∞
by parts
= (—α2e −α|∞
0 +2 αe −α dα
∫0 ∞
by parts −α ∞ −α
= 2(—αe |0 ) + 2 e dα
0
= 2.
Thus, E(X2) = 2. As E(X) = 0, we find that σ2 = 2 and σX =
√ X
2.

3. Problem 3.
The expected value of the random variable
∫ ∞ is:
E(X) = √ αe−(α− dα
1 µ)2 /(2σ 2 )
2πσ∫ ∞−∞
u=(α−µ)/σ 1 −u 2/2
= √ (σu + µ)e dα.
2π −∞
2
Clearly the piece of the integral associated with ue−u /2 is zero. The
remaining integral is just µ times the integral of the PDF of the standard
normal RV—and must be equal to µ as advertised.
Now let us consider the variance of the RV—let us consider E((X—µ)2).
We find that: ∫∞
E((X — µ) )
2
= √ (α — µ)2e−(α− dα
1 µ)2/(2σ 2 )
2πσ ∫−∞ ∞
u=(α−µ)/σ 2 1 2 −u 2/2
= σ √ u e dα.
2π −∞

As this is just σ2 times the variance of a standard normal RV, we find
that the variance here is σ2.

4. Problem 4.

(a) Clearly (β —α)2 ≥ 0. Expanding this and rearranging it a bit we
find that:
β2 ≥ 2αβ — α2.

(b) Because β2 ≥ 2αβ — α2 and e−a is a decreasing function of a, the
inequality must hold.
(c)
∫ ∞ 2
∫ ∞ 2
−β /2
e dβ ≤ e−(2αβ− α )/2 dβ
α α

, Solutions Manual 3
The PDF Function





0

1/2
2





−2 2




−2
1/2
0




FIGURE 1.1
The PDF of Problem 6.
∫ ∞
α2 /2
=e e−2αβ/2 dβ
α
∞
2 e−αβ
= eα /2
—α α
2
e−α
2
= eα /2
α
2
e−α
=
α
The final step is to plug this into the formula given at the beginning
of the problem statement.
5. Problem 5.
If two random variables are independent, then their joint PDF must be
the product of their marginal PDFs. That is, fXY (α, β) = fX (α)fY (β).
The regions in which the joint PDF are non-zero must be the intersection
of regions in which both marginal PDFs are non-zero. As these regions
are strips in the α, β plains, their intersections are rectangles in that
plain. (Note that for our purposes an infinite region all of whose borders
are right angles to one another is also considered a rectangle.)
6. Problem 6.
Consider the PDF given in Figure 1.1. It is the union of two rectangu-
lar regions. Thus, it is at least possible that the two random variables
are independent. In order for the random variables to actually be in-
dependent it is necessary that fXY (α, β) = fX (α)fY (β) at all points.