SOLUTIONS MANUAL for Random Signals and Noise A Mathematical Introduction 1st edition by Shlomo Engelberg

All 11 Chapters Covered




SOLUTIONS

,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



1

,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 k0PDF k0Function





0

1/2
2





−2 2




−2
1/2
0




FIGURE 1.1 k0



The PDF of Problem
k0 k0 k0



6.
k0

∫ k 0
∞ k0
2
α /2 −2αβ/2
= e k0 k 0 e dβ
α
∞
2 e−αβ k 0


= eα k0
k 0 /2 k0

—α k0 k0
α
2
2 e−α
= eα k0
k 0 /2
α
2
e−α
=
α
The final step is to plug this into the formula given at the
k0 k0 k0 k0 k0 k0 k0 k0 k0 k0 k0 k0


beginning of the problem statement.
k0 k0 k0 k0 k0




5. Problem 5. k0




If two random variables are independent, then their joint PDF must
k0 k0 k0 k0 k0 k0 k0 k0 k0 k0



be the product of their marginal PDFs. That is, fXY (α, β) = fX
k0 k0 k0 k0 k0 k0 k0 k0 k0 k0 k0 k0 k0 k0


(α)fY (β). The regions in which the joint PDF are non-zero must
k0 k0 k0 k0 k0 k0 k0 k0 k0 k0 k0 k0



k0be the intersection of regions in which both marginal PDFs are non-
k0 k0 k0 k0 k0 k0 k0 k0 k0 k0 k0



zero. As these regions are strips in the α, β plains, their
k0 k0 k0 k0 k0 k0 k0 k0 k0 k0 k0


intersections are rectangles in that plain. (Note that for our
k0 k0 k0 k0 k0 k0 k 0 k0 k0 k0



purposes an infinite region all of whose borders are right angles to
k0 k0 k0 k0 k0 k0 k0 k0 k0 k0 k0 k0


one another is also considered a rectangle.)
k0 k0 k0 k0 k0 k0 k0




6. Problem 6. k0




Consider the PDF given in Figure 1.1. It is the union of two
k0 k0 k0 k0 k0 k0 k0 k0 k0 k0 k0 k0


rectangu- lar regions. Thus, it is at least possible that the two
k0 k0 k0 k0 k0 k0 k0 k0 k0 k0 k0 k0



random variables are independent. In order for the random
k0 k0 k0 k0 k 0 k0 k0 k0 k0



variables to actually be in- dependent it is necessary that fXY
k0 k0 k0 k0 k0 k0 k0 k 0 k 0 k 0 k0


(α, β) = fX(α)fY (β) at all points.
k0 k0 k0 k0 0
k k0 k 0 k 0 k 0