Solutions for Random Signals and Noise, 1st Edition Engelberg (All Chapters included)
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Module
Analysis of Signals And systems
Institution
Analysis Of Signals And Systems
Complete Solutions Manual for Random Signals and Noise, 1st Edition by Shlomo Engelberg ; ISBN13: 9780849375545.(Full Chapters included Chapter 1 to 11)....1.ELEMENTARY PROBABILITY THEORY
2.AN INTRODUCTION TO STOCHASTIC PROCESSES
3.THE WEAK LAW OF LARGE NUMBERS
4.THE CENTRAL LIMIT THEOREM
5.EXT...
SOLUTIONS MANUAL FOR
Random Signals
and Noise:
A Mathematical
Introduction
Complete Chapter Solutions Manual
are included (Ch 1 to 11)
by
Shlomo Engelberg
** Immediate Download
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** All Chapters included
8861.indd 1 9/8/08 3:39:43
,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:
Z ∞ Z ∞
fX (α) dα = 0.5 e−|α| dα
−∞ −∞
Z0 Z ∞
α −α
= 0.5 e dα + e 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 α2 fX (α) is an even function of α, we find that:
Z ∞ Z ∞
α2 fX (α) dα = 2 α2 fX (α) dα
−∞ 0
1
, 2 Random Signals and Noise: A Mathematical Introduction
Z ∞
= α2 e−α dα
0
Z ∞
by parts
= (−α2 e−α |∞ 0 + 2 αe−α dα
0
Z ∞
by parts
= 2(−αe−α |∞ 0 ) + 2 e−α dα
0
= 2.
Thus, E(X 2 ) = 2. As E(X) = 0, we find that σX
√
2
= 2 and σX =
2.
3. Problem 3.
The expected value of the random variable is:
Z ∞
1 2 2
E(X) = √ αe−(α−µ) /(2σ ) dα
2πσ −∞
Z ∞
u=(α−µ)/σ 1 2
= √ (σu + µ)e−u /2 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:
Z ∞
1 2 2
E((X − µ) )2
= √ (α − µ)2 e−(α−µ) /(2σ ) dα
2πσ −∞
Z ∞
u=(α−µ)/σ 2 1 2
= σ √ u2 e−u /2 dα.
2π −∞
As this is just σ 2 times the variance of a standard normal RV, we find
that the variance here is σ 2 .
4. Problem 4.
(a) Clearly (β − α)2 ≥ 0. Expanding this and rearranging it a bit we
find that:
β 2 ≥ 2αβ − α2 .
(b) Because β 2 ≥ 2αβ − α2 and e−a is a decreasing function of a, the
inequality must hold.
(c)
Z ∞ Z ∞
2 2
e−β /2
dβ ≤ e−(2αβ−α )/2
dβ
α α
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