Solutions Manual: For Random Signals and Noise a Mathematical Introduction (1st Edition) - Graded A+ Comprehensi...

STUDY NOTES & REFERENCE GUIDE




All 11 Chapters Covered




SOLUTIONS




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Solutions Manual


S UMMARY: 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:
∫∞
∫ ∞
fX (α) dα = 0.5 e−|α| dα
−∞ −∞
∫ 0 ∫ ∞
= 0.5 eα dα e−α
−∞ 0
+ dα
= 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:
∫ ∞ ∫ ∞
α2fX(α) dα = 2 α2f (α) dα
X
−∞ 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 d
∫0 ∞ α
by parts −α ∞ −α
= 2(—αe |0 ) + 2 e dα
0
= 2.
Thus, E(X ) = 2. As E(X) = 0, we find that σ2 = 2 and σX =
2
√ X
2.

3. Problem 3.
The expected value of the random variable
∫ ∞ is:
E(X) = √ αe−(α− dα
1 µ)2 /(2σ
2
)
2 σ ∫−∞
u=(α−µ)/σ 1 −u 2/2
∞

= √ (σu + dα.
2π µ)e
−∞
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
)
2 σ ∫−∞∞
u=(α−µ)/σ 2 1 2 −u2 /2
= σ √ u dα.
e 2
−∞

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.
β
(c) /2
α
∫ ∞ 2

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Solutions 3
∫ ∞
Manual
dβ ≤
e−(2αβ− 2
α )/2
dβ
α




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