Solutions Manual for Random Signals and Noise: A Mathematical Introduction (1st Edition by Shlomo Engelberg) – Complete Step-by-Step Solutions

All 11 Chapters Covered
m m m




SOLUTIONS

,Solutions Manual m




SUMMARY: In this chapter we present complete solution to the
m m m m m m m m m




exercises set in the text.
m m m m m




Chapter 1 m




1. Problem 1. As defined in the problem, A —B is composed of the elements
m m m m m m m m m m m m m




min A that are not in B. Thus, the items to be noted are true. Making
m m m m m m m m m m m m m m m




muse of the properties of the probability function, we find that:
m m m m m m m m m m




P (A ∪ B) = P (A) + P (B — A)
m m m m m m m m m m m




and that:m




P (B) = P (B — A) + P (A ∩ B).
m m m m m m m m m m m




Combining the two results, we find that: m m m m m m




P (A ∪ B) = P (A) + P (B) — P (A ∩ B).
m m m m m m m m m m m m m m




2. Problem 2. m




(a) It is clear that fX (α) ≥ 0. Thus, we need only check that
m m m m m m m m m m m




the integral of the PDF is equal to 1. We find that:
m m m m m m m m m m m m



∫∞
∫ ∞
m
m




(α) dα = 0.5 e−|α| dα
fX
m m m m m




−∞ −∞
∫ 0 ∫ ∞ m m m




= 0.5 α
e dα + e−α dα m m m m


−∞ 0
= 0.5(1 + 1)m m m




= 1. m




Thus fX(α) is indeed a PDF. m m m m m m




(b) Because fX(α) is even, its expected value must be zero. Addition-
m m m m m m m m m m m




ally, because α2fX(α) is an even function of α, we find that:
m m m m m m m m m m m m m



∫ ∞ ∫ ∞ m m



α2f X (α) dα = 2 α f2X (α) dα m m m m
m



−∞ 0


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1

,2 Random Signals and Noise: A Mathematical Introduction
m m m m m m




∫∞ m




= α2e−α dα m


0
∫ m
∞
by parts
= (—α2m
m −α|0 ∞ m
m + αe −α dα
e m 2
m
∫0 m
∞ m m

by parts
m
−α ∞ m −α
= 2(—αe |0 ) + 2 m
m m e dα
0
= 2.
Thus, E(X2) = 2. As E(X) = 0, we find that σ2 = 2 and σX =
√
m m m m m m m m m m m m m m m m


X
2.

3. Problem 3. m




The expected value of the random variable
m ∫ ∞ is: m m m m m m
m
m




E(X) = √ αe−(α− dα
1 µ)2 /(2σ2 m


)
2πσ ∫−∞ m




u=(α−µ)/σ 1 −u 2 /2 m m m m
m

m
∞

= √ (σu + µ)e m m dα.
2π −∞
2
Clearly the piece of the integral associated with ue−u /2 is zero. The
m m m m m m m m m m m m




remaining integral is just µ times the integral of the PDF of the
m m m m m m m m m m m m




mstandard normal RV—and must be equal to µ as advertised.
m m m m m m m m m




Now let us consider the variance of the RV—let us consider E((X —µ)2).
m m m m m m m m m m m m




We find that:
m m m ∫ ∞ m
m




E((X — µ) ) 2
= √ m (α — µ)2e−(α−m dα m m


1 µ) 2 /(2σ2 m


)
2πσ −∞
m


∫
u=(α−µ)/σ 2 1 2 −u2 /2 m m m m
m m

m
∞

= σ √ m u dα.
e
m
2π
−∞

As this is just σ2 times the variance of a standard normal RV, we
m m m m m m m m m m m m m




find that the variance here is σ2.
m m m m m m m




4. Problem 4. m




(a) Clearly (β —α)2 ≥ 0. Expanding this and rearranging it a bit we
m m m m m m m m m




find that:
m m




β2 ≥ 2αβ — α2. m m m m




(b) Because β2 ≥ 2αβ — α2 and e−a is a decreasing function of a, the
m m m m m m m m m m




inequality must
m

∫ hold. m m

m ∞ m 2
β
(c) e− m/2
α
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, Solutions Manual 3
∫ ∞
m

m




dβ ≤
m
m m



2
e−(2αβ− α m )/2 m
dβ
α




@@
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isloaltaiotinon