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

All 11Chapters Covered
c c c




SOLUTIONS

,Solutions Manual c




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




exercises set in the text.
c c c c c




Chapter 1 c




c c —c ompose d o f t he e le me nts in A
1. Proble m 1. As define d in the pr oble m, A B is c c c c c c c c c c c c c




that are not in B. Thus, the ite ms to be noted are true. Making use of
c c c c c c c c c c c c c c c c




the properties of the probability function, we find that:
c c c c c c c c c




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




and that:
c




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




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




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




2. Problem 2. c




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




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



∫∞ c

∫ ∞ c




(α) d α = 0 .5 e−|α| dα
fX
c c c c c




−∞ −∞
∫ 0 ∫ ∞ c c c




= 0.5 α
e dα + e−α dα c c c c


−∞ 0
= 0.5(1 + 1 )
c c c




= 1. c




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




(b) Because fX (α) is e ven, its e xpected value must be zero. Addition-
c c c c c c c c c c c




ally, because α2 fX (α) is an even function of α, we find that:
c c c c c c c c c c c c c



∫ ∫
∞ ∞ c c


α2f X (α) dα = 2 α 2f X (α) dα c c c c
c


−∞ 0


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1

,2 Random Signals and Noise: A Mathema tical Introduction c c c c c c




∫ ∞ c



= α2 e−α d α c


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

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


X
2.

3. Problem 3. c




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




E(X) = √ αe−(α− dα
1 µ) 2 /(2σ 2 c


)
2πσ ∫ −∞
c




u=(α−µ)/ σ 1 −u 2/ 2 c c c c
c

c
∞

= √ (σu + µ)e c c dα.
2π −∞

2
Clearly the piece of the integral associated with ue−u /2 is zero. The
c c c c c c c c c c c c



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




cstandard normal RV—and must be equal to µ as advertise d.
c c c c c c c c c




Now let us c onsider the variance of the RV—le t us c onsider E((X µ)—
c c c
2 ). We c c c c c c c c c c




find that:
c c ∫ ∞ c
c




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


1 µ) 2 /(2σ2 c


)
2πσ −∞
c


∫∞ c



u=(α−µ)/ σ 2 1 2 −u2 /2 c c c c




σ √
c c


= u e dα. c c


2π −∞

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




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




4. Problem 4. c




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




find that:c c




β2 ≥ 2αβ— α2 . c c c c




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




inequality must hold.
c c c




(c ) α


∫ c ∞ c 2
∫ c
∞ c


β
e− /2 c
dβ ≤ c c e−(2αβ−
α

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, Solutions Manual
c 3

2
α )/2c c
dβ




@@
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