****INSTANT DOWNLOAD****PDF***Solutions Manual for Random Signals and Noise: A Mathematical Introduction (1st Edition) by Shlomo Engelberg

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




cthat are not in B. Thus, the items to be noted are true. Making use of
c c c c c c c c c c c c c c c




cthe properties of the probability function, we find that:
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)dα = 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 even, its expected value must be zero. Addition-
c c c c c c c c c c c




ally, because α2fX(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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,2 Random Signals and Noise: A Mathematical Introduction c c c c c c




∫ ∞ c




= α2e−α dα c


0
∫ c
∞
by parts
=
c

(—α2e 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 /(2σ2 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 advertised.
c c c c c c c c c




Now let us consider the variance of the RV—let us consider 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) = √ (α — µ)2e−(α−
c dα
c c c


1 µ) 2 /(2σ2 c


)
2πσ ∫−∞
c



∞ c



u=(α−µ)/σ 2 1 2 −u2 /2 c c c c
c c



= σ √ ue 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 )/2
c c
dβ




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