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

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r r r




SOLUTIONS

,Solutions Manual r




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




exercises set in the text.
r r r r r




Chapter 1 r




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




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




rthe properties of the probability function, we find that:
r r r r r r r r




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




and that: r




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




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




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




2. Problem 2. r




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




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



∫∞
∫ ∞
r
r




(α)dα=0.5 e−|α|dα
fX
r r r r r




−∞ −∞
∫0 ∫∞ r r r




= 0.5 α
e dα + e−α dα r r r r


−∞ 0
= 0.5(1 + 1)
r r r




= 1. r




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




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




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



∫ ∞ ∫ ∞ r r



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



−∞ 0


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1

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




∫ ∞ r




= α2e−α dα r


0
∫ r
∞
by parts
=
r

(—α2e r
r −α|0∞ r
r +2 r αe −α dα
∫0 r
∞ r r

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


X
2.

3. Problem 3. r




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




E(X) = √ αe−(α− dα
1 µ)2 /(2σ2 r


)
2πσ ∫ −∞ r




u=(α−µ)/σ 1 −u 2 /2 r r r r
r

r
∞

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




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




rstandard normal RV—and must be equal to µ as advertised.
r r r r r r r r r




Nowletus consider thevariance ofthe RV—letus consider E((X µ)—
r r r
2
). We r r r r r r r r r r




find that:
r r ∫ ∞ r
r




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


1 µ) 2 /(2σ2 r


)
2πσ ∫−∞
r



∞ r



u=(α−µ)/σ 2 1 2 −u2 /2 r r r r
r r



= σ √ ue dα. r r


2π −∞

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




that the variance here is σ2.
r r r r r r




4. Problem 4. r




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




find that: r r




β2 ≥ 2αβ — α2. r r r r




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




inequality must hold.
r r r




(c) α

∫ r ∞ r 2
∫ r
∞ r


β
e− /2 r
dβ ≤ r r e−(2αβ−
α
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, Solutions Manual
r 3

2
α )/2
r r
dβ




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