Solutions Manual for Random Signals and Noise A Mathematical Introduction 1st Edition by Shlomo Engelberg PDF | Complete Step-by-Step Solutions and Explanations | Covers Probability, Random Variables, Expectation, Power Spectral Density, Autocorrelation,

All 11 Chapters Covered




SOLUTIONS

,Solutions Manual


SUMMARY: In this chapter we present complete solution to the
exercises set in the text.




Chapter 1
— A B is composed of the
1. Problem 1. As defined in the problem,
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 f X (α) ≥ 0. Thus, we need only check that
the integral of the PDF is equal to 1. We find that:
∫∞
∫
∞ (α) dα = 0.5 e−|α| dα
fX
−∞ −∞
∫ 0 ∫ ∞
= 0.5 α
e dα + e−α dα
−∞ 0
= 0.5(1 + 1)
= 1.
Thus fX (α) is indeed a PDF.
(b) Because f X (α) is even, its expected value must be zero.
Addition- ally, because α2f X (α) is an even function of α, we
find that:
∫ ∞ ∫ ∞ X
−∞ 0
2
α f X(α) dα = 2 α2f

@LECTJULIESOLUTIONSSTUVIA

,(α) dα

1

,
,2 Random Signals and Noise: A Mathematical Introduction
∫ ∞
= α 2e−α dα
0
∫ ∞
by parts
= (—α2e −α |∞
0 +2 αe −α dα
∫0 ∞
by parts −α ∞ −α
= 2(—αe |0 ) + 2 e dα
0
= 2.
Thus, E(X2) = 2. As E(X) = 0, we find that σ 2 = 2 and σX =
√ X
2.

3. Problem 3.
The expected value of the random ∫ variable
∞ is:
E(X) = √ αe −(α− dα
1 µ)2 /(2σ2 )
2πσ∫ ∞
−∞
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 /(2σ2 )
2πσ ∫−∞∞
u=(α−µ)/σ 2 1 2 −u2 /2
= σ √ u e dα.
2π −∞

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.
(c)
∫ ∫ ∞
∞ 2 2
− β /2
e dβ ≤ e−(2αβ− α )/2
dβ
α α
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,Solutions Manual 3
The PDF Function




0

1/2
2




−2 2




−2
1/2
0




FIGURE 1.1
The PDF of Problem
6. ∫ ∞
2
= eα /2 e −2αβ/2 dβ
α
∞
2 e−αβ
= eα /2
—α α
2
e−α 2
= eα /2
α
2
e−α
=
α
The final step is to plug this into the formula given at the
beginning of the problem statement.
5. Problem 5.
If two random variables are independent, then their joint PDF
must be the product of their marginal PDFs. That is, fXY (α, β)
= fX (α)fY (β). The regions in which the joint PDF are non-zero
must be the intersection of regions in which both marginal PDFs
are non-zero. As these regions are strips in the α, β plains, their
intersections are rectangles in that plain. (Note that for our
purposes an infinite region all of whose borders are right angles to
one another is also considered a rectangle.)
6. Problem 6.
Consider the PDF given in Figure 1.1. It is the union of two
rectangu- lar regions. Thus, it is at least possible that the two
random variables are independent. In order for the random
variables to actually be in- dependent it is necessary that fXY
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,4 Random
(α, β) = f X (α)f Signals and Noise: A Mathematical Introduction
Y (β) at all points.




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,Solutions Manual 5

Let us consider the point (—2.5, 2.5). It is clear that f X (—2.5)
= 0.5 and fY (2.5) = 0.5. Thus if the random variable are
—
independent, f XY (· 2.5, 2.5) = 0.5 0.5. However, the actual
value of the PDF at that point is 0. Thus, the random variables
are not independent.
Are the random variables correlated? Let us consider E(XY ).
Because the probability is only non-zero when either both α and
β are positive or both are negative, it is clear that:
∫ ∫
αβfXY (α, β) dαdβ > 0.

It is also easy to see that the marginal PDFs of X and Y are
even func- tions. Thus, E(X) = E(Y ) = 0. We find that
E(XY ) = E(X)E(Y ) and the random variables are correlated.

7. Problem 7. Making use of the definition of the fact that the Xi
are zero-mean, the fact that the Xi have a common variance,
and the fact that the Xi are mutually uncorrelated, we find
that:

E(Q) = E(R) = E(S) = 0

and that:
σ2 = σ2 = σ2 = E((X3 + X4)2) = E(X2 + 2X3X4 + X2) = 2σ2 .
Q R S 3 4 X

Now let us calculate several important expectations. We find that:

E(QR) = E((X1 + X2)(X2 + X3))
= E(X1X2 + X1X 3 +2 X + X2X3)
2

=0+0+σ X +0
2

= σX
2 ,


and that:

E(QS) = E((X1 + X2)(X3 + X4))
= E(X1X 3 + X1X4 + X2X3 + X2X4)
= 0+0+0+0
= 0,

and that:

E(RS) = E((X2 + X3)(X3 + X4))
= E(X2X3 + X2X 4 +
3 X + X3X4)
2

=0+0+σ X +0
2

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,6 Random Signals and Noise: A Mathematical Introduction
= σ2 .
X




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, Solutions Manual 7

Making use of the preceding calculations and the definition of the
cor- relation coefficient we find that:
ρQR = 1/2, ρQS = 0, ρRS = 1/2.

These results are quite reasonable. If the correlation coefficient
really measures the degree of “sameness,” then as Q and R are “half
the same” and Q and S have no overlap their correlation
coefficients ought to be 1/2 and zero respectively. Similarly, as
R and S overlap in half their constituent parts the degree of
correlation ought to be 1/2.
8. Problem 8.
—
(a) With f X (α) a pulse of unit height that stretches from 1/2
to 1/2, we find that:
∫ 1/2
ϕX (t) = ejαtdα
−1/2
∫ 1/2 ∫ 1/2
= cos(αt)dα + j sin(αt)dα
−1/2 −1/2
1
= (sin(t/2) — sin(—t/2)) + 0
t
2 sin(t/2)
= .
t
(How can this argument be made more precise (correct)
when t = 0?)
(b) We must calculate ϕ′X (t)|t=0 and ϕ′X′ (t)|t=0. The easiest
way to do do this is to calculate the Taylor series associated
with ϕX (t). We find that:
2(t/2 — (t/2)3/3! + · · ·)
ϕX (t) =
t
= 1 — t2/24 + · · ·
= ϕX (0) + ϕ′X (0)t + ϕ′X′ (0)t2/2 + · · · .
By inspection, we find that ϕ ′X (0) = 0 and ϕ′X′ —
(0) =
1/12. We
find that:
jE(X) = 0
—E(X2) = —1/12.
Thus E(X) = 0, and E(X2) = 1/12.
9. Problem 9.
Making use of the definition of the characteristic function, we find that:

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