All 11ChaptersCovered
w w w
SOLUTIONS
,Solutions Manual w
S UMMARY: In this chapter we present complete solution to the exercis
w w w w w w w w w w
es set in the text.
w w w w
Chapter 1 w
1. Problem 1. As defined in the problem, A B—
w is composed of the elements in
w w w w w w w w w w w w w w
A that are not in B. Thus, the items to be noted are true. Making use of
w w w w w w w w w w w w w w w w w
the properties of the probability function, we find that:
w w w w w w w w
P(A ∪ B) = P (A) + P (B — A)
w w w w w w w w w w w
and that:w
P(B) = P (B — A) + P (A ∩ B).
w w w w w w w w w w w
Combining the two results, we find that: w w w w w w
P (A ∪ B) = P (A) + P (B) — P (A ∩ B).
w w w w w w w w w w w w w w
2. Problem 2. w
(a) It is clear that fX (α)
w w w w w ≥
0. Thus, we need only check that the i w w w w w w w w
ntegral of the PDF is equal to 1. We find that: w w w w w w w w w w
∫∞
∫ ∞
w
w
fX (α) dα = 0.5 e−|α| dα w w w w w
−∞ −∞
∫0 ∫∞ w w w
= 0.5 α
e dα + e−α dα w w w w
−∞ 0
= 0.5(1 + 1)w w w
= 1. w
Thus fX(α) is indeed a PDF. w w w w w w
(b) Because fX(α) is even, its expected value must be zero. Addition-w w w w w w w w w w w
ally, because α 2fX(α) is an even function of α, we find that:
w w w w w w w w w w w w w
∫ ∞ ∫ ∞ w w
α2f X (α) dα = 2 α2f X (α) dα w w w w
w
−∞ 0
@@
SeSiesim
smiciiicsiosloaltaiotinon
1
,2 Random Signals and Noise: A Mathematical Introduction
w w w w w w
∫ ∞ w
= α2e−α dα w
0
∫ w
∞
by parts
=
w
(—α2e −α|∞
0 +2 w
w w
w w αe −α dα
∫∞
0 w
w w
by partsw
−α ∞ w −α
= 2(—αe |0 ) + 2
w
w w e dα
0
= 2.
Thus, E(X2 ) = 2. As E(X) = 0, we find that σ2 = 2 and σX =
w w w w w w w w w w w w w w w w
√ X
2.
3. Problem 3. w
The expected value of the random variable
w
∫ ∞ is: w w w w w
w
w
w
E(X) = √ αe−(α− dα
1 µ)2/(2σ 2) w w
2πσ
∫ ∞ −∞ w
u=(α−µ)/σ 1 −u 2/2 w w w w
√
w
= (σu + µ)e dα. w w
2π −∞
2
Clearly the piece of the integral associated with ue−u /2 is zero. The
w w w w w w w w w w w w
remaining integral is just µ times the integral of the PDF of the standard no
w w w w w w w w w w w w w w
rmal RV—and must be equal to µ as advertised.
w w w w w w w w
Now let us consider the variance of the RV—
w w w — w w w w w
let us consider E((X µ)2). We find that:
w w ∫∞ w w w w w w
w
E((X — µ)2) = √ (α — µ)2e−(α−
w w dα w w
1 µ)2/(2σ 2) w w
2πσ ∫ −∞
∞ w
u=(α−µ)/σ 2 1 2 −u 2/2 ww w w
w w
= σ√ ue dα. w w
2π −∞
As this is just σ2 times the variance of a standard normal RV, we find th
w w w w w w w w w w w w w w w
at the variance here is σ2.
w w w w w
4. Problem 4. w
(a) Clearly (β —α)2 ≥ w
0. Expanding this and rearranging it a bit we find that: w w w w w w w w w w
β2 ≥ 2αβ — α2. w w w w
(b) Because β2 ≥ 2αβ — w
α2 and e−a is a decreasing function of a, the inequali w w w w w w w w w w
ty must hold. w w
(c)
∫ w
∞ w
2
∫∞ w
w
2
e − β /2 w
dβ ≤ w w e−(2αβ− α )/2
w w
dβ
α α
@@
SeSie
sm iciiis
i sm co
islo
altaiotinon
, Solutions Manual w 3
The PDF Function
w w
0
1/2
2
−2 2
1/2
−2
0
FIGURE 1.1 w
The PDF of Problem 6.
w w w w
∫ w
∞ w
2
=eα w w
w /2
e −2αβ/2 dβ
α
∞
2 e−αβ w
w
= eα w /2 w
—α α
w
w w
2
2 e−α
= eα /2 w
w
α
2
e−α
=
α
The final step is to plug this into the formula given at the beginning of
w w w w w w w w w w w w w w w
the problem statement.
w w
5. Problem 5. w
If two random variables are independent, then their joint PDF must be th
w w w w w w w w w w w w
e product of their marginal PDFs. That is, fXY (α, β) = fX (α)fY (β). The r
w w w w w w w w w w w w w w w w
egions in which the joint PDF are non-
w w w w w w w
zero must be the intersection of regions in which both marginal PDFs are
w w w w w w w w w w w w w
non-
zero. As these regions are strips in the α, β plains, their intersections ar
w w w w w w w w w w w w w
e rectangles in that plain. (Note that for our purposes an infinite region all o
w w w w w w w w w w w w w w
f whose borders are right angles to one another is also considered a recta
w w w w w w w w w w w w w
ngle.)
6. Problem 6. w
Consider the PDF given in Figure 1.1. It is the union of two rectangu-
w w w w w w w w w w w w w
lar regions. Thus, it is at least possible that the two random variables ar
w w w w w w w w w w w w w w
e independent. In order for the random variables to actually be in-
w w w w w w w w w w w
dependent it is necessary that f XY (α, β) = fX(α)fY (β) at all points.
w w w w w w w w w w w w w w w
@@
SeSie
sm iciiis
i sm co
islo
altaiotinon
w w w
SOLUTIONS
,Solutions Manual w
S UMMARY: In this chapter we present complete solution to the exercis
w w w w w w w w w w
es set in the text.
w w w w
Chapter 1 w
1. Problem 1. As defined in the problem, A B—
w is composed of the elements in
w w w w w w w w w w w w w w
A that are not in B. Thus, the items to be noted are true. Making use of
w w w w w w w w w w w w w w w w w
the properties of the probability function, we find that:
w w w w w w w w
P(A ∪ B) = P (A) + P (B — A)
w w w w w w w w w w w
and that:w
P(B) = P (B — A) + P (A ∩ B).
w w w w w w w w w w w
Combining the two results, we find that: w w w w w w
P (A ∪ B) = P (A) + P (B) — P (A ∩ B).
w w w w w w w w w w w w w w
2. Problem 2. w
(a) It is clear that fX (α)
w w w w w ≥
0. Thus, we need only check that the i w w w w w w w w
ntegral of the PDF is equal to 1. We find that: w w w w w w w w w w
∫∞
∫ ∞
w
w
fX (α) dα = 0.5 e−|α| dα w w w w w
−∞ −∞
∫0 ∫∞ w w w
= 0.5 α
e dα + e−α dα w w w w
−∞ 0
= 0.5(1 + 1)w w w
= 1. w
Thus fX(α) is indeed a PDF. w w w w w w
(b) Because fX(α) is even, its expected value must be zero. Addition-w w w w w w w w w w w
ally, because α 2fX(α) is an even function of α, we find that:
w w w w w w w w w w w w w
∫ ∞ ∫ ∞ w w
α2f X (α) dα = 2 α2f X (α) dα w w w w
w
−∞ 0
@@
SeSiesim
smiciiicsiosloaltaiotinon
1
,2 Random Signals and Noise: A Mathematical Introduction
w w w w w w
∫ ∞ w
= α2e−α dα w
0
∫ w
∞
by parts
=
w
(—α2e −α|∞
0 +2 w
w w
w w αe −α dα
∫∞
0 w
w w
by partsw
−α ∞ w −α
= 2(—αe |0 ) + 2
w
w w e dα
0
= 2.
Thus, E(X2 ) = 2. As E(X) = 0, we find that σ2 = 2 and σX =
w w w w w w w w w w w w w w w w
√ X
2.
3. Problem 3. w
The expected value of the random variable
w
∫ ∞ is: w w w w w
w
w
w
E(X) = √ αe−(α− dα
1 µ)2/(2σ 2) w w
2πσ
∫ ∞ −∞ w
u=(α−µ)/σ 1 −u 2/2 w w w w
√
w
= (σu + µ)e dα. w w
2π −∞
2
Clearly the piece of the integral associated with ue−u /2 is zero. The
w w w w w w w w w w w w
remaining integral is just µ times the integral of the PDF of the standard no
w w w w w w w w w w w w w w
rmal RV—and must be equal to µ as advertised.
w w w w w w w w
Now let us consider the variance of the RV—
w w w — w w w w w
let us consider E((X µ)2). We find that:
w w ∫∞ w w w w w w
w
E((X — µ)2) = √ (α — µ)2e−(α−
w w dα w w
1 µ)2/(2σ 2) w w
2πσ ∫ −∞
∞ w
u=(α−µ)/σ 2 1 2 −u 2/2 ww w w
w w
= σ√ ue dα. w w
2π −∞
As this is just σ2 times the variance of a standard normal RV, we find th
w w w w w w w w w w w w w w w
at the variance here is σ2.
w w w w w
4. Problem 4. w
(a) Clearly (β —α)2 ≥ w
0. Expanding this and rearranging it a bit we find that: w w w w w w w w w w
β2 ≥ 2αβ — α2. w w w w
(b) Because β2 ≥ 2αβ — w
α2 and e−a is a decreasing function of a, the inequali w w w w w w w w w w
ty must hold. w w
(c)
∫ w
∞ w
2
∫∞ w
w
2
e − β /2 w
dβ ≤ w w e−(2αβ− α )/2
w w
dβ
α α
@@
SeSie
sm iciiis
i sm co
islo
altaiotinon
, Solutions Manual w 3
The PDF Function
w w
0
1/2
2
−2 2
1/2
−2
0
FIGURE 1.1 w
The PDF of Problem 6.
w w w w
∫ w
∞ w
2
=eα w w
w /2
e −2αβ/2 dβ
α
∞
2 e−αβ w
w
= eα w /2 w
—α α
w
w w
2
2 e−α
= eα /2 w
w
α
2
e−α
=
α
The final step is to plug this into the formula given at the beginning of
w w w w w w w w w w w w w w w
the problem statement.
w w
5. Problem 5. w
If two random variables are independent, then their joint PDF must be th
w w w w w w w w w w w w
e product of their marginal PDFs. That is, fXY (α, β) = fX (α)fY (β). The r
w w w w w w w w w w w w w w w w
egions in which the joint PDF are non-
w w w w w w w
zero must be the intersection of regions in which both marginal PDFs are
w w w w w w w w w w w w w
non-
zero. As these regions are strips in the α, β plains, their intersections ar
w w w w w w w w w w w w w
e rectangles in that plain. (Note that for our purposes an infinite region all o
w w w w w w w w w w w w w w
f whose borders are right angles to one another is also considered a recta
w w w w w w w w w w w w w
ngle.)
6. Problem 6. w
Consider the PDF given in Figure 1.1. It is the union of two rectangu-
w w w w w w w w w w w w w
lar regions. Thus, it is at least possible that the two random variables ar
w w w w w w w w w w w w w w
e independent. In order for the random variables to actually be in-
w w w w w w w w w w w
dependent it is necessary that f XY (α, β) = fX(α)fY (β) at all points.
w w w w w w w w w w w w w w w
@@
SeSie
sm iciiis
i sm co
islo
altaiotinon
