All11ChaptersCovered
q q q
SOLUTIONS
,Solutions Manual q
SUMMARY: In this chapter we present complete solution to the
q q q q q q q q q q
exercises set in the text. q q q q
Chapter 1 q
1. Problem 1. As defined in the problem, A B—is composed of the elements i
q q q q q q q q q q q q q q
n A that are not in B. Thus, the items to be noted are true. Making use
q q q q q q q q q q q q q q q q q
of the properties of the probability function, we find that:
q q q q q q q q q
P(A ∪ B) = P(A) + P(B — A)
q q q q q q q q q q q
and that: q
P(B) = P(B — A) + P(A ∩ B).
q q q q q q q q q q q
Combining the two results, we find that: q q q q q q
P(A ∪ B) = P(A) + P(B) — P(A ∩ B).
q q q q q q q q q q q q q q
2. Problem 2. q
(a) It is clear that fX(α) q q q q q ≥
0. Thus, we need only check that the i q q q q q q q q
ntegral of the PDF is equal to 1. We find that: q q q q q q q q q q
∫∞
∫ ∞
q
q
(α) d α = 0.5 e−|α|dα
fX
q q q q q
−∞ −∞
∫0 ∫∞ q q q
= 0.5 eα dα + e−α dα q q q q
−∞ 0
= 0.5(1 + 1)q q q
= 1. q
Thus fX(α) is indeed a PDF. q q q q q q
(b) Because fX(α) is even, its expected value must be zero. Addition- q q q q q q q q q q q
ally, because α2fX(α) is an even function of α, we find that:
q q q q q q q q q q q q q
∫ ∞ ∫ ∞ q q
α2f X (α) dα = 2 α2f q q q q
X
−∞ 0
,(α) dα
q
1
,
q q q
SOLUTIONS
,Solutions Manual q
SUMMARY: In this chapter we present complete solution to the
q q q q q q q q q q
exercises set in the text. q q q q
Chapter 1 q
1. Problem 1. As defined in the problem, A B—is composed of the elements i
q q q q q q q q q q q q q q
n A that are not in B. Thus, the items to be noted are true. Making use
q q q q q q q q q q q q q q q q q
of the properties of the probability function, we find that:
q q q q q q q q q
P(A ∪ B) = P(A) + P(B — A)
q q q q q q q q q q q
and that: q
P(B) = P(B — A) + P(A ∩ B).
q q q q q q q q q q q
Combining the two results, we find that: q q q q q q
P(A ∪ B) = P(A) + P(B) — P(A ∩ B).
q q q q q q q q q q q q q q
2. Problem 2. q
(a) It is clear that fX(α) q q q q q ≥
0. Thus, we need only check that the i q q q q q q q q
ntegral of the PDF is equal to 1. We find that: q q q q q q q q q q
∫∞
∫ ∞
q
q
(α) d α = 0.5 e−|α|dα
fX
q q q q q
−∞ −∞
∫0 ∫∞ q q q
= 0.5 eα dα + e−α dα q q q q
−∞ 0
= 0.5(1 + 1)q q q
= 1. q
Thus fX(α) is indeed a PDF. q q q q q q
(b) Because fX(α) is even, its expected value must be zero. Addition- q q q q q q q q q q q
ally, because α2fX(α) is an even function of α, we find that:
q q q q q q q q q q q q q
∫ ∞ ∫ ∞ q q
α2f X (α) dα = 2 α2f q q q q
X
−∞ 0
,(α) dα
q
1
,
