STUDY NOTES & REFERENCE GUIDE
All 11 Chapters Covered
SOLUTIONS
Acadexas Stuvia | Page 1 of 71
, STUDY NOTES & REFERENCE GUIDE
Solutions Manual
SUMMARY: In this chapter we present complete solution to the
exercises set in the text.
Chapter 1
1. Problem 1. As defined in the problem,
— A B is composed of the
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 fX (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 ∫ ∞
α
= 0.5 e dα e−α
−∞ 0
+ dα
= 0.5(1 + 1)
= 1.
Thus fX (α) is indeed a PDF.
(b) Because fX (α) is even, its expected value must be zero.
Addition- ally, because α2fX (α) is an even function of α,
we find that:
∫ ∞ ∫ ∞ @@
SeSiesim
smiciiicsiosloaltaiotinon
2
α f X(α) dα = 2 α2f
X
−∞ 0
Acadexas Stuvia | Page 2 of 71
, STUDY NOTES & REFERENCE GUIDE
(α) dα
1
Acadexas Stuvia | Page 3 of 71
, STUDY NOTES & REFERENCE GUIDE
Acadexas Stuvia | Page 4 of 71
All 11 Chapters Covered
SOLUTIONS
Acadexas Stuvia | Page 1 of 71
, STUDY NOTES & REFERENCE GUIDE
Solutions Manual
SUMMARY: In this chapter we present complete solution to the
exercises set in the text.
Chapter 1
1. Problem 1. As defined in the problem,
— A B is composed of the
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 fX (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 ∫ ∞
α
= 0.5 e dα e−α
−∞ 0
+ dα
= 0.5(1 + 1)
= 1.
Thus fX (α) is indeed a PDF.
(b) Because fX (α) is even, its expected value must be zero.
Addition- ally, because α2fX (α) is an even function of α,
we find that:
∫ ∞ ∫ ∞ @@
SeSiesim
smiciiicsiosloaltaiotinon
2
α f X(α) dα = 2 α2f
X
−∞ 0
Acadexas Stuvia | Page 2 of 71
, STUDY NOTES & REFERENCE GUIDE
(α) dα
1
Acadexas Stuvia | Page 3 of 71
, STUDY NOTES & REFERENCE GUIDE
Acadexas Stuvia | Page 4 of 71
