We are given that the proportion of defectives

We are given that the proportion of defectives follows a uniform distribution on (0, 1), so
the unconditional probability that Y = 2 can be found by
∫ ∫ ∫ ∫
− = − = = = = = = −
1
0
3 2
1
0
1
0
1
0
1 3 2 ) ( 3 ) 1 ( 3 ) ( ) | 2 ( ) , 2 ( ) 2 ( dp p p dp p p dp p f p Y P dp p Y P Y P
= 1/4.
5.42 (Similar to Ex. 5.41) Let Y = # of defects per yard. Then,
( )
1
2
1
0
!
0 0
) ( ) | ( ) , ( ) (
+
λ −
∞
λ
∞ ∞
= λ = λ λ λ = = λ λ = =
∫ ∫ ∫
λ −
y
y
e d e d f y Y P d y Y P y p
y , y = 0, 1, 2, … .
Note that this is essentially a geometric distribution (see Ex. 3.88).
5.43 Assume ). ( ) | (
1 1 2 1
y f y y f = Then, ) ( ) ( ) ( ) | ( ) , (
2 2 1 1 2 2 2 1 2 1
y f y f y f y y f y y f = = so that
Y
1
and Y
2
are independent. Now assume that Y
1
and Y
2
are independent. Then, there
exists functions g and h such that ) ( ) ( ) , (
2 1 2 1
y h y g y y f = so that
∫ ∫ ∫ ∫
× = =
2 2 1 1 2 1 2 1
) ( ) ( ) , ( 1 dy y h dy y g dy dy y y f .
Then, the marginals for Y
1
and Y
2
can be defined by
∫
∫
∫ ∫
=
×
=
1 1
1
2
2 2 1 1
2 1
1 1
) (
) (
) ( ) (
) ( ) (
) (
dy y g
y g
dy
dy y h dy y g
y h y g
y f
, so
∫
=
2 2
2
2 2
) (
) (
) (
dy y h
y h
y f
.
Thus, ) ( ) ( ) , (
2 2 1 1 2 1
y f y f y y f = . Now it is clear that
) ( ) ( / ) ( ) ( ) ( / ) , ( ) | (
1 1 2 2 2 2 1 1 2 2 2 1 2 1
y f y f y f y f y f y y f y y f = = = ,
provided that ) (
2 2
y f > 0 as was to be shown.
5.44 The argument follows exactly as Ex. 5.43 with integrals replaced by sums and densities
replaced by probability mass functions.
5.45 No. Counterexample: P ( Y
1
= 2, Y
2
= 2) = 0 ≠ P ( Y
1
= 2) P ( Y
2
= 2) = (1/9)(1/9).
5.46 No. Counterexample: P ( Y
1
= 3, Y
2
= 1) = 1/8 ≠ P ( Y
1
= 3) P ( Y
2
= 1) = (1/8)(4/8).

102 Chapter 5: Multivariate Probability Distributions
Instructor’s Solutions Manual
5.47 Dependent. For example: P ( Y
1
= 1, Y
2
= 2) ≠ P ( Y
1
= 1) P ( Y
2
= 2).
5.48 Dependent. For example: P ( Y
1
= 0, Y
2
= 0) ≠ P ( Y
1
= 0) P ( Y
2
= 0).
5.49 Note that 1 0 , 3 3 ) (
1
2
1
0
2 1 1 1
1
≤ ≤ = =
∫
y y dy y y f
y
, 1 0 ], 1 [ 3 ) (
2
2
2 2
3
1
1 1 2 2
1
≤ ≤ − = =
∫
y y dy y y f
y
.
Thus, ) ( ) ( ) , (
2 2 1 1 2 1
y f y f y y f ≠ so that Y
1
and Y
2
are dependent.
5.50 a. Note that 1 0 , 1 1 ) (
1
1
0
2 1 1
≤ ≤ = =
∫
y dy y f and 1 0 , 1 1 ) (
2
1
0
1 2 2
≤ ≤ = =
∫
y dy y f . Thus,
) ( ) ( ) , (
2 2 1 1 2 1
y f y f y y f = so that Y
1
and Y
2
are independent.
b. Yes, the conditional probabilities are the same as the marginal probabilities.
5.51 a. Note that 0 , ) (
1
0
2
) (
1 1
1 2 1
> = =
−
∞
+ −
∫
y e dy e y f
y y y
and 0 , ) (
2
0
1
) (
2 2
2 2 1
> = =
−
∞
+ −
∫
y e dy e y f
y y y
.
Thus, ) ( ) ( ) , (
2 2 1 1 2 1
y f y f y y f = so that Y
1
and Y
2
are independent.
b. Yes, the conditional probabilities are the same as the marginal probabilities.
5.52 Note that ) , (
2 1
y y f can be factored and the ranges of y
1
and y
2
do not depend on each
other so by Theorem 5.5 Y
1
and Y
2
are independent.
5.53 The ranges of y
1
and y
2
depend on each other so Y
1
and Y
2
cannot be independent.
5.54 The ranges of y
1
and y
2
depend on each other so Y
1
and Y
2
cannot be independent.
5.55 The ranges of y
1
and y
2
depend on each other so Y
1
and Y
2
cannot be independent.
5.56 The ranges of y
1
and y
2
depend on each other so Y
1
and Y
2
cannot be independent.
5.57 The ranges of y
1
and y
2
depend on each other so Y
1
and Y
2
cannot be independent.
5.58 Following Ex. 5.32, it is seen that ) ( ) ( ) , (
2 2 1 1 2 1
y f y f y y f ≠ so that Y
1
and Y
2
are
dependent.
5.59 The ranges of y
1
and y
2
depend on each other so Y
1
and Y
2
cannot be independent.
5.60 From Ex. 5.36,
2
1
1 1 1
) ( + = y y f , 0 ≤ y
1
≤ 1, and
2
1
2 2 2
) ( + = y y f , 0 ≤ y
2
≤ 1. But,
) ( ) ( ) , (
2 2 1 1 2 1
y f y f y y f ≠ so Y
1
and Y
2
are dependent.
5.61 Note that ) , (
2 1
y y f can be factored and the ranges of y
1
and y
2
do not depend on each
other so by Theorem 5.5, Y
1
and Y
2
are independent.