HW 3 Solutions
4.136
a.
Observing that the exponential distribution is a special case of the gamma
distribution, we can use the gamma moment–generating function with α = 1 and β = θ:
t
t
m
θ

=
1
1
)
(
,
t
< 1/θ.
b.
The first two moments are found by
2
)
1
(
)
(
t
t
m
θ

θ
=
′
,
E
(
Y
) =
θ
=
′
)
0
(
m
.
3
)
1
(
2
)
(
t
t
m
θ

θ
=
′
′
,
E
(
Y
2
) =
2
2
)
0
(
θ
=
′
′
m
.
So, V(Y) = 2θ
2
– θ
2
= θ
2
.
4.140
a.
Gamma with α = 2, β = 4
b.
Exponential with β = 3.2
c.
Normal with μ = –5, σ
2
= 12
6.37
The mass function for the Bernoulli distribution is
y
y
p
p
y
p


=
1
)
1
(
)
(
,
y
= 0, 1.
a.
t
x
ty
tY
Y
pe
p
y
p
e
e
E
t
m
+

=
=
=
∑
=
1
)
(
)
(
)
(
1
0
1
1
.
b.
n
t
n
i
Y
tW
W
pe
p
t
m
e
E
t
m
i
]
1
[
)
(
)
(
)
(
1
+

=
=
=
∏
=
c.
Since the mgf for
W
is in the form of a binomial mgf with
n
trials and success
probability
p
, this is the distribution for
W
.
6.40
Using Theorem 6.4 we know that
is a ChiSquared Random Variable
with 2 degrees of freedom
6.56
Let
U
= total service time for two cars.
Similar to Ex. 6.13,
U
has a gamma
distribution with α = 2, β = 1/2.
Then,
P
(
U
> 1.5) =
du
ue
u
∫
∞

5
.
1
2
4
= .1991.
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 Spring '08
 EduardoMontoya
 Normal Distribution, Probability theory, Exponential distribution, 1 degrees, Chisquare distribution

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