(a)
From the previous problem,
M
x
(
t
) = exp[
λ
(
e
t
 1)].
Suppose
y
is distributed as Poisson with
parameter
μ
.
Then,
M
y
(
t
)=exp[
μ
(
e
t
1)].
The product of these two moment generating functions is
M
x
(
t
)
M
y
(
t
)= exp[
λ
(
e
t
 1)]exp[
μ
(
e
t
 1)] =
exp[(
λ
+
μ
)(
e
t
 1)], which is the moment generating function of the
Poisson distribution with parameter
λ
+
μ
.
Therefore, on the basis of the theorem given in the problem, it
follows that
x
+
y
has a Poisson distribution with parameter
λ
+
μ
.
(b)
The density of the Chisquared distribution with
n
degrees of freedom is [from (C39)]
fx
n
ex
x
n
x
n
()
(/)
,.
/
=>
−−
12
2
0
2
1
22
1
Γ
Let the constant term be
k
for the present.
The moment generating function is
M
(
t
)
=
k
ee
x
d
x
tx
x
n
∞
∫
/(
/
)
1
0
=
k
.
d
x
xt
n
−
∞
∫
(1/
)
( / )
1
0
This is a gamma integral which reduces to
M
(
t
) =
k
(1/2 
t
)

n
/2
Γ
(
n
/2).
Now, reinserting the constant
k
and
simplifying produces the moment generating function
M
(
t
)
=
(1  2
t
)

n
/2
.
Suppose that
x
i
is distributed as
chisquared with
n
i
degrees of freedom.
The moment generating function of
Σ
i
x
i
is
Π
i
M
i
(
t
)
=
/
2
−
−
∑
t
n
i
i
which is the MGF of a chisquared variable with
n
=
Σ
i
n
i
degrees of freedom.
(c)
We let
y
=
σ
z
+
μ
.
Then,
M
y
(
t
) =
E
[exp(
ty
)] =
[ ] [
][
Ee
e Ee
tz
t
t
z
t
t
z
(
)
σμ
μ
μ
+
==
σσ
]
=
[ ]
t
t
tt
μσ
σ
−
=−
/
exp
(
) /
2
2
μ
2
Using the same approach as in part b., it follows that the moment generating function for a sum of random
variables with means
μ
i
and standard deviations
σ
i
is
Mt
x
i
i
i
i
i
i
∑
⎛
⎝
⎜
⎞
⎠
⎟
⎡
⎣
⎢
⎤
⎦
⎥
∑∑
exp
1
2
.
±
171
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View Full DocumentAppendix C
Estimation and Inference
1.
The following sample is drawn from a normal distribution with mean
μ
and standard deviation
σ
:
x
=
1.3, 2.1, .4, 1.3, .5, .2, 1.8, 2.5, 1.9, 3.2.
Compute the mean, median, variance, and standard deviation of the sample.
x
x
n
i
i
n
=
=
∑
1
=
1.52,
s
2
=
(
)
xx
n
i
i
n
−
−
=
∑
2
1
1
=
.9418,
s
=
.97
m
e
d
i
a
n
=
1.55, midway between 1.3 and 1.8.
2.
Using the data in the previous exercise, test the following hypotheses:
(a)
μ
>
2.
(b)
μ
<
.7.
(c)
σ
2
= .5.
(d)
Using a likelihood ratio test, test the following hypothesis
μ
= 1.8,
σ
2
= .8.
(a)
We would reject the hypothesis if 1.52 is too small relative to the hypothesized value of 2.
Since
the data are sampled from a normal distribution, we may use a
t
test to test the hypothesis.
The
t
ratio is
t
[9]
=
(1.52  2) / [.97/
10 ]=
1.472.
The 95% critical value from the
t
distribution for a one tailed test is 1.833.
Therefore, we would not reject
the hypothesis at a significance level of 95%.
(b)
We would reject the hypothesis if 1.52 is excessively large relative to the hypothesized mean of
.7.
The
t
ratio is
t
[9]
=
(1.52  .7) / [.97/
10 ]=
2.673.
Using the same critical value as in the previous
problem, we would reject this hypothesis.
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 Spring '10
 Dr.Fang
 Normal Distribution, σ

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