1)
{X
1
,
....
, X
n
} can be modeled by a Gamma distribution
(
29
(
29
(
29
1
1
exp
/
X
f
x
x
x

α
α
�
�
= β Γ α

β
�
�
with a
fixed
value of the shape parameter
α
.V
Thus you only need to calculate the
value of
β
.
[In environmental applications when n is small,
α
may be computed as an average from
many samples to increase its precision.]
a) What is the MLE of
β
in this case?
We know α so we only need to estimate β. Assuming that the observations are iid, we can
construct the likelihood function as:
1
1
1
exp(
/
)
(
)
n
i
i
i
L
x
x
α
α
β
β
α

=
= Π

Γ
We can simplify the likelihood function to obtain:
1
1
1
1
1
1
1
1
1
exp
[ (
)]
exp
(
)
n
n
n
n
n
n
n
i
i
i
i
i
i
i
i
L
x
x
x
x
α
α
α
α
β
α
β
α
β
β




=
=
=
=
�
�
�
�
�
�
=

Π
=
+
Γ

Π
�
�
�
�
�
�
Γ
�
�
�
�
�
�
�
�
The MLE of β will be the value which maximizes the likelihood function. NOTE: It is
not necessary to compute the log of the likelihood function, but it is generally easier to
find the optimal value of β using the log of the likelihood function.
1
1
1
ln( )
ln(
)
ln[ (
)]
(
1)
ln(
)
n
n
i
i
i
i
L
n
n
x
x
α
β
α
α
β
=
=
= 

Γ

+

�
�
Now to find the MLE of β we take the derivative of the log of the likelihood function
with respect to β, set it equal to zero, and solve for β.
2
1
ln( )
1
0
n
i
i
L
n
x
α
β
β
β
=

=
+
=
FINAL ANSWER
1
1
ˆ
n
i
i
X
x
n
β
α
α
=
=
=
This is our MLE for β. Notice that it is only a function of observable/known values.
(Not Required
: You should also check the second order derivative, to ensure that this is
in fact a maximum.
(
29
0
ˆ
ln
2
2
<
∂
∂
β
β
L
(
29
(
29
0
ˆ
2
1
2
1
2
ˆ
ln
2
2
2
3
2
2
2
<

=

=

=

=
∂
∂
∑
∑
β
α
α
α
β
β
α
β
β
β
α
β
β
n
n
n
x
n
x
n
L
i
i
Because for gamma distribution, we have
α
> 0.
b) What is the MM estimator of
β
in this case?
For the methodofmoments (MOM) estimator of β when α is not known, we simply
equate the sample mean to the expected value of X, and solve for β. We know that X ~
Gamma(α,β), therefore,
FINAL ANSWER
ˆ
[
]
X
E X
X
αβ
αβ
β
α
=
=≈
=
(Not Required)
: If α were not known, then
ˆ
ˆ
X
β
α
=
, and thus the estimator for β is a
function of the estimator for α. But we really need an estimator for β which is only based
on observable values. However, we know that for a Gamma distribution Var[X] = αβ
2
,
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and using MOM we can equate Var[X] to the sample variance. Therefore, using MOM
we have two equations:
2
2
X
S
αβ
αβ
=
=
So we have two equations which we can use to solve for the two unknowns ˆ
α
and
ˆ
β
.
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 Fall '08
 Stedinger
 Derivative, Normal Distribution, probability density function, Logarithm, mle

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