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STAT 710 2006 Second Exam Suggested
Solution
1 Problem 1
(a)
fθ (x) =
axθx
c (θ)
(^1) Z+ (x) =
θx
c (θ)
ax (^1) Z+ (x) ,
where Z+ = { 0 , 1 , · · ·}. Hence
log fθ (x) = x log θ − log c (θ) + log ax (^1) Z+ (x)
l (θ) =
∑^ n
i=
log fθ (xi) = log θ
∑^ n
i=
xi − n log c (θ) +
∑^ n
i=
log axi 1 Z+ (xi).
Hence the likelihood equation is
∂l (θ)
∂θ
∑^ n
i=
xi
θ
nc
′ (θ)
c (θ)
I.e.,
X =
θc
′ (θ)
c (θ)
(b) On the other hand,
EX 1 =
∑^ ∞
x=
xfθ (x) =
∑^ ∞
k=
k
akθk
c (θ)
c (θ)
∑^ ∞
k=
kakθ
k
c (θ)
∑^ ∞
k=
kakθ
k
θ
c (θ)
∑^ ∞
k=
kakθ
k− 1
θ
c (θ)
c
′ (θ) ,
since c′^ (θ) =
k=
kakθk−^1.
Hence the method of moment equation is also X =
θc′(θ) c(θ)
(c) Since
∂l (θ)
∂θ
∑^ n
i=
xi
θ
nc
′ (θ)
c (θ)
n
θ
[
X −
θc
′ (θ)
c (θ)
]
n
θ
[
X − μ (θ)
]
n
θ
X −
n
θ
μ (θ) ,
In (θ) = −E
[
∂^2 l (θ)
∂θ^2
]
= −E
[
n
θ^2
X − n
μ′^ (θ) θ − μ (θ)
θ^2
]
= n
EX − μ (θ)
θ^2
μ′^ (θ) θ
θ^2
= n
μ
′ (θ)
θ
(d) From (c), μ′^ (θ) =
In(θ)θ n
> 0. Hence μ (θ) is monotone. Also observe that the likelihood
equation can be written as n
θ
[
X − μ (θ)
]
i.e.,
X = μ (θ).
Therefore if the likelihood equation has solution, it has unique solution.
(e) Since θ̂n is the unique solution of the likelihood equation, θ̂n is consistent by Theorem
4.17(i), and
√ n
θn − θ
→ N
I 1 (θ)
by Theorem 4.17(ii), with I 1 (θ) =
In(θ) n =^
μ′(θ) θ.
2 Problem 2
(a) By Example 5.10, the asymptotic relative efficiency of sample median with respect to
sample mean is
e = 4 [f (μ)]
2 var (X 1 ).
Observe that f (μ) =
1 2(1−e−c) (NOT 0!).
