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STAT 710 2007 Second Exam Suggested
Solution
1 Problem 1
(a)
fθ (x 1 , · · ·, xn) =
∏^ n
i=
fθ (xi) =
[
θ
2
]
Pn
i=
|xi|
(1 − θ)
n−
Pn i=
|xi| ∏n
i=
(^1) {− 1 , 0 , 1 } (xi)
log fθ (x) = (log θ − log 2)
∑^ n
i=
|xi| + log (1 − θ)
[
n −
∑^ n
i=
|xi|
]
∑^ n
i=
log 1{− 1 , 0 , 1 } (xi)
∂ log fθ (x)
∂θ
∑^ n
i=
|xi|
θ
n −
∑n
i=
|xi|
1 − θ
The likelihood equation is ∑n
i=
|xi|
θ
n −
∑n
i=
|xi|
1 − θ
It is easy to see that the necessary and sufficient condition for this equation to have a
unique solution is
∑^ n
i=
|xi| < n.
(b) If 0 <
∑n
i=
|xi| < n, we get θ̂ =
P^ n i=
|xi|
n.^ And
∂^2 log fθ (x)
∂θ^2
∑^ n
i=
|xi|
θ^2
n −
∑n
i=
|xi|
(1 − θ)
2 <^0 ,
hence θ̂ =
P^ n i=
|xi|
n is the unique global maximum of log^ fθ^ (x) when 0^ <^
∑n
i=
|xi| < n. I.e.,
θ̂ M LE =
P^ n
i=
|xi|
n
when 0 <
∑n
i=
|xi| < n.
If
∑n
i=
|xi| = 0, we must have that each |xi| = 0, hence
fθ (x) =
[
θ
2
]
Pn i=
|xi|
(1 − θ)
n−
Pn i=
|xi| ∏n
i=
(^1) {− 1 , 0 , 1 } (xi)
[
θ
2
] 0
(1 − θ)
n
(1 − θ)
n
Hence in this case ̂θM LE = 0. Similarly we can see that when
∑n
i=
|xi| = n, ̂θM LE = 1.
So
θ̂ M LE =
∑^ n
i=
|Xi|
n
∑^ n
i=
|Xi| ≤ n.
(c) By WLLN, θ̂M LE =
P^ n
i=
|Xi|
n
p → E |Xi| = θ ∈ (0, 1).
Let ε > 0 be small enough such that (θ − ε, θ − ε) ⊂ (0, 1). Then
P
∣̂θM LE −^ θ
∣ < ε
hence
P
0 < ̂θM LE < 1
≥ P
∣̂θM LE − θ
∣ < ε
(d)
In (θ) = −E
[
∂^2 log fθ (x)
∂θ^2
]
= −E
∑^ n
i=
|xi|
θ^2
n −
∑n
i=
|xi|
(1 − θ)
2
E
[
∑n
i=
|xi|
]
θ^2
n − E
[
∑n
i=
|xi|
]
(1 − θ)
2
nθ
θ^2
n − nθ
(1 − θ)
2
n
θ (1 − θ)
Since this is an exponential family and ̂θM LE always exists and is unique, we have that
√ n
θM LE − θ
d → N (0, θ (1 − θ)).
(e) If we restrict θ ∈
1 2 ,^1
, since log fθ (x) is a strictly concave function on (0, 1) , the
maximum of log fθ (x) is θ˜ =
1 2 ∨ ̂θ M LE ,^ with ̂θ M LE =
P^ n i=
|Xi|
n.
