Homework solutions
Math525
Fall 2003
Text book: Bickel-Doksum, 2nd edition
Assignment # 6
Section 3.4.
1. E X̄ = θ. So X̄ is unbiased. p(x, θ) = θ x (1 − θ)1−x , x = 0, 1.
log p(x, θ) = x log θ + (1 − x) log(1 − θ) and
x 1−x
∂
log p(x, θ) = −
∂θ
θ
1−θ
hX
i2
h∂
1 − X i2
log p(X, θ) = E
−
∂θ
θ
1−θ
h X2
hX
2i
X(1 − X) (1 − X)
1−X i 1
1
1
=E 2 −2
+
+
=
=
E
= +
2
2
2
θ
θ(1 − θ)
(1 − θ)
θ
(1 − θ)
θ 1−θ
θ(1 − θ)
I1 (θ) = E
IX~ (θ) = nI1 (θ) = n
The information inequality lower bound is
1
θ(1 − θ)
θ(1 − θ)
1
=
IX~ (θ)
n
On the other hand,
V ar(X̄) =
θ(1 − θ)
n
So X̄ is UMVU estimate of θ.
2.
n o
R(θ, δ) = El θ, δ(X) = E E l θ, δ(X) |T (X)
≥ El θ, E δ(X)|T (X) = El θ, δ ∗ (X) = R(θ, δ ∗ )
12. (a). f (x, θ) = θxθ−1 , 0 ≤ x ≤ 1.
p(~x, θ) = θ
n
n
Y
i=1
xθ−1
i
and log p(~x, θ) = n log θ + (θ − 1)
Solve
n
∂
n X
log p(~x, θ) = +
log xi = 0
∂θ
θ i=1
we have
n
1X
1
=−
log xi
θ
n i=1
1
n
X
i=1
log xi
So the MLE of 1/θ is
n
~ =−
δ(X)
1X
log Xi
n
i=1
Since
~ = −E log X = −
Eδ(X)
Z
1
0
log x · θxθ−1 dx =
1
θ
~ is unbiased.
δ(X)
n
h∂
i2
i2
i2
hn X
h1
~
I(θ) = E
log p(X, θ) = E
+
log Xi = nE + log X
∂θ
θ i=1
θ
Z
n 1
n2
n
1o
1o
n
1o
(log x)2 · θxθ−1 dx − 2 = n 2 − 2 = 2
= n E(log X)2 − 2 = n
θ
θ
θ
θ
θ
0
Write ψ(θ) = θ1 . The information inequality lower bound is
[ψ 0 (θ)]2
1
=
I(θ)
nθ 2
On the other hand,
n
1X
1
1
[ψ 0 (θ)]2
~
V ar δ(X) = V ar −
log Xi = V ar(log X) =
=
n
n
nθ 2
I(θ)
i=1
So the information inequality lower bound is achieved.
(b).
E X̄ = EX =
So X̄ is unbiased. Write ψ(θ) =
lower bound is
θ
1+θ .
Z
1
0
x · θxθ−1 dx =
Then ψ 0 (θ) =
θ
1+θ
1
(1+θ)2 .
So the information inequality
θ2
[ψ 0 (θ)]2
=
I(θ)
n(1 + θ)4
On the other hand,
Z
θ 2 o
θ 2 o
1n
1
1n 1 2
2
V ar(X̄) = V ar(X) =
EX −
=
x · θxθ−1 dx −
n
n
1+θ
n
1+θ
0
n
o
2
θ
1
θ
1
θ
=
=
−
n 2+θ
1+θ
n (2 + θ)(1 + θ)2
By the simple fact that
θ
1
θ2
>
n (2 + θ)(1 + θ)2
n(1 + θ)4
2
the information inequality lower bound is not achieved by X̄.
Section 4.1
2. (a). By Problem B.3.4,
2λ
n
X
i=1
d
2
Xi = X2n
So the power
β(λ) = Pλ {X̄ ≥ µ0 x(1 − α)/2n}
n
n X
o
n
o
2
= Pλ 2λ
Xi ≥ λµ0 x(1 − α) = P X2n ≥ λµ0 x(1 − α)
i=1
When H is true, or λ ≥ 1/µ0 ,
n
o
2
Pλ {X̄ ≥ µ0 x(1 − α)/2n} ≤ P X2n
≥ x(1 − α) = α
and the equality holds as λ = 1/µ0 . So
sup Pλ {X̄ ≥ µ0 x(1 − α)/2n} = α
λ≥1/µ0
Hence, this test has level α.
(b) the power is
n
o
2
beta(λ) = P X2n
≥ λµ0 x(1 − α)
(c). Let Z1 , · · · , Z2n be i.i.d. N (0, 1) random variables. By Theorem B.3.1, p. 491,
d
2
2
= Z12 + · · · + Z2n
X2n
As n is sufficiently large,
n
o
n X 2 − 2nEZ 2
λµ0 x(1 − α) − 2nEZ 2 o
2
p
β(λ) = P X2n
≥ λµ0 x(1 − α) = P p2n
≥
2nV ar(Z 2 )
2nV ar(Z 2 )
λµ x(1 − α) − 2nEZ 2 0
p
≈ 1−Φ
2nV ar(Z 2 )
where the approximation follows from the central limit theorem. Notice that EZ 2 = 1 and
V ar(Z 2 ) = EZ 4 − 1 = 3 − 1 = 2. Therefore,
√
n
o
λµ x(1 − α) − 2n λµ0 x(1 − α) 0
2
√
√
=Φ n−
β(λ) = P X2n
≥ λµ0 x(1 − α) ≈ 1 − Φ
2 n
2 n
3
Take µ = 1/µ0 gives that
λµ0 x(1 − α) − 2n
√
≈ z(1 − α)
2 n
√
Or, x(1 − α) ≈ 2 nz(1 − α) + 2n. Therefore,
√
√n(µ − µ ) µ z
0
0
+
(1 − α)
β(λ) ≈ Φ n(1 − λµ0 ) + λµ0 z(1 − α) = Φ
µ
µ
as λ = 1/µ.
8. (a). For each x, |F̂ (x) − F0 (x)| ≤ Dn . Hence,
|F̂ (x) − F0 (x)| ≥ kα ⊂ {Dn ≥ kα }
Consequently,
P |F̂ (x) − F0 (x)| ≤ kα ≤ P {Dn ≥ kα }
Taking supremum over x on the left hand side gives
sup P |F̂ (x) − F0 (x)| ≤ kα ≤ P {Dn ≥ kα }
x
(b). Omitted
(c). By the fact that F 6= F0 , there is a x such that F (x) 6= F0 (x). As kα <
|F (x) − F0 (x)|, by the law of large numbers,
P |F̂ (x) − F0 (x)| ≤ kα −→ P |F (x) − F0 (x)| ≤ kα = 1
So the conclusion follows from Part (a).
Section 4.2
2. (a).
n
Y
f (xi , θ) = f (1, θ)N1 f (2, θ)N2 f (3, θ)N3 = f (1, θ)N1 f (2, θ)N2 f (3, θ)n−N1 −N2
i=1
= θ 2N1 2θ(1 − θ)
Hence,
L(~x, θ0 , θ1 ) =
N2
n
Y
i=1
(1 − θ)2(n−N1 −N2 ) = 2N2 (1 − θ)2n
f (xi , θ1 )
n
.Y
f (xi , θ0 ) =
i=1
θ 2N1 +N2
1−θ
1 − θ 2n 1 − θ θ 2N1 +N2
1
0 1
1 − θ0
1 − θ 1 θ0
So the conclusion follows from the fact that
1 − θ 0 θ1
>1
1 − θ 1 θ0
(b) Write
1 − θ 2n 1 − θ θ c
0 1
1
k=
1 − θ0
1 − θ 1 θ0
Then L(~x, θ0 , θ1 ) ≥ k if and only if 2N1 + N2 ≥ c. The rest follows from Theorem 4.2.1-(a).
4