Homework solutions
Math526
Spring 2004
Text book: Bickel-Doksum, 2nd edition
Assignment # 1
Section 4.3.
1. (a) The model is a exponential family satisfying condition in Example 4.3.3. So
the sample distribution
Y
n
n
n
o
n
X
Y
θ xi −θ
1
xi − nθ
e =
p(~x, θ) =
exp log θ ·
xi !
xi !
i=1
i=1
i=1
is a monotone likelihood ratio family in T (~x) =
δ(~x) =
i=1
Pn

1

Pn
i=1
Pn
0
i=1
xi . So the UMP test take the form
xi > k
xi < k
Pn
Pn
Notice that i=1 Xi has the Poisson distribution with parameter nθ. When i=1 xi = k,
we take the integer k > 0 to be
∞
X
(nθ0 )m −nθ0
e
≤ α and
m!
∞
X
(nθ0 )m −nθ0
e
≥α
m!
m=k+1
We define
δ(~x) =
m=k
∞
X
(nθ0 )m −nθ0 . (nθ0 )k −nθ0 α−
e
e
m!
k!
m=k+1
~ = 1 − α.
in order that Eδ(X)
(b) For any 0 < α < 1, the α-level UMP test exists if and only if there is a natural
number k such that
∞
X
(nθ0 )m −nθ0
e
=α
m!
m=k
in which case we have α-level UMP test

1
δ(~x) =

0
Pn
i=1
Pn
i=1
xi ≥ k
xi < k
c) We need a table for Poisson distribution with the parameter nθ0 .
1
2. We have α = 0.01.
0.01 = sup Pθ
θ≤10
= P10
n
1
√
10n
n
nX
o
Xi ≥ kα = P10
i=1
n
X
n
nX
i=1
Xi ≥ k α
o
k − 10n kα − 10n o
α
(Xi − 10) ≥ √
≈1−Φ √
10n
10n
i=1
Hence
kα − 10n
√
= z(0.99) = 2.33
10n
On the other hand,
0.99 = inf Pθ
θ≥15
n
= P15 √
Hence
n
nX
o
Xi ≥ kα = P15
i=1
1
15n
n
X
i=1
(Xi − 15) ≥
n
nX
i=1
Xi ≥ k α
o
k − 15n kα − 15n o
α
√
≈1−Φ √
15n
15n
kα − 15n
√
= z(0.01) = −z(0.99) = −2.33
15n
So we have
We have
n=
√
kα − 10n = 2.33 √
10n
kα − 15n = −2.33 15n
√
√
2
2.33 · ( 10 + 15)
≈ 10.75
5
So we take n = 11.
6. (a).
µ=
Z
∞
c
x · cθ θx−(1+θ) dx = c
θ
θ−1
The probelm is to test H: θ = θ0 versus K: θ > θ0 , where θ0 =
The sample density
p(~x, θ) = cnθ θ n
n
Y
−(1+θ)
xi
= cn exp
i=1
is a MLR family in T (~x) = −
Pn
n
− (1 + θ)
log xi . Hence,

T (~x) ≥ k
 1,
δ(~x) =

0,
T (~x) < k
i=1
2
n
X
i=1
µ0
µ0 −c
(notice that µ0 > c).
log xi + n log θ
o
where k satisfies
~ ≥k =α
Pθ0 T (X)
To compute its value, we use central limit theorem that claims that
q
~ − ET (X)
~ / V ar(T (X)
~
T (X)
converges to N (0, 1) in distribution as n → ∞.
Let η = 1 + θ
n
o
~ + n log(η − 1)
p(~x, θ) = cn exp ηT (X)
By Theorem 1.6.2,
~ = n log(η − 1)
ET (X)
0
=
n
n
=
η−1
θ
n 0
n
n
~
= 2
V ar T (X) =
=
2
η−1
(η − 1)
θ
Hence,
α = P θ0
So we take
Therefore,
n T (X)
k− n ~ − n
k − nθ o
θ0
θ0
√
≥
≈
1
−
Φ
√
−1
−1 √
θ −1 n
θ0
n
θ0
n
k − θn0
√ = z(1 − α)
θ0−1 n
i µ − ch
i
h
√
√
0
n + nz(1 − α)
k = θ0−1 n + nz(1 − α) =
µ0
3