THE UNIVERSITY OF NEW SOUTH WALES
DEPARTMENT OF STATISTICS
MID SESSION TEST – 2021 – Wednesday, 31st March (Week 7)
Solutions
MATH5905
Time allowed: 135 minutes
1. Let X = (X1, X2, . . . , Xn) be a sample of i.i.d. Geometric(θ) random variables with
density function
f(x; θ) = θ(1 θ)x, x = {0, 1, 2, . . . }; θ ∈ (0, 1)),
with
E(X) =
1 θ
θ
and Var(X) =
1 θ
θ2
.
a) Provide justification why the statistic T (X) =
∑n
i=1Xi is complete and sufficient
for θ
The Geometric distribution belongs to the one-parameter exponential
family since
f(x, θ) = θ exp(x log(1 θ))
with d(x) = x and hence T (X) =
∑n
i=1Xi is complete and (minimal)
sufficient for θ.
b) Derive the UMVUE of h(θ) = θ. You must justify each step in your answer.
Hint: Use the interpretation that P(X1 = 0) = θ and the fact that
T =
n∑
i=1
Xi ～ Negative Binomial
(
n, θ
)
with probability mass function
P (T = t) =
(
n+ t 1
t
)
θn(1 θ)t, t = 0, 1, 2, . . .
1) T is sufficient and complete for θ.
2) Let W = I{X1=0}(X) and
then using the fact that P (X1 = 0) = θ we
have
E(W ) = E
(
I{X1=0}(X)
)
= P (X1 = 0) = θ,
which is unbiased for h(θ) = θ.
1
3) Now, apply the Theorem of Lehmann-Scheffe, to obtain
τ (T ) = E(W |T = t)
= E
(
I{X1=0}(X)
∣∣∣ n∑
i=1
Xi = t
)
= P
(
X1 = 0
∣∣∣ n∑
i=1
Xi = t
)
=
P
(
X1 = 0,
∑n
i=1Xi = t
)
P
(∑n
i=1Xi = t
)
The events in the numerator must be satisfied simultaneously to
have a non-zero probability and hence this reduces to
τ (T ) =
P
(
X1 = 0,
∑n
i=2Xi = t
)
P
(∑n
i=1Xi = t
)
=
P (X1 = 0)P
(∑n
i=2Xi = t
)
P
(∑n
i=1Xi = t
)
=
θ
(
n 1+t 1
t
)
θn 1(1 θ)t(
n+t 1
t
)
θn(1 θ)t
=
(n 2 + t)!
t!(n 1 + t 1 t)! ·
t!(n+ t 1 t)!
(n+ t 1)!
=
(n+ t 2)!
(n t 1)! ·
(n 1)!
(n 2)!
=
n 1
n 1 t
Therefore the UMVUE for h(θ) = θ is:
humvue(θ) =
n 1
n 1 +
n∑
i=1
Xi
c) Calculate the Cramer-Rao lower bound for the minimal variance of an unbiased
estimator of h(θ) = θ.
2
First we calculate the Fisher Information as follows:
log f(x; θ) = log θ + x log(1 θ)