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DEPARTMENT OF STATISTICS
SOLUTIONS
MATH5905
Time allowed: 135 minutes
1. Let X = (X1, X2, . . . , Xn) be i.i.d. Poisson(θ) random variables with density function
f(x, θ) =
e−θθx
x!
, x = 0, 1, 2, . . . , and θ > 0.
a) The statistic T (X) =
∑n
i=1Xi is complete and sufficient for θ. Provide justifi-
cation for why this statement is true.
Using the property of the one-parameter exponential family, we ob-
serve that
f(x, θ) =
e−θθx
x!
= e−θ
1
x!
exp
(
x log θ
)
Thus, the Poisson distribution belongs to the one-parameter expo-
nential family. Then this implies that T (X) =
∑n
i=1Xi is complete and
(minimal) sufficient for θ.
b) Derive the UMVUE of h(θ) = e−kθ where k = 1, 2, . . . , n is a known integer.
You must justify each step in your answer. Hint: Use the interpretation that
P (X1 = 0) = e
−θ and therefore P (X1 = 0, . . . , Xk = 0) = P (X1 = 0)k = e−kθ.
Since T is sufficient and complete for θ, we first need to find an unbi-
ased estimator of h. Let W = I{X1=0,...,Xk=0}(X) and then using the fact
that P (Xi = 0) = e
−θ we see that
E(W ) = P (X1 = 0, . . . , Xk = 0) = P (X1 = 0)
k =
[
e−θ
]k
= e−kθ,
which is unbiased for h(θ). Now, we apply the Theorem of Lehmann-
Scheffe, and obtain the following
1
τˆ(T ) = E(W |T = t) = E
(
I{X1=0,...,Xk=0}|
n∑
i=1
Xi = t
)
= P
(
X1 = 0, . . . , Xk = 0|
n∑
i=1
Xi = t
)
=
P
(
X1 = 0, . . . , Xk = 0,
∑n
i=1Xi = t
)
P
(∑n
i=1Xi = t
)
Now we observe that the events in the numerator must be satisfied
simultaneously to have a non-zero probability and hence this reduces
to
τˆ(T ) =
P
(
X1 = 0, . . . , Xk = 0,
∑n
i=k+1Xi = t
)
P
(∑n
i=1Xi = t
)
= e−kθ
P
(∑n
i=k+1Xi = t
)
P
(∑n
i=1Xi = t
)
= e−kθ
e−θ(n−k)(θ(n− k))t
t!
· t!
e−nθ(nθ)t
=
(
n− k
n
)t
=
(
1− k
n
)nX¯
which is the UMVUE for h(θ) = e−kθ.
c) Calculate the Cramer-Rao lower bound for the minimal variance of an unbiased
estimator of h(θ) = e−kθ.
First we calculate the fisher information as follows. We have that
log f(x) = −θ + x log θ − log x!
∂
∂θ
log f(x) = −1 + x
θ
∂2
∂θ2
log f(x) = − x
θ2
and so the Fisher information in a single sample is
2
IX1(θ) = −E
[
∂2
∂θ2
log f(x)
]
=
E(X)
θ2
=
θ
θ2
=
1
θ
Hence the Fisher information for the whole sample is
IX = nIX1(θ) =
n
θ
.
Then notice that
∂
∂θ
h(θ) = −ke−kθ.
Hence the Cramer-Rao lower bound is(
∂
∂θ
h(θ)
)2
IX(θ)
=
θ
n
k2e−2kθ
d) Show that there does not exist an integer k for which the variance of the UMVUE
of h(θ) attains this bound.
To show that the Cramer-Rao bound is attainable we will look at the
score function:
V (X, θ) = −n+ 1
θ
n∑
i=1
Xi
= −n+ nX¯
θ
=
n
e−kθ
(
X¯e−kθ
θ
− e−kθ
)
Therefore, no matter what integer k we choose the term X¯e
−kθ
θ
cannot
be a statistic as it will still depend on θ.
e) Determine the MLE hˆ of h(θ).
The MLE for θ is simply θˆ = X¯ and hence the
ĥ(θ)mle = h(θˆmle) = e
−kX¯ .
f) Suppose that n = 5, T = 10 and k = 1 compute the numerical values of the
UMVUE in part (b) and the MLE in part (e). Comment on these values.
The UMVUE is
ĥ(θ)umvue =
(
1− 1
5
)5×2
= 0.107
The MLE is
ĥ(θ)mle = e
−2 = 0.135
As n → ∞ the UMVUE would converge to the MLE while for finite
sample size these values are slightly different.
3
g) Consider testing H0 : θ ≤ 2 versus H1 : θ > 2 with a 0-1 loss in Bayesian setting
with the prior τ(θ) = 4θ2e−2θ. What is your decision when n = 5 and T = 10.
You may use: ∫ 2
0
x12e−7xdx = 0.00317
Note: The continuous random variable X has a gamma density f with param-
eters α > 0 and β > 0 if
f(x;α, β) =
1
Γ(α)βα
xα−1e−x/β
and
Γ(α + 1) = αΓ(α) = α!
First we need to compute the posterior by observing that it propor-
tional to the likelihood times the prior. The likelihood is
L(X|θ) = e
−nθθ
∑n
i=1Xi∏n
i=1Xi!
∝ e−5θθ10
therefore
h(θ|x) ∝ 4θ2e−2θ × e−5θθ10 = θ12e−7θ
which implies that
θ|X ∼ gamma(13, 1
7
).
Hence we are interested in computing the posterior probability
P (θ < 2|X) =
∫ 2
0
1
Γ(13)
(
1
7
)13 θ12e−7θdθ
= 202.27× 0.00317
= 0.64
We compare this posterior probability with 0.5 since we are dealing
with a 0-1 loss. Since this probability is greater than 0.5 we must
NOT reject H0.
4
2. Let X1, X2, . . . , Xn be independent random variables, with a density
f(x; θ) =
{
e−(x−θ), x > θ,
0 else
where θ ∈ R1 is an unknown parameter. Let T = min{X1, . . . , Xn} = X(1) be the
minimal of the n observations.
a) Show that T is a sufficient statistic for the parameter θ.
First calculate the likelihood as follows
L(X, θ) =
n∏
i=1
e−(Xi−θ)I(−∞,Xi)(θ) = exp
(
−
n∑
i=1
Xi + nθ
)
I(−∞,X(1))(θ)
Therefore the likelihood can be written as L(X, θ) = g(T, θ)h(X) where
g(T, θ) = exp(nθ)I(−∞,T )(θ) and h(X) = exp
(
−
n∑
i=1
Xi
)
Hence, T = X(1) is sufficient by the Neyman Fisher Factorization Cri-
terion.
b) Show that the density of T is
fT (t) =
{
ne−n(t−θ), t > θ,
0 else
Hint: You may find the CDF first by using
P (X(1) < x) = 1− P (X1 > x ∩X2 > x · · · ∩Xn > x).
First note that for x < θ we have P (X1 ≥ x) = 1 and for x ≥ θ we have,
P (X1 ≥ x) =
∫ ∞
x
e−(y−θ)dy =
[
− e−(y−θ)
]y=∞
y=x
= e−(x−θ)
Hence,
FT (t, θ) = P (T ≤ t)
= 1− P (X1 ≥ t,X2 ≥ t, . . . , Xn ≥ t)
= 1− P (X1 ≥ t)n
=
{
1− e−n(t−θ) if t ≥ θ
0 if t < θ.
Then by differentiation
fT (t, θ) = ne
−n(t−θ), t ≥ θ,
otherwise zero.
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c) Find the maximum likelihood estimator of θ and provide justification.
The MLE for θ is calculated by maximizing over all θ values
L(X, θ) = exp
