STAT 610 Final Exam
7:00pm-8:30pm, May 6, 2021
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1. Let X1, …, Xn be independent and identically distributed random variables with a
common pdf fθ(x) =
1
2θ
e |x|/θ, where θ is an unknown parameter and θ > 0.
(a) (2 points) Derive the UMVUE of θ.
(b) (2 points) Derive the MLE θ of θ and prove that your solution θ is the unique
MLE.
(c) (2 points) Obtain the asymptotic distribution of
√
n(θ θ) as n→∞.
(d) (2 points) Let = 1/θ and the prior for to be the gamma distribution with pdf
1
γαΓ(α)
α 1e /γI( > 0), where α and γ are known positive constants. Derive
the Bayes estimator of θ = 1/ under the squared error loss (i.e., the posterior
mean of θ = 1/ ).
(e) (2 points) Obtain the asymptotic relative efficiency of the MLE θ in (b) w.r.t. the
Bayes estimator of θ in (d).
2. Let X1, …, Xn be independent and identically distributed positive random variables
with a common pdf
fθ(x) =
θθ
Γ(θ)
xθ 1e θxI(x > 0),
where θ > 0 is unknown.
(a) (2 points) Let ψ(θ) = dΓ(θ)
dθ
/Γ(θ). Show that the Fisher information contained in
X1, …, Xn is
In(θ) = n
(
dψ(θ)
dθ
1
θ
)
(b) Consider the hypotheses H0 : θ = θ0 versus H1 : θ 6= θ0, where θ0 is a fixed
positive constant.
i. (2 points) Derive the Wald test with a given significance level α (do not try
to find an explicit formula for the MLE of θ).
ii. (2 points) Derive the score test with a given significance level α.
(c) Consider the hypotheses H0 : θ ≤ θ0 versus H1 : θ > θ0, where θ0 is a fixed
positive constant.
i. (2 points) Show that the family of joint pdf’s of X1, …, Xn has a monotone
likelihood ratio in T =
∑n
i=1(logXi Xi).
ii. (1 point) Show that the UMP test of size α ∈ (0, 1) rejects H0 if and only
if T > c with a constant c satisfying α = Pθ0(T > c), where Pθ0 is the
probability under θ = θ0 (you do not need to find an explicit form of c).
(d) (2 points) Let Fθ(t) be the cdf of T . Show that Fθ(t) is non-increasing in θ for each
fixed t and, using this result, obtain a confidence interval for θ with confidence
coefficient 1 α.
3. Let X1, …, Xn be independent and identically distributed random variables with a
common pdf θ 1f(x/θ), where θ > 0 is unknown and f is a known pdf satisfying
f(x) = f( x).
(a) (2 points) Show that (
∑n
i=1X
2
i /cα)
1/2 is a lower confidence bound for θ with
confidence coefficient 1 α, where cα is the 1 α quantile of the cdf of θ 2∑ni=1X2i .
(b) (3 points) Suppose that f(x) is the pdf of the standard normal. Sho
w that the
lower confidence bound in (a) is (0, θ)-UMA, where θ is the true parameter value.
Compared with other lower confidence bounds for θ, what can you say why the
bound in (a) is better
(c) (4 points) Suppose that f(x) = x when 0 < x < 1 and f(x) = 0 when x ≥ 1. Let
T = maxi=1,...,n |Xi|. Show that among all confidence intervals of θ in the class{(
T
b
,
T
a
)
: 0 < a < b ≤ 1 are constants satisfying P
(
T
b
≤ θ ≤ T
a
)
= 1 α
}
the one with b = 1 and a = α1/(2n) has the shortest length.
4. Let X1, ..., Xn be independent and identically distributed random variables with a
common pdf φ 1e x/φI(x > 0) and Y1, …, Yn be independent and identically distributed
random variables with a common pdf 1e y/ I(y > 0), where both φ > 0 and > 0
are unknown parameters and Xi’s and Yi’s are independent. Let θ = (φ, )
′. Consider
testing
H0 : R(θ) = 0 versus H1 : R(θ) 6= 0
where R(θ) = φ c and c is a known positive constant.
(a) (2 points) Under the hypothesis H0 : R(θ) = 0, obtain the MLE’s of φ and .
(b) (3 points) Derive the Wald test statistic
Wn = [R(θ )]
′{[C(θ )]′[In(θ )] 1C(θ )} 1R(θ )
where θ is the MLE of θ, C(θ) = R(θ)/ θ, and In(θ) is the Fisher information
contained in X1, …, Xn, Y1, …, Yn.
(c) (2 points) Derive the score test statistic Rn.
(d) (3 points) Obtain a 1 α asymptotic confidence interval for φ/ by inverting the
acceptance regions of Wald’s tests in (b) with different c’s.
2
Solution
1. (a) The family of pdf’s is an exponential family with complete and sufficient statistic
T =
∑n
i=1 |Xi|. Since E|Xi| = θ, T/n is unbiased for θ and hence it is the UMVUE
of θ.
(b) The log-likelihood is n log θ T/θ, T = ∑ni=1 |Xi|. The score function S(θ) =
T/θ2 n/θ, and S(θ) = 0 has a unique solution θ = T/n. Since S(θ) = n
θ
( θ
θ
1),
which is negative if θ < θ and positive if θ > θ , θ is the unique MLE.
(c) We apply Theorem 10.1.12. Since S ′(θ) = n
θ2
2T
θ3
, the Fisher information is
E{S ′(θ)} = 2E(T )
θ3
n
θ2
= n
θ2
. Hence the asymptotic distribution of
√
n(θ θ) is
N(0, θ2).
(d) The posterior of is proportional to
α 1e /γ ne T = n+α 1e (T+γ
1)
which is the gamma distribution with shape parameter n+α and scale parameter
(T + γ 1) 1. The Bayes estimator of θ = 1/ is
θ B =
(T + γ 1)n+α
Γ(n+ α)
∫ ∞
0
n+α 2e (T+γ
1)d
=
(T + γ 1)n+α
Γ(n+ α)
Γ(n+ α 1)
(T + γ 1)n+α 1
=
T + γ 1
n+ α 1
(e) Note that
√
n(θ θ B) =
√
n
(
θ T + γ
1
n+ α 1
)
=
√
n
θ θ + 1γn
1 + α 1
n