The University of of Mathematics and Statistics
Solutions to Tutorial Week 6
STAT3023: Statistical Inference
Lecturers: and
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Semester 2, 2021
1. Complete any unfinished exercises from the week 5 tutorial.
2. In the week 5 tutorial we found that if X1,…,Xn are iid geometric with P(X1 = x) = (1−p)x−1p,
x = 1,2,… then
• the Cram ́er-Rao lower bound for estimating the parameter θ = p is θ2(1−θ);
• the Cram ́er-Rao lower bound for estimating the parameter θ = 1/p is θ(θ−1) and the estimator ̄n
X attains the lower bound.
Thus the sample mean is optimal (in this sense) for estimating 1/p. We might then think that
1/X ̄ is a “good” estimator for p. Show (using the delta method) that 1/X ̄ is AMVU. Solution: NotefirstlythatE(X)=μ=1/pandVar(X)=σ2=1−p andsowehavethat
1 1p2 √nX ̄−μ→d N(0,σ2)
or in other words, X ̄ ∼ AN μ, σ2 . By the delta method, for any function g(·) which is differen- n
tiable at μ,
√ng(X ̄)−g(μ)→d N0,g′(μ)2σ2
or in other words, g(X ̄) ∼ AN g(μ), g′(μ)2σ2 . Taking g(μ) = 1/μ, we have that g′(μ) = −1/μ2
andso1/X ̄∼AN1,1 σ2. Butifwewritethisallintermsofpweget1/μ=pforthe
μ μ4 n asymptotic mean and
1 σ2 (1−p) p2(1−p) = p4 =
for the asymptotic variance. Replacing p with θ we see that the asymptotic variance coincides
with the Cram ́er-Rao lower bound for estimating p, so 1/X ̄ is AMVU.
3. Suppose X1, . . . , Xn are iid exponential random variables with rate θ, so the common CDF is
1 − e−θx for x > 0, Pθ(X1 ≤ x) = 0 otherwise
In the week 5 tutorial we saw that the sample mean X ̄ is MVU for estimating the population mean Eθ(X1) = 1/θ.
(a) Determine the form of the common PDF.
∂Fθ(x) θe−θx for x > 0 fθ(x)= ∂x = 0 forx≤0.
(b) Write down the log-likelihood lθ(X) based on X = (X1,…,Xn). Solution: lθ(X) = ni=1 log fθ(Xi) = n log θ − θ ni=1 Xi.
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(c) Write down the score function ∂ lθ(X). Can it be written in the form ∂θ
C θˆ(X)−θ? θ
′ nn ̄1
It cannot be written in the desired form. However this suggests that we should change
(d) Determine the Cram ́er-Rao lower bound for estimating the rate parameter θ.
parameters to η = η(θ) = θ1 .
Solution: Note that the population variance is σ2 = 1/θ2, so
so for any unbiased estimator θˆ(X), we have
Varθ θ(X)≥Varθ(l′θ(X))=n,. (e) Use the delta method to show that the estimator 1/X ̄ is AMVU.
nn Var (l′(X))=Var X =nσ2 = ,
θθ θ i θ2 i=1
Solution: Firstly note that the population mean is μ = 1/θ and the population variance is σ2 = 1/θ2. For any function g(·) differentiable at μ, using the same argument as the
previous question, g(X ̄) ∼ AN g(μ), g′(μ)2σ2 . Expressing everything in terms of θ, since n
g′(μ) = −1/μ2 we get that the asymptotic mean is g(μ) = 1/μ = θ and the asymptotic variance is
g′(μ)2σ2 1 1 θ2 n = μ4 θ2n = n
which coincides with the Cram ́er-Rao lower bound. 4. The Pareto distribution with mean θ > 1 has CDF
Fθ(x)= 1−x−(θ ) forx>1
fθ(x)= θ x−(2θ−1) forx>1.
Suppose X1, . . . , Xn are iid with common PDF fθ(·).
(a) The sample mean X ̄ = n1 ni=1 Xi is an unbiased estimator of θ. Derive a formula for its variance as a function of θ.
Solution: The mean square of X1 is given by ∞θ2θ−1 θ∞ 1
x2 x−(θ−1 )dx=
1 θ−1 θ−11
for x ≤ 1;
for x ≤ 1;
x−(θ−1)dx θ−2∞
θ xθ−1 = θ =θ−2 12−θ
for 1 < θ < 2, for θ ≥ 2.
Varθ(X1) = 2 − θ − θ
for 1 < θ < 2, otherwise Varθ(X1) = ∞; dividing by n gives the variance of X ̄.
θ − 2θ2 + θ3 2 − θ
θ(θ − 1)2 = 2 − θ
(b) Write down the log-likelihood lθ(X) based on X = (X1,...,Xn). Solution:
n lθ (X) =
i=1 (c) Write down the score function ∂ lθ(X). Can it be written in the form
log fθ (Xi ) = n log
2 θ − 1 n
θ − 1 log Xi
− =nlogθ−nlog(θ−1)− 2+θ−1
This is not in the special form, but it suggests that we should change parameters to η =
η(θ)=θ−1 =1−1. θθ
(d) Show using the distribution function method that log(X1) has an exponential distribution with a rate depending on θ. Hence determine the CRLB for estimating θ (question 3 above may prove useful for this).
Solution: For y > 0,
1 1 θ−1 θ−1
= (θ − 1)2 n
The score function is
C θˆ(X)−θ? θ
′ n n 1 n
lθ(X)= θ − θ−1 + (θ−1)2 −n 1n
= θ(θ−1) + (θ−1)2
log(Xi) log(Xi)
log(Xi) − θ .
P{log(X )≤y}=P{X ≤ey}=1−(ey)−( θ )=1−e−( θ )y
which is the CDF of an exponential random variable with rate θ and mean η = θ−1 =
θ−1 1−1. ThusVarθ[log(X1)]=η2 =θ−12,sothevarianceofthescoreis
θ − 1 2 n
θ = θ2(θ−1)2 .
θ2(θ − 1)2 n.
1 (θ−1)4 n
Thus the CRLB for estimating θ is
(e) Identify a change of parameters from θ to η = η(θ) such that a MVU estimator ηˆ(X) exists for η. Determine the form of this ηˆ(X).
Solution: Ifwechangetoη(θ)=θ−1 =1−1,θ= 1 ,θ−1= η θ θ 1−η 1−η
2θ−1=2+ 1 =2+1−η=1+1=η+1. θ−1θ−1ηηη
Thus the density is
ηη f(x)=1x−η+1 ,
the log-likelihood is
lη(X)=−nlogη− 1+η
1n i=1
and the score function is
′ n1n n1n
lη(X)=−η + η2 log(Xi)j = η2 n logXi −η . i=1 i=1
This is in “the magic form” which shows us that the logarithmic average ηˆ(X) = n1 ni=1 log(Xi) is the MVU estimator of η, which is the expectation of the exponential log(Xi)’s, which (un- surprisingly) is optimally estimated using their sample mean.
(f) Showusingthedeltamethodthattheestimatorθˆ(X)=θ(ηˆ(X))(whereη→θ(η)isthe inverse transformation of θ → η(θ)) is asymptotically normal with asymptotic variance equal to the CRLB for estimating θ.
Solution: We have that ηˆ(X) ∼ AN η, η2 . Now defining θˆ(X) = g [ηˆ(X)] = 1 as n 1−ηˆ(X)
the “reverse transformation”, by the delta method we have that g′(η) = 1 , so (1−η)2
g′(η)2η2 g(η) = θ, n
η2 θ−1 2 θ
θ2(θ − 1)2 n ;
= n(1 − η)4 = n 1 = θ4
Note that the asymptotic variance coincides with the CRLB. Thus the estimator θˆ(X) is AMVU.
5. Suppose that X has a geometric distribution with probability mass function (PMF) fθ(x)=Pθ(X=x)=(1−θ)x−1θ, forx=1,2,…
Determine the most powerful test for testing H0 : θ = 0.5 against H1 : θ = 0.25 at level α = 0.05. Solution: The likelihood ratio is
f1(X) (3/4)X−1(1/4) 3X 1 Y = f0(X) = (1/2)X−1(1/2) = 2 3
which is an increasing function of X; we need to thus “reject for large X”. The PMF and upper tail of X under H0 are given by
x123456···
P0(X = x) 0.5 0.25 0.125 0.0625 0.03125 0.015625 P0(X ≥ x) 1 0.5 0.25 0.125 0.0625 0.03125
Because P(X ≥ 5) < α < P(X ≥ 6) the test function will be of the form 1 forX≥6
δ(X)= γ forX=5 0 forX≤4
· · · · · ·
where the constant γ is chosen so that the level of the test is α, that is
E0 [δ(X)] = γP (X = 5) + P (X ≥ 6) = 0.03125γ + 0.03125 = 0.03125(γ + 1) = α = 0.05 ,
0.05 1 32 123
γ= −1= 20 −1= 0.03125 1
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