The University of of Mathematics and Statistics
Solutions to Tutorial Week 10
STAT3023: Statistical Inference Semester 2, 2021
Lecturers: and
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1. Suppose X1,…,Xn are iid N(θ,1) random variables and X ̄ = n1 ni=1 Xi. If θ0 < θ < θ1 and 0 < C < ∞ show that
as n → ∞. Solution:
Pθ θ0+√n
corresponds to obtaining an interval estimate of θ, with risk equal to the non-coverage probability of the interval; the midpoint of the interval is still regarded as an “estimator” of θ though.
Consider using the ordinary maximum likelihood estimator θˆML as the estimator, giving the interval θˆML ± √C .
(a) Write down the likelihood and derive θˆML as a function of the Xi’s.
Solution: The likelihood is
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fθ(X) = θe−θXi = θne−Tθ
where T = ni=1 Xi; taking logs and differentiating gives l ′ ( θ ; X ) = nθ − T ;
setting equal to zero and solving gives
θˆML = n = 1 ̄ ,
TX where, as usual, X ̄ = T/n denotes the sample mean.
(b) Since X ̄ has a gamma distribution with shape n and rate nθ, the product Yn = θX ̄ has a gamma distribution with shape n and rate n (i.e. its distribution is free of θ). Show that the risk function
R(θ|θˆML)=Gn
C −1 C −1
1+θ√n + 1− (y) = P(Yn ≤ y)
is the CDF of Yn.
The risk function is ˆˆCˆC
R(θ|θML)=Pθ θ<θML−√n +Pθ θ>θML+√n
= P θ = P θ
11 X ̄ < θ + √C + P θ X ̄ > θ − √C
θθ θ X ̄ < θ + √C + P θ θ X ̄ > θ − √C
=P Yn<1+ C +P Yn>1− C √√
C −1 C −1 =Gn 1+θ√n + 1−Gn 1−θ√n .
(c) Determine, for any 0 < a < b < ∞, the maximum risk max R(θ|θˆML).
Solution: Each of the two terms making up the risk is an increasing function of θ and so
the sum of them is also an increasing function of θ. Thus
C −1 C −1 1+ √ +1−Gn 1− √
max R(θ|θˆML)=R(b|θˆML)=Gn
a≤θ≤b bn bn
(d) Determine, for any 0 < a < b < ∞, the limiting maximum risk lim maxR(θ|θˆML).
You may use the facts that E(Yn) = 1, Var(Yn) = n1 and Yn is asymptotically normal.
Solution: Note firstly that for any sequence zn → z, since √n(Yn − 1) →d N (0, 1) we have P√n(Yn −1)≤zn→Φ(z)
where Φ(·) is the N(0,1) CDF. Consider the first term making up the maximum risk: C−1 1
this tends to
Φ−b=1−Φb. C−1√ √1
Gn 1+√ =PYn≤
=P n(Yn−1)≤n1+C−1 √
bn −C
n(Yn − 1) ≤ b . √
In a similar way,
1−Gn1−√ =Pn(Yn−1)>n C−1
n(Yn − 1) > b 1−C
Therefore the limiting maximum risk is
lim max R(θ|θML)=2 1−Φ .
C →1−Φb.
ˆ C n→∞ a≤θ≤b b
3. Suppose X1, . . . , Xn are iid U[0, θ] random variables and that we wish to estimate θ using squared error loss (so the risk is the mean-squared error). Assume that n ≥ 3.
(a) The maximum likelihood estimator of θ is the sample maximum X(n) = maxi=1,…,n Xi. In
last week’s tutorial, using the fact that E (X ) = nθ and Var X = nθ2
θ (n) n+1 showed that the exact risk of this estimator is
θ (n) (n+1)2(n+2)
2 Eθ X(n) −θ
= (n+2)(n+1) .
Determine, for 0 ≤ a < b < ∞, the limiting maximum (rescaled) risk over [a, b]:
lim max n2E X
−θ2= lim max θ2 2n2 n→∞ a≤θ≤b (n + 2)(n + 1)
lim maxn2EX −θ2. n→∞ a≤θ≤b θ (n)
= lim b n→∞
= lim b n→∞
n2 1 + 2 1 + 1 nn
1 + 2 1 + 1 nn
(b) In last week’s tutorial we also showed that the unbiased estimator θˆunb = n+1 X(n) has n
exact risk
For 0 ≤ a < b < ∞ find
ˆ 2 θ2
lim maxnEθ n→∞ a≤θ≤b
= lim max θ n→∞ a≤θ≤b
ˆ 2 θunb−θ
= n(n+2). ˆ 2
lim maxnEθ n→∞ a≤θ≤b
1 + 2 n
The likelihood is (assuming
(c) Show that the Bayes procedure using the “flat
= lim b n→∞
= lim b n→∞
n2 1 + 2 n
n(n + 2) n2
prior” weight function w(θ) ≡ 1 is given by θflat(X)= n−2 X(n).
n 1{0 ≤ Xi ≤ θ} 1X(1) ≥ 01X(n) ≤ θ
(where X(1) is the sample minimum). Assuming X(1) ≥ 0, when viewed this is a multi- ple of a Pareto density (with shape n − 1 and scale X(n)) so since the weight function is just w(θ) ≡ 1 this Pareto distribution is also the posterior distribution. Thus the Bayes procedure/estimator is the posterior mean, which is
shape × scale = (n − 1)X(n) . shape−1 (n−2)
(d) Using the expressions for the expectation and variance of X(n) given in part (a) above, determine the variance, bias and thus exact risk of θˆflat.
ˆ n−1n−1nθ Eθ θflat(X) = n−2 Eθ X(n) = n−2 n+1.
Thus the bias is
Biasθ θflat = n−2 Eθ X(n) −θ= (n−2)(n+1)
ˆ n − 1 (n − 1)nθ − θ(n − 2)(n + 1) n2 −n−[n2 −n−2]
The variance is given by
= θ (n − 2)(n + 1) = 2θ .
ˆ n−12
θflat = n − 2 Varθ X(n)
n−12 nθ2
= n−2 (n+1)2(n+2)
(n−2)(n+1)
Thus the exact risk is
ˆ 2 θflat − θ
ˆ ˆ 2 = Varθ θflat + Biasθ θflat
θ2 n(n−1)2 = (n−2)2(n+1)2 n+2 +4
= θ2(n−1)2 n + 4 (n−2)2(n+1)2 n+2 (n−1)2
(e) Determine,for0≤a