The University of Sydney School of Mathematics and Statistics

Solutions to Tutorial Week 11

STAT3023: Statistical Inference Semester 2, 2023 1. Suppose X1,...,Xn are iid N(θ,1) random variables and X ̄ = n1 Pni=1 Xi. If θ0 < θ < θ1 and

 0 < C < ∞ show that

as n → ∞. Solution:

?C ̄C?

Pθ θ0+√n<X<θ1−√n →1

Since X ̄ ∼ N (θ, n1 ), √n(X ̄ − θ) ∼ N (0, 1). Therefore, so long as n is big enough that θ 0 + √C ≤ θ 1 − √C

nn

?√? C ? √? ̄ ? √? C ?? n θ0+√n−θ < n X−θ < n θ1−√n−θ

n 2C o2

(i.e. n ≥ θ1−θ0 ), we have

? C  ̄ C?

Pθ θ0+√n<X<θ1−√n =Pθ

Since

also, since

= P θ ? C − √ n ( θ − θ 0 ) < √ n ? X ̄ − θ ? < √ n ( θ 1 − θ ) − C = Φ ?√n (θ1 − θ) − C? − Φ ?C − √n (θ − θ0)? .

√

n (θ1 − θ) − C → +∞

Φ?√n(θ1 −θ)−C?→1; √

C− n(θ−θ0)→−∞, Φ ?C − √n (θ − θ0)? → 0 .

Therefore the difference tends to 1, as required.

2. Interval estimation of an exponential rate parameter

Suppose X = (X1, . . . , Xn) consists of iid random variables whose common distribution is expo-

nential with rate θ ∈ Θ = (0, ∞) unknown. Consider the formal decision problem with decision space D = Θ = (0, ∞) and loss function (sequence) L(d|θ) = Ln(d|θ) = 1 n|d − θ| > √C o. This

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 .

Solution: The likelihood is

n

fθ(X) = Y?θe−θXi? = θne−Tθ

i=1

where T = Pni=1 Xi; taking logs and differentiating gives

l ′ ( θ ; X ) = nθ − T ; Copyright © 2023 The University of Sydney 1

n

(a) Write down the likelihood and derive θˆML as a function of the Xi’s.

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

Solution:

The risk function is ˆ?ˆC??ˆC?

ˆ R(θ|θML)=Gn

? C ?−1! " 1+θ√n + 1−Gn

Gn(y) = P(Yn ≤ y)

? C ?−1!# 1−θ√n

 where

is the CDF of Yn.

R(θ|θML)=Pθ θ<θML−√n +Pθ θ>θML+√n

= P θ = P θ

(1)(1) X ̄ < θ + √C + P θ X ̄ > θ − √C

nn

(θ)(θ) θ X ̄ < θ + √C + P θ θ X ̄ > θ − √C

nn

   (1)(1)

=P Yn<1+ C +P Yn>1− C √√

θn θn

? 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).

a≤θ≤b

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

ˆ ˆ ?C?−1! ?C?−1!

max R(θ|θML)=R(b|θML)=Gn 1+ √ +1−Gn 1− √ . a≤θ≤b bn bn

(d) Determine, for any 0 < a < b < ∞, the limiting maximum risk lim maxR(θ|θˆML).

n→∞ a≤θ≤b

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)

Gn 1+√ =PYn≤