CS代写 STAT3023: Statistical Inference Semester 2, 2021

The University of of Mathematics and Statistics
Solutions to Tutorial Week 12
STAT3023: Statistical Inference Semester 2, 2021
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1. Suppose X1,…,Xn are iid U(0,θ) random variables and that 0 < θ0 < θ < θ1. Suppose 0 < C < ∞ is fixed and known and write X(n) = maxi=1,...,n Xi for the sample maximum. (a) Determine a value n0 = n0(C, θ, θ1) such for n ≥ n0, 􏰌C􏰍 Pθ X(n)>θ1−n =0.
Solution: We have (for all n) that
Pθ􏰄X(n) ≤θ􏰅=1.
Thusfornsolargethatθ≤θ1−Cn wealsohave 􏰌C􏰍
This occurs when
Pθ X(n)≤θ≤θ1−n =1. Cn ≤ θ 1 − θ
n≥n0= C . θ1 − θ
So by considering the complementary event we see that for this definition of n0, and all n ≥ n0,
Pθ X(n)>θ1−n =0.
(b) Write down the CDF of X(n) and show that as n → ∞,
Hence determine
We have already seen (see, for example, Tutorial 10) that the CDF of X(n) is
Foralln≥ C ,θ0+C ≤θ,so θ−θ0 n
C􏰍 􏰙θ0+C􏰚n = n
􏰌 C 􏰍 C/θ0 X(n) ≤ θ0 + n → e .
􏰌CC􏰍 lim Pθ θ0 + < X(n) < θ1 − . 0 Pθ􏰄X(n) ≤x􏰅=􏰄x􏰅n for x ≤ 0, for0 0, so the common density is
fθ(x) = xθ+1 1{x > m} .
It is desired to estimate θ under squared-error loss.
(a) Determine dflat(X), the Bayes procedure using the “flat prior” w(θ) = 1 {θ > 0}.
Solution: The likelihood is n􏰐􏰢n􏰐θ􏰢n
fθ(X) = θ+1 1 {Xi > m}
􏰝 􏰌m􏰍 1 i=1 Xi Xi
i=1 log(Xi/m)
􏰝 1 i=1 Xi
assuming all Xis exceed the known lower endpoint m.
i=1 Xi Thus the product
w(θ)fθ(X) = const. θne−θ 􏰑ni=1 log(Xi/m)
and thus we see the posterior is the gamma density with shape n+1 and rate 􏰑ni=1 log(Xi/m).
The Bayes procedure is thus the posterior mean, which is shape n + 1
rate = 􏰑ni=1 log(Xi/m) .
(b) By deriving the distribution of Y1 = log(X1/m), show that dflat(X) has an inverse gamma
distribution.
Solution: The CDF of X1 is (for x > m),
Pθ{Y1 ≤y}=Pθ{log(X1/m)≤y}=Pθ(X1 ≤m−1ey =1−e−θy
so Y1 is exponential with rate θ. Consequently, 􏰑ni=1 log(Xi/m) is gamma with shape n
and rate θ, the ratio
􏰑ni=1 log(Xi/m) n+1
is gamma with shape n and rate (n+1)θ, thus dflat(X) is inverse gamma with shape n and rate (n + 1)θ.
􏰟x 􏰖−θ􏰗x 􏰊􏰋 θt θ􏰆−θ −θ􏰇 x−θ
Pθ(X1 ≤x)= fθ(t)dt=θm −θ =m m −x =1− m . mm

(c) Show that the exact risk of dflat (X ) is
θ2(n + 7) .
(n−1)(n−2)
Solution: We have
Eθ [dflat(X)] = rate = (n + 1)θ ,
shape−1 n−1 Biasθ[dflat(X)]=Eθ[dflat(X)]−θ=(n+1)θ−(n−1)θ= 2θ ,
Varθ [dflat(X)] = (shape − 1)2(shape − 2) = (n − 1)2(n − 2) Thus the risk is
n−1 n−1 rate2 θ2(n + 1)2
R(θ|dflat) = Varθ [dflat(X)] + {Biasθ [dflat(X)]}2
θ2(n + 1)2 4θ2(n − 2) = (n−1)2(n−2) + (n−1)2(n−2) =θ2n2 +2n+1+4n−8
(n−1)2(n−2) =θ2 n2+6n−7
(n−1)2(n−2) =θ2 (n+7)(n−1)
(n−1)2(n−2) = θ2 (n + 7) .
(n−1)(n−2)
(d) It is known that if d(X) is the Bayes procedure for this problem using a U(θ0,θ1) prior (for
0<θ0 <θ1 <∞)thenforallθ0 <θ<θ1, lim nR(θ|d)= lim nR(θ|dflat). Use this to prove that dflat(X) is asymptotically minimax over any interval [a,b] with 0 < a < b < ∞. Solution: Firstly, for any θ0 < θ < θ1 lim nR(θ|d)= lim nR(θ|dflat)= lim θ2 n(n+7) =θ2 lim n(n+7) =θ2 =S(θ). n→∞ 􏰓 n→∞ n→∞ (n−1)(n−2) n→∞ (n−1)(n−2) So for any procedure (sequence) {dn(·)}, according to the asymptotic minimax lower bound theo- rem, for any 0 < a < b, lim max nR(θ|dn) ≥ max S(θ) = b2 . n→∞ a≤θ≤b a≤θ≤b Now, focusing on dflat(X), for any 0 < a < b < ∞, max nR(θ|dflat)= max θ2n(n+7) =b2 n(n+7) →b2 a≤θ≤b a≤θ≤b (n−1)(n−2) (n−1)(n−2) as n → ∞, and so since this coincides with the lower bound derived above, this proves dflat(X) is asymptotically minimax over [a, b]. The “shifted” unit-scale exponential distribution 3. Suppose X1, . . . , Xn are iid with common distribution a unit-scale exponential “shifted” by θ, with density given by fθ(x) = 1 {x ≥ θ} e−(x−θ) and unknown “location” parameter θ ∈ Θ = R. (a) Write down the common CDF. Solution: Forx≤θ,Pθ{X1 ≤x}=0. Forx>θ, 􏰟x􏰟xx
Pθ{X1 ≤x}= e−(y−θ)dy=eθ e−ydy=eθ􏰆−e−y􏰇 =eθ􏰀e−θ −e−x􏰁=1−e−(x−θ).
(b) Write down the likelihood function.
fθ(X) = 􏰝 􏰔1 {Xi ≥ θ} e−(Xi−θ)􏰕 = 1 􏰀X(1) ≥ θ􏰁 enθ−􏰑ni=1 Xi
where X(1) is the sample minimum.
(c) Determine the maximum-likelihood estimator of θ.
Solution: The likelihood is positive and increasing for −∞ < θ ≤ X(1), and then is zero for θ > X(1). So the MLE is simply the sample minimum.
(d) Show that this family has the monotone likelihood ratio property in a statistic T(X); what
is the statistic T(X)?
Solution: Fix θ0 < θ1. The likelihood ratio f (X) 1􏰀X(1) ≥ θ1􏰁enθ1−􏰑ni=1 Xi θ1 =􏰀 􏰁nθ0−􏰑nXi= 􏰐en(θ1−θ0) 0 if θ < θ ≤ X 01(1) if θ0 θ0 under either θ0 or θ1 so this is not a problem. This is a non-decreasing function of X(1) so this family has the monotone likelihood ratio property in the sample minimum.
THIS TEXT IN BLUE IS NOT EXAMINABLE, IT IS JUST IN RESPONSE TO A QUESTION FROM AN INTERESTED STUDENT.
Note that the Neyman- Ratio Test for simple H0 : θ = θ0 versus H1 : θ = θ1, for θ0 < θ1 depends on the alternative value θ1 and so this does not provide a strategy for finding a UMP test directly, we must instead use the monotone likelihood ratio property as in part (e) below. The Neyman- RT in this case can be found as follows: note that the discrete LR = LR(θ0,θ1) = fθ1(X)/fθ0(X) only takes two possible values, en(θ1−θ0) and 0, with probabilities (under H0) given by 􏰂LR=en(θ1−θ0)􏰃=P 􏰀X >θ 􏰁 θ0 (1) 1
= Pθ0 {X1 > θ1}n
= e−n(θ1−θ0) (using (1) above from part (a)) =1−Pθ0 {LR=0}.
􏰐e−n(θ1−θ0) for x = en(θ1−θ0) Pθ0 {LR=x}= 1−e−n(θ1−θ0) forx=0.
According to the discrete version of the Neyman- emma, the most powerful test of H0:θ=θ0 againstH1:θ=θ1 isgivenby
1 forLR>C; 
δ(X)= γ forLR=C; 0 forLR C) ≤ P(LR > C) + γP(LR = C) = α ≤ P(LR ≥ C)
so since there are only two possible values for C, we know which it is when we know p0 =P(LR=0).

• If1−p0=P(LR>0)=e−n(θ1−θ0)≤α(i.e.θ1≥θ0+n1log(1/α)),thenC=0,and furthermore
in other words,
α − P(LR > 0) γ= P(LR=0)
δ(X)= α−e−n(θ1−θ0)
α − e−n(θ1−θ0) = 1−e−n(θ1−θ0) ;
for X(1) ≥ θ1
forθ 0)=e−n(θ1−θ0)>α(i.e.θ1<θ0+n1log(1/α)),C=en(θ1−θ0)and γ = α = αen(θ1−θ0) ; in other words, 􏰐αen(θ1−θ0) if X(1) > θ1
0 ifθ0 θ0 + n log(1/α) .
(e) Determine the UMP level-α test for testing H0 : θ = θ0 against H1 : θ > θ0.
Solution: Since the family of distributions has the monotone likelihood ratio property, the UMP test is of the form
δ(X)=1􏰀X(1) >C􏰁 where C is chosen so that Eθ0 [δ(X)] = α.
Pθ􏰀X(1) >C􏰁=Pθ{X1 >C,…,Xn >C} =Pθ{X1 >C}···Pθ{Xn >C}
= [Pθ {X1 > C}]n by independence. By part (a) above, we need to solve
􏰀 􏰁 􏰔 −(C−θ0)􏰕n α=Pθ0 X(1)>C = e =e
C = θ0 − log α = θ0 + [− log α] . nn
(f) Consider a decision problem with decision space D = Θ = R. 5

(i) Determine the posterior distribution using the “flat prior” weight function w(θ) ≡ 1. Solution: The product of the prior and likelihood function is
1 􏰀X(1) ≥ θ􏰁 enθ−􏰑ni=1 Xi = const. enθ1 􏰀θ ≤ X(1)􏰁 .
As a function of θ, this is (a multiple of) a reversed, shifted exponential distribution with rate n, supported on the interval (−∞,X(1)).
Renormalising, the posterior is (the factor e− 􏰑ni=1 Xi may be dropped)
􏰀 􏰁nθ 􏰀 􏰁nθ 􏰀 􏰁nθ
p(θ|X)= 1 θ≤X(1) e = 1 θ≤X(1) e = 1 θ≤X(1) e =1􏰀θ≤X 􏰁nen(θ−X(1)).
􏰨X(1) enθdθ 􏰆1enθ􏰇X(1) n1enX(1) −0 (1) −∞ n−∞
(ii) Determine the Bayes procedure/estimator when the loss function is
(A) squared error: L(d|θ) = (d − θ)2;
Solution: The Bayes procedure is the posterior mean which is given by
θp(θ|X)dθ =
θnen(θ−X(1)) dθ
= e−nX(1) = e−nX(1)
θ nenθ dθ 􏰐X􏰟X(1) 􏰢
=e (1) X(1)e (1) −0−n e (1) −0 = X(1) − n1 .
(B) absolute error: L(d|θ) = |d − θ|;
Solution: The Bayes procedure is the posterior median which is that value −∞ < m < X(1) satisfying Solving for m we get e−n(X(1)−m) . log 21 = −n(X(1) − m) p(θ|X)dθ nen(θ−X(1)) dθ 􏰆θenθ􏰇 (1) − −∞ X(1)enX(1) − 0 − nenθ −nX􏰎nX 1􏰆nX􏰇􏰏 nenθ dθ e−nX(1) 􏰆enθ􏰇m −∞ e−nX(1) [enm − 0] m = X(1) − log 2 ≈ X(1) − 0.69 . nn (C) non-coverage (or “zero-one”) : L(d|θ) = 1{|d−θ| > C/n} for some given C > 0.
Solution: The Bayes procedure is the midpoint of the interval of length 2C/n which has the highest probability under the posterior distribution. Since the posterior density increases up to X(1) and then drops to zero, the corresponding interval is 􏰆X(1) − 2C , X(1)􏰇. The midpoint is then just
X(1) − Cn . 6

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