46 Section 8.3.Efficiency Suppose we want to estimate a real-valued function of 0,say g(0), g():RR.Assume that g has continuous partial derivatives. Consider estimating g(0o)by g((Xn)),where (Xn)is the MLE. Assuming that the 8 regularity conditions hold,we know that √元((Xn)-do)巳N(0,I-l(do).By the multivariate delta method,we know that √m(g((xn)-g(0o)2N(0,a(0o)'I-1(0o)a(0o) where a(0o)is the gradient of g(0)at 00
46 Section 8.3. Efficiency Suppose we want to estimate a real-valued function of θ, say g(θ), g(·) : Rk → R. Assume that g has continuous partial derivatives. Consider estimating g(θ0) by g( ˆ θ(Xn)), where ˆ θ(Xn) is the MLE. Assuming that the 8 regularity conditions hold, we know that √n( ˆ θ(Xn) − θ0) D→ N(0, I−1(θ0)). By the multivariate delta method, we know that √n(g(ˆθ(Xn)) − g(θ0)) D→ N(0, a(θ0)I−1(θ0)a(θ0)) where a(θ0) is the gradient of g(θ) at θ0
47 By the information inequality,we know that among all unbiased estimators,6(Xn),of g(0), Varoo [6(Xn)]=Eoo[(8(Xn)-g(0o))2]>a(00)'In (0o)a(0o)(1) where In(00)=nI(00). Consider an estimator,T(Xn),of g(0o)which is asymptotically normal and asymptotically unbiased,i.e., n(T(Xn)-g(0o))2 N(0,V(0o)) (2) It turns out that under some additional regularity conditions on T(Xn),we can show that V(0o)≥a(0o)'I-1(0o)a(0o) (3)
47 By the information inequality, we know that among all unbiased estimators, δ(Xn), of g(θ), V arθ0 [δ(Xn)] = Eθ0 [(δ(Xn) − g(θ0))2] ≥ a(θ0)I−1 n (θ0)a(θ0) (1) where In(θ0) = nI(θ0). Consider an estimator, T(Xn), of g(θ0) which is asymptotically normal and asymptotically unbiased, i.e., √n(T(Xn) − g(θ0)) D→ N(0, V (θ0)) (2) It turns out that under some additional regularity conditions on T(Xn), we can show that V (θ0) ≥ a(θ0)I−1(θ0)a(θ0) (3)
48 Definition:A regular estimator T(Xn)of g(0o)which satisfies (2) with V(0o)=a(0o)'I1(00)a(0o)is said to be asymptotically efficient. If g((X))is regular,then we know that it is asymptotically efficient. Remarks about Lower Bounds (1)and (3) .(1)is attained only under exceptional circumstances (i.e., usually need completeness),while (3)is obtained under quite general regularity conditions. The UMVUE tends to be unique,while asymptotically efficient estimators are not.If T(Xn)is asymptotically efficient,then so is T(Xn)+Rn;provided nRn 0. In (1),the estimator must be unbiased,whereas in(3),the estimator must be consistent and asymptotically unbiased
48 Definition: A regular estimator T(Xn) of g(θ0) which satisfies (2) with V (θ0) = a(θ0)I−1(θ0)a(θ0) is said to be asymptotically efficient. If g( ˆ θ(Xn)) is regular, then we know that it is asymptotically efficient. Remarks about Lower Bounds (1) and (3) • (1) is attained only under exceptional circumstances (i.e., usually need completeness), while (3) is obtained under quite general regularity conditions. • The UMVUE tends to be unique, while asymptotically efficient estimators are not. If T(Xn) is asymptotically efficient, then so is T(Xn) + Rn, provided √nRn P→ 0. • In (1), the estimator must be unbiased, whereas in (3), the estimator must be consistent and asymptotically unbiased
49 .V(0o)in (3)is an asymptotic variance,whereas (1)refers to the actual variance of 6(Xn). For a long time,it was believed that the regularity conditions needed to make (3)hold involved regularity conditions on the density p(x;0).This belief was exploded by Hodges
49 • V (θ0) in (3) is an asymptotic variance, whereas (1) refers to the actual variance of δ(Xn). For a long time, it was believed that the regularity conditions needed to make (3) hold involved regularity conditions on the density p(x; θ). This belief was exploded by Hodges
50 Hodges'Example of Super-Efficiency Suppose thatXn=(X1,...,Xn)where the Xi's are i.i.d. Normal(uo,1).We can show that I(uo)=1 for all uo.Consider the following estimator of uo, n|xn≥n-1/4 )|xml<n-1/4 Hodges showed that: VHECN. N(0,1)ifo≠0 The latter variance makes the asymptotic distribution of the MLE inadmissible
50 Hodges’ Example of Super-Efficiency Suppose that Xn = (X1,...,Xn) where the Xi’s are i.i.d. Normal(µ0, 1). We can show that I(µ0) = 1 for all µ0. Consider the following estimator of µ0, µ ˆ(Xn) = ⎧⎨⎩ X¯ n |X¯ n| ≥ n−1/4 0 |X¯ n| < n−1/4 Hodges showed that: √n(ˆµ(Xn) − µ0) →D ⎧⎨⎩ N(0, 1) if µ0 = 0 0 if µ0 = 0 The latter variance makes the asymptotic distribution of the MLE inadmissible
51 Remarks about Super-Efficiency The problem here is not due to irregularities of the density function,but due to the partialness of our estimator to uo =0. This example shows that no regularity conditions on the density can prevent an estimator from violating (3).This possibility can only be avoided by placing restrictions on the sequence of estimators. LeCam (1953)showed that for any sequence of estimators satisfying (2),the set of points in violating (3)has Lebesgue measure zero. ●Superefficiency shows that“parametric”models can be useful and justified when checking them in appropriate ways
51 Remarks about Super-Efficiency • The problem here is not due to irregularities of the density function, but due to the partialness of our estimator to µ0 = 0. • This example shows that no regularity conditions on the density can prevent an estimator from violating (3). This possibility can only be avoided by placing restrictions on the sequence of estimators. • LeCam (1953) showed that for any sequence of estimators satisfying (2), the set of points in Θ violating (3) has Lebesgue measure zero. • Superefficiency shows that “parametric” models can be useful and justified when checking them in appropriate ways
52 Regular Estimator When we first discussed estimators we wanted to rule out partial estimators,i.e.,estimators which favored some values of the parameters over others.In an asymptotic sense,we may want our sequence of estimators to be impartial so that we rule out estimators like the one presented by Hodges.Toward this end,we may restrict ourselves to regular estimators.A regular sequence of estimators is one whose asymptotic distribution remains the same in shrinking neighborhoods of the true parameter value
52 Regular Estimator When we first discussed estimators we wanted to rule out partial estimators, i.e., estimators which favored some values of the parameters over others. In an asymptotic sense, we may want our sequence of estimators to be impartial so that we rule out estimators like the one presented by Hodges. Toward this end, we may restrict ourselves to regular estimators. A regular sequence of estimators is one whose asymptotic distribution remains the same in shrinking neighborhoods of the true parameter value
53 More formally.consider a sequence of estimators {T(Xn)}and a sequence of parameter values {on}so that vn(n-00)is bounded. T(Xn)is a regular sequence of estimators if Pan[v元(T(Xn)-g(0n)）≤ad converges to the same limit as Pa[v元(T(Xn)-g(0o)≤ad for all a.For estimators that are regular and satisfy (2),(3)holds. When we talk about influence functions,we will formally establish this result
53 More formally. consider a sequence of estimators {T(Xn)} and a sequence of parameter values {θn} so that √n(θn − θ0) is bounded. T(Xn) is a regular sequence of estimators if Pθn [√n(T(Xn) − g(θn)) ≤ a] converges to the same limit as Pθ0 [√n(T(Xn) − g(θ0)) ≤ a] for all a. For estimators that are regular and satisfy (2), (3) holds. When we talk about influence functions, we will formally establish this result
54 Is Hodge's Estimator Regular? Suppose that uo =0.We know that the limiting distribution of vn((n)-Lo)is a degenerate distribution with point mass at zero.Consider the sequence un=T/√元，where r is some positive constant.Note that vn(un-uo)T.Now,we can show that ()u).To see this,not that Pμn【元(a(xn)-hn)=-]=Pμn【v元(n(Xn)-T/√元)=-T] Pμn[m(Xn)=o] PunllXnl<n-1/4] = Pun【-n-1/4<xn<n-1/4] Pun元(-n-1/4-un)<V元(xn-un)<V元(n-1/4-u】 = Φ(W元(n-1/4-n)》-Φ（元(-n-1/4-un) Φ(n1/4-r)-Φ(-n1/4-r) 1
54 Is Hodge’s Estimator Regular? Suppose that µ0 = 0. We know that the limiting distribution of √n(ˆµ(Xn) − µ0) is a degenerate distribution with point mass at zero. Consider the sequence µn = τ /√n, where τ is some positive constant. Note that √n(µn − µ0) → τ. Now, we can show that √n(ˆµ(Xn) − µn) P (µn) → −τ. To see this, not that Pµn [√n(ˆµ(Xn) − µn) = −τ] = Pµn [√n(ˆµ(Xn) − τ/√n) = −τ] = Pµn [ˆµ(Xn) = 0] = Pµn [|X ¯ n| < n−1/4] = Pµn [−n−1/4 < X¯ n < n−1/4] = Pµn [√n(−n−1/4 − µn) < √n(X¯ n − µn) < √n(n−1/4 − µn)] = Φ(√n(n−1/4 − µn)) − Φ(√n(−n−1/4 − µn)) = Φ(n1/4 − τ) − Φ(−n1/4 − τ) → 1
55 So,the limiting distribution of vn((Xn)-un)is also degenerate, but with point mass at -7.This is a different limiting distribution than that of vn((Xn)-uo).Therefore,(Xn)is not regular
55 So, the limiting distribution of √n(ˆµ(Xn) − µn) is also degenerate, but with point mass at −τ. This is a different limiting distribution than that of √n(ˆµ(Xn) − µ0). Therefore, ˆµ(Xn) is not regular