Section 5.2

Admissibility and Minimaxity in Exponential Families

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Problem 5.2.1

Lemma 2.1 has been extended by Berger (1990a) to include the case where the estimand need not be restricted to a finite interval, but, instead, attains a maximum or minimum at a finite parameter value.

Lemma 8.1 Let the estimand g(θ)g(\theta ) be nonconstant with global maximum or minimum at a point θ∗∈Ω\theta^{*} \in \Omega for which f(x∣θ∗)>0f\left(x \mid \theta^{*}\right)>0 a.e. (with respect to a dominating measure μ)\mu ), and let the loss L(θ,d)L(\theta , d) satisfy the assumptions of Lemma 2.1. Then, any estimator δ\delta taking values above the maximum of g(θ)g(\theta ), or below the minimum, is inadmissible.

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(a)

Show that if θ∗\theta^{*} minimizes g(θ)g(\theta ), and if g^(x)\hat{g}(x) is an unbiased estimator of g(θ)g(\theta ), then there exists ϵ>0\epsilon >0 such that the set Aϵ={x∈X:g^(x)<g(θ)−ϵ}A_{\epsilon }=\left\{ x \in \mathcal{X}: \hat{g}(x)<g(\theta )-\epsilon \right\} satisfies P(Aϵ)>0\mathbb {P}\left(A_{\epsilon }\right)>0. A similar conclusion holds if g(θ∗)g\left(\theta^{*}\right) is a maximum.

(b)

Suppose g(θ∗)g\left(\theta^{*}\right) is a minimum. (The case of a maximum is handled similarly.) Show that the estimator

δ(x)={g^(x) if g^(x)≥g(θ∗)g(θ∗) if g^(x)<g(θ∗) \delta (x)= \begin{cases} \hat{g}(x) & \text{ if } \hat{g}(x) \geq g\left(\theta ^{*}\right) \\ g\left(\theta ^{*}\right) & \text{ if } \hat{g}(x)<g\left(\theta ^{*}\right)\end{cases}

satisfies R(δ,θ)−R(g^(x),θ)<0R(\delta , \theta )-R(\hat{g}(x), \theta )<0.

(c)

For the situation of Example 2.3, apply Lemma 8.1 to establish the inadmissibility of the UMVU estimator of σA2\sigma_{A}^{2}. Also, explain why the hypotheses of Lemma 8.1 are not satisfied for the estimation of σ2\sigma^{2}

Problem 5.2.2

Determine the Bayes risk of the estimator (2.4) when θ\theta has the prior distribution N(μ,τ2)N\left(\mu , \tau^{2}\right).

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Problem 5.2.3

Prove part (d) in the second proof of Example 2.8, that there exists a sequence of values θi→−∞\theta_{i} \rightarrow -\infty with b′(θi)→0b^{\prime }\left(\theta_{i}\right) \rightarrow 0.

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Problem 5.2.4

Show that an estimator aX+b(0≤a≤1)a X+b(0 \leq a \leq 1) of Eθ[X]\mathbb {E}_{\theta }\left[X\right] is inadmissible (with squared error loss) under each of the following conditions:

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(a)

if Eθ[X]≥0\mathbb {E}_{\theta }\left[X\right] \geq 0 for all θ\theta, and b<0b<0;

(b)

if Eθ[X]≤k\mathbb {E}_{\theta }\left[X\right] \leq k for all θ\theta, and ak+b>ka k+b>k.

Problem 5.2.5

Show that an estimator [1/(1+λ)+ε]X[1 /(1+\lambda )+\varepsilon ] X of Eθ[X]\mathbb {E}_{\theta }\left[X\right] is inadmissible (with squared error loss) under each of the following conditions:

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(a)

if var⁡θ(X)/Eθ2(X)>λ>0\operatorname {var}_{\theta }(X) / E_{\theta }^{2}(X)>\lambda >0 and ε>0\varepsilon >0,

(b)

if var⁡θ(X)/Eθ2(X)<λ\operatorname {var}_{\theta }(X) / E_{\theta }^{2}(X)<\lambda and ε<0\varepsilon <0.

Problem 5.2.6

Show that if var⁡θ(X)/Eθ2(X)>λ>0\operatorname {var}_{\theta }(X) / E_{\theta }^{2}(X)>\lambda >0, an estimator [1/(1+λ)+ε]X+b[1 /(1+\lambda )+\varepsilon ] X+b is inadmissible (with squared error loss) under each of the following conditions:

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(a)

if Eθ[X]>0\mathbb {E}_{\theta }\left[X\right]>0 for all θ,b>0\theta , b>0 and ε>0\varepsilon >0;

(b)

if Eθ[X]<0\mathbb {E}_{\theta }\left[X\right]<0 for all θ,b<0\theta , b<0 and ε>0\varepsilon >0 (Gupta 1966).

Problem 5.2.7

Brown (1986a) points out a connection between the information inequality and the unbiased estimator of the risk of Stein-type estimators.

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(a)

Show that (2.7) implies

R(θ,δ)≥[1+b′(θ)]2n+b2(θ)≥1n+2b′(θ)n+b2(θ) R(\theta , \delta ) \geq \frac{\left[1+b^{\prime }(\theta )\right]^{2}}{n}+b^{2}(\theta ) \geq \frac{1}{n}+\frac{2 b^{\prime }(\theta )}{n}+b^{2}(\theta )

and, hence, if R(θ,δ)≤R(θ,Xˉ)R(\theta , \delta ) \leq R(\theta , \bar{X}), then 2b′(θ)n+b2≤0\frac{2 b^{\prime }(\theta )}{n}+b^{2} \leq 0.

(b)

Show that a nontrivial solution b(θ)b(\theta ) would lead to an improved estimator x−g(x)x-g(x), for p=1p=1, in Corollary 4.7.2.

Problem 5.2.8

A density function f(x∣θ)f(x \mid \theta ) is variation reducing of order n+1(VRn+1)n+1\left(V R_{n+1}\right) if, for any function g(x)g(x) with k(k≤n)k(k \leq n) sign changes (ignoring zeros), the expectation Eθ[g(X)]=∫g(x)f(x∣θ)dx\mathbb {E}_{\theta }\left[g(X)\right]= \int g(x) f(x \mid \theta ) d x has at most kk sign changes. If Eθ[g(X)]\mathbb {E}_{\theta }\left[g(X)\right] has exactly kk sign changes, they are in the same order. Show that f(x∣θ)f(x \mid \theta ) is VR2V R_{2} if and only if it has monotone likelihood ratio. (See TSH2, Lemma 2, Section 3.3 for the "if" implication). Brown et al. (1981) provide a thorough introduction to this topic, including VRV R characterizations of many families of distributions (the exponential family is VR∞V R_{\infty }, as is the χv2\chi_{v}^{2} with ν\nu the parameter, and the noncentral χν2(λ)\chi_{\nu }^{2}(\lambda ) in λ\lambda ). There is an equivalence between VRnV R_{n} and TPnT P_{n}, Karlin's (1968) total positivity of order nn, in that VRn=TPnV R_{n}=T P_{n}.

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Problem 5.2.9

For the situation of Example 2.9, show that:

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(a)

without loss of generality, the restriction θ∈[a,b]\theta \in [a, b] can be reduced to θ∈[−m,m]\theta \in [-m, m], m>0m>0.

(b)

If Λ\Lambda is the prior distribution that puts mass 1/2 on each of the points ±m\pm m, then the Bayes estimator against squared error loss is

δΛ(xˉ)=memnxˉ−e−mnxˉemnxˉ+e−mnxˉ=mtanh⁡(mnxˉ). \delta ^{\Lambda }(\bar{x})=m \frac{e^{m n \bar{x}}-e^{-m n \bar{x}}}{e^{m n \bar{x}}+e^{-m n \bar{x}}}=m \tanh (m n \bar{x}).
(c)

For m<1/nm<1 / \sqrt{n},

max⁡θ∈[−m,m]R(θ,δ(Xˉ))=max⁡{R(−m,δΛ(Xˉ)),R(m,δΛ(Xˉ))} \max _{\theta \in [-m, m]} R(\theta , \delta (\bar{X}))=\max \left\{ R\left(-m, \delta ^{\Lambda }(\bar{X})\right), R\left(m, \delta ^{\Lambda }(\bar{X})\right)\right\}

and hence, by Corollary 1.6, δΛ\delta^{\Lambda } is minimax.

(d)

For m>1.05/n,δΛm>1.05 / \sqrt{n}, \delta^{\Lambda } of part (b) is no longer minimax. Explain why this is so and suggest an alternate estimator in this case.

Problem 5.2.10

For the situation of Example 2.10, show that:

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(a)

max⁡θ∈[−m,m]R(θ,aXˉ+b)=max⁡{R(−m,aXˉ+b),R(m,aXˉ+b)}\max_{\theta \in [-m, m]} R(\theta , a \bar{X}+b)=\max \left\{ R(-m, a \bar{X}+b), R(m, a \bar{X}+b)\right\}.

(b)

The estimator a∗Xˉa^{*} \bar{X}, with a∗=m2/(1n+m2)a^{*}=m^{2} /\left(\frac{1}{n}+m^{2}\right), is the linear minimax estimator for all mm with minimax risk a∗/na^{*} / n.

(c)

Xˉ\bar{X} is the linear minimax estimator for m=∞m=\infty.

Problem 5.2.11

Suppose XX has distribution FξF_{\xi } and YY has distribution GηG_{\eta }, where ξ\xi and η\eta vary independently. If it is known that η=η0\eta =\eta_{0}, then any estimator δ(X,Y)\delta (X, Y) can be improved upon by

δ∗(x)=EY[δ(x,Y)]=∫δ(x,y)dGη0(y) \delta ^{*}(x)=\mathbb {E}_{Y}\left[\delta (x, Y)\right]=\int \delta (x, y) d G_{\eta _{0}}(y)
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Problem 5.2.12

In Example 2.13, prove that the estimator aY+ba Y+b is inadmissible when a>1/(r+1)a>1 /(r+1).

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Problem 5.2.13

Let X1,…,XnX_{1}, \ldots , X_{n} be iid according to a N(0,σ2)N\left(0, \sigma^{2}\right) density, and let S2=∑Xi2S^{2}=\sum X_{i}^{2}. We are interested in estimating σ2\sigma^{2} under squared error loss using linear estimators cS2+dc S^{2}+d, where cc and dd are constants. Show that:

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(a)

admissibility of the estimator aY+ba Y+b in Example 2.13 is equivalent to the admissibility of cS2+dc S^{2}+d, for appropriately chosen cc and dd.

(b)

the risk of cS2+dc S^{2}+d is given by R(cS2+d,σ2)=2nc2σ2+[(nc−1)σ2+d]2R\left(c S^{2}+d, \sigma^{2}\right)=2 n c^{2} \sigma^{2}+\left[(n c-1) \sigma^{2}+d\right]^{2}

(c)

for d=0,R(cS2,σ2)<R(0,σ2)d=0, R\left(c S^{2}, \sigma^{2}\right)<R\left(0, \sigma^{2}\right) when c<2/(n+2)c<2 /(n+2), and hence the estimator aY+ba Y+b in Example 2.13 is inadmissible when a=b=0a=b=0.

Remark.
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This exercise illustrates the fact that constants are not necessarily admissible estimators.

Problem 5.2.14

For the situation of Example 2.15, let Z=Xˉ/SZ=\bar{X} / S.

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(a)

Show that the risk, under squared error loss, of δ=φ(z)s2\delta =\varphi (z) s^{2} is minimized by taking

φ(z)=φμ,σ∗(z)=E[S2/σ2∣z]/E[(S2/σ2)2∣z]. \varphi (z)=\varphi _{\mu , \sigma }^{*}(z)=\mathbb {E}\left[S^{2} / \sigma ^{2} \mid z\right] / \mathbb {E}\left[\left(S^{2} / \sigma ^{2}\right)^{2} \mid z\right].
(b)

Stein (1964) showed that φμ,σ∗(z)≤φ0,1∗(z)\varphi_{\mu , \sigma }^{*}(z) \leq \varphi_{0,1}^{*}(z) for every μ,σ\mu , \sigma. Assuming this is so, deduce that φs(Z)S2\varphi_{s}(Z) S^{2} dominates [1/(n+1)]S2[1 /(n+1)] S^{2} in squared error loss, where

φs(z)=min⁡{φ0,1∗(z),1n+1}. \varphi _{s}(z)=\min \left\{ \varphi _{0,1}^{*}(z), \frac{1}{n+1}\right\} .
(c)

Show that φ0,1∗(z)=(1+z2)/(n+2)\varphi_{0,1}^{*}(z)=\left(1+z^{2}\right) /(n+2), and, hence, φs(Z)S2\varphi_{s}(Z) S^{2} is given by (2.31).

(d)

The best equivariant estimator of the form φ(Z)S2\varphi (Z) S^{2} was derived by Brewster and Zidek (1974) and is given by

φBZ(z)=E[S2∣Z≤z]E[S4∣Z≤z], \varphi _{B Z}(z)=\frac{\mathbb {E}\left[S^{2} \mid Z \leq z\right]}{\mathbb {E}\left[S^{4} \mid Z \leq z\right]},

where the expectation is calculated assuming μ=0\mu =0 and σ=1\sigma =1. Show that φBZ(Z)S2\varphi_{B Z}(Z) S^{2} is generalized Bayes against the prior

π(μ,σ)=1σ∫0∞u−1/2(1+u)−1e−unμ2/σ2dudμdσ \pi (\mu , \sigma )=\frac{1}{\sigma } \int _{0}^{\infty } u^{-1 / 2}(1+u)^{-1} e^{-u n \mu ^{2} / \sigma ^{2}} d u d \mu d \sigma
Remark.
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Brewster and Zidek did not originally derive their estimator as a Bayes estimator, but rather first found the estimator and then found the prior. Brown (1968) considered a family of estimators similar to those of Stein (1964), which took different values depending on a cutoff point for z2z^{2}. Brewster and Zidek (1974) showed that the number of cutoff points can be arbitrarily large. They constructed a sequence of estimators, with decreasing risks and increasingly dense cutoffs, whose limit was the best equivalent estimator.

Problem 5.2.15

Show the equivalence of the following relationships:

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(a)

(2.26) and (2.27),

(b)

(2.34) and (2.35) when c=(n−1)/(n+1)c=\sqrt{(n-1) /(n+1)}, and

(c)

(2.38) and (2.39).

Problem 5.2.16

In Example 2.17, show that the estimator aX/n+ba X / n+b is inadmissible for all (a,b)(a, b) outside the triangle (2.39).

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Problem 5.2.17

Prove admissibility of the estimators corresponding to the interior of the triangle (2.39), by applying Theorem 2.4 and using the results of Example 4.1.5.

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Problem 5.2.18

Use Theorem 2.14 to provide an alternative proof for the admissibility of the estimator aXˉ+ba \bar{X}+b satisfying (2.6), in Example 2.5.

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Problem 5.2.19

Determine which estimators aX+ba X+b are admissible for estimating E[X]\mathbb {E}\left[X\right] in the following situations, for squared error loss:

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(a)

XX has a Poisson distribution.

(b)

XX has a negative binomial distribution (Gupta 1966).

Problem 5.2.20

Let XX have the Poisson⁡(λ)\operatorname {Poisson}(\lambda ) distribution, and consider the estimation of λ\lambda under the loss (d−λ)2/λ(d-\lambda )^{2} / \lambda with the restriction 0≤λ≤m0 \leq \lambda \leq m, where mm is known.

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(a)

Using an argument similar to that of Example 2.9, show that XX is not minimax, and a least favorable prior distribution must have a set w∧w_{\wedge } [of (1.5)] consisting of a finite number of points.

(b)

Let Λa\Lambda_{a} be a prior distribution that puts mass ai,i=1,…,ka_{i}, i=1, \ldots , k, at parameter points bi,i=1,…,kb_{i}, i=1, \ldots , k. Show that the Bayes estimator associated with this prior is

δΛa(x)=1E[λ−1∣x]=∑i=1kaibixe−bi∑i=1kaibix−1e−bi. \delta ^{\Lambda a}(x)=\frac{1}{\mathbb {E}\left[\lambda ^{-1} \mid x\right]}=\frac{\sum _{i=1}^{k} a_{i} b_{i}^{x} e^{-b_{i}}}{\sum _{i=1}^{k} a_{i} b_{i}^{x-1} e^{-b_{i}}}.
(c)

Let m0m_{0} be the solution to m=e−m(m0≈.57)m=e^{-m}\left(m_{0} \approx .57\right). Show that for 0≤λ≤m,m≤m00 \leq \lambda \leq m, m \leq m_{0} a one-point prior (ai=1,b1=m)(a_{i}=1, b_{1}=m) yields the minimax estimator. Calculate the minimax risk and compare it to that of XX.

(d)

Let m1m_{1} be the first positive zero of (1+δΛ(m))2=2+m2/2\left(1+\delta^{\Lambda }(m)\right)^{2}=2+m^{2} / 2, where Λ\Lambda is a two-point prior (a1=a,b1=0;a2=1−a,b2=m)\left(a_{1}=a, b_{1}=0 ; a_{2}=1-a, b_{2}=m\right). Show that for 0≤λ≤m,m0<m≤m10 \leq \lambda \leq m, m_{0}<m \leq m_{1}, a two-point prior yields the minimax estimator (use Corollary 1.6). Calculate the minimax risk and compare it to that of XX.

Remark.
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As mm increases, the situation becomes more complex and exact minimax solutions become intractable. For these cases, linear approximations can be quite satisfactory. See Johnstone and MacGibbon 1992, 1993.

Problem 5.2.21

Show that the conditions (2.41) and (2.42) of Example 2.22 are not only sufficient but also necessary for admissibility of (2.40).

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Problem 5.2.22

Let XX and YY be independently distributed according to Poisson distributions with E[X]=ξ\mathbb {E}\left[X\right]=\xi and E[Y]=η\mathbb {E}\left[Y\right]=\eta, respectively. Show that aX+bY+ca X+b Y+c is admissible for estimating ξ\xi with squared error loss if and only if either 0≤a<1,b≥0,c≥00 \leq a<1, b \geq 0, c \geq 0 or a=1,b=c=0a=1, b=c=0 (Makani 1972).

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Problem 5.2.23

Let XX be distributed with density 12β(θ)eθxe−∣x∣,∣θ∣<1\frac{1}{2} \beta (\theta ) e^{\theta x} e^{-\left|x\right|},\left|\theta \right|<1.

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(a)

Show that β(θ)=1−θ2\beta (\theta )=1-\theta^{2}.

(b)

Show that aX+ba X+b is admissible for estimating Eθ[X]\mathbb {E}_{\theta }\left[X\right] with squared error loss if and only if 0≤a≤1/20 \leq a \leq 1 / 2.

Problem 5.2.24

Let XX be distributed as N(θ,1)N(\theta , 1) and let θ\theta have the improper prior density π(θ)=eθ\pi (\theta )=e^{\theta } (−∞<θ<∞)(-\infty <\theta <\infty ). For squared error loss, the formal Bayes estimator of θ\theta is X+1X+1, which is neither minimax nor admissible. (See also Problem 2.15.) Conditions under which the formal Bayes estimator corresponding to an improper prior distribution for θ\theta in Example 3.4 is admissible are given by Zidek (1970).

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Problem 5.2.25

Show that the natural parameter space of the family (2.16) is (−∞,∞)(-\infty , \infty ) for the normal (variance known), binomial, and Poisson distribution but not in the gamma or negative binomial case.

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