Extensions

Show that

where $S=\sum_{j}\left(X_{j}-\overline{\mathbf{X}}\right)^{2}$.

Use integration by parts to show

Use the results of parts (a) and (b) to establish (6.2).

The estimator $\delta^{L}$ is minimax if $r \geq 4$ and $c \leq 2$.

The risk of $\delta^{L}$ is infinite if $r \leq 3$

The minimum risk is equal to $3 / r$, and is attained at $\theta_{1}=\theta_{2}=\cdots =\theta$.

The estimator $\delta^{L}$ is dominated in risk by its positive-part version

Show that $k \mathbf{x}$ is the MLE if $\boldsymbol {\theta } \in \mathcal{L}_{k}$.

Show that $\delta^{k}(\mathbf{x})$ of (6.8) is minimax under squared error loss.

Verify that $\theta_{i}$ of the form (6.4) satisfy $T\left(T^{\prime } T\right)^{-1} T^{\prime } \boldsymbol {\theta }=\boldsymbol {\theta }$ for $T$ of (6.5), and construct a minimax estimator that shrinks toward this subspace.

Find the MLE of $\boldsymbol {\theta }$, say $\hat{\boldsymbol {\theta }}_{R}$, if $\boldsymbol {\theta }$ is assumed to be in the linear subspace

Show that $\mathcal{L}$ can be written in the form (6.7), and find $K$.

Construct a Stein estimator that shrinks toward the MLE of part (a) and prove that it is minimax.

Show that $\delta^{c}(\mathbf{x}, \mathbf{y})$ is minimax if $0 \leq c \leq 2$.

Show that if $\boldsymbol {\xi }=0, R\left(\boldsymbol {\theta }, \delta^{1}\right)=1-\frac{\sigma^{2}}{\sigma^{2}+\tau^{2}} \frac{r-2}{r}, R\left(\boldsymbol {\theta }, \delta^{\text{comb }}\right)=1-\frac{\sigma^{2}}{\sigma^{2}+\tau^{2}}$, and, hence, $R\left(\boldsymbol {\theta }, \delta^{1}\right)>R\left(\boldsymbol {\theta }, \delta^{\text{comb }}\right)$.

For $\boldsymbol {\xi } \neq 0$, show that $R\left(\boldsymbol {\theta }, \delta^{\text{comb }}\right)=1-\frac{\sigma^{2}}{\sigma^{2}+\tau^{2}}+\frac{|\boldsymbol {\xi }|^{2} \sigma^{2}}{r\left(\sigma^{2}+\tau^{2}\right)}$ and hence is unbounded as $\left|\boldsymbol {\xi }\right| \rightarrow \infty$.

For $X\left|\boldsymbol {\theta } \sim N_{r}\left(\boldsymbol {\theta }, \sigma^{2} I\right), Y\right| \boldsymbol {\theta }, \boldsymbol {\xi } \sim N_{r}\left(\boldsymbol {\theta }+\boldsymbol {\xi }, \tau^{2} I\right), \boldsymbol {\xi } \sim N\left(0, \gamma^{2} I\right)$, and $\theta_{i} \sim$ Uniform $(-\infty , \infty )$, with $\sigma^{2}$ and $\tau^{2}$ assumed to be known, show how to derive $\delta^{r-2}(\mathbf{x}, \mathbf{y})$ as an empirical Bayes estimator.

Calculate the Bayes estimator, $\delta^{\pi }$, against squared error loss.

Compare $r\left(\pi , \delta^{\pi }\right)$ and $r\left(\pi , \delta^{r-2}\right)$.

Verify the risk function (6.13).

Verify that for unknown $\sigma^{2}$, the risk function of the estimator (6.14) is given by (6.15).

Show that the minimum risk of the estimator (6.14) is $1-\frac{v}{v+2} \frac{r-2}{r}$.

Show that $R\left(\boldsymbol {\theta }, \delta^{L}\right)=1-\frac{v}{v+2} \frac{(r-3)^{2}}{r} \mathbb {E}_{\theta }\left[\frac{\sigma^{2}}{\Sigma \left(x_{i}-\bar{x}\right)^{2}}\right]$.

Show that both $\delta^{L}$ and $\delta$ of (6.14) can be improved by using their positive-part versions.

Show that the positive-part Lindley estimator can be written as

The null hypothesis is rejected if $F$ is large. Show that this corresponds to using the MLE under $H_{0}$ if $F$ is small, and a Stein estimator if $F$ is large.

The null hypothesis is rejected at level $\alpha$ if $F>F_{s-1, s(n-1), \alpha }$. For $s=8$ and $n=6$: what is the level of the test that corresponds to choosing $c=1$, the optimal risk choice?

For the same $s=8$ and $n=6$: what values of $c$ correspond to choosing $\alpha =.05$ or $\alpha =.01$, typical $\alpha$ levels? Are the resulting estimators minimax?

Show that $\mathbb {E}_{\sigma }\left[\frac{1}{\sigma^{2}}\right]=\mathbb {E}_{0}\left[\frac{r-2}{|\mathrm{x}|^{2}}\right]$.

If $1 / \sigma^{2} \sim \chi_{v}^{2} / v$, then $f(\left|\mathbf{x}-\boldsymbol {\theta }\right|)$ of (6.19) is the multivariate $t$-distribution, with $v$ degrees of freedom and $\mathbb {E}_{0}\left[|\mathbf{X}|^{-2}\right]=(r-2)^{-1}$.

If $1 / \sigma^{2} \sim Y$, where $\chi_{v}^{2} / v$ is stochastically greater than $Y$, then $\delta (\mathbf{x})$ of (6.20) is minimax for this mixture as long as $0 \leq c \leq 2(r-2)$.

Verify that the estimator (6.25) is minimax if $0 \leq c \leq 2$. (Theorem 5.5 will apply.)

Referring to (6.27), show that

where $Y \sim \chi_{r}^{2}$.

If $\chi_{v}^{2}$ denotes a chi-squared random variable with $v$ degrees of freedom, establish the identity

to show that

where, now, $Y \sim \chi_{r-2}^{2}$.

Verify (6.29), hence showing that $r\left(\pi , \delta^{R}\right) \leq r\left(\pi , \delta^{c}\right)$.

Show that $\mathbb {E}\left[(Y-a)^{2} I(Y>a)\right]$ is a decreasing function of $a$, and hence the maximal Bayes risk improvement, while maintaining minimaxity, is obtained at $c=2$.

For what values of $a, \alpha$, and $\beta$ are the estimators of (4.6.29) minimax? Are they also proper Bayes for these values?

Let $\Lambda =\Sigma \lambda_{i}$ and define $\theta_{i}=\lambda_{i} / \Lambda , i=1, \ldots , r$. For the prior distribution $\pi (\boldsymbol {\theta }, \Lambda )= m(\Lambda ) d \Lambda \prod_{i=1}^{r} d \theta_{i}$, show that the Bayes estimator is

where $z=\Sigma x_{i}$ and

Show that the choice $m(\Lambda )=1$, yields the estimator $\delta (\mathbf{x})=[1-(r-1) /(z+r-1)] \mathbf{x}$, which is minimax.

Show that the choice $m(\Lambda )=(1+\Lambda )^{-\beta }, 1 \leq \beta \leq r-1$ yields an estimator that is proper Bayes minimax for $r>2$.

The estimator of part (d) is difficult to evaluate. However, for the prior choice

show that the generalized Bayes estimator is

and determine conditions for its minimaxity. Show that it is proper Bayes if $\beta >1$.

Show that the usual estimator $\mathbf{x} / p$ has constant risk $r(1-p) / p$.

For $r \geq 2$, show that the estimator

dominates $\mathbf{x} / p$ in risk, where $z=\Sigma x_{i}$ and $0<a<2(r-1)(1-p)$.

Extend the argument from part (b) and find conditions on the function $c(\cdot )$ and constant $b$ so that

dominates $\mathbf{x} / p$ in risk.

Prove Lemma 6.9.

Prove that for $X \sim p_{i}(x \mid \theta )$, where $p_{i}(x \mid \theta )$ is given by (6.36), $\delta^{0}(x)=h_{i}(x- 1) / h_{i}(x)$ is the UMVU estimator of $\theta$ (Roy and Mitra 1957).

Prove Theorem 6.10.

Establish that $\mathbf{x}+g(\mathbf{x})$, where $g(\mathbf{x})$ is given by (6.42), satisfies $\mathcal{D}(\mathbf{x}) \leq 0$ for the loss $L_{0}(\boldsymbol {\theta }, \delta )$ of (6.38), and hence dominates $\mathbf{x}$ in risk.

Derive $\mathcal{D}(\mathbf{x})$ for $X_{i} \sim \operatorname {Poisson}\left(\lambda_{i}\right)$, independent, and loss $L_{-1}(\lambda , \delta )$ of (6.38). Show that $\mathbf{x}+g(\mathbf{x})$, for $g(\mathbf{x})$ given by (6.43), satisfies $\mathcal{D}(\mathbf{x}) \leq 0$ and hence is a minimax estimator of $\lambda$.

Show that the estimator $\delta_{0}(\mathbf{x})+g(\mathbf{x})$, for $g(\mathbf{x})$ of (6.45) dominates $\delta^{0}$ in risk under the loss $L_{-1}(\boldsymbol {\theta }, \delta )$ of (6.38) by establishing that $\mathcal{D}(\mathbf{x}) \leq 0$.

For the loss $L_{0}(\boldsymbol {\theta }, \delta )$ of (6.38), show that the estimator $\delta^{0}(\mathbf{x})+g(\mathbf{x})$, where

with $k_{i}(x)=\Sigma_{\ell =1}^{x}\left(t_{i}-1+\ell \right) / \ell$ and $c(\cdot )$ nondecreasing with $0 \leq c(\cdot ) \leq 2\left[\left(\# x_{i} s>1\right)-2\right]$ has $\mathcal{D}(\mathbf{x}) \leq 0$ and hence dominates $\delta^{0}(\mathbf{x})$ in risk.

Show that the MLE of $\mu$ is $X$, and the risk of an estimator $\delta (\mathbf{x})=\mathbf{x}+g(\mathbf{x})$ can be written

where

and

so that a sufficient condition for domination of the MLE is $\mathcal{D}_{1}(\mathbf{x})+\mathcal{D}_{2}(\mathbf{x}) \leq 0$ for all $\mathbf{x}$.

Show that if $X_{i}$ are $\operatorname {Poisson}\left(\theta_{i}\right)$ (instead of negative binomial), then $\mathcal{D}_{2}(\mathbf{x})=0$. Thus, any estimator that dominates the MLE in the negative binomial case also dominates the MLE in the Poisson case.

Show that the Clevenson-Zidek estimator

satisfies $\mathcal{D}_{1}(\mathbf{x}) \leq 0$ and $\mathcal{D}_{2}(\mathbf{x}) \leq 0$ and, hence, dominates the MLE under both the Poisson and negative binomial model.