Minimax and Bayes rule Could anyone provide some examples that a decision rule which is minimax but not Bayes, and a decision rule which is Bayes but not minimax? Thanks!
 A: In the event the decision problem "has a value", i.e., when minimax meets maximin,
$$\min_\delta\,\max_\theta\,\mathbb E_\theta[L(\theta,\delta(X)]=
\max_\pi\min_\delta\mathbb \int \mathbb E_\theta[L(\theta,\delta(X)] \pi(\text{d}\theta)$$
there exists a Bayes procedure or a limit of Bayes procedures that is minimax. (This is actually a constructive method for producing minimax estimators in difficult settings: seek a Bayesian estimator with minimax risk, as eg in 1981 Casella's and Strawderman's derivation of the [unique] minimax estimator of a bounded Normal mean.)  However, this intersection between both concepts does not mean they are confounded.
For examples of minimax procedures that are not Bayes or limits of Bayes procedures, a classical example is the class of James-Stein estimators: when estimating a Normal mean vector $\theta\in\mathbb R^p$ with $p\ge 3$ and a quadratic loss function$$L(\theta,\delta)=(\delta-\theta)^\text{T}A(\delta-\theta)$$with $A$ a $p\times p$ symmetric positive definite matrix, estimators of the form$$\delta\,: x \longmapsto \delta(x)=\left\{ 1 - {a}||x||^{-2}\right\}x$$are minimax if $0\le a\le \bar{a}(p,A)$ where the upper bound is a function of the dimension $p$ and of the matrix $A$. For instance, $\bar{a}(p,\mathbf I_p)=2(p-2)$.
Conversely, most Bayes procedures are not minimax. Indeed, any prior $\pi$ such that$$\min_\delta\,\max_\theta\,\mathbb E_\theta[L(\theta,\delta(X)]=
> \min_\delta\mathbb \int \mathbb E_\theta[L(\theta,\delta(X)] \pi(\text{d}\theta)$$will not produce a minimax Bayes procedure.
Here is an illustration from my book:

Consider a Bernoulli observation, $x\sim \mathcal B e(\theta)$  with
$\theta\in\{0.1,0.5\}$. Four nonrandomized \est s of $\theta$ are
available, \begin{eqnarray*} \delta_1(x) & = & 0.1,\qquad \qquad
  \delta_2(x) = 0.5, \\ \delta_3(x) & = & 0.1\, \mathbb I_{x = 0} +
  0.5\, \mathbb I_{x = 1}, \quad
     \delta_4(x)  = 0.5\, \mathbb I_{x = 0} + 0.1\, \mathbb I_{x = 1}. \end{eqnarray*} Assume that the penalty for a wrong answer is $2$ when
$\theta = 0.1$ and $1$ when $\theta = 0.5$. The risk vectors
$(R(0.1,\delta),R(0.5,\delta))$ of the four estimators are then,
respectively, $(0,1)$, $(2,0)$, $(0.2,0.5)$, and $(1.8,0.5)$. It is
straightforward to see that the risk vector of any randomized
estimator is a convex combination of these four vectors or,
equivalently, that the risk set, $\mathfrak R$, is the convex hull of
the above four vectors, as represented by the following Figure



In this case, the minimax estimator is obtained at the intersection of
the diagonal of $\mathbb R^2$ with the lower boundary of $\mathfrak
  R$. As shown by this Figure, this estimator $\delta^*$ is randomized
and takes the value $\delta_3(x)$ with probability $\alpha = 0.87$ and
$\delta_2(x)$ with probability $1-\alpha$. The weight $\alpha$ is
actually derived from the equation
$$0.2 \alpha + 2(1-\alpha)  =  0.5 \alpha. $$ This estimator $\delta^*$ is also a (randomized) Bayes estimator with respect to the prior
$$\pi(\theta)  =  0.22\, \mathbb I_{0.1}(\theta) + 0.78\, \mathbb
  I_{0.5} (\theta); $$ the prior probabilities $\pi_1 = 0.22$
corresponds to the slope between $(0.2,0.5)$ and $(2,0)$, i.e., $$
  {\pi_1\over 1-\pi_1}  =  {0.5 \over 2-0.2}. $$ Notice that every
randomized estimator that is a combination of $\delta_2$ and of
$\delta_3$ is a Bayes estimator for this distribution, but that
$\delta^*$ only is also a minimax estimator.

