# Prove/counter example: A minimax decision rule is always Bayes wrt some proper prior

Not sure whether the claim is true or false.

If claim is true, intuitively, it might have something to do with "least favorable priors", but am not able to figure out the connection.

If claim is false, one example is when $X_i|\theta \sim$ Poisson$(\theta)$, then $\bar{X}$ is minimax. But a Gamma$(\alpha, \beta)$ prior fails since, that would indicate $\beta = 0$, which is improper. But, how do we know there is no other prior that gives $\bar{X}$ as a Bayes rule?

• The connection between least favorable priors and minimax procedures is described in Kiefer, Introduction to Statistical Inference, and E. L. Lehmann, Theory of Point Estimation. The former discusses (without specific attribution) the existence of counterexamples, referring to "work by C. Stein and others since the 1950s." – whuber Oct 28 '13 at 15:33
• Thank you. Will look up Kiefer's book and see what I find. – Greenparker Nov 3 '13 at 14:48

$X_i \overset{iid}\sim N(\theta, 1)$. Then $\bar{X}$ is minimax. (This can be shown by using a sequence of priors). However, $\bar{X}$ being unbiased for $\theta$, can only be Bayes for any proper prior if the Bayes Risk is 0.
$R(\theta,\bar{X}) =\dfrac{1}{n} \implies \int_{\Theta} R(\theta,\bar{X}) \pi(\theta) d\theta = \dfrac{1}{n} \int_{\Theta} \pi(\theta) d\theta = \dfrac{1}{n}$
Thus, as long $\pi$ is proper, Bayes risk of $\bar{X}$ cannot be 0. Hence $\bar{X}$ cannot be Bayes with respect to any proper. The same reasoning can be made for admissible estimators.