# Bayes Minimax Estimation

Let $$S\sim B(n,\theta), l(\theta, a) = (\theta - a)^2,\delta=\bar{X} = S/n,$$ and $$\delta^*(S)=\left(S+\frac{1}{2}\sqrt{n}\right)/\left(n+\sqrt{n}\right)$$ where $$B(n,\theta)$$ is Bernoulli distribution, $$l(\theta, a)$$ is loss function

How can I show that $$\delta^*$$ has constant risk and is Bayes for the beta, $$\beta(\sqrt{n}/2, \sqrt{n}/2)$$, prior. So, $$\delta^*$$ would be minimax. Also if we have $$\theta \ne \frac{1}{2}$$, would $$lim_{n\rightarrow \infty}\left[R(\theta,\delta^*)/R(\theta,\delta)\right] > 1$$? Otherwise, if $$\theta = \frac{1}{2}$$, would that make $$lim_{n\rightarrow \infty}\left[R(\theta,\delta^*)/R(\theta,\delta)\right] = 1$$? How do I show this?

• This is covered in details in Berger (1985) and my book. – Xi'an May 3 '19 at 7:23

Lemma 2.25 If $$\delta_0$$ is a Bayes estimator with respect to $$\pi_0$$ and if $$R(\theta,\delta_0) \le r(\pi_0)$$ for every $$\theta$$ in the support of $$\pi_0$$, $$\delta_0$$ is minimax and $$\pi_0$$ is the least favorable distribution.
Example 2.26 (Berger, 1985a) Consider $$x\sim{\mathcal B}(n,\theta)$$ when the probability $$\theta$$ is to be estimated under the quadratic loss, $$\mathrm L (\theta,\delta) = (\delta-\theta)^2.$$ Bayes estimators are then given by posterior expectations and, when $$\theta \sim{\cal B}e \left({\sqrt{ n} \over 2}, {\sqrt{ n} \over 2} \right)$$, the posterior mean is $$\delta^ \ast (x) = {x+ \sqrt{n}/2 \over n+ \sqrt{ n}}.$$ Moreover, this estimator has constant risk, $$R(\theta,\delta^*) = 1/4(1+\sqrt{n})^2$$. Therefore, integrating out $$\theta$$, $$r(\pi) = R(\theta,\delta^*)$$ and $$\delta^*$$ is minimax according to Lemma 2.25. Notice the difference with the MLE, $$\delta_0(x) = x/n$$, for the small values of $$n$$, and the unrealistic concentration of the prior around $$0.5$$ for larger values of $$n$$.[