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?