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?

  • $\begingroup$ This is covered in details in Berger (1985) and my book. $\endgroup$ – 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 $n$.[

[The Bayesian Choice, 2007, Chapter 2]

  • $\begingroup$ Thank you, this is very helpful! $\endgroup$ – John Ling May 5 '19 at 1:09

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