# Showing $\delta'(X)$ is a Bayes estimator of $\theta^k$ for specified prior

Problem: Define $$\delta(X) = \frac{\sqrt{n}}{1+\sqrt{n}}\bar{X}_n + \frac{1}{2(1+\sqrt{n})}$$

We assume $$\bar{X}_n|\theta \sim Bin(\theta,n)$$.

It is known that $$\delta(X)$$ is the Bayes estimator of $$\theta$$ with respect to the $$Beta(\sqrt{n}/2,\sqrt{n}/2)$$ prior for squared error loss. Since its risk is constant, it is also minimax.

The problem at hand is to show that

$$\delta'(X) = \frac{\sqrt{n}}{1+\sqrt{n}}\frac{1}{n}\sum_i X_i^k + \frac{1}{2(1+\sqrt{n})}$$

Is Bayes for $$\theta^k$$ with respect to the same prior.

Since it's squared error loss, the Bayes estimate is given by $$E[\theta^k |X]$$ and we know that $$[\theta|X] \sim Beta(\sqrt{n}/2+X,\sqrt{n}+n-X)$$.

However, the $$k^{th}$$ moment for a Beta distribution seems to be (according to Wikipedia) a nasty product of $$k-1$$ terms. So I assume there must be an easier way to show the above is the Bayes estimator of the $$k^{th}$$ moment.

Is there an easier way forward?

• $\delta^\prime$ cannot be Bayes since it does not factorise through a sufficient statistic. Can you provide the source of this exercise? May 23, 2019 at 7:55
• It is from Lehman & Romano theory of point estimation, under the problems of chapter 4. I will include the problem number when I get home! Perhaps I have misunderstood what it was asking. Thanks for your help May 23, 2019 at 7:59

In Lehmann's book, Example 2.5, Chapter 4, under $$\mathfrak F_1$$, the $$X_i$$'s are either zero or one (while under $$\mathfrak F_0$$, the $$X_i$$'s are iid Bernoulli $$\mathcal B(\theta)$$) and therefore $$X_i^k=X_i\qquad k>0$$ Unless I miss something from this exercise it is rather anti-climactic.
A side remark: Let $$n \bar X_n=Z_1+\cdots+Z_n\qquad Z_i\stackrel{\text{iid}}{\sim}\mathcal B(\theta)$$and assume $$k\le n$$. (Otherwise, I think there exists no unbiased estimator of $$\theta^k$$.) An unbiased estimator of $$\theta^k$$ based on the $$Z_i$$'s is then$$\delta_1(Z_1,\ldots,Z_n)=Z_1\times\cdots\times Z_k$$and, by Rao-Blackwell, an improved estimator [under squared error loss] is $$\delta_2(\bar X_n)=\Bbb E[\delta_1(Z_1,\ldots,Z_n)\mid\bar X_n]=\Bbb E[Z_1\cdots Z_k\mid \bar X_n]$$