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

Question

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?

Answer 1

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.

Original text from Lehmann's Theory of Point Estimation (1983): Example 2.5 (left) and Exercise 2.15 (right)

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 $$\delta_2(\bar X_n)=\Bbb E[\delta_1(Z_1,\ldots,Z_n)\mid\bar X_n]=\Bbb E[Z_1\cdots Z_k\mid \bar X_n]$$