3
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ $\delta^\prime$ cannot be Bayes since it does not factorise through a sufficient statistic. Can you provide the source of this exercise? $\endgroup$
    – Xi'an
    May 23, 2019 at 7:55
  • $\begingroup$ 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 $\endgroup$
    – Xiaomi
    May 23, 2019 at 7:59

1 Answer 1

2
$\begingroup$

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.

enter image description here enter image description here

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

$\endgroup$
1
  • 2
    $\begingroup$ Apologies! I have found Lehmann's book very difficult so I misunderstood what it was asking I guess. Thanks for clarifying $\endgroup$
    – Xiaomi
    May 23, 2019 at 14:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.