# UMVUE of $p^4$ when $X_1,\ldots,X_n$ is a sample from Bernoulli$(p)$

Let $$X_1,\ldots,X_n$$ be a random sample from $$\text{Bernoulli}(p)$$. For $$n\geq 4$$ show that the product $$X_1X_2X_3X_4$$ is a unbiased estimator for $$p^4$$, and use this fact for find the best unbiased estimator of $$p^4$$.

For the first part I did $$E[X_1X_2X_3X_4]=E[X_1]E[X_2]E[X_3]E[X_4]=p^4$$

I find that Cramer Rao Lower Bound for $$p^4$$ is $$\frac{16p^7(1-p)}{n}$$

I even managed to find some estimators, but neither had variance equal to Cramer-Rao lower bound.

Define $T=\sum X_i$

T is a complete sufficient statistic for $p$.

Now, consider indicator $I_{X_1=1,X_2=1,X_3=1,X_4=1}$ which is an unbiased estimator of $p^4$(As you proved in the first part)

Rao-Blackwellising:

\begin{align}\phi(T) &= E[I_{X_1=1,X_2=1,X_3=1,X_4=1}|T] \\&=P(X_1=1,X_2=1,X_3=1,X_4=1|T=t) \\&=\frac{P(X_1=1,X_2=1,X_3=1,X_4=1,X_1+X_2+X_3+X_4+\dots X_n=t)}{P(X_1+X_2+\dots +X_n=t)} \\&= \frac{P(X_1=1,X_2=1,X_3=1,X_4=1)\times P(X_5+X_6+\dots X_n=t-4)}{P(X_1+X_2+\dots +X_n=t)} \\&= \frac{p^4\times \binom{n-4}{t-4}p^{t-4}(1-p)^{n-t}}{\binom{n}{t}p^{t}(1-p)^{n-t}} \\&= \frac{\binom{n-4}{t-4}}{\binom{n}{t}} \\&=\frac{t(t-1)(t-2)(t-3)}{n(n-1)(n-2)(n-3)} \end{align}

Check:

$E[\phi(T)] = 0$ using moments from here