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.


1 Answer 1


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)


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


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


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