# Find UMVUE of $p^3$

Let $$X_1, X_2, ..., X_n$$ be a random sample from $$Binom(1, p)$$. I'm trying to find the UMVUE of $$p^3$$.

Some thoughts:

1. Apparently, $$\bar{X}^3$$ is not the answer, although it's the MLE of $$p^3$$.
2. For distinct $$i$$, $$j$$, and $$k$$, the distribution of $$X_iX_jX_k$$ is $$Binom(1, p^3)$$, but what if $$n$$ is not divisible by three?
• What is an unbiased estimator of $p^3$? – StubbornAtom Apr 20 '19 at 15:48
• See math.stackexchange.com/q/2687375. And if the answers there are helpful you might answer this question yourself. – StubbornAtom Apr 20 '19 at 16:14
• @StubbornAtom Thanks for the information; they are really helpful! Wasn't aware of the Lehmann–Scheffé theorem. +1 – nalzok Apr 20 '19 at 16:32
• Please provide an answer to your question so that it goes off the unanswered list. – StubbornAtom Apr 23 '19 at 10:33
• @StubbornAtom Done! – nalzok Apr 23 '19 at 12:16

If a statistic that is unbiased, complete and sufficient for some parameter $$\theta$$, then it is the UMVUE for $$\theta$$.
Here $$\theta$$ is $$p^3$$, and $$T = \sum_{i=1}^n X_i$$ is a sufficient and complete statistics for $$p^3$$, so we simply need to construct a unbiased estimator of $$p^3$$ with $$T$$. In other words, we need to find $$\phi$$ such that $$E(\phi(T)) = p^3$$. For example, you can readily verify
$$\phi(T) = \frac{T(T-1)(T-2)}{n(n-1)(n-2)}$$
And that's the UMVUE of $$p^3$$.