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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?
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  • $\begingroup$ What is an unbiased estimator of $p^3$? $\endgroup$ – StubbornAtom Apr 20 '19 at 15:48
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    $\begingroup$ See math.stackexchange.com/q/2687375. And if the answers there are helpful you might answer this question yourself. $\endgroup$ – StubbornAtom Apr 20 '19 at 16:14
  • $\begingroup$ @StubbornAtom Thanks for the information; they are really helpful! Wasn't aware of the Lehmann–Scheffé theorem. +1 $\endgroup$ – nalzok Apr 20 '19 at 16:32
  • $\begingroup$ Please provide an answer to your question so that it goes off the unanswered list. $\endgroup$ – StubbornAtom Apr 23 '19 at 10:33
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    $\begingroup$ @StubbornAtom Done! $\endgroup$ – nalzok Apr 23 '19 at 12:16
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Actually, this problem is a classic example of the Lehmann–Scheffé theorem. The theorem states

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 $p^3$.

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