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Let $x_1, x_2, \cdots, x_n$ be a random sample from the Bernoulli ($\theta$).

The question is to find the UMVUE of $\theta^k$.

I know the $\sum_1^nx_i$ is the complete sufficient statistics for $\theta$.

Is $(\frac{\sum_1^nx_i}{n})^k$ the estimator or any other possible estimator?

Could someone just help me?

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  • $\begingroup$ Is the thing you wrote down even unbiased? $\endgroup$ – spaceisdarkgreen Mar 12 '18 at 4:29
  • $\begingroup$ I just guess. no idea how to do it $\endgroup$ – Nicolas Mar 12 '18 at 5:02
  • $\begingroup$ Considering this estimator is a good start as the expectation is a polynomial of $\theta$ with degree $k$. I am not sure how to do this general $k$ case, but if $k$ is just a small natural number, it is possible to find an unbiased estimator in terms of a polynomial of sample mean. The key feature is notice that $\displaystyle E\left[\prod_{i=1}^k X_i\right] = \theta^k$ $\endgroup$ – BGM Mar 12 '18 at 6:38
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    $\begingroup$ Look at this answer: math.stackexchange.com/a/2673087 For your case change notations $Y_1=\sum x_i$, $p(1)=\theta$, $l=k$. $\endgroup$ – NCh Mar 12 '18 at 15:25
  • $\begingroup$ I don't think it's possible. Since Bernoulli is an exponential family with sufficient statistic $\sum_{i=1}^n x_i$ and $E_\theta(\sum_{i=1}^n x_i) = n\theta$, only the affine functions of $\theta$ have an umvue. $\endgroup$ – Gabriel Romon Mar 14 '18 at 9:08
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  • You have shown that $T=\sum_{i=1}^n X_i$ is the complete sufficient statistic for $\theta$.

  • It is easily seen that $\displaystyle\text{E}_{\theta}\left(\frac{T}{n}\right)=\theta$ and $\displaystyle\text{E}_{\theta}\left(\frac{T(T-1)}{n(n-1)}\right)=\theta^2$ for all $\theta\in(0,1)$.

  • Inductively (or otherwise) one can show that

$\quad\quad\displaystyle\text{E}_{\theta}\left(\frac{T_{(k)}}{n_{(k)}}\right)=\text{E}_{\theta}\left(\frac{T(T-1)(T-2)...(T-k+1)}{n(n-1)(n-2)...(n-k+1)}\right)=\theta^k$ for all $\theta\in(0,1)$ and $k\in\mathbb N$.

Being a function of the complete sufficient statistic $T$, $\dfrac{T_{(k)}}{n_{(k)}}$ is the UMVUE for $\theta^k$ for all $k\in\mathbb N$.

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Having that

$$\theta^m=P\{ X_1=x_1,X_2=x_2,...,X_m=x_m\}$$

An unbiased estimator for $\theta^m$ is

$$T= \begin{cases} 1, & if \ \ X_1=X_2= \, ... \,=X_m =1 \\ 0, & in \ other \ case \end{cases}$$

But $$\begin{align} E[T|S=s] & = P\{X_1=1,X_2=1,...,X_m=1|S=s\}=\frac{P\{X_1=1,X_2=1,...,X_m=1,S=s\}}{P\{S=s\}} = \\\\ & = \begin{cases} 0, & if \ \ m>s \\ \frac{\theta^m\binom{n-m}{s-m}\theta^{s-m}(1-\theta)^{n-s}}{\binom{n}{s}\theta^s(1-\theta)^{n-s}}, & if \ \ m\leq s \end{cases} \end{align}$$

By the theorem of Lehmann-Scheffé, the UMVUE for $\theta^m$ is, after operating the latter expression:

$$E[T|S=s]=\begin{cases} 0, & if \ \ m>s \\ \frac{s!(n-m)!}{n!(s-m)!}, & if \ \ m\leq s \end{cases}$$

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  • $\begingroup$ Note that the estimators T and S are not independent random variables $\endgroup$ – Akerbeltz Apr 18 '18 at 21:43
  • $\begingroup$ Thank you so much.... $\endgroup$ – Nicolas Apr 18 '18 at 21:45
  • $\begingroup$ I forgot to mention that $S$ represents the estimator $\sum_{i=1}^{n}X_i$, although it could be interpreted from the context. I will edit this post later in the day. $\endgroup$ – Akerbeltz Apr 19 '18 at 17:49
  • $\begingroup$ Can you explain your first line? And what do you mean by $S$? $\endgroup$ – Hendrra Jan 9 at 17:55
  • $\begingroup$ $S$ is the sum of the given variables, as I explained in my last comment. $\endgroup$ – Akerbeltz Jan 24 at 10:16

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