Let $X_1, ..., X_n$ be iid from the Poisson ($\theta$) distribution.

I have proven that $T = \sum_{i=1}^{n} x_i$ is the complete and sufficient statistic and it has a Poisson($n\theta$) distribution. I need to find the UMVUE of $\theta^k$, k>0.

Attempts: I have noticed that $E(x_1x_2...x_k) = \theta^k$ due to the iid property and they are the unbiased estimator.

I tried using the Lehmann Scheffe Theorem

$\begin{equation} \sum_{t=0}^{n} g(t)\frac{e^{-n\theta}{(n\theta)}^t}{t!} = \theta^k\end{equation}$ but arrived with the result of $g(t) = \theta^k$, which does not make sense. I tried using factorial moments, but think it is too complicated when k is not specified.

Can anyone point me in the right direction?

Thank you.


2 Answers 2


As you mentioned, $$P(T=t)=\frac{e^{-n\theta} (n\theta)^t}{t!} $$

Let $I(T)$ be a function such that

\begin{align} I(T)=\begin{cases}\displaystyle\frac{T!}{(T-k)!}&,\text{ when }T\geq k\\0 &,\text{ elsewhere }\end{cases} \end{align}


$$E\left[ I(T) \right]=\sum_{t=k}^{\infty} \frac{e^{-n\theta} (n\theta)^{t-k}}{(t-k)!} n^k \theta^k =n^k\theta^k$$

So, $$E\left[ \frac{I(T)}{n^k} \right]=\theta^k$$

And we know that $T$ is the complete and sufficient statistic, so the function of $T$ given by $I(T)/n^k$ is the UMVUE of $\theta^k$.

  • $\begingroup$ It would be great if someone proofread my answer correcting grammar or mathematical errors. $\endgroup$
    – KDG
    Commented Sep 10, 2018 at 1:38
  • $\begingroup$ We can check this answer for a simple case when $k=1$. In this case, $I(T)=T/n$. This is the sample mean of $X$. $\endgroup$
    – KDG
    Commented Sep 10, 2018 at 1:45
  • 2
    $\begingroup$ (+1) Your answer is correct. A comparison of coefficients indeed leads to your $I(T)/n^k$ as an unbiased estimator of $\theta^k$ for a positive integer $k$. $\endgroup$ Commented Sep 10, 2018 at 18:43

Try comparing power series coefficients (warning: this can lead to very complicated calculations).

Specifically, assuming that there is a UMVUE $\delta$, we know it is unbiased and a function of $T$, so we can write that $\delta(X) = \eta(T(X))$, and that $\mathbb{E}_{\theta}[\eta(T(X))] = \theta^k$, i.e. that $$\sum_{t=0}^{\infty} \frac{\theta^t n^t e^{-n\theta}}{t!}\eta(t) = \sum_{t=0}^{\infty}\mathbb{P}_{\theta}(T = t) \eta(t) = \theta^k \,.$$

Alternatively, (by Theorem 4.4. in Keener, Theoretical Statistics: Topics for a Core Course - proof involves Rao-Blackwell - is this called Lehmann-Scheffe in the literature?) we also can say that: $$\eta(T(X)) = \mathbb{E}_{\theta}[\delta(X)|T] $$ This method is even more general, but for me at least is even more difficult to use in practice.


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