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