Find a function so that for integer $k>1$ and $\lambda>0$: $ \sum_{t=0}^{\infty} h(t) \frac{e^{-n\lambda}(n\lambda)^t}{t!}=\lambda^k, $

Question

If I want to find a function $h(\cdot): \mathbb R\to \mathbb R$ so that for integer $k>1$ and $\lambda>0$: $$ \sum_{t=0}^{\infty} h(t) \frac{e^{-n\lambda}(n\lambda)^t}{t!}=\lambda^k, $$

The answer is that $h(t)=t(t-1)\cdots (t-k+1)/n^k$. But I have no idea how to get this and I got a different result: note that this is same as

$$ \sum_{t=0}^{\infty} h(t)\lambda^{-k} \frac{e^{-n\lambda}(n\lambda)^t}{t!}=1 $$

Since we have $$ \sum_{t=0}^{\infty}\frac{e^{-n\lambda}(n\lambda)^t}{t!}=1, $$ then we can just take $$ h(t)\lambda^{-k}=1 $$ That means $h(t)=\lambda^k$... Where am I wrong?

Answer 1

You have

$$\sum_{t=0}^{\infty} h(t) \frac{e^{-n\lambda}(n\lambda)^t}{t!}=\lambda^k$$

Or, $$\sum_{t=0}^{\infty} \frac{h(t) n^t}{t!}\lambda^t=\color{blue}{e^{n\lambda}}\lambda^k=\color{blue}{\sum_{j=0}^\infty \frac{(n\lambda)^j}{j!}}\lambda^k=\sum_{j=0}^\infty \frac{n^j}{j!}\lambda^{j+k}$$

That is,

$$\sum_{t=0}^{\infty} \frac{h(t) n^t}{t!}\lambda^t=\sum_{t=k}^\infty \frac{n^{t-k}}{(t-k)!}\lambda^t$$

Comparing coefficients of $\lambda^t$ on both sides, you get

$$h(t)=\begin{cases} \frac{t!}{n^t}\cdot\frac{n^{t-k}}{(t-k)!}&,\text{ if }t\ge k \\ 0 &,\text{ else } \end{cases}$$

Or,

$$h(t)=\frac{t!}{n^k(t-k)!}\mathbf1_{\{t\ge k\}}=\frac{t(t-1)(t-2)\cdots(t-k+1)}{n^k}\mathbf1_{\{t\ge k\}}$$

This is just one way of finding the best unbiased estimator of $\lambda^k$ when $X_1,X_2,\ldots,X_n$ are i.i.d $\text{Poisson}(\lambda)$. There are other ways of course, like using the Rao-Blackwell theorem.