For what values of $k \in \mathbb{Z}\setminus\{0\}$ does there exist an unbiased estimator of $e^{k \lambda}$?

Question

This is inspired by Finding an unbiased estimator of $e^{-2\lambda}$ for Poisson distribution, reminding me of a qualifying exam question that I was frustrated with.

Suppose $X_1, \dots, X_n \overset{\text{iid}}{\sim}\text{Poisson}(\lambda)$.

For some subset of size $k \leq n$, it can be seen that if $\mathbf{I}$ is the indicator function that $$\mathbf{I}(X_{i_1}+X_{i_2}+\cdots+X_{i_k}=0)$$ is an unbiased estimator of $e^{-k\lambda}$ for $k \geq 1$.

The OP of the linked question above asked an interesting question: does there exist an unbiased estimator of $e^{2\lambda}$? More generally,

Is there an unbiased estimator of $e^{k\lambda}$ for $k > 0$?

In the comments to the link above, I told the OP I wouldn't know where to begin with this. My first thought was to try finding the distribution of the reciprocal of a Poisson distribution. Say $X \sim \text{Poisson}(\lambda)$, then $$f_{1/X}(y)=\dfrac{e^{-\lambda}\lambda^{1/y}}{(1/y)!}$$ (what is even the support of this thing?) and I suppose $(1/y)!$ would have to be extended in cases where $1/y$ isn't an integer... so basically, every case except $y = 1$, and then we'd have to use the Gamma function. But even with this, using indicator functions (as above) will still give $e^{-k \lambda}$ for $k \geq 1$; it doesn't help the problem much.

Answer 1

Here are some unbiased estimators of $e^{a\lambda}$, for every nonnegative $a$, based on any i.i.d. sample $(X_1,\ldots,X_n)$ of Poisson distribution with parameter $\lambda$:

• Using $X_1$ only, $$(1+a)^{X_1}$$
• Using $(X_1,\ldots,X_n)$, $$\left(1+\frac an\right)^{X_1+X_2+\cdots+X_n}$$
• If asymmetry is allowed, $$\prod_{i=1}^na_i^{X_i}$$ for every real numbers $(a_i)$ such that $$\sum_{i=1}^na_i=n+a$$

And so on.