# Unbiased estimator of $\lambda(1 - e^\lambda)$ when $x_1,\ldots,x_n$ are i.i.d Poisson($\lambda$)

Suppose $$x_1, x_2, x_3,\ldots, x_n$$ are i.i.d. random variables with a common Poisson$$(\lambda)$$ distribution.

I was trying to find an unbiased estimator for $$\lambda(1 - e^\lambda)$$, but I could not find any. Is there a way to prove that there is not any unbiased estimator for $$\lambda(1 - e^\lambda)$$ ?

Actually, an unbiased estimator does exist. Let us define $$\tau = \lambda e^\lambda$$ so that $$\lambda(1-e^\lambda) = \lambda - \tau$$ Since the sample mean $$\bar{X}$$ is unbiased for $$\lambda$$, really all we need is an unbiased estimator for $$\tau$$. An obvious starting place is to use the invariance property of the MLE. $$\hat\tau_\text{mle} = \bar{X}e^\bar{X}$$ For reasons which will shortly become clear, let's adjust this estimator by introducing a quantity $$m$$ in the exponential term.
$$\hat\tau_m = \frac{T}{n}e^{T/m}$$ where $$T = \sum_{i=1}^n X_i$$ has a $$\text{Poisson}(n\lambda)$$ distribution. The expected value of $$\hat\tau_m$$ can be found directly.
\begin{aligned} E(\hat\tau_m) &= \sum_{t=0}^\infty \left(\frac{t}{n}e^{t/m}\right)\left(\frac{e^{-n\lambda}(n\lambda)^t}{t!}\right) \\[1.2ex] &= \cdots && \text{show this on your own} \\[1.2ex] &= \lambda\left(e^{\lambda(e^{1/m} - 1)n}\right)e^{1/m} \end{aligned}
This estimator is clearly biased (for now). To make this an unbiased estimator, we need $$(e^{1/m}-1)n = 1$$ for all $$n$$. Solving this equation, we obtain $$m_\star = (\log(1+1/n))^{-1}$$. Using this value for $$m$$ yeilds, $$E(\hat\tau_{m_\star}) = \frac{\lambda e^\lambda}{1+1/n}$$ I'll leave the rest of the details up to you, but this estimator can now be adjusted so that it is unbiased for $$\tau$$.
• In your first comment, you are nearly there. When you set $y=t-1$, the sum in your comment becomes $\sum_{y=0}^\infty e^{(y+1)/m}(n\lambda)^y/y!$. You can pull out a $e^{1/m}$ and obtain $(e^{1/m}n\lambda)^y$ in the numerator. – knrumsey Mar 12 '19 at 16:32