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 $\tau$.
self-studytag. $\endgroup$