Unbiased estimator for a parameter in a Poisson distribution

Question

Suppose that $X_1,X_2,\ldots , X_n$ for a random sample from a Poisson distribution with parameter $\lambda$. Propose an unbiased estimator for $\theta = e^{-\lambda}$

My attempt was to substitute this estimator in the Poisson distribution $f(x|\lambda)$ = $\frac{\lambda^{x} e^{-\lambda}}{x!}$ obtaining $\frac{\log(\theta^{x}) \theta}{x!}$. I don't know if this is the right approach and I wouldn't know how to find an unbiased estimator afterwards. I thought about calculating the MLE of this new distribution, but I don't know if it was correct.

Any help/hint would be appreciated.

Answer 1

Notice that $P(X_n = 0) = e^{-\lambda}$. This motivates the following estimator. Set $Y_n = 1(X_n = 0)$, so that $Y_n \sim \mathrm{Bern}(e^{- \lambda})$ and $\mathbb{E}(Y_n) = P(X_n = 0) = e^{-\lambda}$. The estimator $$\overline{Y}_n = \frac{1}{n}(Y_1 + \ldots + Y_n) = \frac{1}{n} (1(X_1 = 0) + \ldots + 1(X_n = 0))$$ is then an unbiased estimator for $\theta = e^{- \lambda}$.