# Unbiased estimator for a parameter in a Poisson distribution

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.

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 $$\theta = e^{- \lambda}$$.
• $1(X_1 = 0)$ and $1(X_n = 0)$ etc. are also unbiased estimators, and yours is the average of these so also unbiased Jan 31, 2022 at 16:11