# Unbiased estimator for $e^\lambda$ in Poisson distribution

How to generate an unbiased estimator for $$e^{-\lambda}$$ in Poisson distribution: $$\frac{\lambda^k}{k!}{e^{-\lambda}}$$

I tried: $$E[a^x]=\sum_{x=0}^\infty a^x\frac{1}{e^{\lambda}}\frac{\lambda^x}{x!}=\frac{1}{e^{\lambda}}\sum_{x=0}^\infty \frac{(a\lambda)^x}{x!}=e^{a\lambda-\lambda}=e^{\lambda(a-1)}$$ But here I cannot just let a=0. So I have to find other ways.

Furthermore, how to generate the UMVUE for it if possible? Thank you.

• Possible duplicate of Is this an unbiased estimator? – Xi'an Apr 17 '19 at 7:12
• Hint: the probability of a poisson variable equal to zero is $e^{-\lambda}$. – B.Liu Apr 17 '19 at 7:15
• @Xi'an Sorry that doesn't solve my problem. He/She can fit (-2) in that problem, but I cannot fit 0 in my problem. – CuteCat Apr 17 '19 at 10:14
• Since $P(X_i=0) = e^{-\lambda}$ for $X_i \sim Po(\lambda)$, if you can construct an unbiased estimator for the former, then you automatically get one for the later. Can you construct one using e.g. empirical probability? Do see @Ben's answer below if you need more inspiration. – B.Liu Apr 17 '19 at 10:36
• – B.Liu Apr 17 '19 at 10:46

Hint: Given observed data $$x_1,...,x_n$$, consider the estimator of the form:
$$\widehat{e^{-\lambda}} = \sum_{k=0}^\infty w_k \Bigg[ \frac{1}{n} \sum_{i=1}^n \mathbb{I}(x_i = k) \Bigg],$$
where $$w_0,w_1,w_2,...$$ are a series of estimator weights corresponding to the possible outcomes of a Poisson random variable. You can see that this estimator estimates the target value as a weighted sum of the proportions of values equal to each possible outcome of a Poisson random variable.
Try to find an expression for the expected value of this estimator, and then see if there is any choice of values you could make for $$w_0,w_1,w_2,...$$ that would lead the expected value of this estimator to be equal to the quantity you are trying to estimate.