Finding an estimator of $e^{-\theta}$ for Poisson distribution with parameter $\theta$ using the methods of moments

Question

Let $X_1, \ldots, X_n \stackrel{i . i . d}{\sim} \mathcal{P}(\theta)$
My aim: Let's set $\lambda=e^{-\theta}$. Propose an estimator $\hat{\lambda}_1$ of $\lambda$ using the method of moments.
My attempt: I link the $\lambda$ with the probability event. I got $P(X=0) =\lambda$. But i don't know how to continue my solution.

Answer 1

Indeed $\lambda = P(X=0)$.

So if you think of probabilities as asymptotic frequencies, you should have $\lambda \approx \dfrac {\text{number of } i, 1\leq i \leq n \text{ such that } X_i = 0}{n}$.

I suggest you take this last quantity as your estimator. Call it $\hat \lambda_n$ maybe because it depends on $n$.

I think the method of moments comes in to show that $\hat \lambda_n\to \lambda$ in probability, by computing the first two moments of $\hat \lambda_n - \lambda$.