Finding moment estimator and its asymptotic distribution I got a question:
We let $X$ and $Y$ be independent random variables with $X$ Poisson distributed with mean $\lambda$ and $Y$ exponentially distributed with rate $\lambda>0$ and we let $(X_1,Y_1),\ldots,(X_n,Y_n)$ be a sample from this distribution.

*

*I have to find the moment estimator $\hat{\lambda}$ based on the statistic $t(x,y)=x-y$ and the sample $(X_1,Y_1),\ldots,(X_n,Y_n)$ and find the moment estimator's asymptotic distribution. Can anyone help me?

My thoughts so far is that: To find the asymtotic distribution I think I can use that $V(t(x,y))/m'(\lambda)^2$, but how do I find $m(\lambda)$? And can I then find moment estimator by solving $\lambda$ in $m(\lambda)$?

First, the mean of $X_1-Y_1$ is $\lambda-\lambda^{-1}$. Thus, the MM estimator of $\lambda$ is
$$
\hat{\lambda}_n=\frac{1}{2}\left(m_n+\sqrt{m_n^2+4}\right),
$$
where $m_n:=n^{-1}\sum_{i=1}^{n}(X_i-Y_i)$. For the asymptotic distribution of $\hat{\lambda}_n$, note that
$$
\sqrt{n}\left(m_n-(\lambda-\lambda^{-1})\right)\xrightarrow{d}N\!\left(0,\lambda+\lambda^{-2}\right).
$$
Thus, using delta method with $g(x)=(x+\sqrt{x^2+4})/2$,
$$
\sqrt{n}\left(\hat{\lambda}_n-\lambda\right)\xrightarrow{d}N\!\left(0,(\lambda+\lambda^{-2})[g'(\lambda-\lambda^{-1})]^2\right).
$$
