Suppose $X_1,X_2, \cdots$ are i.i.d. observations from a $Poisson(\lambda)$ distribution. Define $\bar{X}_n=\sum_{i=1}^nX_i/n$. What will be the asymptotic distribution of $\left(1-\frac{1}{n}\right)^{n\bar{X}_n}$?
I have solved this problem, however I need to use the facts that $\left(1-\frac{1}{n}\right)^{n\bar{X}_n}$ and $e^{-\bar{X}_n}$ are the UMVUE and MLE for $e^{-\lambda}$ respectively. I don't how to use this fact. Please tell me how to do this problem using this interesting fact.
No idea why these (not so) "interesting facts" should be used (in this context) since, considering the random variables $Y_n=\left(1-\frac{1}{n}\right)^{n\bar{X}_n}$, the fact that $\mathrm e^{-(n+1)/n^2}\leqslant1-1/n\leqslant\mathrm e^{-1/n}$ for every $n\geqslant2$ yields readily $$ \mathrm e^{-(1+1/n)\bar X_n}\leqslant Y_n\leqslant\mathrm e^{-\bar X_n}. $$ Thus, the almost sure convergence $\bar X_n\to\lambda$ implies that $Y_n\to\mathrm e^{-\lambda}$ almost surely (and in probability and in distribution and in every $L^p$, $0\lt p\lt\infty$).
