Let $X_1,X_2,\dots$ be identically distributed and independent random variables. If distribution of $X_1$ is Poisson($\lambda$).
Let $\bar{X_n} =\frac{\sum_{i=1}^{n}X_n}{n}$ and $Y_n=(1-\frac{1}{n})^{n\bar{X_n}}$
Find suitable sequences $a_n$ and $b_n$ such that $a_n(Y_n-b_n)$ converges to non-degenerated distribution (i.e converges to a random variable whose distribution, having the distribution of constant random variable).
My attempt: I have tried to find a continuous function $f$ such that $Y_n = f(\bar{X_n})$ and then use continuous mapping theorem. But I couldn't find such a function. I think the solution will use central limit theorem and continuous mapping theorem but I can't figure out how.
Best Answer
Note that $$\ln Y_n = -\bar X_n\frac{\ln(1-\frac 1n)}{-\frac 1n}.$$
By the SLLN, $\bar X_n$ converges a.s. to $\lambda$, and $\frac{\ln(1-\frac 1n)}{-\frac 1n}$ converges deterministically to $1$, thus $\ln Y_n$ converges a.s. to $-\lambda$.
By the continuous mapping theorem, $Y_n$ converges a.s. to $e^{-\lambda}$, hence $Y_n-e^{-\lambda}$ converges a.s. to $0$.
Note that $$\sqrt n (\ln Y_n + \lambda) = -\sqrt n (\bar X_n - \lambda) - \bar X_n\sqrt n \left(\frac{\ln(1-\frac 1n)}{-\frac 1n}-1 \right) = -\sqrt n (\bar X_n - \lambda) - \bar X_n(\frac 1{2\sqrt n} + o(\frac 1{\sqrt n}))$$
The second summand $-\bar X_n(\frac 1{2\sqrt n} + o(\frac 1{\sqrt n}))$ converges a.s to $0$, while by the CLT $-\sqrt n (\bar X_n - \lambda)$ converges in distribution to $\mathcal N(0,\lambda)$.
By Slutsky's theorem, $\sqrt n (\ln Y_n - (- \lambda))$ converges in distribution to $\mathcal N(0,\lambda)$.
By the Delta method, $\sqrt n (Y_n - e^{- \lambda})$ converges in distribution to $\mathcal N(0,\lambda e^{\lambda})$.