Limit of distribution function of Poisson distributed random variables

Question

Let $\left\{X_n\right\}_{n \geq 1}$ be i.i.d. random variables with Poisson$(\lambda)$ distribution, $\lambda>0$. For each $n \geq 1$, let $F_n$ denote the distribution function of $$ \frac{1}{\sqrt{n}} \sum_{i=1}^n\left(X_{2 i-1}-X_{2 i}\right) . $$ How can I calculate the limit $\lim _{n \rightarrow \infty} F_n(x), x \in \mathbb{R}$?

---

I thought I could apply the weak law of large numbers? Let $\left\{X_{n}\right\}$ be a sequence of random variables with finite expected values. We will say that $\left\{X_{n}\right\}$ satisfies the weak law of large numbers if $$ \frac{1}{n} \sum_{i=1}^{n} X_{i}-\frac{1}{n} \sum_{i=1}^{n} \mathbb{E}\left[X_{i}\right] \stackrel{P}{\rightarrow} 0. $$ The problem here is that this converges in probability but above I just take the limit. Any hints?

Answer 1

The Central Limit Theorem seems more appropriate. If $Y_i$ denotes $X_{2i-1}-X_{2i}$, then $F_n$ is the distribution function of $$ \frac1{\sqrt n}(Y_1+\cdots+Y_n)\tag{$\ast$}$$ where the $Y$'s are IID (why?). You are looking for the limiting distribution of ($\ast$) as $n\to\infty$.

Answer 2

Let $Y_i=X_{2i-1}-X_{2i}$ and notice that the $Y_i$ are iid, with $\mathbb E[Y_i]=\lambda-\lambda=0$ and $\mathrm{Var}(Y_i)=\lambda+\lambda=2\lambda$, using basic properties of the variance and the variance of the Poisson distribution. The central limit theorem shows that $S_n=\frac{1}{\sqrt{n}}\sum_{i=1}^n(X_{2i-1}-X_{2i})$ converges to a normal $S^*\sim\mathcal N(0,2\lambda)$ in distribution, so $\lim_{n\rightarrow\infty}\mathbb P(S\le s)=\mathbb P(S^*\le s)$ for all $s$ and thereby $F_n$ converges pointwise to the cumulative distribution function $F^*$ of $S^*$.