Convergence in Distribution of a Normalized Sum of Independent Poisson($k$) Random Variables

Question

Let $\{X_k \}_{k=1}^{\infty}$ be a sequence of independent random variables such that $X_k \sim$ Poisson($k$) for each $k.$ Consider the sequence of random variables $\{Y_n \}_{n=1}^{\infty}$ such that $$Y_n = \frac{1}{n^2} \sum_{k=1}^{n} X_k.$$ Prove that $Y_n$ converges in distribution as $n \to \infty,$ and find the limiting distribution.

On first glance, this appears to be a question that can be modified so as to be handled using the Central Limit Theorem; however, this has proven quite difficult thus far. Consider a sequence of independent and identically distributed Poisson(1) random variables $\{W_i \}_{i = 1}^{\infty}.$ We note that each of our original Poisson($k$) random variables is equal in distribution to a sum of independent Poisson(1) random variables, particularly $X_k \stackrel{d}{=} \sum_{i=1}^{k} W_i,$ from which it follows that $$Y_n = \frac{1}{n^2} \sum_{k=1}^{n} X_k \stackrel{d}{=} \frac{1}{n^2} \sum_{k=1}^{n} \bigg( \sum_{i=1}^{k} W_i \bigg) = \frac{n-1}{n^2} \sum_{k=1}^{n} W_k - \frac{1}{n^2} \sum_{k=1}^{n} k W_k,$$ where the last equality follows from writing out the double sum and doing some rearranging. Unfortunately, it is here that I get stuck. I am uncertain how to deal with either of these sums as we take the limit as $n$ approaches infinity. Can anyone provide any useful information or hints?

Ultimately, I would like to use the fact that $$\frac{1}{\sqrt{n}} \sum_{k=1}^{n} (W_k - 1) \stackrel{d}{\to} \mbox{n}(0,1)$$ by the Central Limit Theorem, noting that $\mu = \mathbb{E} (W_1) = 1$ and $\sigma = \sqrt{\mbox{Var} (W_1)} = 1.$

Answer 1

Each $X_k$ can be thought of as the sum of $k$ iid copies of a Poisson(1) random variable. Let $\{W_i\}_{i=1}^\infty$ be an iid sequence of Poisson(1) random variables. Then $\sum_{k=1}^n X_k$ is the sum of $1+2+\cdots+n=n(n+1)/2$ iid Poisson(1) random variables. Let $m=n(n+1)/2$.

By the Law of Large Numbers,

$$\frac 1m \sum_{j=1}^m W_j \xrightarrow{d}E[W_1]=1. $$

Now note that because $\lim_{n\to\infty} \frac m{n^2}=\lim_{n\to\infty}\frac{n(n+1)/2}{n^2}=\frac12$

$$Y_n=\frac1{n^2}\sum_{k=1}^n X_n = \frac{m}{n^2}\cdot \frac1m \sum_{j=1}^m W_j\xrightarrow{d}\frac12 \cdot 1=\frac12.$$

Answer 2

Your $n^2 Y_n$ is Poisson distributed with expectation $1+2+\cdots+n = \binom{n+1}{2} \sim n^2/2$. So $(n^2 Y_n - n^2/2)/\sqrt{n^2/2}$ converges in distribution to a standard normal.

Answer 3

Since each $X_k$ is Poisson distributed with parameter $k$, it has mean and variance equal to $k$. Hence $$ \mathbb{E}[Y_n]=\frac{1+2+\dots+n}{n^2}=\frac{n+1}{2n}$$ and $$ \mathrm{var}(Y_n)=\frac{1+2+\dots+n}{n^4}=\frac{n+1}{2n^3}\leq \frac{1}{n^2}$$

For a given $\varepsilon>0$, if we choose $n$ sufficiently large that $\frac{n+1}{2n}-\frac{1}{2}<\varepsilon$, then it follows from Chebyshev's inequality that $$ \mathbb{P}\Big(\Big|Y_n-\frac{1}{2}\Big|>2\varepsilon\Big)\leq \mathbb{P}(|Y_n-\mathbb{E}[Y_n]|>\varepsilon)\leq \frac{\mathrm{var}(Y_n)}{\varepsilon^2}\leq \frac{1}{\varepsilon^2n^2}\to 0$$ as $n\to\infty$.

Therefore $Y_n\to\frac{1}{2}$ in probability, hence in distribution.