Central Limit Theorem (Question)

Question

Use the central limit theorem to verify that,

\[\begin{equation*} \begin{array}{rclll} \lim\limits_{n \to \infty}e^{-n}\displaystyle\sum^{n}_{k=0}\dfrac{2n^{k}}{k!} & = & 1. \end{array} \end{equation*}\]

Answer:

Let \(X_{1}, X_{2}, \cdots\), i.i.d. with \(\sim Pois(1)\). Then, \(\mathbb{E}(X_{i}) = 1\) and \(Var(X_{1}) = 1\).

Furthermore, we know that \(S_{n} = X_{1} + X_{2} + \cdots + X_{n}\), will have \(S_{n} \sim Pois(n)\), \(\mathbb{E}(X_{i}) = n\) and \(Var(X_{1}) = n\).

Now, using the central limit theorem we get

\[\begin{equation*} \begin{array}{rclll} \displaystyle\sum^{n}_{k=0}e^{-n}\dfrac{n^{k}}{k!} & = & \displaystyle\sum^{n}_{k=0}\mathbb{P}(S_{n} = k) & = & \mathbb{P}(S_{n}\leq n) \end{array} \end{equation*}\]

in which,

\[\begin{equation*} \begin{array}{rclll} \mathbb{P}\left( \dfrac{S_{n} - n}{\sqrt{n}} \leq \dfrac{n - n}{\sqrt{n}}\right) & {\underset{n \to \infty}{\longrightarrow}} & \mathbb{P}(Z\leq 0) & = & \dfrac{1}{2} \end{array} \end{equation*}\]

\(Z \sim N(0,1)\). Therefore, the proof is done.

\[\begin{equation*} \begin{array}{rclll} \lim\limits_{n \to \infty}\displaystyle\sum^{n}_{k=0}e^{-n}\dfrac{n^{k}}{k!} & = & \dfrac{1}{2}. \end{array} \end{equation*}\]