$\lim_{n\to \infty} \exp\{\exp\{it/\sqrt n\}-1-it\sqrt{n}\}=\exp\{-t^2/2\}$

Question

Let $X_1,\dots ,X_n$ be an iid sample from $\text{Poi}(1)$.

1. Using characteristic function of $\text{Poi}(1)$, find the coefficient of skewness $\gamma:=\frac{\mu_3}{\mu_2^{3/2}}$

2. Hence, or otherwise, show that $$\lim_{n\to \infty} \exp\{\exp\{it/\sqrt n\}-1-it\sqrt{n}\}=\exp\{-t^2/2\}$$

I have been able to solve the first part and the answer is $1$. But, I can't figure out how to use that (or otherwise) to calculate the given limit.

Answer 1

The original expression is definitely wrong. At the very least at each $t \in \{(2a + 1)\pi/b : a, b \in \mathbb Z_+\}$, $\exp\{\exp\{it/\sqrt{n}\} - 1 -it\sqrt{n} \}$ fails to converge to any value (let alone $\exp(-t^2/2)$). I demonstrate that later at the end. However the following slight modification of the original expression: $$ \lim_{n \to \infty} \exp\{n(\exp\{it/\sqrt{n}\} - 1) -it\sqrt{n} \} = \exp(-t^2/2) $$ is true and is appropriate considering that the question uses $n$ i.i.d. $\text{Poisson}(1)$ variables.

Here is a short "statistical flavored" proof of it. By the central limit theorem $\sqrt{n}(\overline{X}_n - 1) \to_d N(0, 1)$. As convergence in distribution $\implies$ pointwise convergence of characteristic functions, $\lim\limits_{n \to \infty}\phi_{\sqrt{n}(\overline{X}_n - 1)}(t) = \phi_{N(0, 1)}(t) = \exp(-t^2/2)$. To finish, let's verify that $\phi_{\sqrt{n}(\overline{X}_n - 1)}(t)$ has the necessary form: for each $n \in \mathbb{N}$, \begin{align*} & \phi_{\sqrt{n}(\overline{X}_n - 1)}(t) = E[e^{it\sqrt{n}(\overline{X}_n - 1)}] =e^{-it\sqrt{n}} \ \phi_{\sum_{m = 1}^n X_m}\Big(\frac{t}{ \sqrt{n} }\Big) \\ = & e^{-it\sqrt{n}} [\phi_{\text{Poisson}(1)}\Big(\frac{t}{ \sqrt{n} }\Big)]^n = \exp\{n(\exp\Big\{\frac{it}{\sqrt{n}}\Big\} - 1) -it\sqrt{n} \} \end{align*}

Non-Convergence of Original Expression: Fix some $t = (2a + 1)\pi/b$ for $a, b \in \mathbb Z_+$. Consider the two subsequences $n_k^{(1)} = ((2k + 1)b)^2$ and $n_k^{(2)} = (2kb)^2$ of $(n)_{n = 1}^\infty$. \begin{align*} \exp\{\exp\{it/\sqrt{n_k^{(1)}}\} - 1 -it\sqrt{n_k^{(1)}} \} = e^{\exp\{it/\sqrt{n_k^{(1)}} \} - 1} \underbrace{ e^{-i(2a + 1)(2k + 1)\pi} }_{e^{-i(2k' + 1)\pi} =-1} = -e^{\exp\{it/\sqrt{n_k^{(1)}} \} - 1} \end{align*} and this converges to $-e^{\exp(0) - 1} = -1$ as $k \to \infty$. But note \begin{align*} \exp\{\exp\{it/\sqrt{n_k^{(2)}}\} - 1 -it\sqrt{n_k^{(2)}} \} = e^{\exp\{it/\sqrt{n_k^{(2)}} \} - 1} \underbrace{ e^{-i(2a + 1)2k\pi} }_{e^{-i(2k')\pi} = +1} = +e^{\exp\{it/\sqrt{n_k^{(2)}} \} - 1} \end{align*} and this converges to $+e^{\exp(0) - 1} = +1$ as $k \to \infty$. So evaluating the original expression on two different subsequences led to two different limits and so it can't converge at any $t$ with the form $(2a + 1)\pi/b$.