# Proof of Poisson theorem using characteristic function

Suppose, $\lbrace X_n\rbrace_{n=1}^{\infty}$ is a sequence of random variables with distribution $B(n, p_n)$, and $\lim_{n\to\infty} np_n = \lambda > 0$, then $X_n \rightarrow X$ in distribution, where $X$ has $Poi(\lambda)$ distribution.

I want to prove it using characteristic function, so for each $n$

$\phi_{X_n}(t) = (p_ne^{it} + (1-p_n))^n$, but I don't know how to prove, that

$\lim_{n\to\infty}\phi_{X_n}(t) = \exp{\lambda(e^{it} - 1)}$.

So fix any $t_0 \in \mathbb{R}$. For any $n \geq 1$, we have $$\phi_{X_n}(t_0) = (p_ne^{it_0} + (1-p_n))^n = e^{n \ln \left( p_ne^{it_0} + (1-p_n) \right)} = e^{n \ln \left( 1+ p_n(e^{it_0}-1) \right)}$$ Now, since by assumption $p_n\xrightarrow[n\to\infty]{} 0$ (as $n p_n \to \lambda$), then $p_n(e^{it_0}-1)\xrightarrow[n\to\infty]{} 0$ and we can do a first-order Taylor expansion of the logarithm. Specifically, when $n\to\infty$ we have $$\phi_{X_n}(t_0) = e^{n\left( p_n(e^{it_0}-1) + o(p_n) \right)} = e^{n p_n(e^{it_0}-1) + o(n p_n)} \underbrace{=}_{(\ast)} e^{\lambda(e^{it_0}-1 + o(1))} \xrightarrow[n\to\infty]{} e^{\lambda(e^{it_0}-1)}$$ where for $(\ast)$ we used the fact that by assumption $np_n = \lambda + o(1)$ when $n\to\infty$.
• @Maciej By the way, if this fully answered your question, consider accepting the answer by clicking on the green $\color{green}\checkmark$ on the left. (If it did not, don't -- but please let me know how to improve my answer). – Clement C. Sep 3 '16 at 21:41