Use Levy's Continuity Theorem to show Poisson Approximation of the Binomial Distribution

Question

Let $(X_{n})_{n \in \mathbb N}$ be random variables where $X_{n}$~$\operatorname{Bin}(n,p_{n})$ where $np_{n} \xrightarrow{n \to \infty}\lambda>0$

Show, using Levy's Continuity Theorem, that $X_{n} \xrightarrow{d} \operatorname{Poi}(\lambda)$ while $n \to \infty$

Only idea thus far:

$F_{X_{n}}(c)=P(X_{n} \leq c)=\sum_{k=0}^{c}\binom{n}{k}P(X_{n}=k)=\sum_{k=0}^{c}\binom{n}{k}(1-p_{n})^{n-k}(p_{n})^{k}$

I know that characteristic function is $\phi_{X_{n}}(t)=\mathbb E[e^{itX_{n}}]=\sum_{k=0}^{n}\binom{n}{k}e^{itk}(1-p)^{n-k}p^k=\sum_{k=0}^{n}\binom{n}{k}(1-p_{n})^{n-k}(e^{it}p_{n})^k$

How can I go on to use Levy's Continuity Theorem to show anything here?

Answer 1

By the binomial theorem the last expression (with $p=p_n$) is $(e^{it}p_n+1-p)^n =\left(1-\frac{np_n(1-e^{it})}{n}\right)^n \to \exp(-\lambda(1-e^{it}))$ which is the characteristic function of the Poisson distribution.

Answer 2

The theorem states that if the sequence of characteristic functions $\phi_{X_n}(t)$ converges to a function which is the characteristic function of some other r.v., say $\phi_Y(t)$, then the sequence $X_1,\ldots,X_n$ converges in distribution to $Y$, that is $$F_{X_n}(t)\to F_Y(t)$$ for every $t$.

Since $X_n\sim Bin(n,p_n)$, you can prove that $$\phi_{X_n}(t)=(1-p_n+p_ne^it)^n,$$ so if you prove that for $n\to \infty$ this expression goes to $$e^{\lambda (e^{it}-1)},$$ which is the characteristic function of any variable distributed as $\mathcal P(\lambda)$, then you can say that $$X_n\to \mathcal P(\lambda),$$ as desired.