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

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?

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.
• I only have the following version of Levy's Continuity Theorem: Weak convergence between two finite measures $\mu_{1}$ and $\mu{2}$ $\iff$ pointwise convergence of the characteristic functions of $\mu_{1}$ and $\mu{2}$. But in this proof we are using convergence in distribution rather than weak convergence. Does weak convergence simply imply convergence in distribution? Jan 29, 2019 at 13:54
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.