Example. Sequence of binomially distributed random variables
Let $ \left\{ \mathbf{X}_{n}\right\} _{n\geq1} $ be a sequence of binomially distributed random variables, with the $ n_{th} $ random variable $ \mathbf{X}_{n} $ having pmf
$ P_{\mathbf{X}_{n}}(k)=P\left(\left\{ \mathbf{X}_{n}=k\right\} \right)=\left(\begin{array}{c} n\\ k \end{array}\right)p_{n}^{k}\left(1-p_{n}\right)^{n-k}\;,\; k=0,1,\cdots,n,\; p_{n}\in\left(0,1\right). $ Show that, if the $ p_{n} $ have the property that $ np_{n}\rightarrow\lambda $ as $ n\rightarrow\infty $ , where $ \lambda $ is a positive constant, then the sequence $ \left\{ \mathbf{X}_{n}\right\} _{n\leq1} $ converges in distribution to a Poisson random variable $ \mathbf{X} $ with mean $ \lambda $ .
Hint:
You may find the following fact useful:
$ \lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^{n}=e^{x}. $
Solution
If $ \mathbf{X}_{n} $ converges to $ \mathbf{X} $ in distribution, then $ F_{\mathbf{X}_{n}}(x)\rightarrow F_{\mathbf{X}}(x) $$ \forall x\in\mathbf{R} $ , where $ F_{\mathbf{X}}(x) $ is continuous. This occurs iff $ \Phi_{\mathbf{X}_{n}}(\omega)\rightarrow\Phi_{\mathbf{X}}(\omega) $$ \forall x\in\mathbf{R} $ . We will show that $ \Phi_{\mathbf{X}_{n}}(\omega) $ converges to $ e^{-\lambda\left(1-e^{i\omega}\right)} $ as $ n\rightarrow\infty $ , which is the characteristic function of a Poisson random variable with mean $ \lambda $ .
$ \Phi_{\mathbf{X}_{n}}(\omega)=E\left[e^{i\omega\mathbf{X}_{n}}\right]=\sum_{k=0}^{n}e^{i\omega k}\left(\begin{array}{c} n\\ k \end{array}\right)p_{n}^{k}\left(1-p_{n}\right)^{n-k}=\sum_{k=0}^{n}\left(\begin{array}{c} n\\ k \end{array}\right)\left(p_{n}e^{i\omega}\right)^{k}\left(1-p_{n}\right)^{n-k} $$ =\left(p_{n}e^{i\omega}+1-p_{n}\right)^{n}=\left(1+p_{n}\left(e^{i\omega}-1\right)\right)^{n}. $
Now as $ n\rightarrow\infty $ , $ np_{n}\rightarrow\lambda\Rightarrow p_{n}\rightarrow\frac{\lambda}{n} $ .
$ \lim_{n\rightarrow\infty}\Phi_{\mathbf{X}_{n}}(\omega)=\lim_{n\rightarrow\infty}\left(1+p_{n}\left(e^{i\omega}-1\right)\right)^{n}=\lim_{n\rightarrow\infty}\left(1+\frac{\lambda}{n}\left(e^{i\omega}-1\right)\right)^{n}=e^{\lambda\left(e^{i\omega}-1\right)}=e^{-\lambda\left(1-e^{i\omega}\right)}, $
which is the characteristic function of Poisson random variable with mean $ \lambda $ .
c.f.
The problem 2 of the August 2007 QE is identical to this example.
