View source for ECE 600 Exams Sequence of binomially distributed random variables - Rhea

=Example. Sequence of binomially distributed random variables=

Let <math class="inline">\left\{ \mathbf{X}_{n}\right\} _{n\geq1}</math>  be a sequence of binomially distributed random variables, with the <math class="inline">n_{th}</math>  random variable <math class="inline">\mathbf{X}_{n}</math>  having pmf

<math class="inline">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).</math> Show that, if the <math class="inline">p_{n}</math>  have the property that <math class="inline">np_{n}\rightarrow\lambda</math>  as <math class="inline">n\rightarrow\infty</math> , where <math class="inline">\lambda</math>  is a positive constant, then the sequence <math class="inline">\left\{ \mathbf{X}_{n}\right\} _{n\leq1}</math>  converges in distribution to a Poisson random variable <math class="inline">\mathbf{X}</math>  with mean <math class="inline">\lambda</math> . 

Hint: 

You may find the following fact useful:

<math class="inline">\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^{n}=e^{x}.</math> 

Solution

If <math class="inline">\mathbf{X}_{n}</math>  converges to <math class="inline">\mathbf{X}</math>  in distribution, then <math class="inline">F_{\mathbf{X}_{n}}(x)\rightarrow F_{\mathbf{X}}(x)</math><math class="inline">  \forall x\in\mathbf{R}</math> , where <math class="inline">F_{\mathbf{X}}(x)</math>  is continuous. This occurs iff <math class="inline">\Phi_{\mathbf{X}_{n}}(\omega)\rightarrow\Phi_{\mathbf{X}}(\omega)</math><math class="inline">  \forall x\in\mathbf{R}</math> . We will show that <math class="inline">\Phi_{\mathbf{X}_{n}}(\omega)</math>  converges to <math class="inline">e^{-\lambda\left(1-e^{i\omega}\right)}</math>  as <math class="inline">n\rightarrow\infty</math> , which is the characteristic function of a Poisson random variable with mean <math class="inline">\lambda</math> .

<math class="inline">\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}</math><math class="inline">=\left(p_{n}e^{i\omega}+1-p_{n}\right)^{n}=\left(1+p_{n}\left(e^{i\omega}-1\right)\right)^{n}.</math> 

Now as <math class="inline">n\rightarrow\infty</math> , <math class="inline">np_{n}\rightarrow\lambda\Rightarrow p_{n}\rightarrow\frac{\lambda}{n}</math> .

<math class="inline">\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)},</math> 

which is the characteristic function of Poisson random variable with mean <math class="inline">\lambda</math> .

c.f.

The problem 2 of the [[ECE600 QE 2007 August|August 2007 QE]] is identical to this example.

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