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=Example. Sequence of binomially distributed random variables= | =Example. Sequence of binomially distributed random variables= | ||
| − | Let <math>\left\{ \mathbf{X}_{n}\right\} _{n\geq1}</math> be a sequence of binomially distributed random variables, with the <math>n_{th}</math> random variable <math>\mathbf{X}_{n}</math> having pmf | + | 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>P_{\mathbf{X}_{n}}(k)=P\left(\left\{ \mathbf{X}_{n}=k\right\} \right)=\left(\begin{array}{c} | + | <math class="inline">P_{\mathbf{X}_{n}}(k)=P\left(\left\{ \mathbf{X}_{n}=k\right\} \right)=\left(\begin{array}{c} |
n\\ | n\\ | ||
k | 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>p_{n}</math> have the property that <math>np_{n}\rightarrow\lambda</math> as <math>n\rightarrow\infty</math> , where <math>\lambda</math> is a positive constant, then the sequence <math>\left\{ \mathbf{X}_{n}\right\} _{n\leq1}</math> converges in distribution to a Poisson random variable <math>\mathbf{X}</math> with mean <math>\lambda</math> . | + | \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: | Hint: | ||
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You may find the following fact useful: | You may find the following fact useful: | ||
| − | <math>\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^{n}=e^{x}.</math> | + | <math class="inline">\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^{n}=e^{x}.</math> |
Solution | Solution | ||
| − | If <math>\mathbf{X}_{n}</math> converges to <math>\mathbf{X}</math> in distribution, then <math>F_{\mathbf{X}_{n}}(x)\rightarrow F_{\mathbf{X}}(x)</math><math> \forall x\in\mathbf{R}</math> , where <math>F_{\mathbf{X}}(x)</math> is continuous. This occurs iff <math>\Phi_{\mathbf{X}_{n}}(\omega)\rightarrow\Phi_{\mathbf{X}}(\omega)</math><math> \forall x\in\mathbf{R}</math> . We will show that <math>\Phi_{\mathbf{X}_{n}}(\omega)</math> converges to <math>e^{-\lambda\left(1-e^{i\omega}\right)}</math> as <math>n\rightarrow\infty</math> , which is the characteristic function of a Poisson random variable with mean <math>\lambda</math> . | + | 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>\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} | + | <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\\ | n\\ | ||
k | k | ||
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n\\ | n\\ | ||
k | k | ||
| − | \end{array}\right)\left(p_{n}e^{i\omega}\right)^{k}\left(1-p_{n}\right)^{n-k}</math><math>=\left(p_{n}e^{i\omega}+1-p_{n}\right)^{n}=\left(1+p_{n}\left(e^{i\omega}-1\right)\right)^{n}.</math> | + | \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>n\rightarrow\infty</math> , <math>np_{n}\rightarrow\lambda\Rightarrow p_{n}\rightarrow\frac{\lambda}{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>\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> | + | <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>\lambda</math> . | + | which is the characteristic function of Poisson random variable with mean <math class="inline">\lambda</math> . |
c.f. | c.f. | ||
| − | The problem 2 of the August 2007 QE | + | The problem 2 of the [[ECE600 QE 2007 August|August 2007 QE]] is identical to this example. |
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Latest revision as of 12:14, 30 November 2010
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.
