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=Example. Addition of two independent Poisson random variables= | =Example. Addition of two independent Poisson random variables= | ||
| − | Let <math>\mathbf{Z}=\mathbf{X}+\mathbf{Y}</math> where <math>\mathbf{X}</math> and <math>\mathbf{Y}</math> are independent Poisson random variables with means <math>\lambda</math> and <math>\mu</math> , respectively. | + | Let <math class="inline">\mathbf{Z}=\mathbf{X}+\mathbf{Y}</math> where <math class="inline">\mathbf{X}</math> and <math class="inline">\mathbf{Y}</math> are independent Poisson random variables with means <math class="inline">\lambda</math> and <math class="inline">\mu</math> , respectively. |
(a) | (a) | ||
| − | Find the pmf of <math>\mathbf{Z}</math> . | + | Find the pmf of <math class="inline">\mathbf{Z}</math> . |
According to the characteristic function of Poisson random variable | According to the characteristic function of Poisson random variable | ||
| − | <math>\Phi_{\mathbf{X}}(\omega)=e^{-\lambda\left(1-e^{i\omega}\right)},\Phi_{\mathbf{Y}}(\omega)=e^{-\mu\left(1-e^{i\omega}\right)}</math>. | + | <math class="inline">\Phi_{\mathbf{X}}(\omega)=e^{-\lambda\left(1-e^{i\omega}\right)},\Phi_{\mathbf{Y}}(\omega)=e^{-\mu\left(1-e^{i\omega}\right)}</math>. |
| − | <math>\mathbf{X}</math> and <math>\mathbf{Y}</math> are independent <math>\Longrightarrow \mathbf{X}</math> and <math>\mathbf{Y}</math> are uncorrelated <math>\Longrightarrow e^{i\omega\mathbf{X}}</math> and <math>e^{i\omega\mathbf{Y}}</math> are uncorrelated. | + | <math class="inline">\mathbf{X}</math> and <math class="inline">\mathbf{Y}</math> are independent <math class="inline">\Longrightarrow \mathbf{X}</math> and <math class="inline">\mathbf{Y}</math> are uncorrelated <math class="inline">\Longrightarrow e^{i\omega\mathbf{X}}</math> and <math class="inline">e^{i\omega\mathbf{Y}}</math> are uncorrelated. |
| − | <math>\Phi_{\mathbf{Z}}(\omega)=E\left[e^{i\omega\mathbf{Z}}\right]=E\left[e^{i\omega\left(\mathbf{X}+\mathbf{Y}\right)}\right]=E\left[e^{i\omega\mathbf{X}}e^{i\omega\mathbf{Y}}\right]=E\left[e^{i\omega\mathbf{X}}\right]\cdot E\left[e^{i\omega\mathbf{Y}}\right]</math><math>=e^{-\lambda\left(1-e^{i\omega}\right)}\cdot e^{-\mu\left(1-e^{i\omega}\right)}=e^{-\left(\lambda+\mu\right)\left(1-e^{i\omega}\right).}</math> | + | <math class="inline">\Phi_{\mathbf{Z}}(\omega)=E\left[e^{i\omega\mathbf{Z}}\right]=E\left[e^{i\omega\left(\mathbf{X}+\mathbf{Y}\right)}\right]=E\left[e^{i\omega\mathbf{X}}e^{i\omega\mathbf{Y}}\right]=E\left[e^{i\omega\mathbf{X}}\right]\cdot E\left[e^{i\omega\mathbf{Y}}\right]</math><math class="inline">=e^{-\lambda\left(1-e^{i\omega}\right)}\cdot e^{-\mu\left(1-e^{i\omega}\right)}=e^{-\left(\lambda+\mu\right)\left(1-e^{i\omega}\right).}</math> |
| − | Now, we know that \mathbf{Z} is a Poisson random variable with mean <math>\lambda+\mu</math> . | + | Now, we know that \mathbf{Z} is a Poisson random variable with mean <math class="inline">\lambda+\mu</math> . |
| − | <math>\therefore p_{\mathbf{Z}}(k)=\frac{e^{-\left(\lambda+\mu\right)}\left(\lambda+\mu\right)^{k}}{k!}.</math> | + | <math class="inline">\therefore p_{\mathbf{Z}}(k)=\frac{e^{-\left(\lambda+\mu\right)}\left(\lambda+\mu\right)^{k}}{k!}.</math> |
(b) | (b) | ||
| − | Show that the conditional pmf of <math>\mathbf{X}</math> conditioned on the event <math>\left\{ \mathbf{Z}=n\right\}</math> is binomially distributed, and determine the parameters of binomial distribution (<math>n</math> and <math>p</math> ). | + | Show that the conditional pmf of <math class="inline">\mathbf{X}</math> conditioned on the event <math class="inline">\left\{ \mathbf{Z}=n\right\}</math> is binomially distributed, and determine the parameters of binomial distribution (<math class="inline">n</math> and <math class="inline">p</math> ). |
| − | <math>P_{\mathbf{X}}\left(\mathbf{X}|\left\{ \mathbf{Z}=n\right\} \right)=P\left(\left\{ \mathbf{X}=k\right\} |\left\{ \mathbf{Z}=n\right\} \right)=\frac{P\left(\left\{ \mathbf{X}=k\right\} \cap\left\{ \mathbf{Z}=n\right\} \right)}{P\left(\left\{ \mathbf{Z}=n\right\} \right)}=\frac{P\left(\left\{ \mathbf{X}=k\right\} \cap\left\{ \mathbf{Y}=n-k\right\} \right)}{P\left(\left\{ \mathbf{Z}=n\right\} \right)}</math><math>=\frac{\frac{e^{-\lambda}\lambda^{k}}{k!}\cdot\frac{e^{-\mu}\mu^{n-k}}{\left(n-k\right)!}}{\frac{e^{-\left(\lambda+\mu\right)}\left(\lambda+\mu\right)^{n}}{n!}}=\left(\frac{n!}{k!\left(n-k\right)!}\right)\left(\frac{\lambda}{\lambda+\mu}\right)^{k}\left(\frac{\mu}{\lambda+\mu}\right)^{n-k}</math><math>=\left(\begin{array}{c} | + | <math class="inline">P_{\mathbf{X}}\left(\mathbf{X}|\left\{ \mathbf{Z}=n\right\} \right)=P\left(\left\{ \mathbf{X}=k\right\} |\left\{ \mathbf{Z}=n\right\} \right)=\frac{P\left(\left\{ \mathbf{X}=k\right\} \cap\left\{ \mathbf{Z}=n\right\} \right)}{P\left(\left\{ \mathbf{Z}=n\right\} \right)}=\frac{P\left(\left\{ \mathbf{X}=k\right\} \cap\left\{ \mathbf{Y}=n-k\right\} \right)}{P\left(\left\{ \mathbf{Z}=n\right\} \right)}</math><math class="inline">=\frac{\frac{e^{-\lambda}\lambda^{k}}{k!}\cdot\frac{e^{-\mu}\mu^{n-k}}{\left(n-k\right)!}}{\frac{e^{-\left(\lambda+\mu\right)}\left(\lambda+\mu\right)^{n}}{n!}}=\left(\frac{n!}{k!\left(n-k\right)!}\right)\left(\frac{\lambda}{\lambda+\mu}\right)^{k}\left(\frac{\mu}{\lambda+\mu}\right)^{n-k}</math><math class="inline">=\left(\begin{array}{c} |
n\\ | n\\ | ||
k | k | ||
\end{array}\right)\left(\frac{\lambda}{\lambda+\mu}\right)^{k}\left(\frac{\mu}{\lambda+\mu}\right)^{n-k}\;,\; k=0,\,1,\,2,\,\cdots</math> | \end{array}\right)\left(\frac{\lambda}{\lambda+\mu}\right)^{k}\left(\frac{\mu}{\lambda+\mu}\right)^{n-k}\;,\; k=0,\,1,\,2,\,\cdots</math> | ||
| − | This is a binomial pmf <math>b(n,p)</math> with parameters <math>n</math> and <math>p=\frac{\lambda}{\lambda+\mu}</math> . | + | This is a binomial pmf <math class="inline">b(n,p)</math> with parameters <math class="inline">n</math> and <math class="inline">p=\frac{\lambda}{\lambda+\mu}</math> . |
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Latest revision as of 11:58, 30 November 2010
Example. Addition of two independent Poisson random variables
Let $ \mathbf{Z}=\mathbf{X}+\mathbf{Y} $ where $ \mathbf{X} $ and $ \mathbf{Y} $ are independent Poisson random variables with means $ \lambda $ and $ \mu $ , respectively.
(a)
Find the pmf of $ \mathbf{Z} $ .
According to the characteristic function of Poisson random variable
$ \Phi_{\mathbf{X}}(\omega)=e^{-\lambda\left(1-e^{i\omega}\right)},\Phi_{\mathbf{Y}}(\omega)=e^{-\mu\left(1-e^{i\omega}\right)} $.
$ \mathbf{X} $ and $ \mathbf{Y} $ are independent $ \Longrightarrow \mathbf{X} $ and $ \mathbf{Y} $ are uncorrelated $ \Longrightarrow e^{i\omega\mathbf{X}} $ and $ e^{i\omega\mathbf{Y}} $ are uncorrelated.
$ \Phi_{\mathbf{Z}}(\omega)=E\left[e^{i\omega\mathbf{Z}}\right]=E\left[e^{i\omega\left(\mathbf{X}+\mathbf{Y}\right)}\right]=E\left[e^{i\omega\mathbf{X}}e^{i\omega\mathbf{Y}}\right]=E\left[e^{i\omega\mathbf{X}}\right]\cdot E\left[e^{i\omega\mathbf{Y}}\right] $$ =e^{-\lambda\left(1-e^{i\omega}\right)}\cdot e^{-\mu\left(1-e^{i\omega}\right)}=e^{-\left(\lambda+\mu\right)\left(1-e^{i\omega}\right).} $
Now, we know that \mathbf{Z} is a Poisson random variable with mean $ \lambda+\mu $ .
$ \therefore p_{\mathbf{Z}}(k)=\frac{e^{-\left(\lambda+\mu\right)}\left(\lambda+\mu\right)^{k}}{k!}. $
(b)
Show that the conditional pmf of $ \mathbf{X} $ conditioned on the event $ \left\{ \mathbf{Z}=n\right\} $ is binomially distributed, and determine the parameters of binomial distribution ($ n $ and $ p $ ).
$ P_{\mathbf{X}}\left(\mathbf{X}|\left\{ \mathbf{Z}=n\right\} \right)=P\left(\left\{ \mathbf{X}=k\right\} |\left\{ \mathbf{Z}=n\right\} \right)=\frac{P\left(\left\{ \mathbf{X}=k\right\} \cap\left\{ \mathbf{Z}=n\right\} \right)}{P\left(\left\{ \mathbf{Z}=n\right\} \right)}=\frac{P\left(\left\{ \mathbf{X}=k\right\} \cap\left\{ \mathbf{Y}=n-k\right\} \right)}{P\left(\left\{ \mathbf{Z}=n\right\} \right)} $$ =\frac{\frac{e^{-\lambda}\lambda^{k}}{k!}\cdot\frac{e^{-\mu}\mu^{n-k}}{\left(n-k\right)!}}{\frac{e^{-\left(\lambda+\mu\right)}\left(\lambda+\mu\right)^{n}}{n!}}=\left(\frac{n!}{k!\left(n-k\right)!}\right)\left(\frac{\lambda}{\lambda+\mu}\right)^{k}\left(\frac{\mu}{\lambda+\mu}\right)^{n-k} $$ =\left(\begin{array}{c} n\\ k \end{array}\right)\left(\frac{\lambda}{\lambda+\mu}\right)^{k}\left(\frac{\mu}{\lambda+\mu}\right)^{n-k}\;,\; k=0,\,1,\,2,\,\cdots $
This is a binomial pmf $ b(n,p) $ with parameters $ n $ and $ p=\frac{\lambda}{\lambda+\mu} $ .
