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=Example. A sum of a random number of i.i.d. Gaussians= | =Example. A sum of a random number of i.i.d. Gaussians= | ||
| − | Let <math>\left\{ \mathbf{X}_{n}\right\}</math> be a sequence of i.i.d. Gaussian random variables, each having characteristic function | + | Let <math class="inline">\left\{ \mathbf{X}_{n}\right\}</math> be a sequence of i.i.d. Gaussian random variables, each having characteristic function |
| − | <math>\Phi_{\mathbf{X}}\left(\omega\right)=e^{i\mu\omega}e^{-\frac{1}{2}\sigma^{2}\omega^{2}}</math>. Let <math>\mathbf{N}</math> be a Poisson random variable with pmf | + | <math class="inline">\Phi_{\mathbf{X}}\left(\omega\right)=e^{i\mu\omega}e^{-\frac{1}{2}\sigma^{2}\omega^{2}}</math>. Let <math class="inline">\mathbf{N}</math> be a Poisson random variable with pmf |
| − | <math>p(n)=\frac{e^{-\lambda}\lambda^{n}}{n!},\; n=0,1,2,\cdots,\;\lambda>0,</math> and assume <math>\mathbf{N}</math> is statistically independent of <math>\left\{ \mathbf{X}_{n}\right\}</math> . Define a new random variable | + | <math class="inline">p(n)=\frac{e^{-\lambda}\lambda^{n}}{n!},\; n=0,1,2,\cdots,\;\lambda>0,</math> and assume <math class="inline">\mathbf{N}</math> is statistically independent of <math class="inline">\left\{ \mathbf{X}_{n}\right\}</math> . Define a new random variable |
| − | <math>\mathbf{Y}=\mathbf{X}_{1}+\mathbf{X}_{2}+\cdots+\mathbf{X}_{N}.</math> | + | <math class="inline">\mathbf{Y}=\mathbf{X}_{1}+\mathbf{X}_{2}+\cdots+\mathbf{X}_{N}.</math> |
Note | Note | ||
| − | If <math>\mathbf{N}=0</math> , then <math>\mathbf{Y}=0</math> . | + | If <math class="inline">\mathbf{N}=0</math> , then <math class="inline">\mathbf{Y}=0</math> . |
| − | (a) Find the mean of <math>\mathbf{Y}</math> . | + | (a) Find the mean of <math class="inline">\mathbf{Y}</math> . |
| − | • Probability generating function of <math>\mathbf{N}</math> is <math>P_{\mathbf{N}}\left(z\right)=E\left[z^{\mathbf{N}}\right]=\sum_{n=0}^{\infty}z^{n}\frac{e^{-\lambda}\lambda^{n}}{n!}=e^{-\lambda}\sum_{n=0}^{\infty}\frac{\left(z\lambda\right)^{n}}{n!}=e^{-\lambda}e^{z\lambda}=e^{-\lambda\left(1-z\right)}.</math> | + | • Probability generating function of <math class="inline">\mathbf{N}</math> is <math class="inline">P_{\mathbf{N}}\left(z\right)=E\left[z^{\mathbf{N}}\right]=\sum_{n=0}^{\infty}z^{n}\frac{e^{-\lambda}\lambda^{n}}{n!}=e^{-\lambda}\sum_{n=0}^{\infty}\frac{\left(z\lambda\right)^{n}}{n!}=e^{-\lambda}e^{z\lambda}=e^{-\lambda\left(1-z\right)}.</math> |
| − | • The characteristic function of <math>\mathbf{Y}</math> is <math>\Phi_{\mathbf{Y}}\left(\omega\right)=P_{\mathbf{N}}\left(z\right)\Bigl|_{z=\Phi_{\mathbf{X}}\left(\omega\right)}=e^{-\lambda\left(1-z\right)}\Bigl|_{z=e^{i\mu\omega}e^{-\frac{1}{2}\sigma^{2}\omega^{2}}}=e^{-\lambda\left(1-e^{i\mu\omega}e^{-\frac{1}{2}\sigma^{2}\omega^{2}}\right)}.</math> | + | • The characteristic function of <math class="inline">\mathbf{Y}</math> is <math class="inline">\Phi_{\mathbf{Y}}\left(\omega\right)=P_{\mathbf{N}}\left(z\right)\Bigl|_{z=\Phi_{\mathbf{X}}\left(\omega\right)}=e^{-\lambda\left(1-z\right)}\Bigl|_{z=e^{i\mu\omega}e^{-\frac{1}{2}\sigma^{2}\omega^{2}}}=e^{-\lambda\left(1-e^{i\mu\omega}e^{-\frac{1}{2}\sigma^{2}\omega^{2}}\right)}.</math> |
| − | • Now, we can get the mean of <math>\mathbf{Y}</math> using the characteristic function. <math>E\left[\mathbf{Y}\right]=\frac{d}{d\left(i\omega\right)}\Phi_{\mathbf{Y}}\left(\omega\right)\Bigl|_{i\omega=0}=e^{-\lambda}\cdot\frac{d}{d\left(i\omega\right)}e^{\lambda e^{\mu\left(i\omega\right)+\frac{1}{2}\sigma^{2}\left(i\omega\right)^{2}}}\Bigl|_{i\omega=0}</math><math>=e^{-\lambda}\cdot e^{\lambda e^{\mu\left(i\omega\right)+\frac{1}{2}\sigma^{2}\left(i\omega\right)^{2}}}\cdot\lambda e^{\mu\left(i\omega\right)+\frac{1}{2}\sigma^{2}\left(i\omega\right)^{2}}\cdot\left(\mu+\sigma^{2}\left(i\omega\right)\right)\Bigl|_{i\omega=0}</math><math>=e^{-\lambda}\cdot\lambda\cdot\mu=\lambda\mu e^{-\lambda}.</math> | + | • Now, we can get the mean of <math class="inline">\mathbf{Y}</math> using the characteristic function. <math class="inline">E\left[\mathbf{Y}\right]=\frac{d}{d\left(i\omega\right)}\Phi_{\mathbf{Y}}\left(\omega\right)\Bigl|_{i\omega=0}=e^{-\lambda}\cdot\frac{d}{d\left(i\omega\right)}e^{\lambda e^{\mu\left(i\omega\right)+\frac{1}{2}\sigma^{2}\left(i\omega\right)^{2}}}\Bigl|_{i\omega=0}</math><math class="inline">=e^{-\lambda}\cdot e^{\lambda e^{\mu\left(i\omega\right)+\frac{1}{2}\sigma^{2}\left(i\omega\right)^{2}}}\cdot\lambda e^{\mu\left(i\omega\right)+\frac{1}{2}\sigma^{2}\left(i\omega\right)^{2}}\cdot\left(\mu+\sigma^{2}\left(i\omega\right)\right)\Bigl|_{i\omega=0}</math><math class="inline">=e^{-\lambda}\cdot\lambda\cdot\mu=\lambda\mu e^{-\lambda}.</math> |
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Latest revision as of 07:15, 1 December 2010
Example. A sum of a random number of i.i.d. Gaussians
Let $ \left\{ \mathbf{X}_{n}\right\} $ be a sequence of i.i.d. Gaussian random variables, each having characteristic function
$ \Phi_{\mathbf{X}}\left(\omega\right)=e^{i\mu\omega}e^{-\frac{1}{2}\sigma^{2}\omega^{2}} $. Let $ \mathbf{N} $ be a Poisson random variable with pmf
$ p(n)=\frac{e^{-\lambda}\lambda^{n}}{n!},\; n=0,1,2,\cdots,\;\lambda>0, $ and assume $ \mathbf{N} $ is statistically independent of $ \left\{ \mathbf{X}_{n}\right\} $ . Define a new random variable
$ \mathbf{Y}=\mathbf{X}_{1}+\mathbf{X}_{2}+\cdots+\mathbf{X}_{N}. $
Note
If $ \mathbf{N}=0 $ , then $ \mathbf{Y}=0 $ .
(a) Find the mean of $ \mathbf{Y} $ .
• Probability generating function of $ \mathbf{N} $ is $ P_{\mathbf{N}}\left(z\right)=E\left[z^{\mathbf{N}}\right]=\sum_{n=0}^{\infty}z^{n}\frac{e^{-\lambda}\lambda^{n}}{n!}=e^{-\lambda}\sum_{n=0}^{\infty}\frac{\left(z\lambda\right)^{n}}{n!}=e^{-\lambda}e^{z\lambda}=e^{-\lambda\left(1-z\right)}. $
• The characteristic function of $ \mathbf{Y} $ is $ \Phi_{\mathbf{Y}}\left(\omega\right)=P_{\mathbf{N}}\left(z\right)\Bigl|_{z=\Phi_{\mathbf{X}}\left(\omega\right)}=e^{-\lambda\left(1-z\right)}\Bigl|_{z=e^{i\mu\omega}e^{-\frac{1}{2}\sigma^{2}\omega^{2}}}=e^{-\lambda\left(1-e^{i\mu\omega}e^{-\frac{1}{2}\sigma^{2}\omega^{2}}\right)}. $
• Now, we can get the mean of $ \mathbf{Y} $ using the characteristic function. $ E\left[\mathbf{Y}\right]=\frac{d}{d\left(i\omega\right)}\Phi_{\mathbf{Y}}\left(\omega\right)\Bigl|_{i\omega=0}=e^{-\lambda}\cdot\frac{d}{d\left(i\omega\right)}e^{\lambda e^{\mu\left(i\omega\right)+\frac{1}{2}\sigma^{2}\left(i\omega\right)^{2}}}\Bigl|_{i\omega=0} $$ =e^{-\lambda}\cdot e^{\lambda e^{\mu\left(i\omega\right)+\frac{1}{2}\sigma^{2}\left(i\omega\right)^{2}}}\cdot\lambda e^{\mu\left(i\omega\right)+\frac{1}{2}\sigma^{2}\left(i\omega\right)^{2}}\cdot\left(\mu+\sigma^{2}\left(i\omega\right)\right)\Bigl|_{i\omega=0} $$ =e^{-\lambda}\cdot\lambda\cdot\mu=\lambda\mu e^{-\lambda}. $
