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}. $
