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Part 3

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Show that the sum of two jointly distributed Gaussian random variables that are not necessarily statistically independent is a Gaussian random variable.

Solution 1:

Suppose $ \mathbf{X} $ and $ \mathbf{Y} $ are jointly distributed Gaussian random variables with jointly pdf

$ f_{X,Y}(x,y)=\frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-r^2}}\text{exp}\left\{\frac{-1}{2(1-r^2)}\left(\frac{(x-\eta_X)^2}{\sigma_X^2}-2r\frac{(x-\eta_X)(y-\eta_Y)}{\sigma_X\sigma_Y}+\frac{(y-\eta_Y)^2}{\sigma_Y^2}\right)\right\} $

Then, compute the characteristic function of $ \mathbf{X}+\mathbf{Y} $:

$ \begin{align} \Phi_{X+Y}(\omega) &= E[e^{i\omega(X+Y)}]\\ &=\int_{-\infty}^\infty\int_{-\infty}^\infty e^{i\omega(x+y)}f_{X,Y}(x,y)dxdy \\ &=\int_{-\infty}^\infty e^{i\omega y}\left(\int_{-\infty}^\infty e^{i\omega x}f_{X,Y}(x,y)dx\right)dy \\ &=\int_{-\infty}^\infty e^{i\omega y}\left(\int_{-\infty}^\infty e^{i\omega x} \frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-r^2}}\text{exp}\left\{ \frac{-1}{2(1-r^2)}\left(\frac{(x-\eta_X)^2}{\sigma_X^2}-2r\frac{(x-\eta_X)(y-\eta_Y)}{\sigma_X\sigma_Y}+\frac{(y-\eta_Y)^2}{\sigma_Y^2}\right)\right\}dx\right)dy \\ &=\int_{-\infty}^\infty \frac{e^{i\omega y}}{\sqrt{2\pi}\sigma_Y}\left(\int_{-\infty}^\infty e^{i\omega x} \frac{1}{\sqrt{2\pi}\sigma_X\sqrt{1-r^2}}\text{exp}\left\{-\frac{(x-(\eta_X+\frac{\sigma_X}{\sigma_Y}r(y-\eta_Y)))^2}{2(1-r^2)\sigma_X^2}\right\}dx\right)\text{exp}\left\{-\frac{(y-\eta_Y)^2}{2\sigma_Y^2}\right\}dy \end{align} $

We know that the characteristic function of a Gaussian random variable with mean $ \mu $ and variance $ \sigma^2 $ is $ (e^{i\omega\mu-\frac{1}{2}\sigma^2\omega^2}) $. Then,

$ \begin{align} \Phi_{X+Y}(\omega) &= \int_{-\infty}^\infty \dfrac{e^{i\omega y}}{\sqrt{2\pi}\sigma_Y}\left(e^{i\omega(\eta_X+\frac{\sigma_X}{\sigma_Y}r(y-\eta_Y))-\frac{1}{2}(1-r^2)\sigma_X^2\omega^2}\right)\text{exp}\left\{-\dfrac{(y-\eta_Y)^2}{2\sigma_Y^2}\right\}dy \\ &= e^{i\omega(\eta_X-\frac{\sigma_X}{\sigma_Y}r\eta_Y)-\frac{1}{2}(1-r^2)\sigma_X^2\omega^2}\int_{-\infty}^\infty e^{i\omega(1+\frac{\sigma_X}{\sigma_Y}r)y}\cdot\dfrac{1}{\sqrt{2\pi}\sigma_Y}\text{exp}\left\{-\dfrac{(y-\eta_Y)^2}{2\sigma_Y^2}\right\}dy \\ &= e^{i\omega(\eta_X-\frac{\sigma_X}{\sigma_Y}r\eta_Y)-\frac{1}{2}(1-r^2)\sigma_X^2\omega^2}\cdot e^{i\omega(1+\frac{\sigma_X}{\sigma_Y}r)\eta_Y-\frac{1}{2}\sigma_Y^2\omega^2(1+\frac{\sigma_X}{\sigma_Y}r)^2} \\ &= e^{i\omega(\eta_X+\eta_Y)-\frac{1}{2}\omega^2(\sigma_X^2+2\sigma_X\sigma_Yr+\sigma_Y^2)} \end{align} $

So, $ \mathbf{X}+\mathbf{Y} $ is a Gaussian random variable with mean $ (\eta_X+\eta_Y) $ and variance $ (\sigma_X^2+2\sigma_X\sigma_Yr+\sigma_Y^2) $

Solution 2:

Let $ Z=X+Y $, where $ X\sim(\mu_1,\sigma_1^2) $ and $ Y\sim(\mu_2,\sigma_2^2) $.

The characteristic function

$ E(e^Z)=E(e^{X+Y})=e^{i(\mu_1+\mu_2)t-\frac{1}{2}\left(\sigma_1^2+\sigma_2^2+2\text{cov}(x,y)\right)t^2} $

according to the property of jointly distributed Gaussian random variable.

$ \therefore Z\sim\left(\mu_1+\mu_2, sigma_1^2+\sigma_2^2+2\text{cov}(x,y)\right) $ according to uniqueness of characteristic function.

Comments:

Related Problems:

$ \mathbf{X} $ and $ \mathbf{Y} $ are two exponential random variables with different means, $ \lambda_X $ and $ \lambda_Y $, respectively. Show that $ \text{Min}(X,Y) $ is also an exponential random variable.

"Communication, Networks, Signal, and Image Processing" (CS)- Question 1, August 2011

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