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Random Variables and Signals

Topic 13: Functions of Two Random Variables


One Function of Two Random Variables

Given random variables X and Y and a function g:R$ ^2 $→R, let Z = g(X,Y). What is f$ _Z $(z)?

We assume that Z is a valid random variable, so that ∀z ∈ R, there is a D$ _z $ ∈ b(R$ ^2 $) such that

$ \{Z\leq z\} = \{(X,Y)\in D_z\} $

i.e.

$ D_z = \{(x,y)\in\mathbb R^2:g(x,y)\leq z\} $

Then,

$ F_Z(z) = P((X,Y)\in D_z)=\int\int_{D_z}f_{XY}(x,y)dxdy $

and we can find f$ _Z $ from this.

Example $ \qquad $ g(x,y) = x + y. So Z = X + Y. Here,

$ D_z = \{(x,y)\in\mathbf R^2:x+y\leq z\} $

and

$ F_Z(z) = \int_{-\infty}^{\infty}\int_{-\infty}^{z-x}f_{XY}(x,y)dydx $


Fig 1: The shaded region indicates D$ _z $.


If we now assume that X and Y are independent, then

$ \begin{align} F_Z(z) &= \int_{-\infty}^{\infty}\int_{-\infty}^{z-x}f_X(x)f_Y(y)dydx \\ &=\int_{-\infty}^{\infty}f_X(x)F_Y(z-x)dx \end{align} $

Then,

$ \begin{align} f_Z(z) &= \frac{d}{dz}[\int_{-\infty}^{\infty}f_X(x)F_Y(z-x)dx] \\ &=\int_{-\infty}^{\infty}f_X(x)\frac{dF_Y(z-x)}{dz}dx \\ &=\int_{-\infty}^{\infty}f_X(x)f_Y(z-x)dx \\ &=(f_X\ast f_Y)(z) \end{align} $

So if X and Y are independent, and Z = X + Y, we can find f$ _Z $ by convolving f$ _X $ and f$ _Y $.

Example $ \qquad $ X and Y are independent exponential random variables with means $ \mu_X $ = $ \mu_Y $ = $ \mu $. So,

$ \begin{align} f_Z(z)&=\int_{-\infty}^{\infty}\frac{1}{\mu}e^{-\frac{x}{\mu}}u(x)\frac{1}{\mu}e^{-\frac{(z-x}{\mu}}u(z-x)dx \\ &=\int_0^z\frac{1}{\mu^2}e^{-\frac{z}{\mu}}dx\;\;\mbox{for}\;z=0 \\ &=\frac{z}{\mu^2}e^{-\frac{z}{\mu}}u(z) \end{align} $

So the sum of two independent exponential random variables is not an exponential random variable.


Two Functions of Two Random Variables

Given two random variables X and Y, cobsider random variables Z and W defined as follows:

$ \begin{align} Z &= g(X,Y) \\ W &= h(X,Y) \end{align} $

where g:R$ ^2 $→R and h:R$ ^2 $→R.

For example, if we have a linear transformation, then

$ \begin{bmatrix}Z\\W\end{bmatrix} =A \begin{bmatrix}X\\Y\end{bmatrix} $
where A is a 2x2 matrix. In this case,
$ \begin{align} g(x,y) &= a_{11}x+a_{12}y \\ h(x,y) &= a_{21}x+a_{22}y \end{align} $

where

$ A=\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} $

How do we find f$ _{ZW} $?

If we could find D$ _{z,w} $ = {(x,y) ∈ R$ ^2 $: g(x,y) ≤ z and h(x,y) ≤ w} for every (z,w) ∈ R$ ^2 $, then we could compute

$ \begin{align} F_{ZW}(z,w) &= p(Z\leq z, W\leq w) \\ &=P((X,Y)\in D_{z,w}) \\ &=\int\int_{D_{z,w}}f_{XY}(x,y)dxdy \end{align} $

It is often quite difficult to find D$ _{z,w} $, so we use a formula for f$ _{ZW} $ instead.


Formula for Joint pdf f$ ZW $

Assume that the functions g and h satisfy

  • z = g(x,y) and w = h(x,y) can be solved simultaneously for unique x and y. We will write x = g$ ^{-1} $(z,w) and y = h$ ^{-1} $(z,w) where g$ ^{-1} $:R$ ^2 $→R and h$ ^{-1} $:R$ ^2 $→R.
For the linear transformation example, this means we assume A$ ^{-1} $ exists (i.e. it is invertible). In this case
$ \begin{align} g^{-1}(z,w) &= b_{11}z+b_{12}w \\ h^{-1}(z,w) &= b_{21}z+b_{22}w \end{align} $
where
$ A^{-1}=\begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix} $
  • The partial derivatives
$ \frac{\partial x}{\partial z},\;\frac{\partial x}{\partial w},\;\frac{\partial y}{\partial z},\;\frac{\partial y}{\partial w},\;\frac{\partial z}{\partial x},\;\frac{\partial z}{\partial y},\;\frac{\partial w}{\partial x},\;\frac{\partial w}{\partial y} $
exist.

Then it can be shown that

$ f_{ZW}(z,w) = \frac{f_{XY}(g^{-1}(z,w),h^{-1}(z,w))}{\left|\frac{\partial\left(z,w\right)}{\partial\left(x,y\right)}\right|} $

where

$ \begin{align} \left|\frac{\partial(z,w)}{\partial(x,y)}\right| &\equiv \left|\begin{bmatrix} \frac{\partial z}{\partial x} & \frac{\partial z}{\partial y} \\ \frac{\partial w}{\partial x} & \frac{\partial w}{\partial y} \end{bmatrix}\right| \\ &= \left|\frac{\partial z}{\partial x}\frac{\partial w}{\partial y}-\frac{\partial z}{\partial y}\frac{\partial w}{\partial x} \right| \end{align} $
$ z=g(x,y)\qquad x=h(x,y) $

Note that we take the absolute value of the determinant. This is called the Jacobian of the transformation. For more on the Jacobian, see here.

The proof for the above formula is in Papoulis.


Example X and Y are iid (independent and identically distributed) Gaussian random variables with $ \mu_X=\mu_Y=\mu = 0 $, $ \sigma^2 $$ _X $=$ \sigma^2 $$ _Y $=$ \sigma^2 $ and r = 0. Then, br/>

$ f_{XY}(x,y)=\frac{1}{2\pi\sigma^2}e^{-\frac{x^2+y^2}{2\sigma^2}} $

Let

$ R = \sqrt{X^2+Y^2}\qquad \Theta=\tan^{-1}\left(\frac{Y}{X}\right) $

So,

$ \begin{align} g(x,y) &= \sqrt{x^2+y^2} \\ h(x,y) &= \tan^{-1}\left(\frac{y}{x}\right) \\ g^{-1}(r,\theta)&=r\cos\theta \\ h^{-1}(r,\theta)&=r\sin\theta \end{align} $

Now use

$ f_{R\Theta}(r,\theta)=f_{XY}(g^{-1}(r,\theta),h^{-1}(r,\theta))\left|\frac{\partial(x,y)}{\partial(r,\theta)}\right| $

where

$ \begin{align} \left|\frac{\partial(x,y)}{\partial(r,\theta)}\right| &= \left|\begin{bmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial\theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial\theta} \end{bmatrix}\right| \\ \\ &=\left|\mbox{det}\begin{bmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{bmatrix}\right| \\ \\ &=|r\cos^2\theta+r\sin^2\theta| \\ &=r \end{align} $


$ \begin{align} f_{R\Theta}(r,\theta) &=\frac{1}{2\pi\sigma^2}e^{-\frac{r^2\cos^2\theta+r^2\sin^2\theta}{2\sigma^2}}r\cdot u(r) \\ \\ &=\frac{r}{2\pi\sigma^2}e^{-\frac{r^2}{2\sigma^2}}u(r)\qquad -\pi\leq\theta\leq\pi \end{align} $

We can then find the pdf of R from

$ \begin{align} f_R(r) &= \int_{-\infty}^{\infty}f_{R,\Theta}(r,\theta)d\theta \\ &= \frac{r}{\sigma^2}e^{-\frac{r^2}{2\sigma^2}}u(r) \end{align} $

This is the Rayleigh pdf.


Auxiliary Variables

If Z=g(X,Y), we can define another random variable W, find f$ _{ZW} $, and then integrate to get f$ _Z $. We call W an "auxiliary" or "dummy" variable. We should select h(x,y) so that the solution for f$ _Z $ is as simple as possible when we let W=h(X,Y). For the polar coordinate problem in the previous example, h(x,y) = tan$ ^{-1} $(y/x) works well for finding f$ _R $. Often, W = X works well since

$ \frac{\delta w}{\delta x}=1 \qquad \frac{\delta y}{\delta w}=0 $

References


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