ECE600 F13 Functions of Two Random Variables mhossain - Rhea

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\mathbb 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}\left[\int_{-\infty}^{\infty}f_X(x)F_Y(z-x)dx\right] \\ &=\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} =\mathbf 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

\mathbf 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

\mathbf 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|\mbox{det}\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 w=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,

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|\mbox{det}\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

M. Comer. ECE 600. Class Lecture. Random Variables and Signals. Faculty of Electrical Engineering, Purdue University. Fall 2013.

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ECE600 F13 Functions of Two Random Variables mhossain - Rhea

Contents

One Function of Two Random Variables

Two Functions of Two Random Variables

Formula for Joint pdf f $_{ZW}$

Auxiliary Variables

References

Questions and comments

Alumni Liaison

ECE600 F13 Functions of Two Random Variables mhossain - Rhea

Contents

One Function of Two Random Variables

Two Functions of Two Random Variables

Formula for Joint pdf f$ _{ZW} $

Auxiliary Variables

References

Questions and comments

Alumni Liaison

Formula for Joint pdf f $_{ZW}$