Difference between revisions of "ECE600 F13 Functions of Two Random Variables mhossain" - Rhea

Revision as of 00:06, 4 November 2013

Random Variables and Signals

Topic 13: Functions 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 $\qquad$

@@ Line 48: / Line 48: @@
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-So if X and Y are independent, and Z = X + Y, we can find f<math>_Z</math> by convolving f<math>_X,/math> and f<math>_Y</math>.
+So if X and Y are independent, and Z = X + Y, we can find f<math>_Z</math> by convolving f<math>_X</math> and f<math>_Y</math>.
 '''Example''' <math>\qquad</math>