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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 $

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