# Thread: P.D.F of X-Y when X Y are independent

1. ## P.D.F of X-Y when X Y are independent

In general, how do you solve for X-Y?

e.g X and Y have p.d.f $\displaystyle f(x)=3x^2$ for 0<x<1
I know how to solve for X+Y
but for X-Y, C.D.F is effectively the region above a line Y=X-Z bounded by the square 0 to 1.

I think I am stuck with a non standard multiple integral from top left to right bottom.

any idea?

2. ## Re: P.D.F of X-Y when X Y are independent

Hello,

X-Y=X+(-Y)
you can easily get the pdf of -Y, with the region that is different. You can then use the convolution or the transform of variables method.

3. ## Re: P.D.F of X-Y when X Y are independent

Note that $\displaystyle -1\le a\le 1$

where $\displaystyle P(Z\le a)=P(X-Y\le a)$, which has two cases.

If $\displaystyle -1\le a\le 0$ use $\displaystyle \int_{-a}^1\int_0^{a+y}9x^2y^2dxdy$

and if $\displaystyle 0< a\le 1$ use $\displaystyle 1-\int_{a}^1\int_0^{x-a}9x^2y^2dydx$

It's 2am, I hope my bounds are correct, but the idea is certainly right.