# transformation function of random variables

• Sep 29th 2009, 06:14 AM
oscaralive
transformation function of random variables
If X and Y are exponentially distributed with the same parameter and they are also independent:

To compute the density function of :

Case 1) Z = X-Y with the constraint that X is strictly higher than Y, (X>Y)

Case 2) Z = X+Y without any constraint

If I want to do it by computing the cdf first:

Can someone explain which should be the limits in the following integral in both cases:

$\displaystyle$
$\displaystyle F_Z(z) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}f_X(x)f_Y(y) dx dy$

• Sep 29th 2009, 05:44 PM
matheagle
Case 2) Z = X+Y without any constraint

is easy, you're adding two independent gammas, use the MGF technique here

Case 1) Z = X-Y with the constraint that X is strictly higher than Y
Here you can draw the region.
I would get the joint density and just integrate the infinite triangle in the first quadrant
where X>Y
• Sep 30th 2009, 01:45 AM
oscaralive
First of all thank you!

For the first case is OK. It is just the convolution, however in the second case I do not reach the rigth solution, I get lost with the integral limits in getting the pdf of Z= X-Y...
• Sep 30th 2009, 11:02 AM
matheagle
$\displaystyle F_Z(a)=P(Z\le a)=P(X-Y\le a)$

NOW draw this region.
It's in the first quadrant with the boundary x-y=a or y=x-a.
You have the region that has vertex (a,0) in the first quadrant.

So it's the region between x-y=a and x-y=0 (that's y=x)

If that's true, dxdy is preferable

$\displaystyle {1\over \beta^2}\int_0^{\infty}\int_y^{y+a}e^{-(x+y)/\beta}dxdy$
• Oct 1st 2009, 02:05 AM
oscaralive
Thank you very much. I finally got the idea!!