# Thread: Finding a PDF with Two Variables Changing to One

1. ## Finding a PDF with Two Variables Changing to One

I have been stumped with this problem as well. I have tried several methods, but cannot seem to get the right answer.

I have two variables $X_1$ and $X_2$ which have the joint PDF $f(x_1,x_2) = e^{-x_1}e^{-x_2}$ if $x_1,x_2 > 0$ and $0$ otherwise. Now, we let $W = w_1X_1 + w_2X_2$ with $w_1,w_2 > 0$, and we need to confirm that the PDF is

$f(w) = \frac{1}{w_1-w_2} \left(e^{-\frac{w}{w_1}} - e^{-\frac{w}{w_2}}\right)$ when $w>0$ and $0$ otherwise.

Any help is appreciated! I can only seem to get one of the exponential terms and not both.

2. ## Re: Finding a PDF with Two Variables Changing to One

toying with this I found the easiest way to reproduce this answer is via first finding the distributions for the weighted variables $w_1 X_1$ and $w_2 X_2$ and then just doing the convolution.

It's well known that given two random variables the distribution of the sum is the convolution of the individual distributions.

it's quickly seen that

$w_1 X_1 \sim \dfrac 1 {w_1} f_{X_1}\left(\dfrac{x}{w_1}\right)$

and so the convolution integral we are after is

$\displaystyle{\int_0^x}\dfrac{e^{-t/w_1}}{w_1}~\dfrac{e^{-(x-t)/w_2}}{w_2}~dt$

$f_W(w) = \dfrac{e^{-\frac{w}{w_1}}-e^{-\frac{w}{w_2}}}{w_1-w_2}$