# Statistics Problem: finding a PDF using the CDF technique

1. Oct 30, 2012

### _Steve_

Hey guys, I'm stuck on a question in my homework assignment and I was wondering if you could push me in the right direction? So here's the question:

X and Y are continuous random variables with joint pdf f(x,y)= 4xy (0<x<1, 0<y<1, and otherwise 0). Find the pdf of T=X+Y using the CDF technique.

So this is how I started off, I first let G(t) be the CDF of T, then I look at three different cases:
t<=0: G(t) = P[X+Y<=t] = 0
t>=2: G(t) = P[X+Y<=t] = 1
0<t<2: G(t) = P[X+Y<=t] = ?
So here I'm a little confused, I'm trying to figure out the limits of the double integral I'm supposed to take, I think I might have to take one integral from 0<t<1, then another from 1<=t<2, but then I'm stuck with two "functions" (one from (0,1), one from [1,2)) for my g(t). Are there any possible limits for this double integral that would save me from having to separate 0<t<2 into two double integrals?

2. Oct 30, 2012

### Ray Vickson

You are correct in that you do need different integrals for t > 1 and t < 1. If you use the cdf method this is unavoidable. (Other, completely different, methods are easier, but you are told not to use them.)

RGV

3. Oct 30, 2012

### _Steve_

Oh, okay, so how do I get a g(t) with only one function from 0<t<2 instead of having:
g(t)={f(t), t<1
h(t), t>1)
?

Does it make sense to write:
G(t)={0, t<0
F(t), t<1
H(t), t>1
1, t>2}
as:
G(t)={0, t<0
F(t)+H(t), 0<t<2
1, t>2}
?

EDIT: Nevermind, I think I was doing the other part of the question wrong.. thanks for your help!

Last edited: Oct 30, 2012