# Conditional Distribution Functions

1. Jul 10, 2013

### Brandon1994

1. The problem statement, all variables and given/known data
If X1 is uniform on [0,1], and, conditional on X1, X2, is uniform on [0,X1], find the joint and marginal distributions of X1 and X2

2. Relevant equations

conditional joint distribution

3. The attempt at a solution

f(x1|x2) = 1/x1 (for 0<x2<x1)
f(x1) = 1 ( for 0<x1<1)

then
F(x1,x2) = Integrate (1/x1) from {x2,0,x2}{x1,0,x1}
I get an ln(0) when i try to integrate however

for the marginal distribution of x2, i get X2~[0,X1] //i am not sure if thats the answer they are looking for, if i try to write an explicit density function for X2 i get that the density is infinity, again due to the ln (0) term.

Thanks

2. Jul 10, 2013

### Ray Vickson

You have it exactly backwards: you are given $f(x_2|x_1)$, not $f(x_1|x_2).$

Also: what does the notation "Integrate(1/x1) from {x2,0,x2}{x1,0,x1}" mean? I have never seen that before.

Anyway, the first thing to do is to answer the question "what is the joint distribution of $(X_1,X_2)?$ You have not done that.

3. Jul 10, 2013

### Brandon1994

wouldn't the joint density be:
f(X1,X2) = 1/X1

and I was saying to find the marginal density f(X2) you would integrate the above expression

4. Jul 10, 2013

### Ray Vickson

Yes, f(x1,x2) = 1/x1, bit only on an appropriate region in (x1,x2)-space. You need to spell out the details.

Of course you need to do an integral to get the marginal distibution of x2, but that was not the point. I asked what you meant by the weird notation "Integrate(1/x1) from {x2,0,x2}{x1,0,x1}". This looks like something you invented that nobody else knows about, but surely you must have in mind some meaning for it. I am asking you to explain that meaning---in detail, not just saying that you need to integrate. That is: what is the integration variable, and what are the limits of integration?Even better, what is the final answer you get?