# Math Help - Joint density function

1. ## Joint density function

Let X and Y be independent, iid uniform random variables on [0,1]. Compute the joint density of:
a) U = X + Y, V = X/Y
b) U = X, V = X/Y
c) U = X + Y, V = X/(X+Y)

I know how to approach this problem, but I am having trouble with some of the details. I rearranged U and V in part (a) to get X=UV/(V+1) and Y=U/(V+1). I know you are supposed to integrate the function and somehow use the jacobian determinant, which I calculated to be (-X/Y^2) - (1/Y). However I am not sure how to actually calculate the joint distribution function.

2. I'm interested in your problem. Please, give me a simple example as to understand the details. As a first step, I would infere the support for U and V. For a, 0<=U<=1 and 0<Z<Inf. Then, I would construct the marginal for U and V. At last, construct f(u,v). But, I don't exactly know how.

Do you have any web page were the issue is discussed?

Thank you.

Federico.

3. Originally Posted by almono

Let X and Y be independent, iid uniform random variables on [0,1]. Compute the joint density of:
a) U = X + Y, V = X/Y
[snip]
$X = UV/(V+1) \text{ and } Y = U/(V+1)$

so the Jacobian matrix is

$J =
\begin{pmatrix}
V/V+1 & U/(V+1)^2\\
1/(V+1) & -U/(V+1)^2
\end{pmatrix}$

and $|\det(J)| = \frac{U}{(V+1)^2}$

So by the change-of-variables formula, the joint pdf of U and V is
$
f(U,V) =
\begin{cases}
\frac{U}{(V+1)^2} &\text{ if } 0 \leq U/(V+1) \leq 1 \text{ and } 0 \leq UV/(V+1) \leq 1\\
0 &\text{ otherwise}
\end{cases}$

The only part that is a little tricky is sketching the region where the pdf is non-zero, and I'm not good at posting images, but it's the region of the first quadrant in the U-V plane bounded by the V axis, the U axis, the line $V = U-1$ and the curve $V = 1/(U-1)$. The region stretches off to infinity in the direction of positive V.

jw