# Transform f(U,V) to f(X,Y)

1. Sep 1, 2006

### island-boy

given
$$f(U,V) = \lambda^{2} e^{-(u+v)\lambda}$$

How do I get:
f(X,Y)
where
X = U+V
Y =UV

all I'm able to get is
$$f(X,Y) = \lambda^{2} \e^{-(x)\lambda} |J|$$

where J is the Jacobian.
But the Jacobian is too complicated since I was able to solve that:
$$U = \frac{X + \sqrt{X^{2} - 4Y}}{2}$$
and
$$V = \frac{2Y}{X + \sqrt{X^{2} - 4Y}}$$

2. Sep 1, 2006

### AKG

Why do you need the Jacobian? You're not finding df, or anything like that.

3. Sep 2, 2006

### island-boy

I neglect to mention above that f(U,V) and f(X,Y) are density functions

that is
$$f(U,V) = \lambda^{2} e^{-(u+v)\lambda}$$
for
$$u \geq 0, v\geq 0$$

isn't the Jacobian needed when you are transforming variables for a distribution?