# Conditional expectation (w/ transformation)

1. Aug 29, 2006

### island-boy

Any hints on how to solve for E(Y|X) given the ff:

Suppose U and V are independent with exponential distributions
$$f(t) = \lambda \exp^{-\lambda t}, \mbox{ for } t\geq 0$$

Where X = U + V and Y = UV.

I am having difficulty finding f(Y|X)...
Also, solving for f(X,Y), I am also having difficulty transforming U and V to X and Y. I was able to define U and V to X and Y, but the terms are so complicated that its difficult to get the Jacobian.

So maybe, there's no need for transformation?

Last edited: Aug 29, 2006
2. Aug 29, 2006

### island-boy

ETA:
here's what I was able to get so far.

Since U and V are independent,
then f(U,V) = f(U)f(V)
thus
$$f(U,V) = \lambda^{2} \exp^{-\lambda (u+v)}, \mbox{ for } u\geq 0 v\geq 0$$

To solve for E(Y|X), I would need to find f(Y|X)
$$f(Y|X) = \frac{f(X,Y)}{f(X)}$$

to get f(X,Y), I need to transform U and V to X and Y.
thus
$$f(X,Y) = \lambda^{2} \exp^{-\lambda (x)} |J|$$
where J is the jacobian.

my rpoblem is in solving for the Jacobian.

Since X = U+V
and Y =UV

then either
$$U = \frac{2Y}{X+ \sqrt{X^{2} - 4Y}}$$
$$V = \frac{X+\sqrt{X^{2} - 4Y}}{2}$$

or
$$U = \frac{2Y}{X- \sqrt{X^{2} - 4Y}}$$
$$V = \frac{X-\sqrt{X^{2} - 4Y}}{2}$$

problem is, getting the Jacobian of U and V would result in a very complicated and long expresion.

so I was thinking, maybe, I don't need to do the transformation. If not. What should I do?

Thanks