1. Originally Posted by mr fantastic
where $f(x) = \frac{1}{\lambda} \, e^{-x/\lambda}$ ....
Yea, but I was trying to convert from (s,t) -> to (x,y).
By right, the answers should tally, but they don't , so I am guessing the original question was wrong.

Shouldn't it be :

$
f_{ST}(s,t) = \frac{\lambda^{-2} s}{(1+t)^2} e^{- \lambda s}
$

2. Originally Posted by chopet
Yea, but I was trying to convert from (s,t) -> to (x,y).
By right, the answers should tally, but they don't , so I am guessing the original question was wrong. Mr F asks: What original question??

I have no idea what question you're trying to do but there's certainly nothing wrong with any of the questions asked in this thread.

Shouldn't it be :

$
f_{ST}(s,t) = \frac{\lambda^{-2} s}{(1+t)^2} e^{- \lambda s}
$
..

3. Originally Posted by mr fantastic
..
I'm trying to do a transformation of variables from (s,t) -> (x,y) without using the fact that f(x) and f(y) are exponential distributions.

4. Originally Posted by chopet
I'm trying to do a transformation of variables from (s,t) -> (x,y) without using the fact that f(x) and f(y) are exponential distributions.
S = X + Y .... (1)
T = X/Y .... (2)

$f_{S, T} (s,t) = \frac{\lambda^2 s}{(1 + t)^2} \, e^{-\lambda \, s}$.

From (1) and (2):

X = ST/(1 + T)

Y = S/(1 + T)

$|J| = \frac{s}{(1 + t)^2}$.

Therefore $f_{X,Y} (x,y) = f_{S, T} (s,t)/|J| = \lambda^2 e^{-\lambda s} = \lambda^2 e^{-\lambda (x + y)}$.

This distribution is obviously equivalent to $f_{X,Y} (x,y) = \frac{1}{\lambda^2} e^{-(x + y)/\lambda}$ and there's no problem that I can see.

Edit: I see that you get it now. Excellent. Now the trick is to derive the given joint distribution for S and T from the distributions for X and Y lol!

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