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This question already has an answer here:

Suppose $X$ and $Y$ are independent and identically distributed uniform r.v on $(0,1)$. Compute the joint density of $U=X+Y$ and $V = \dfrac{X}{X+Y}$. How about if $X$ and $Y$ are iid exponential with $\lambda=1$?

Try

Notice that $f_{X,Y}(x,y)=1$ and $X= V(X+Y)= UV $ and so $Y = U -X = U - UV = U(1-V)$. We compute the Jacobian of map $(U,V)$ and obtain

$$ J = 1 \cdot \left( - \frac{X}{(X+Y)^2} \right) - 1 \cdot \left( \frac{Y}{(X+Y)^2} \right) = - \frac{1}{X+Y} = \frac{-1}{U} $$

And since

$$ f_{X,Y}(X,Y) = 1 $$

then

$$ f_{U,V}(u,v) = f_{X,Y} (x,y) \frac{1}{|J|} = u $$

Since $0 \leq UV \leq 1$ and $0 \leq U(1-V) \leq 1$, we have $0 \leq U \leq \frac{1}{V}$ and $1 - V \leq \frac{1}{U}$ and so $1 - \frac{1}{U} \leq V \leq 1$

is this correct?

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marked as duplicate by StubbornAtom, Lord Shark the Unknown, Namaste, Paul Frost, user10354138 Nov 15 '18 at 0:04

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  • $\begingroup$ Although you have followed the method correctly, you have overlooked some technicalities. See this post. The second question has been asked here numerous times. Here is a relevant post. $\endgroup$ – StubbornAtom Nov 14 '18 at 15:41

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