# Computing joint density [duplicate]

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?