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If $U=X+Y$ and $V=\frac {X} {X+Y}$ I have my Jacobian: $\dfrac {\partial U} {\partial X}=1,\dfrac {\partial U} {\partial Y}=1 ,\dfrac {\partial V} {\partial X}=\dfrac{Y} {(X+Y)^2}, \dfrac {\partial V} {\partial Y}=\dfrac{-X} {(X+Y)^2}$ And Then I solve the determinant and get $J=-\dfrac{X+Y} {(X+Y)^2}$ Solving in terms of $U,V$ I get $-\dfrac{1} {U}$. Thus $\left| J^{-1}\right| =U$. I also have $Y=U(1-V)$ and $X=UV$. So then I know I have to integrate according to the new intervals $\int \int \left| J^{-1}\right| f\left( u,v\right)dVdU $ Which gives me $\int \int U dUdV $, I think. What I'm having trouble resolving are the intervals, could someone give me a hand, please?

Did
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Jabernet
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    The general approach is to write $(x,y)$ in terms of $(u,v)$ and to deduce the $(u,v)$-domain from these formulas. Here, $x=uv$ and $y=u-uv=u(1-v)$ and the $(x,y)$-domain is $$0<x<1\qquad0<y<1,$$ hence the $(u,v)$-domain is $$0<uv<1\qquad0<u(1-v)<1.$$ Since $u=uv+u(1-v)$, this can be rewritten as $$0<u<2,$$ and $$\max(0,1-1/u)<v<\min(1,1/u).$$ Finally, the $v$-inequality is clearer if one splits it into some inequalities valid for $0<u<1$, then $$0<v<1,$$ and for $1<u<2$, then $$1-\frac1u<v<\frac1u.$$ – Did Nov 21 '15 at 11:44
  • https://math.stackexchange.com/questions/541322/cdf-of-x-xy-where-x-and-y-are-i-i-d-uniform – StubbornAtom Jan 09 '20 at 13:39

1 Answers1

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If all you want is the joint density function, you don't need to integrate: you have $f_{U,V}(u,v)=u$, which I think is right.

You do still need the ranges to complete the function definition though.

It's clear that with $U=X+Y,$ we have $0\lt U\lt 2$. We need to split this, however, as follows:

Case $0\lt U\leq 1$:

It's easy to see that we need $0\lt V\lt U$ here ($V$ is minimised when $X=0$ and maximised when $X=U$).

Case $1\lt U\lt 2$:

Here, $X$ (and also $Y$) must be at least $U-1$. So $V$ is minimised when $X=U-1$ and maximised when $X=1$. This gives us $\dfrac{U-1}{U}\lt V\lt \dfrac{1}{U}$.

Mick A
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