# Jacobian Transformation p.d.f

Suppose $X$ and $Y$ are continuous random variables with joint p.d.f.

$$f(x,y) = e^{-y},\,\, 0<x<y <\infty$$

(a) Find the joint p.d.f. of $U=X+Y$ and $V=X$. Be sure to specify the support of $(U,V)$.

(b) Find the marginal p.d.f. of $U$ and the marginal p.d.f. of $V$. Be sure to specify their support.

I can't figure out what I am doing wrong with this question. So far, I have gotten that the support for $U$ and $V$ is $0<v<u<\infty$, the Jacobean matrix has determinant $-1$ and that the joint p.d.f for part a) is $e^{v-u}$ but this p.d.f doesn't make sense when I try to find the marginals. Could someone help guide me in the right direction?

• To start, maybe you or @A.D should repair the TeX code for your PDF. When something is botched in an edit, it is difficult to discern original intent. Just a guess, but check the limits on integrals, or tell us what limits you used. – BruceET Jun 25 '15 at 17:51
• The support of (U,V) is not the set of (u,v) such that u>v>0 since U>2V almost surely. – Did Jun 28 '15 at 20:10

First I checked if it is indeed a density $$\int_0^\infty\int_0^ye^{-y}dxdy=1$$

and $X,Y$ are not independent.So I proceeded as you already did

$(1)U=g_1(X,Y)=X+Y$ and $(2)V=g_2(X,Y)=X$ then $J=\begin{bmatrix}\frac{\partial u}{\partial x}&&\frac{\partial u}{\partial y}\\\frac{\partial v}{\partial x}&&\frac{\partial v}{\partial y}\end{bmatrix}=-1$ hence $|J|^{-1}=1$

You know that $Y=U-V$ from $(1)$ and $(2)$ $$f_{U,V}(u,v)=f_{X,Y}(g_1^{-1}(u,v),g_2^{-1}(u,v))|J|=e^{v-u}$$

You know that $$0<x<y\iff0<v<u-v\iff0<2v<u$$ hence $$f_U(u)=\int_0^{u/2} e^{v-u}dv=(e^{-u/2}-e^{-u})\,I_{[0,\infty)}(u)$$ $$f_V(v)=\int_{2v}^\infty e^{v-u}du=e^{-v}I_{[0,\infty)}(v)$$

• The density of V cannot depend on some parameter u. Also, see my comment to the question. – Did Jun 28 '15 at 20:11
• @Did I was trying something here, and to me it seems $u>0$ and $v>0$, it's wrong? – Roland Jun 28 '15 at 20:20
• But now your f_V is not a density. – Did Jun 28 '15 at 20:22
• @Did I think now it's fine. – Roland Jun 28 '15 at 20:51
• Seems perfect to me. +1. – Did Jun 28 '15 at 21:11