Let X,Y be independent $Exp(\lambda)$ random variables. Let $U=X+Y$ and $V=X-Y$.
Find the joint pdf of U,V then use it to find the marginal pdf for $V$.
My attempt:
Using $f_{U,V}(u,v)=f_{X,Y}(\gamma(u,v))|det(J(u,v))|$, I found the joint pdf
$$f_{U,V}(u,v)=\frac{\lambda^2e^{-\lambda u}}{2}$$
where $u+v\geq 0$ and $u-v\geq 0$
I know it is correct because when I integrate it, it equals $1$.
When finding the marginal pdf, I integrated this wrt $u$, from $v$ to $\infty$, and got
$$f_V(v)=\int^\infty_v\frac{\lambda^2e^{-\lambda u}}{2}du=\frac{\lambda e^{-\lambda v}}{2}$$
However, when I integrate this from $0$ to $\infty$, I get $\frac{1}{2}$ instead of 1. I tried tweaking it so that I have $|v|$ instead of just $v$, then integrated from $-\infty$ to $\infty$, and this gave me $1$, but I don't think this is the right approach. Can someone point out what part I did wrong? Thank you.
Best Answer
$u+v \geq 0$ and $u-v \geq 0$ iff $u \geq |v|$. So $f_{V} (v)=\frac 1 2\int_{|v|}^{\infty} \lambda^{2} e^{-\lambda u}du=\frac 1 2 \lambda e^{-\lambda |v|}$ for $-\infty <v<\infty$.