Question:
Let $X$ and $Y$ be independent random variables, each with probability density function $$f(x) = \lambda e^{−\lambda x} $$for $x > 0$. Let $U = X − Y$ and $V = X + Y$ . Find the joint probability density function $$ f_{U,V} (u, v) $$ of $(U,V)$, describing carefully the region of the $(u, v)$-plane in which $$f_{U,V} (u, v)$$ is non-zero.
Find also the marginal probability density functions $$f_U(u)$$ and $$ f_V(v)$$ of U and V , and the conditional probability density function $$f_{U|V} (u|v)$$.
I am having trouble finding the marginal distribution.
My work so far: Jacobian = 1/2 $$f_{U,V}(u,v) =\frac{\lambda^2}2 exp(-\lambda v)$$ Now I thought then $U$ would go from $-\infty$ to $\infty$ since it can take any of these values? Then $$f_U(u) =\int_{0}^{\infty} \frac{\lambda^2}2 exp( {-\lambda v}) \, dv$$ $$f_U(u) =\int_{-\infty}^{\infty} \frac{\lambda^2}2 exp( {-\lambda v}) \, du$$
This is clearly wrong since the second integral doesn't make sense. I think it's to do with my limits and when $f(u,v)$ satisfies the expression I gave it and is zero otherwise?
Thanks in advance.
Edited typo.
First, notice that $f_{X,Y}(x,y) = f(x)f(y)$. We also have that
\begin{align*} X &= \frac{U+V}{2}, \\ Y &= \frac{V-U}{2}. \end{align*}
Hence,
$$ f_{U,V}(u,v) = \frac{1}{2} f_{X,Y}\left(\frac{u+v}{2},\frac{v-u}{2}\right), $$
where the factor of $\frac{1}{2}$ comes from the determinant of the Jacobian of the transformation described above.
Given that $X,Y\in[0,\infty)$, what can we say about what possible values $U$ and $V$ can take? This will give you the regions where $f_{U,V}$ is zero, and should follow straightforwardly from the possible values for $X$ and $Y$, and the definitions $U=X-Y$ and $V=X+Y$.
I'll let you take the rest from here.