Problem: Let $f$ be the pdf of a positive rv and write $g(x,y) = \frac{f(x+y)}{x+y}$, if $x>0,y>0$. Show that $g$ is a density function in the plane.
$g$ is a pdf if $\int_{0}^{\infty}\int_{0}^{\infty}g(x,y)dxdy = 1$, or equivalently if $\int_{0}^{\infty}\int_{0}^{\infty}\frac{f(x+y)}{x+y}dxdy = 1$. Since $f$ is a pdf, we know that $\int_{0}^{\infty}f(a)da = 1$
I don't know how to show this. Please help
The integrand only depends on $x+y$, so it is natural to try to change variables so that one of the independent variables is $x+y$. For instance we can use $u=x+y,v=y$. This corresponds to the transformation matrix $A=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$, whose determinant is $1$. So by the multivariable change of variables formula, you have a double integral of $\frac{f(u)}{u}$ in a certain region of the $(u,v)$ plane. Can you work out what region this is? (Hint: since $x \geq 0$, $u \geq v$...)
