Let $f$ be the pdf of a positive random variable and write
$$g(x, y) = \frac{f(x + y)}{x + y} , \text{ if } x, y > 0.$$
Show that $g$ is the density function in the plane.
Clearly, $g(x, y) \geq 0$, but I need help evaluating the integral
$$\int_{0}^{\infty}\int_{0}^{\infty}\frac{f(x + y)}{x + y} \mathop{dx dy}. $$
One substitution that looks pretty clear is $u = x + y$. Then, we get the $f(u)$ term, and I know $\int_{0}^{\infty} f(u) \mathop{du} = 1$, but I don't know when I would utilize that.
Does anyone have any suggestions?
Best Answer
You're in the right direction, but recall that change of variables in dimension 2 requires one more variable $v = v(x,y)$ as well as $u = x+y$. One of the simplest choice is $v=y$. Then, $$du = dx+dy,$$ $$dv = dy,$$ and hence $$dudv = dxdy. $$ Also, the support of the integral is mapped to $x=u-v>0, y=v>0$. Therefore, the integral has value $$ \int_0^\infty\int_0^\infty \frac{f(x+y)}{x+y}dxdy =\iint_{0<v<u} \frac{f(u)}{u}dudv = \int_0^\infty\left(\int_0^u \frac{f(u)}{u}dv\right)du = \int_0^\infty f(u)du =1. $$ $\textbf{Note:}$ A jacobian matrix is involved in the calculation. What $du = dx+dy, ,dv=dy $ means is exactly $$\left( {\begin{array}{c} du \\ dv\\ \end{array} } \right) = \left( {\begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} } \right)\left( {\begin{array}{c} dx \\ dy \\ \end{array} } \right), $$ and $\frac{dudv}{dxdy} = 1$ is a jacobian determinant.