I was reading an article and I found this equivalence:
$$ \int_0^\infty\,dy\int_0^\infty dx\,p(x+y)\ =\ \langle X \rangle $$ where $p(x)$ is a p.d.f of a positive random variable $X$ and $\langle X \rangle = \int_0^\infty xp(x)\,dx$.
Is it true? Changing the variables I have only that the l.h.s is equal to $$ \int_0^\infty dy\int_y^\infty p(x)\,dx = \int_0^\infty P(X\geq y)\,dy $$
On
One should be careful to note that this identity holds only when $X$ is a nonnegative random variables, i.e. $\Pr(X\ge0) = 1.$
You have $$ \iint\limits_{x\,\ge\,0 \\ y\,\ge\, 0} p(x+y) \, d(x,y) $$ Let \begin{align} u & = x+y \\ v & = x/(x+y) \\[10pt] \text{So that } x & = uv \\ y & = u(1-v) \\[10pt] d(x,y) & = \left| \frac{\partial(x,y)}{\partial(u,v)} \right| \, d(u,v) = u \, d(u,v) \end{align} As $u$ remains constant while $x$ goes from $0$ to $u,$ then $v$ goes from $0$ to $1$, and we have $$ \iint\limits_{x\,\ge\,0 \\ y\,\ge\, 0} p(x+y) \, d(x,y) = \int_0^\infty \left(\int_0^1 p(u) \cdot u \, dv\right) \, du = \int_0^\infty up(u)\,du. $$
On
Here's a slightly different approach, maybe simpler than my earlier posting. \begin{align} u & = x+y \\ v & = x-y \\[10pt] \text{Therefore } x & = (u+v)/2 \\ y & = (u-v)/2 \\[10pt] d(x,y) & = \left| \frac{\partial(x,y)}{\partial(u,v)} \right| \, d(u,v) = \frac 1 2 \, d(x,y) \end{align} As $x$ goes from $0$ up to $u$ with $u$ fixed, $y$ goes from $u$ to $0$ with $u$ fixed, and $v=x-y$ goes from $-u$ to $u.$ Thus we have $$ \iint\limits_{x\,\ge\,0 \\ y\,\ge\,0} p(x+y) \, d(x,y) = \frac 1 2 \int_0^\infty \left( \int_{-u}^u p(u) \, dv \right) \, du = \frac 1 2 \int_0^\infty 2u p(u) \, du. $$
Yes, this is true, and your transformation is already a first step in showing it (and a useful result in its own right). Now integrate $1\cdot P$ by parts, or alternatively integrate $x\cdot p$ by parts (using $P-1$ as the antiderivative).