How to prove positivity of $\int_{\Bbb R_+\times\Bbb R_+} x(x-y)P(x,y) dx dy $

Question

I need to show that $\int_{\Bbb R_+\times\Bbb R_+}x(x-y)P(x,y) dx dy \geq0$ where $P(x,y)=P(y,x)$ symmetric (potentially $P(x,y)\geq0$ everywhere).

Intuitively this seems obvious by splitting the integral along the diagonal. But so far I failed to prove it.

Answer 1

The additional hypothesis $P\ge0$ a.e. is necessary.

$$\begin{align}\int_{\Bbb R_+\times\Bbb R_+}x(x-y)P(x,y)dxdy&=\int_{y,h>0}(y+h)hP(y+h,y)dhdy+\int_{x,k>0}x(-k)P(x,x+k)dxdk\\&=\int_{y,h>0}\left((y+h)hP(y+h,y)-yhP(y+h,y)\right)dhdy\\&=\int_{y,h>0}h^2P(y+h,y)dhdy\\&\ge0. \end{align}$$