Show $E(Y)-E(X) = \int_{\mathbb R} P[X<t\le Y] - P[Y< t \le X] dt$

Question

Suppose X and Y are integrable random variables on the measure space $(\Omega,\mathcal F, P)$. Im trying to show that $E(Y)-E(X) = \int_{\mathbb R} P[X<t\le Y] - P[Y< t \le X] dt$

but I got that $2(E(Y)-E(X)) = \int_{\mathbb R} P[X<t\le Y] - P[Y< t \le X] dt$

Here is what I did: $\begin{align} \int_{\mathbb R} P[X<t\le Y] - P[Y< t \le X] dt &= \int_{\mathbb R} \int_{\Omega}1_{[X<t \le Y]}dP\; dt -\int_{\mathbb R}\int_{\Omega} 1_{[Y<t \le X]}dP\; dt \\ &= \int_{\Omega} \int_{X}^Ydt\; dP - \int_{\Omega} \int_{Y}^Xdt\; dP \\ &= \int_{\Omega}Y-X \; dP - \int_{\Omega} X - Y\; dP \\ &= E(Y-X) - E(X-Y) \\ &= E(Y)-E(X)-E(X)+E(Y) \\ &= 2\left(E(Y)-E(X)\right) \end{align} $

Can someone let me know what I'm doing wrong?

Answer 1

In the first integral of the third line, you should write $(Y-X)^+$, since you only integrate when $Y\geq X$. Similarly, then the integrand in the second integral becomes $(X-Y)^+=(Y-X)^-$, and then you have $E(Y-X)^+-E(Y-X)^-=E(Y-X)$.

Answer 2

The second equal sign should be followed by $\int_{\{X<Y\}}\int_X^Y\,dt\,dP -\int_{\{Y<X\}}\int_Y^X\,dt\,dP$---these two expectations should not be over all of the sample space.