There are two random variables $X$ and $Y$ PDF's $f(x)$ and $g(y)$. Both distributions functions have the same support [a,b], where $0<a<b<\infty$. Realizations of the variables are observed "ex-post." "Ex-iterim" the following holds: $E[X] = E[Y]$. I wish to find "ex-ante" $\Pr[X-Y<0]$ .
My attempt: $\Pr[X<Y] = \int_a^b F(y)g(y)dy$. Is it true ex-ante?
The respective cdf of $X$ and $Y$ are $$ F(x) = \int_{-\infty}^x f(t)\,dt \\G(y) = \int_{-\infty}^y g(u)\,du $$ For $x+y\leq z$ we need $y\leq z-x$. Therefore, the cdf of $Z\equiv X+Y$ is $$H(z) = \int_{x=-\infty}^x f(t) \int_{y=-\infty}^{z-x}g(u)\,du \,dt $$ The expectation value of $Z$ is $$ E[Z]\,\, = \int_{t=-\infty}^x f(t) \int_{u=-\infty}^{z-t}g(u) (t+u) \,du \,dt $$ The trick to show that $E[Z] = E[X]+ E[Y]$ is to note that you can reverse the order to say that also $$ E[Z]\,\, = \int_{u=-\infty}^x g(u) \int_{t=-\infty}^{z-u}f(t) (t+u) \,dt \,du $$ and a bit of integration by parts gives the result.
However, this trick and the IBP is valid only if $f(x)$ and $g(y)$ meet certain fairly reasonable conditions. In particular, it is valid if $f(x)$ and $g(y)$ each have finite second moments.