Does anyone know how to prove that if
$$\int_{-\infty}^\infty f(x) \, dF(x) \geq \int_{-\infty}^\infty f(x) \, dG(x) $$
$\forall f: \mathbb{R} \rightarrow \mathbb{R}$ non-decreasing
Then
$$E(F) = \int_{-\infty}^\infty x \, dF(x) \geq \int_{-\infty}^\infty x \, dG(x) =E(G)$$
For two cdfs $F$ and $G$? And does the converse hold?
Thanks
The claim isn't true.
For a counterexample, let $F$ be the cdf of a random variable with uniform distribution over $[0,1]$. Let $G$ be the cdf of a random variable uniform over $[1,2]$. Pick $f:=F$. Then clearly $\int fdF >0$ while $\int fdG=0$. But $E(F)<E(G)$.
The converse assertion is also false; just swap $F$ and $G$ in the above counterexample.
EDIT: If the assertion $ \int fdF\ge \int fdG $ holds for all nondecreasing $f$, then the assertion $E(F)\ge E(G)$ is true: just take $f(x):=x$.