Value of a integral under a non-decreasing transformation

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

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 $f(x):=x$.
• That is strange. I tought the converse would not be true. But by the definition of first-order stochastic dominance, we have: "Let F and G be two cdfs in \mathbb{R}. We say that F dominates G in the first-order stochastic sense if $$\int_{-\infty}^\infty f(x) \, dF(x) \geq \int_{-\infty}^\infty f(x) \, dG(x)$$ for all non-decreasing $f: \mathbb{R} \rightarrow \mathbb{R}"$. Then it follows: "Notice that E[F] $\geq$ E[G] if F dominates G in first-order stochastic sense. The converse is not true. – jpugliese Apr 21 '18 at 0:30
• I think the problem in your example is that I can find $f: \mathbb{R} \rightarrow \mathbb{R}$ such that this isn't true. Let $f:=G$. Then we can see that F doesn't first-order stochastic dominate G. I should make myself clearer in my question. – jpugliese Apr 21 '18 at 0:34
• @jpugliese Ah, you want the relation to hold for all $f$, not just one $f$. If the relation is assumed true for all $f$, then take $f(x):=x$ to see that $E(F)\ge E(G)$. – grand_chat Apr 21 '18 at 1:09