second order stochastic dominance without the same mean

Question

Let $F$ and $G$ be two distributions with the same mean. $F$ is said to second order stochastically dominate (SOSD) $G$ if $$\int u(x)\mathrm dF(x)\ge \int u(x)\mathrm dG(x)\tag{1}$$ for all increasing and concave $u(\cdot)$.

This above definition is equivalent to
$$\int_{-\infty}^x F(t)\mathrm dt\le \int_{-\infty}^xG(t)\mathrm dt,\qquad\forall x\in\mathbb R.\tag{2}$$

I was told that the requirement for $F$ and $G$ to have the same mean is not really necessary. Suppose $F$ and $G$ do not have the same mean. Can we then still have the equivalence between $(1)$ and $(2)$?

N.B. I was able to show $(2)\Rightarrow (1)$ without the same mean condition, but not the other way around.

Answer 1

Let $u(x) = x$ which is increasing and concave. Then the defining condition of SOSD reads

$$\int x\mathrm dF(x)\ge \int x\mathrm dG(x) \implies E_F(X) \geq E_G(X) \tag{1}$$

..which would contradict the case $E_F(X) < E_G(X)$ that would be permissible under a general "different means" postulate. On the other hand, we see that the "same mean" condition can be violated by the defining condition for SOSD itself. What does that tell us?

1) That $E_F(X) \geq E_G(X)$ is a necessary condition for $F$ to SOSD $G$.

2) ...And so that the requirement "$F$ and $G$ have the same mean" wrongly restricts the application of the concept of SOSD.