Let $X$ and $Y$ be random variables
with cumulative distribution functions $F\_X(t)$ and $F\_Y(t)$ respectively.


Suppose that 
$\forall t \in \mathbb{R} \ F\_X(t) \geq F\_Y(t)$.
Does it imply that $E(X) \leq E(Y)$?


Note that $X \leq Y$ implies $F\_X(t) \geq F\_Y(t)$ but not vise versa:


1. $X \leq Y$ $\ \Rightarrow \ $ $F\_Y(t) = P(Y \leq t) \leq P(X \leq t) = F\_X(t)$.
2. If $X$ has the standard normal distribution and $Y=-X$, then $F\_X(t) = F\_Y(t)$ but $X$ can be greater than $Y$.



"Generalized" monotonicity of the expected value
Note that $E(X) = \int\_{-\infty}^\infty x f(x) d x = -x (1-F(x))\vert\_{-\infty}^\infty +\int\_{-\infty}^\infty (1-F(x))d x.$


From this you should be able to deduce conditions under which the answer to your question is in the affirmative.


Using monotone convergence for the integrals on $(-\infty, 0)$ and $(0, \infty)$, we obtain:


$$
\int\_{-\infty}^\infty x f(x) dx = \lim\_{R \to \infty} \left( -x (1-F(x))\vert\_{-R}^R +\int\_{-R}^R (1-F(x))d x \right)
$$


By monotony of limits, we obtain the result if a probability density function and the first moment exist.


