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:

- $X \leq Y$ $\ \Rightarrow \ $ $F_Y(t) = P(Y \leq t) \leq P(X \leq t) = F_X(t)$.
- If $X$ has the standard normal distribution and $Y=-X$, then $F_X(t) = F_Y(t)$ but $X$ can be greater than $Y$.

anyrandom variables X, Y with the distributions $F_X$ and $F_Y$ such that $E(X) \le E(Y)$, then the statement must be true forallrandom variables with these distributions. And James Martin's comment shows how to construct such $X, Y$. $\endgroup$