If $F_X(z)>F_Y(z)$ for all $z$, then $P[X0$. If $F_X(z)>F_Y(z)$ for all $z$, then $P[X<Y]>0$.
$$P[X\leq{}z] > P[Y\leq{}z]\Rightarrow{} P[X<z] > P[Y<z] \rightarrow{} (1)$$ 
By (1)
$$P[X<z] > P[Y<z]$$
$$P[X-Y<z] > P[Y-Y<z]$$
$$P[X-Y<{}0] > P[0<0]$$
$$P[X-Y<0] > 0$$
$$P[X<Y] > 0$$
I have two question
Is correct say that $P[0<0]=0$??? and is correct (1)??
Thanks for your help :D
 A: I think it might be of some value for you to look up the concept of stochastic dominance. http://en.wikipedia.org/wiki/Stochastic_dominance
Try and picture the two Cumulative Distribution Functions simultaneously. To say that one CDF stochastically dominates another is to say that
In terms of the cumulative distribution functions of RVs, Y dominating X means that $F_Y(z) \le F_X(z) \forall z,$ with strict inequality at some $z$. This might seem backward at first, but think about what the CDF evaluated at some $z$ is saying. Specifically:
$F_Y(z)=P(Y<z)$
If that probability is smaller than
$F_X(z)=P(X<z)$
then at some point we must have $P(X<Y)>0$
Try to draw two CDFs on paper with one of them "always at least as high" as the other one. The $z$ here represents where you can draw a vertical line through both CDFs simultaneously. One of those cumulative curves will have more area to the left of your line at $z$ than the other one.
Think about what that means for the underlying distributions of $X$ and $Y$.
Hope that helps.
