Random variables and distribution functions relation Suppose I have two random variables $X(\omega) \leq Y(\omega)$ for any $\omega$. Does it mean the probability distribution functions relation $F_x(x) \leq F_y(x)$ still the same for any $x$?
Well, I found out it is not true, but I've failed to get a strict counterexample.
 A: Observe that the inequality $X(\omega) \le Y(\omega)$ for all $\omega$ implies a relationship among events.  In particular, letting $x$ be any real number, suppose the value of $Y$ at $\omega$ does not exceed $x$:
$$\omega\in Y^{-1}(x) = \{\omega\,|\,Y(\omega)\le x\}$$
The inequalities $X(\omega) \le Y(\omega) \le x$ imply $\omega\in X^{-1}(x) =  \{\omega\,|\,X(\omega)\le x\}.$ By definition of $\subset$, this has demonstrated
$$Y^{-1}(x) \subset X^{-1}(x),$$
because every element in the left hand set lies in the right hand set.
The axioms of probability imply the probability of a subset of an event cannot be greater than the probability of the event itself, so
$$F_Y(x) = \Pr(Y^{-1}(x)) \le \Pr(X^{-1}(x)) = F_X(x).$$
The equalities are the very definitions of the distribution functions $F_Y$ and $F_X$.
This chain of implications is reversible.  We may conclude

$X(\omega) \le Y(\omega)$ for all $\omega$ is equivalent to $F_Y(x) \le F_X(x)$ for all real numbers $x$.

A: If you mean the cumulative distribution function, then I believe the opposite must be true.  For any given value, there must be fewer instances (or less cumulative probability) for y to take on that value than for x.  The cumulative probability of x must max out (reach 1) before that happens for y.
