Suppose $X$ and $Y$ are jointly distributed real-valued random variables and for all outcomes $\omega_1$, $\omega_2$, we have $$ X(\omega_1)\le X(\omega_2)\quad\Longrightarrow\quad Y(\omega_1)\le Y(\omega_2). $$
Edit: As Louigi Addario-Berry's answer below shows, it may be better to consider the following variation: $$ X(\omega_1)< X(\omega_2)\quad\Longrightarrow\quad Y(\omega_1)\le Y(\omega_2). $$
Does this property have a name?
This is along the lines of Tom's answer. $X$ induces a partial order on $\Omega$. In fact, it induces a total order on a partition of $\Omega$ into sets $X^{-1}(x)$, $x \in \mathbb{R}$); simply say $X^{-1}(x) < X^{-1}(y)$ if $x < y$.
By your property, there is then some non-decreasing function $y:\mathbb{R} \to \mathbb{R}$ such that for $x \in \mathbb{R}$, if $\omega_1, \omega_2 \in X^{-1}(x)$ then $Y(\omega_1)=Y(\omega_2)=y(x)$.
But then we can write $Y(\omega)=y(X(\omega))$. In other words, $Y$ is just a non-decreasing (measurable) function of $X$.
$$ \Omega = \{ (0,0), (0,1), (1,0), (1,1)\}. $$ $$ X(i,j) = i $$ $$ Y(i,j) = 10i + j $$ If $X=0$ then $Y$ may be either 0 or 1; if $X=1$ then $Y$ may be either 10 or 11. A larger $X$ value entails a larger $Y$ value. But $X$ does not determine $Y$.
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Commented
Nov 9, 2010 at 21:29
Suppose that $\Omega$ is partially ordered. If $$\omega_1 \le \omega_2 \qquad \mathrm{implies} \qquad X(\omega_1) \le X(\omega_2),$$ we say that $X$ is an increasing random variable.
This comes up naturally in percolation theory. In this setting, $\Omega = \{0,1\}^{\mathbb Z^2},$ where $\omega(z) = 0$ if a site $z$ is closed, and $\omega(z) = 1$ if it is open. See, for example, the beginning of Chapter 2 of Grimmett's Percolation.
Your property is useful in settings where $\Omega$ doesn't have a natural ordering, but one may wish to impose an ordering using some random variable $X$. I would say that $Y$ is increasing relative to $X$, though I don't know a standard terminology.
