1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Isn't it the case that $X$ and $Y$ are monotonic functions of $X+Y$? $\endgroup$ – James Martin Nov 9 '10 at 18:56
  • $\begingroup$ @James Martin: I think you're right, I'll update the question... $\endgroup$ – Bjørn Kjos-Hanssen Nov 9 '10 at 19:05
  • $\begingroup$ now my comment looks like a bit of a non-sequitur :) $\endgroup$ – James Martin Nov 9 '10 at 19:38
0
$\begingroup$

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 $X$.

$\endgroup$
  • $\begingroup$ But $Y$ is not in general determined by $X$. Simplest case: $$ \Omega = \{ (0,0), (0,1), (1,0), (1,1)\}. $$ $$ X(i,j) = i $$ $$ Y(i,j) = 10i + j $$ 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 $Y$. $\endgroup$ – Michael Hardy Nov 9 '10 at 21:29
  • $\begingroup$ Correct. A STRICTLY bigger X entails a strictly bigger Y. I was hasty. $\endgroup$ – Michael Hardy Nov 9 '10 at 21:59
  • $\begingroup$ So.... It seems that the answer given was correct, but maybe only shows why a different question should have been asked. $\endgroup$ – Michael Hardy Nov 9 '10 at 22:03
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.