# What does it mean to say that an event is determined by certain random variables?

So I was reading Moser and Tardos' paper on their algorithmic proof for the Lovász Local Lemma and came across the following:

Let $\mathcal{P}$ be a finite collection of mutually independent random variables in a fixed probability space $\Omega$. We will consider events $A$ that are determined by the values of some subset $S \subseteq \mathcal{P}$ of these variables.

(This excerpt was taken from the second paragraph on page 2 of the above mentioned paper.)

What does it mean to say that an event is determined by (the values of) certain random variables? I think it would make sense, for instance, to say that the event $A$ = [$X \neq Y$] is determined by the r.v.s $X$ and $Y$, but I don't know how one would formally write the definition in general.

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Probably that $A$ is in the sigma-algebra generated by those random variables, i.e. $A\in\sigma(\{X\mid X\in S\})$. – Stefan Hansen Jul 19 '13 at 18:22
@StefanHansen Thank you. It seems your guess is correct. The definition you provided makes sense and it also implies the following: if $A$ is determined by a set $S = \{X_1, X_2, ..., X_n\}$ of random variables, then for all real numbers $a_1, a_2, ..., a_n$, we have that the event $[X_1 = a_1, ..., X_n = a_n]$ is either contained in $A$ or in $A$'s complement. What this means is: if we know the values of each random variable, then we know whether $A$ happens or not. – TuringMachine Jul 28 '13 at 22:26
[...] This fact is used in Moser and Tardos' algorithm and it is consistent with what comes to mind when I think of something being determined by something else. I wish I could give you some rep for helping out, but I guess all I can do is say thanks! – TuringMachine Jul 28 '13 at 22:30

$A = f(X, Y, Z ...)$ with $X, Y, Z \in S$
Your $A$ is a random variable, not an event. – Dilip Sarwate Jul 19 '13 at 18:20
@user86828 Yes, but your example is not an event. You should probably define $A$ as the event that $f(X,Y,Z,\ldots)=a$ for some value $a$. – augurar Jul 19 '13 at 18:58