5
$\begingroup$

When teaching random variables for the first time, most of us say that it is a function $$X : \Omega \to \mathbb{R}$$ without any further restriction. Of course, a more general definition is to say that it is a function $$X : \Omega \to \mathbb{R}$$ where $\Omega$ is equipped with a sigma-algebra $\mathcal{A}$ and where $X^{-1}(I) \in \mathcal{A}$ for all interval $I \subset \mathbb{R}.$ I would like to switch from the first definition to the second with my students. In order to justify the relevance of doing so, I would like to have interesting examples of random variables defined with a sigma-algebra $\mathcal{A}$ which is strictly smaller than $\mathcal{P}(\Omega)$. The classical examples using coins and dices are using $\mathcal{P}(\Omega)$. Thanks for any suggestion.

$\endgroup$
4
  • $\begingroup$ I edited the title a bit, since random variables are not "defined on a $\sigma$-algebra" (they are defined on $\Omega$, no matter what the $\sigma$-algebra is), but are "measurable with respect to a $\sigma$-algebra". $\endgroup$ Oct 20, 2022 at 17:18
  • 2
    $\begingroup$ Re the question: It is very easy to come up with examples of functions that are measurable wrt to a non-trivial $\sigma$-algebra (e.g., every continuous function $X: [0,1] \to \mathbb{R}$ is Borel-measurable). The point is rather: Why does one need this measurability assumption? The most straightforward answer to this is: one needs it in order to talk about integrals, which are in turn needed to define expected values. (There's also another, more sophisticated, point about filtrations which are, e.g., related to the amount of information that is available at a certain point in time.) $\endgroup$ Oct 20, 2022 at 17:30
  • $\begingroup$ @Jochen Glueck : Thanks for your answer. Does it mean that, in the discrete case, this assumption is not really relevant ? $\endgroup$
    – user700974
    Oct 20, 2022 at 20:00
  • $\begingroup$ In the discrete case all integrals are sums (or series), so measurability become essentially irrelevant for most probabilistic concepts in the discrete setting. It still makes some sense to talk about conditional expectations (and maybe even filtrations, although I don't know whether anybody in probability really does this seriously for discrete spaces) - but if you're solely interested in the discrete case, I would recommend against describing this by using $\sigma$-algebra terminology, since those things can then be worded much more naturally in terms of partitions of $\Omega$. $\endgroup$ Oct 20, 2022 at 20:24

1 Answer 1

4
$\begingroup$

(this answer is meant as an extension of Jochen Glueck's comment)

As far as I know, there are two fundamental uses for the notion of measurability:

  • If we want to integrate functions (in the context of probability: take expectations of random variables), we have to make sure these functions are measurable with respect to the $\sigma$-algebra on which our measure is defined. For example, the Lebesgue measure on $[0,1]$ cannot be meaningfully defined on all subsets of $[0,1]$, and this means that we cannot Lebesgue-integrate all functions on $[0,1]$.
  • In probability, $\sigma$-algebras can represent information (in a sense). In this context, saying that $X$ is measurable with respect to some $\sigma$-algebra $\mathcal{F}$ means that in we only need certain knowledge (represented by $\mathcal{F}$) to determine the value of $X$.

The first use is more due to a technical obstruction (our inability to define the desired measure on the whole $\mathcal{P}(\Omega)$), and it's actually irrelevant when $\Omega$ is discrete. Even if $\Omega$ is not discrete, this issue of measurability is not very interesting from the point of view of probability (although that's just my opinion).

The second use is quite interesting and meaningful, so let me write a few words about it, using the simplest possible example.


Let $\Omega$ be the space corresponding to two dice throws: $$ \Omega = \{ (a,b) : a,b \in [6] \}, $$ where I used $[6]$ to denote the set $\{ 1,2,\ldots,6 \}$. The $\sigma$-algebra $\mathcal{P}(\Omega)$ corresponds to knowing the outcomes of both throws, and any $X \colon \Omega \to \mathbb{R}$ is measurable with respect to $\mathcal{P}(\Omega)$.

But let us consider the following two $\sigma$-algebras: $$ \mathcal{F}_1 = \{ A \times [6] : A \subseteq [6] \}, \quad \mathcal{F}_2 = \{ [6] \times B : B \subseteq [6] \}. $$ Here, $\mathcal{F}_1$ corresponds to knowing the outcome of the first throw, and $\mathcal{F}_2$ - the second throw. For example, the random variable $$ Y(a,b) = \begin{cases} 0 & 2 \mid a \\ 1 & 2 \nmid a \end{cases} $$ which yields the result of the first throw modulo $2$, is measurable with respect to $\mathcal{F}_1$ but not with respect to $\mathcal{F}_2$. Formally, this is because the sets $Y^{-1}(0) = \{2,4,6\} \times [6]$ and $Y^{-1}(1) = \{1,3,5\} \times [6]$ lie in $\mathcal{F}_1$ but not $\mathcal{F}_2$. Intuitively the reason is that $Y$ is determined by the first throw, but it's not determined by the second throw.

Consider also the example $Z(a,b) = a+b$, which is measurable with respect to $\mathcal{P}(\Omega)$, but not $\mathcal{F}_1$ nor $\mathcal{F}_2$ (as can be checked easily). This corresponds to the fact that knowing one of the throws is not enough to determine $Z$.

There's a nice construction of a $\sigma$-algebra generated by a random variable: $$ \sigma(X) = \{ X^{-1}(U) : U \in \mathcal{B} \}, $$ where I used $\mathcal{B}$ to denote the $\sigma$-algebra of Borel sets in $\mathcal{R}$. Now, when $W$ is $\sigma(X)$-measurable it means that $W$ is determined by $X$ (i.e. it can be represented as a deterministic function of $X$). For example $$ W(a,b) = \begin{cases} 0 & 2 \mid a+b \\ 1 & 2 \nmid a+b \end{cases} $$ is determined by $Z$ and indeed, it is $\sigma(Z)$-measurable.


In the applications, it is usual to think of $(\Omega,\mathcal{F})$ as representing all possible outcomes of everything in the world (I'm exaggerating a bit) and introduce smaller $\sigma$-algebras corresponding to partial knowledge. Usually the time is involved, so for example $S_t$ can represent stock prices at time $t$, and then $$ \mathcal{F}_T = \sigma(S_t : 0 \le t \le T) $$ represents the knowledge of all things that happened on the market up to time $T$.

$\endgroup$
1
  • 1
    $\begingroup$ +1 Reaaly great answer (and much better than my comments)! $\endgroup$ Oct 21, 2022 at 9:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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