3
$\begingroup$

I thought I understood the measure theoretic concept of a probability space, but yesterday I realized that I really don't. Here's what I mean:

  1. I thought we could think of $\Omega$ as the set of outcomes of a certain experiment. So before we can even define the triple $(\Omega, \mathcal{F}, P)$, we would need to think about an experiment. Now, each $\omega \in \Omega$ was supposed to be a possible outcome of the experiment, and the randomness thus comes from the fact that different outcomes can occur with different frequencies over time. These long term frequencies define $P$ for us.

  2. So, if we consider the experiment of flipping a coin once, then $\Omega = \{ H, T\}$, since we only have those possible outcomes.

  3. Ok, but in the discussion of Brownian motion (or the diffusion of a particle across some environment like dye in water), we are told to think of each $\omega \in \Omega$ as a particle. For example, each $\omega$ is a particle of pollen. And the stochastic process $X_{t}(\omega)$ gives us, for each fixed $\omega$, the path that that particle $\omega$ follows. But then what is the experiment we started with in order to get the set $\Omega$? How are the particles that each $\omega$ represents the outcomes of some experiment? And where does the randomness come from? It seems like the randomness comes from the path the particle takes after we've already chosen it, so how can each particle be considered the outcome of some experiment? Which experiment?

Finally, after reading the above (and in addition to the above questions), I'd really like to understand why in many times, the probability space $(\Omega, \mathcal{F}, P)$ is often "hidden" or not explicit. Can anyone explain this intuitively or give me some useful examples where the probability space is hidden? Shouldn't it always be explicitly defined (since math is always explicit)?

$\endgroup$
  • $\begingroup$ We pick a particle, and record it's trajectory as times go by; this is your experiment. The $X_t(\omega)$ for fixed $\omega \in \Omega$ is this recorded trajectory and it represents the result of your experiment ($X_t(\omega)$, of course, represents the position of particle $\omega$ at time $t$). $\endgroup$ – Zoran Loncarevic Jan 14 '16 at 13:40
  • $\begingroup$ Alternatively, you can have just a single particle, and $\Omega$ will be the set of functions $X: \mathbb{R} \to \mathbb{R}$ representing possible trajectories this particle might trace. $\endgroup$ – Zoran Loncarevic Jan 14 '16 at 13:43
  • 1
    $\begingroup$ @ZoranLoncarevic So we are defining our probability triple, and in particular, our $\Omega$, before picking an experiment, which shatters any intuition I had that $\omega$ is an outcome. $\endgroup$ – layman Jan 14 '16 at 13:43
  • 3
    $\begingroup$ Surprise: the choice of $\Omega$ is, in 99% of cases, completely irrelevant. $\endgroup$ – Did Jan 14 '16 at 18:38
  • 5
    $\begingroup$ @Did I wish you would write a full answer out explaining what you mean. I have no idea what the motivation for your remark is, or what you actually mean. $\endgroup$ – layman Jan 14 '16 at 19:10
3
$\begingroup$

You should think of $\Omega$ more as a set of all possible futures. So for the coin flipping, there are some futures where the coin comes up heads and some where it comes up tails. You could let $\Omega$ be $\{H,T\}$, but you could also let it be the half-open interval $(0,1]$ with Lebesgue measure and say that $(0,1/2]$ is the set where the coin turns up heads and $(1/2,1]$ is the set where the coin turns up tails.

The reason for preferring $(0,1]$ to ${H,T}$ is that it allows us to add in new random variables easily. If we wanted to add a separate, independent coin toss to the first model, we'd be stuck. But with $(0,1]$ we can do it (even though it's a bit messy):

$$\begin{array}{c|c|c|} & \text{First coin heads} & \text{First coin tails} \\ \hline \text{Second coin heads} & (0,1/4] & (1/4,1/2] \\ \hline \text{Second coin tails} & (1/2,3/4] & (3/4,1] \\ \hline \end{array}$$

The choice of the probability space itself doesn't really matter. Normally, we set it to be $(0,1]$ (since a theorem tells us that $(0,1]$ is big enough to be the underlying measure space of most things we want to model, including countably infinite sequences of continuous random variables) and then forget about it. It just happens that the language of measures captures our intuition for probability really well, and in order to use measures we need a measure space. Unfortunately, there's no intuitive interpretation of that measure space, which is why it's often glossed over when we're talking about probability.

$\endgroup$
  • $\begingroup$ So basically you are saying that we usually take $\Omega$ to be $(0,1)$ because it can serve as the "outcome" space for many different types of experiments that we can run? And to use this space as a model, we simply take an interval of length $\frac{p}{q}$ if we want the associated probability of an event to be $\frac{p}{q}$? How does this tie into Brownian Motion though? $\endgroup$ – layman Jan 15 '16 at 2:51
  • $\begingroup$ Basically, where is the randomness present with Brownian Motion? And how is this randomness represented in the set $\Omega$? $\endgroup$ – layman Jan 15 '16 at 2:51
  • 1
    $\begingroup$ To show that we can consistently define Brownian motion using measure theory, we use something called the Kolmogorov existence theorem. In this case, the $\Omega$ we end up with is the set of all functions from $[0,\infty)$ to $\mathbb R$. Each possible Brownian motion gives us such a function and the measure of a subset of $\Omega$ corresponds to the probability that we end up with that particular function. $\endgroup$ – John Gowers Jan 15 '16 at 17:07

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.