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Why aren't probability spaces just defined as $(\Omega, p)$ pairs with $\Omega$ as the sample space, $\sum_{\omega \in \Omega}p(\omega) = 1$, and for a subset $A \subseteq \Omega$, $\Pr(A) := \sum_{\omega \in A}p(\omega)$?

Said another way, why aren't all $(\Omega, \mathcal{A}, p)$ probability spaces of the form $(\Omega, \mathcal{P}(\Omega), p)$? What do we gain by giving ourselves the freedom to exclude certain subsets of $\Omega$ from $\mathcal{A}$ ?

alecbz
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    How would you deal with uniform probability on $[0,1]$? – copper.hat Jul 02 '13 at 05:10
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    In addition to what is already posted on the page, note that it is in fact often desirable to consider several sigma-algebras on the same set. As soon as one deals with processes instead of a single random variable, this opens up the possibilities to describe the relations between different random variables (obvious keywords Markov chains, martingales, conditional expectations). – Did Jul 02 '13 at 08:47

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That's a good question. An answer is that there are many probability spaces $(\Omega,\mathcal{A},p)$ where the probability function $p$ cannot be extended to all of $\mathcal{P}(\Omega)$. For example, consider the probability space where $\Omega=[0,1]$, $\mathcal{A}$ is the Lebesgue $\sigma$-algebra on $[0,1]$, and $p=\lambda$ is the Lebesgue measure. Then there is no way of extending $p$ to have a value when given the Vitali set (the standard example of a non-Lebesgue measurable subset of $[0,1]$).

Zev Chonoles
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If $\Omega$ is countable, then we gain nothing. Otherwise, a sum over $\Omega$ is meaningless, and we need to replace that with integration. I don't know too much about the details, but in order for that (Lebesgue) integration to be meaningful, it needs to be done over a measure space, and a probability space is just a measure space where $p(\Omega)=1$. In ZFC theory (and some weaker set theories), there are non-measurable subsets of, say, the reals, so it's useful to be able to deal with that situation.

dfeuer
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  • We can even gain something when $\Omega$ is countable. The $\sigma$-algebra can also present imperfect information, which is equivalent to a partition of $\Omega$. in this case. – Michael Greinecker Jul 02 '13 at 08:55