1
$\begingroup$

On a measurable space $(\Omega, F)$, where $\Omega$ is a set of outcomes and $F$ is a $\sigma -$field, what exactly is the domain of a probability measure $P$? If it's a specific $\sigma -$field such as $F$, then how can we define the independence on sets from multiple spaces? And it can't be the whole power set because we can construct sets that have no probability. I thought the whole point of using a $\sigma -$field was to make the domain of probability measures better to define.

Edit:

So in the textbook $Probability: \ Theory \ and \ Examples$, by Rick Durrett, he says "... $\sigma -$fields $\mathcal{F}_1, \ldots , \mathcal{F}_n$ are independent if whenever $A_i \in \mathcal{F}_i$ for $i=1, \ldots, n$, we have $$P(\cap_{i=1}^{n} A_i)= \prod_{i=1}^n P(A_i)$$..."

How can $P$ be defined on all these $\sigma-$fields? I thought a probability measure was always w/respect to a given $\sigma-$field.

$\endgroup$
  • $\begingroup$ I'm afraid I don't see the question about independence and other probability spaces: what does the domain of $P$ have to do with that? Clarify a bit more? Give an example? $\endgroup$ – Henno Brandsma Feb 18 at 22:28
  • $\begingroup$ Apologies, I'll edit my question $\endgroup$ – JasonM Feb 18 at 22:40
  • $\begingroup$ Ok I edited my question $\endgroup$ – JasonM Feb 18 at 22:58
  • 1
    $\begingroup$ @JasonM the sigma fields $\mathscr{F}_i$ are contained in $\mathscr{F}$ (so $\mathscr{F}$ is at least as large as $\sigma(\cup \mathscr{F}_i)$). $\endgroup$ – Will M. Feb 18 at 23:00
  • $\begingroup$ Without more context (I don't have the book) it also seems confusing to me. One should expect that $P$ is defined on the $\sigma$-field generated by all $\mathcal{F}_i$. $\endgroup$ – Henno Brandsma Feb 18 at 23:03
2
$\begingroup$

The domain of $P$ is definitely $F$, the $\sigma$-field on $\Omega$ (it is in some cases the power set of $\Omega$, e.g. for discrete measures on at most countable $\Omega$).

$P$ a function from $F$ to $[0,1]$ obeying certain axioms. I don't see a relation to independence.

$\endgroup$
  • $\begingroup$ Yes this was my understanding. I hope my edit clarifies my question $\endgroup$ – JasonM Feb 18 at 22:59
  • 2
    $\begingroup$ The relation is that the sigma fields $\mathscr{F}_i$ are subset of $\mathscr{F},$ (otherwise the independence is a nonsensical issue). $\endgroup$ – Will M. Feb 18 at 23:03
1
$\begingroup$

Thanks to Will M. it appears the most sensible $\sigma-$field for $P$ to be defined on is a $\sigma-$field containing $\sigma(\cup_{i=1}^n \mathcal{F}_i)$. Without the experience, I had difficulty seeing this, so thanks to everyone who commented.

$\endgroup$

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.