2
$\begingroup$

Suppose we have the probability space $(\Omega,\mathcal{A},P)$. Which of the following are right?

  1. $P$ is the probability measure defined on the events $\mathcal{A}$ as follows: $P:\mathcal{A}\rightarrow[0,1]$
  2. $P$ is the probability measure defined on the outcome space $\Omega$ as follows: $P:\Omega\rightarrow[0,1]$
  3. $X$ is a function $X:\mathcal{A}\rightarrow(E,\mathcal{E})$, where $(E,\mathcal{E})$ is a measurable space.
  4. $X$ is a function $X:\Omega\rightarrow(E,\mathcal{E})$, where $(E,\mathcal{E})$ is a measurable space.

Basically, I am unsure whether probability measures and random variables are defined on the state space $\Omega$, or the $\sigma-algebra$ $\mathcal{A}$, or both?

$\endgroup$
3
$\begingroup$

Probability measures assign values (probabilities) to sets in the $\sigma$-algebra $\mathcal{A}$. On the other hand, random variables are functions $f\colon \Omega\to E$ that are measurable in this sense: If $B \in \mathcal{E}$, then $f^{-1}(B) \in \mathcal{A}$.

$\endgroup$
  • $\begingroup$ If random variables are functions $f\colon \Omega\to E$, why is it not sufficient for $f^{-1}(B) \in \Omega$ ($B \in \mathcal{E}$) for them to be measurable? Could we construct a random variable on $\mathcal{A}$ instead of $\Omega$, or does that make no sense? I guess I'm a bit confused on $\sigma$-algebras: why are they needed, why are probability measures defined on them instead of on $\Omega$? $\endgroup$ – lodhb Mar 26 '12 at 4:56
  • $\begingroup$ $f^{-1}(B) \subseteq \Omega$ makes sense, but not $f^{-1}(B)\in \Omega$. For any function, the preimage of a set is a subset of the domain, not an element of the domain. $\endgroup$ – Patrick Mar 26 '12 at 5:38
  • $\begingroup$ But $\Omega\subset\mathcal{A}$, so $f^{-1}(B) \in \mathcal{A}$ doesn't imply $f^{-1}(B) \subset \Omega$, right? $\endgroup$ – lodhb Mar 26 '12 at 9:02
  • $\begingroup$ Not quite. $\mathcal{A}$ is a bunch of subsets of $\Omega$, and $f^{-1}(B)\in \mathcal{A}$ means that $f^{-1}(B)$ is one of those subsets. $\endgroup$ – Patrick Mar 26 '12 at 22:03
  • $\begingroup$ We often want to know the probability that some particular "group of outcomes" will occur. For example, suppose we choose a number $x$ from $[0,1]$, and want to know the probability that $\frac{7}{16} < x < \frac23$. The probability is the measure of the set $(\frac{7}{16}, \frac23)$ which is a subset of the sample space $[0,1]$. $\endgroup$ – Patrick Mar 26 '12 at 22:14
-1
$\begingroup$

1 and 4 are right. 2 and 3 are incorrect.

$\endgroup$
  • $\begingroup$ Without an explanation this isn't contributing much. $\endgroup$ – user147263 Jul 1 '15 at 14:20
  • $\begingroup$ @1999 I explicitly answered OP's question. $\endgroup$ – bcf Jul 1 '15 at 14:24
  • $\begingroup$ The question being from 2012, it's safe to assume the OP isn't struggling with that multiple choice question any more. $\endgroup$ – user147263 Jul 1 '15 at 14:27

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.