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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?

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2 Answers 2

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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 $f^{-1}(B) \in \mathcal{A}$.

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  • $\begingroup$ 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, 2012 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, 2012 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, 2012 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, 2012 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, 2012 at 22:14
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1 and 4 are right. 2 and 3 are incorrect.

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  • $\begingroup$ Without an explanation this isn't contributing much. $\endgroup$
    – user147263
    Jul 1, 2015 at 14:20
  • $\begingroup$ @1999 I explicitly answered OP's question. $\endgroup$
    – bcf
    Jul 1, 2015 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, 2015 at 14:27

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