This is a notational issue: Let's say we have a partition of $\mathcal X$ given by $\{\Omega,\Sigma\}$ and we define

  • $P(\omega\in\Omega) = p$,
  • $P(X(\omega)=x|\omega\in\Omega) = \pi$ and
  • $P(X(\omega)=x|\omega\in\Sigma)=\rho$.

Then $P(X(\omega) = x) = \pi\times p + \rho\times(1-p)$ by Bayes theorem.

This is all clear but I cannot formalize this in a proper way. What is the underlying probability space? I guess it is $\mathcal X$. I know that $X$ is a random variable from $\mathcal X$ to $\mathbb R$ but, strictly speaking, isn't this a contradiction to the first bullet point?

While the first bullet point states that $P$ is defined on $\mathcal X$, the second and third (as well as the conclusion using Bayes theorem) suggest that $P$ is defined on $\mathbb R$...

What also confuses me is the notation "$\omega\in\Omega$". Shouldn't it be just $\omega$, or more precisely, $\{\omega\}$?

Can someone help clarify?

  • 1
    $\begingroup$ The first bullet neither states nor implies that $P$ is defined on (the sample space) $\mathcal X.$ $P$ is a probability function, which means it is defined on the (implicitly given) sigma algebra over $\mathcal X.$ Notation like "$x\in\Omega$" is a supererogatory mathematical solecism: it stands for the event "$\Omega$." $\endgroup$ – whuber Jan 13 at 22:58
  • $\begingroup$ that is, I can rewrite the conditions as $p = P(\Omega)$, $P(X(\omega) =x |\Omega) = \pi$ and $\rho=P(X(\omega)=x|\Sigma)$? $\endgroup$ – Syd Amerikaner Jan 13 at 23:16
  • 1
    $\begingroup$ The conditional probability notation doesn't usually work that way, unfortunately. It can be written as you did originally or it can be written as $$\Pr(X(\omega)=x\mid\omega\in\Omega)=\Pr(X^{-1}(x)\mid\Omega)$$ if you like. $\endgroup$ – whuber Jan 13 at 23:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.