# Writing down a conditional probability / finding the underlying probability space

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?

• 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 "$\Omega$." – whuber Jan 13 at 22:58
• that is, I can rewrite the conditions as $p = P(\Omega)$, $P(X(\omega) =x |\Omega) = \pi$ and $\rho=P(X(\omega)=x|\Sigma)$? – Syd Amerikaner Jan 13 at 23:16
• 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. – whuber Jan 13 at 23:25