Understanding the dependence of conditional probability on filtrations

Question

Given a sequence of iid random variables $\{X_n\}$ (not necessarily discrete), and defining the conditional probability $\mathbb{P}_{\mathcal{F}_n}(A)(\omega)$ where the filtration is the natural one and $A$ is an event in the probability space. I am interested in understanding the connection between the conditional probability and $\omega$. Usually one provides the following example, tipically using the discrete case (or more in general a partitioning): for example if $\omega\in\{X_1=x_1,\ldots,X_n=x_n\}$, $\mathbb{P}_{\mathcal{F}_n}(A)(\omega)=\mathbb{P}(A|X_1=x_1,\ldots,X_n=x_n)$, because the events $\{X_1=x_1,\ldots,X_n=x_n\}$ are a partitioning generating $\mathcal{F}_n$. My question is essentially if more generally (that is not just in the case of a partitioning) this relationship extends to some measurable events $H\in\mathcal{F}_n$: when do we have that $\mathbb{P}_{\mathcal{F}_n}(A)(\omega)=\mathbb{P}(A|H)$ for all $\omega\in H$? Thanks for any insight provided. EDIT: As an example, from a paper I read, the following relationship, reminiscent of the above, holds: if $\{X_i\}_{i=1}^N$ are uniform on $(0,1)$, $\mathbb{P}_{\mathcal{F}_n}(X_n=\max\{X_1,\ldots,X_N\})=1_{\{X_n=\max\{X_1,\ldots,X_n\}\}}X_n^{N-n}$. What properties have been used to derive it?

Answer 1

This is generally true if $H$ is an atom of $\mathcal F_n$, i.e. if $G \in \mathcal F_n$ and $G$ is a strict subset of $H$ implies $\mathbb{P}(G)=0$. Otherwise, if $H = H_1 \cup H_2$, it would imply $\mathbb{P}(A|H)=\mathbb{P}(A|H_1)=\mathbb{P}(A|H_2)$. The reason this isn't a problem for atoms is because conditional probabilities aren't defined for events with probability zero.

For the specific example you asked about, we have \begin{align} \mathbb{P}_{\mathcal F_n}(X_n = \max(X_1,...,X_N)) &= \mathbb{E}_{\mathcal F_n}[1_{X_n = \max(X_1,...,X_N)}] \\ &= \mathbb{E}_{\mathcal F_n}[1_{X_n = \max(X_1,...,X_n)} 1_{X_{n+1}\le X_n} \cdots 1_{X_N \le X_n}] \\ &= 1_{X_n = \max(X_1,...,X_n)}\mathbb{E}_{\mathcal F_n}[1_{X_{n+1}\le X_n}] \cdots \mathbb{E}_{\mathcal F_n}[ 1_{X_N \le X_n}] \\ &= 1_{X_n = \max(X_1,...,X_n)} X_n \cdots X_n \\ &= 1_{X_n = \max(X_1,...,X_n)} X_n^{N-n}, \end{align} where the third equality follows from independence and taking out what is known, and the fourth from the fact that $X_{n+1},...,X_N \sim U(0,1)$.