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Given probability space $(\Omega, \mathcal{F}, \mathbb{P})$, I can understand the definition of conditional expectation $\mathbb{E}[X\mid \mathcal{G}]$, where $\mathcal{G}$ is a sub $\sigma$-algebra of $\mathcal{F}$.

If we define $\mathbb{P}(A\mid\mathcal{G}) := \mathbb{E}[\mathbb{I}_A\mid\mathcal{G}]$, we get $\mathbb{P}(A\mid\mathcal{G})$ as a conditional probability.

So, why do we need to define regular conditional probability?


Can I explain the reason like this?

The conditional probability $\mathbb{P}(A\mid\mathcal{G})$ satisfies:

  • $\mathbb{P}(\Omega\mid\mathcal{G})=1$ a.s.;
  • $\mathbb{P}(A\mid\mathcal{G})\ge 0$ a.s.;
  • $\mathbb{P}(\sum_n A_n\mid\mathcal{G}) = \sum_n \mathbb{P}(A_n\mid\mathcal{G})$ a.s. for non-intersecting sets $\{A_n: n\ge 1\}$.

Taking $\mathbb{P}(\omega, A)$ as a representative of $\mathbb{P}(A\mid\mathcal{G})$ (since it is an equivalence class of random variable), we hope that $\mathbb{P}(\omega, \cdot):\mathcal{F}\to\mathbb{R}$ is almost surly a probability measure. It easily satisfies $\mathbb{P}(\omega, \Omega)=1$ and $\mathbb{P}(\omega, A)\ge 0$ a.s., but has difficulty on countable-additivity.

To satisfy $\mathbb{P}(\omega, \sum_n A_n) = \sum_n\mathbb{P}(\omega, A_n)$ a.s., we must exclude a series of null sets $\{N_n: n\ge 1\}$; and for another series of $\{B_n: n\ge 1\}$, to satisfy $\mathbb{P}(\omega, \sum_n B_n) = \sum_n\mathbb{P}(\omega,B_n)$ a.s., we must exclude another series of null sets $\{M_n: n\ge 1\}$... Then totally we need to exclude a big set $$ \left(\bigcup_{n=1}^\infty N_n\right) \cup \left(\bigcup_{n=1}^\infty M_n\right) \cup \cdots $$ which may no longer be a null set.

To deal with this difficult, we introduce the concept of regular conditional probability $\mathbb{P}(\cdot, \cdot): \Omega\times\mathcal{G}\to[0,1]$, such that

  • $\mathbb{P}(\omega,\cdot)$ is a probability measure on $\mathcal{G}$ for every $\omega\in\Omega$.
  • $\mathbb{P}(\cdot,A)$ is a measurable function on $(\Omega,\mathcal{G})$ for every $A\in\mathcal{G}$, and $\mathbb{P}(\omega,A)=\mathbb{P}(A\mid\mathcal{G})$ a.s.

Since we can tolerance difference between $\mathbb{P}(\omega,A)$ and $\mathbb{P}(A\mid\mathcal{G})$ on null sets, we can just directly remove the "almost surly" suffix of $\mathbb{P}(\omega,\Omega)=1$ and $\mathbb{P}(\omega,A)\ge 0$.

Please point out my mistakes.

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    $\begingroup$ Roughly speaking, one knows that each map $\omega\mapsto P(A\mid\mathcal G)(\omega)$ is well behaved and one wants that (almost) every map $A\mapsto P(A\mid\mathcal G)(\omega)$ is well behaved. $\endgroup$ – Did Sep 3 '18 at 6:29
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Conditional probabilites and conditional expectations are only defined up to null sets. In many problems you would like to put many null sets involved in conditional expectations inside one null set, but uncountable unions of null sets need not be null. When we are dealing with Borel measures on 'nice spaces' (say complete separable metric spaces) it is possible to handle these null sets using regular conditional probabilities.

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