Suppose $(\Omega, \mathcal{A}, P)$ is the sample space and $X: (\Omega, \mathcal{A}) \rightarrow (\mathbb{R}, \mathcal{B})$ is a random variable.
- Using language of measure theory, $P(A \mid X)$, the conditional probability of an event given a random variable, is defined from conditional expectation as $E(I_A \mid X)$. So $P(\cdot \mid X)$ is in fact a mapping $\mathcal{A} \times \Omega \rightarrow [0,1]$.
- In elementary probability, we learned that $$P(A \mid X \in B): = \frac{P(A \cap \{X \in B\})}{P(X \in B)}.$$ If I understand correctly, this requires and implies $P(X \in B) \neq 0$. So $P(\cdot \mid X \in \cdot)$ is in fact a mapping $\mathcal{A} \times \mathcal{B} \rightarrow [0,1]$.
My questions are:
When will $P(\cdot \mid X)$ in the first definition and $P(\cdot \mid X \in \cdot)$ in the second coincide/become consistent with each other and how?
Is there some case when they can both apply but do not agree with other? Is the first definition a more general one that include the second as a special case?
Similar questions for conditional expectation.
- In elementary probability, $E(Y \mid X \in B)$ is defined as expectation of $Y$ w.r.t. the p.m. $P(\cdot \mid X \in B)$. So $E(Y \mid X \in \cdot)$ is a mapping $\mathcal{B} \rightarrow \mathbb{R}$.
- In measure theory, $E(Y \mid X )$ is a random variable $\Omega \rightarrow \mathbb{R}$.
I was also wondering how $E(Y \mid X \in \cdot)$ in elementary probability and $E(Y \mid X )$ in measure theory can coincide/become consistent? Is the latter a general definition which includs the former as a special case?
Thanks and regards!
