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Suppose $(\Omega, \mathcal{A}, P)$ is the sample space and $X: (\Omega, \mathcal{A}) \rightarrow (\mathbb{R}, \mathcal{B})$ is a random variable.

  1. 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 $\mathcal{A} \times \Omega \rightarrow [0,1]$.
  2. 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:

  1. 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?

  2. 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?

  3. 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!

1 Answers 1

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$\int_C P(A|X)(\omega)dP(\omega)=\int_C E(I_A|X)(\omega)dP(\omega)=\int_C I_A(\omega)dP(\omega)=P(A\cap C)$ for $C$ in the sigma algebra generated by $X$. So, for $C=\{X\in B\}$, $\frac{\int_{\{X\in B\}} P(A|X)(\omega)dP(\omega)}{P(X\in B)}=P(A\mid X\in B).$

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    @Rasmus: Thanks! Although it has been a while, I was wondering, as $P(⋅∣X \in⋅)$ can be written in terms of $P(⋅∣X) $ as in your reply, if the reverse is possible, i.e. if $P(⋅∣X) $ can be written in terms of $P(⋅∣X \in⋅)$? Thanks!2011-03-19