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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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    @Didier: Right, maybe I should change it to $\omega$.2011-03-13
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    @Rasmus: Thanks! I was wondering if the two definitions of conditional expectation are related via this relation between the two definitions of conditional probability?2011-03-13
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    @Tim: Yes, they are -- in a similar way that I have indicated for conditional probability.2011-03-13
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    see http://en.wikipedia.org/wiki/Conditional_expectation#Conditioning_as_factorization2011-03-13
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    @Rasmus: It seems to me that $P(\cdot \mid X)$ in the first definition is only connected to the denominator in the definition of $P(\cdot \mid X \in \cdot)$. $P(\cdot \mid X)$ actually does not require the conditioning event in $P(\cdot \mid X \in \cdot)$ to be related to $X$, so $P(\cdot \mid X)$ can in the same way be connected to $P(\cdot \mid C)$ for $C$ seemingly unrelated to $X$. Can I say $P(\cdot \mid X)$ in the first definition and $P(\cdot \mid X \in \cdot)$ in the second are actually totally unrelated in concept, instead of the former is the generalization of the latter?2011-03-13
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    @Tim: I think the answer to your question is no, but I do not understand your reasoning that $P(\cdot\mid X)$ is unrelated to $X$.2011-03-13
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    @Rasmus: By no, do you mean I cannot say that? I am saying the two are unrelated in concept, although they are connected via $P(A \cap \{ X \in B \})$ as in your reply. This is rather remote connection, since $P(\cdot \mid X)$ can be connected to $P(\cdot \mid C)$ in exactly the same way. That is what I mean when saying they are unrelated in concept. If my understanding is wrong, please feel free to correct me. Thanks!2011-03-13
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    @Tim. Yes, by no, I meant you cannot say that. I am sure that the two definitions of conditional expectation are not unrelated in concept. The relation I wrote down feels very natural to me. How is your relation of $P(\cdot\mid X)$ to $P(\cdot\mid C)$ supposed to look like?2011-03-14
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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