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  1. In elementary probability,

    $E(Y \mid X =x)$ is defined as expectation of $Y$ w.r.t. the p.m. $P(A \mid X =x): = \frac{P(A \cap \{X=x \})}{P( X=x)}$ when $P( X=x) \neq 0$.

    when $P(X =x) =0$ , define $$P(A\mid X =x): = \lim_{\epsilon \rightarrow 0} \frac{P(A \cap \{|X-x|< \epsilon\} )}{P(|X-x|< \epsilon)}$$ and $$E(Y \mid X =x): = \lim_{\epsilon \rightarrow 0} \frac{E(Y \times 1_{\{|X-x|< \epsilon\}})}{P(|X-x|< \epsilon)}.$$

    If define $f(x):=P(A \mid X =x)$ and $h(x):=E(Y \mid X =x)$, then $f(X)$ and $h(X)$ are both random variables $\Omega \rightarrow \mathbb{R}$.

  2. In probability theory, $E(Y \mid X )$ and $P(A \mid X )$ are both random variables $\Omega \rightarrow \mathbb{R}$.

I was wondering

  1. if $h(X)$ and $E(Y \mid X )$ are the same a.s.?

  2. Similarly, if $f(X)$ and $P(A \mid X )$ are the same a.s.? Is $f(X=x)$ the so called transition probability distribution?

Thanks and regards!

Tim
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    (i) P(Y|X=x) and P(Y|X) do not exist. (ii) Ever tried a book? – Did May 04 '11 at 14:15
  • @Didier: Thanks! My mistake. I have replaced Y in P(Y|X=x) and P(Y|X) with a measurable set A. How about now? – Tim May 04 '11 at 16:24
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    I should not do this but... here we go: $P(A|X=x)$ is not a probability measure but a number; the limits you use to define $P(A|X=x)$ and $E(Y|X=x)$ do not exist in general, so these definitions are false; $f(X)$ and $h(X)$ are not defined on $\mathbb{R}$; $f(X=x)$ does not exist. Note: I find peculiar that you did not even answer my question (ii) but that you continue to try to use my expertise. – Did May 07 '11 at 13:32
  • @Didier: Thanks for following up! (1) Yes, I have tried some books. (2) f and h are functions defined on the codomain of r.v. X. So f(X) and h(X) are r.v.s. I made a typo by saying its domain is R. Now I corrected it. (3) Some more clarification: when I said P(A|X=x) (when it exists) is a p.m., I wanted to say it is when x is fixed and A is varying; when I said P(A|X=x) (when it exists) is a r.v., I wanted to say P(A|X=) is a function of x when A is fixed, so P(A|X) is a r.v.. – Tim May 07 '11 at 15:43
  • @Didier: (4) I know P(A|X=x) and E(Y|X=x) do not exist in general, but I consider cases when they exist. However ill-posed my questions are, my intention is to find connection between, when exist, conditional probabilities/expectations defined in non-measure theoretical courses/books and those in measure-theoretical courses/books. – Tim May 07 '11 at 15:50

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Answer to both questions: Yes if the limit exists.

ps. don't forget to replace t by small x or x by t in your formulas.

Roah
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