# Conditional distribution question

When we talk about conditional distribution we mean $F_{Y|X}(y|x) = \mathbb{P}(Y\leq y|X=x)$. Does there exists the following object: $F_{Y|X}(Y|X)$? I'm refering to conditional expectations when we have $\mathbb{E}[Y|X=x]$ a number and $\mathbb{E}[Y|X] = \mathbb{E}[Y|\sigma(X)]$ a random variable. Sorry if the question sounds stupid or naive.

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$\sigma(X)$ is the sigma algebra generated by $X$? – Learner Jan 12 '13 at 13:16
Yes. Sorry for confusion. – arkadiy Jan 12 '13 at 13:18
What do you want $F_{Y\mid X}(Y\mid X)$ to mean? $P(Y\leq Y\mid X)$? But this is $1$ since $\{Y\leq Y\}=\Omega$. – Stefan Hansen Jan 12 '13 at 13:25

I'm not entirely sure what you are looking for, but here is an attempt to answer your question.

The relationship between conditional expectations $E[Y\mid X]$ versus $E[Y\mid X=x]$, is that if $\varphi(x)=E[Y\mid X=x]$, then $E[Y\mid X]=\varphi(X)$. Note that $P(Y\leq y\mid X=x)$ is just the conditional expectation of the indicator variable $1_{\{Y\leq y\}}$, i.e. $$P(Y\leq y\mid X=x)=E[1_{\{Y\leq y\}}\mid X=x],$$ so if for fixed $y$, we denote this function by $\psi_y(x)$, then the above shows that $$P(Y\leq y\mid X)=E[1_{\{Y\leq y\}}\mid X]=\psi_y(X).$$

It does not really make sense to take this a step further and investigate $\psi_Y(X)$, because this is equal to $1$ a.s., as the following shows: $$\psi_Y(X)=E[1_{\{Y\leq Y\}}\mid X]=E[1\mid X]=1 \quad \text{a.s.}$$

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The abstract definition of conditional expectation (due to Kolmogorov) is actually one where the conditioning is on a $\sigma$-algebra. See this link for the actual definition. More precisely:

Call $\sigma \left( X \right) =\mathcal{B}$. Let $\left( \Omega, \mathcal{A}, P \right)$ be the probability space so that $\mathcal{B} \subset A$. The conditional expectation $E \left[ Y|\mathcal{B} \right]$ is a random variable (measurable function) from $\Omega$ into $\mathbb{R}$ such that $\forall B \in \mathcal{B}$ $$\int_B E \left[ Y|\mathcal{B} \right] \left( \omega \right) \mathrm{d} P\left( \omega \right) = \int_B X \left( \omega \right) \mathrm{d} P\left( \omega \right)$$

Most graduate level probability textbooks contain a detailed treatment. (See for instance Section 34 in Billingsley's "Probability and Measure" (p.445 of the third edition). Rosenthal's treatment in chapter 13 of his book "A First look at rigorous probability theory" is another possibility. Rao's "Conditional Measures and Applications" is a book-length treatment of just that question.

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Thank you, especially for references. – arkadiy Jan 12 '13 at 16:02