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Suppose $(X,Y)$ are random variables that are continuously distributed.

Conditional probabilities are just conditional expectations: $P[Y \leq y | X = x] = E[1(Y \leq y) | X = x]$, and conditional expectations are only a.e. unique.

I am wondering how this a.e. uniqueness relates to the elementary probability formula, i.e. $P[Y \leq y | X = x] = P[Y \leq y, X = x]/P[X = x] = (\frac{\partial}{\partial x}P[Y \leq y, X \leq x])/P[X = x]$.

Specifically, the elementary formula fails if $P[X = x] = 0$. However, it also fails if $P[Y \leq y, X \leq x]$ is not differentiable with respect to $x$, which can only happen on a set of measure zero, because the function is monotonic.

I had thought that the a.e. uniqueness of the conditional expectation resulted from needing a definition of conditional expectation that made sense when conditioning on negligible events. Does it also relate to this non-differentiability in some way?

EDIT: Here is a related question I guess. Are there sufficient conditions under which the conditional expectation will be everywhere unique?

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Perhaps you should evaluate $P[X=x]$ also as $\frac{\partial}{\partial x}P[X \leq x]$ in your formula for $$P[Y \leq y | X = x] = \cdots = (\frac{\partial}{\partial x}P[Y \leq y, X \leq x])/P[X = x]?$$ –  Dilip Sarwate Jan 5 '12 at 16:35
So really both problems are the same: $\frac{\partial}{\partial x}P[X \leq x]$ and $\frac{\partial}{\partial x}P[Y \leq y, X \leq x]$ both only exist a.e. (at points where the underlying density is continuous)? Is there any reason that the sets where these derivatives exist should coincide? If $\frac{\partial}{\partial x}P[Y \leq y, X \leq x]$ exists for a given $y$, that wouldn't seem to imply that $\frac{\partial}{\partial x}P[X \leq x] = \frac{\partial}{\partial x}\int P[Y = y, X \leq x] dy$ would need to exist... –  evencoil Jan 5 '12 at 17:04

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