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In David Williams's Probability with Martingales, there is a remark regarding conditional expectation of a random variable conditional on a $\sigma$-algebra:

The 'a.s.' ambiguity in the definition of conditional expectation is something one has to live with in general, but it is sometimes possible to choose a canonical version of $E(X| \mathcal{Q})$.

What is "canonical version of $E(X| \mathcal{Q})$", and what are some cases when it is possible to choose it?

I don't want to be misleading, but is it referring to elementary definitions of conditional distribution and conditional expectation when they exist i.e. when the denominators are not zero?

Thanks and regards!

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    @GeorgeLowther: Thanks! Why is "you only truly have a canonical version of the conditional expectation if the sigma-algebra Q is finite, and every nonempty element has positive probability"? In the finite $\mathcal{Q}$ case, is the canonical version a continuous function of $Y$?2011-11-09

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Assume that $\mathcal Q=\sigma(Z)$ for some real valued random variable $Z$, then $E(X\mid\mathcal Q)=u(Z)$ almost surely, for a given measurable function $u:\mathbb R\to\mathbb R$, as well as for every other measurable function $v$ such that $u=v$ $P_Z$-almost everywhere. If one of these functions $v$ is, say, continuous, then $v(Z)$ might be called a canonical version of $E(X\mid\mathcal Q)$.

Unfortunately, this is a dubious denomination since it may well happen that \mathcal Q=\sigma(Z') for a quite different real valued random variable Z'. Even if E(X\mid\mathcal Q)=v'(Z') almost surely, for a given continuous function v', nothing ensures that v(Z)=v'(Z') everywhere. One only knows that v(Z)=v'(Z') almost surely and one is back at square one, which is that there is no way to decide which random variable $v(Z)$ or v'(Z') is more canonical than the other...

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    See edit. Thanks for the constructive comments.2011-11-08
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If $E(X|\mathcal{Q})$ is equal (a.s.) to a continuous function, then the continuous function would be a canonical version.

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    Who knows... $ $2011-11-14