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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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For example if $E(X|\mathcal{Q})$ is equal (a.s.) to a continuous function, then the continuous function would be a canonical version. –  Quinn Culver Nov 8 '11 at 3:46
    
@QuinnCulver: Thanks! Are you saying if a random variable $Y$ equals $E(X|\mathcal{Q})$ a.e., then $Y$ is a canonical version of $E(X|\mathcal{Q})$? In other words, any version of $E(X|\mathcal{Q})$ is a canonical one? –  Tim Nov 8 '11 at 3:54
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No. I'm saying that if a continuous random variable $Y$ equals $E(X|\mathcal{Q})$ a.e., then $Y$ Is a canonical version of $E(X|\mathcal{Q})$. In general, there won't be a continuous random variable that is equal to $E(X|\mathcal{Q})$ a.e., but when there is, it is certainly a canonical version. –  Quinn Culver Nov 8 '11 at 4:21
    
Thanks! Why is a continuous version canonical? How is "canonical" defined? –  Tim Nov 8 '11 at 6:11
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You only truly have a canonical version of the conditional expectation if the sigma-algebra $\mathcal{Q}$ is finite, and every nonempty element has positive probability. Otherwise, you can ask for a canonical continuous version with respect to some topology on $\Omega$. More generally, given a (Borel) measurable map $Y\colon\Omega\to E$ to a topological space $E$ with full support, then you could ask for the conditional expectation to be a continuous function of $Y$. This is more natural if $\mathcal{Q}=\sigma(Y)$, and is only canonical with respect to $Y$ and the topology on $E$. –  George Lowther Nov 9 '11 at 0:42

2 Answers 2

up vote 2 down vote accepted

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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What's $Z$? Is it a real valued random variable, or can it lie in some more general space? Also, when you say $u=v$ almost everywhere, is this with regards to the probability distribution of $Z$? Also, I think you need the support of the distribution of $Z$ to be the whole space ($\mathbb{R}$ or whichever space $Z$ lies in) in order to conclude that $v$ is unique. –  George Lowther Nov 8 '11 at 23:18
    
See edit. Thanks for the constructive comments. –  Did Nov 8 '11 at 23:48

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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Continuous function of what? This would assume that $\Omega$ is a topological space... –  Did Nov 8 '11 at 21:36
    
@DidierPiau I used the word "if" for that exact reason. Of course $E(X|\mathcal{Q})$ won't be continuous in general (and in general that might not even make sense since $\Omega$ need not be a topological space), but if it is, it's canonical. Notice that the quote was that it is "sometimes possible". –  Quinn Culver Nov 9 '11 at 1:48
    
Are you talking about standard spaces, à la Rokhlin? In the present state of your answer, this is not clear to me. –  Did Nov 9 '11 at 6:31
    
No, I'm just giving an instance of when it is possible to choose a canonical representative: when the sample space is also a topological space and $E(X|\mathcal{Q})$ is a.s. equal to a continuous function. –  Quinn Culver Nov 9 '11 at 15:08
    
@DidierPiau Is your point that if, for example, the topology is trivial, then any function is continuous, so there's still not a canonical representative? –  Quinn Culver Nov 10 '11 at 17:10

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