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MathematicsWhat is a canonical version of conditional expectation?
[+4] [2] Tim
[2011-11-08 02:51:28]
[ probability-theory ]
[ https://math.stackexchange.com/questions/80074/what-is-a-canonical-version-of-conditional-expectation ]

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!

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
@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
(2) 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
Thanks! Why is a continuous version canonical? How is "canonical" defined? - Tim
"Canonical" is not defined. Rather it is used to denote a form of a given object that is intuitively natural, well-behaved, or best in some other sense, among other similar objects. In this case, since the relation $W=Z$ a.s. is an equivalence relation on the collection of random variables, asking for a canonical version of $E(X|\mathcal{Q})$ is asking for a best representative of its equivalence class. If there's a continuous representative, then a) it's unique, b) it behaves well and c) a random variable can't get much better than that. - Quinn Culver
@QuinnCulver: Thanks! But generally, the domain space of $E(X|\mathcal{Q})$ is a general probability space which may not have a topology. So discussing continuous mapping defined on it may well not apply. - Tim
Good point. But the quote was that "...it is sometimes possible to choose a canonical version..." (emphasis mine). - Quinn Culver
(2) 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
@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 $Y$? - Tim
[+2] [2011-11-08 21:41:11] Did [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...


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
See edit. Thanks for the constructive comments. - Did
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[+1] [2011-11-08 21:29:00] Quinn Culver

If $E(X|\mathcal{Q})$ is equal (a.s.) to a continuous function, then the continuous function would be a canonical version.


Continuous function of what? This would assume that $\Omega$ is a topological space... - Did
@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
Are you talking about standard spaces, à la Rokhlin? In the present state of your answer, this is not clear to me. - Did
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
@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
No. Since the beginning, I am referring to the fact that you assumed that $(\Omega,\mathcal F)$ is a topological space (with its Borel sigma-algebra), without saying so, although in the canonical probabilistic setting, $(\Omega,\mathcal F)$ is just any measurable space. By the way, to mention continuous random variables like you did in comments is misleading (and I think it did mislead the OP) because continuous here often refers to the distribution of the random variable having no discrete part, and not to the random variable itself, as a function on $\Omega$, being continuous. - Did
@DidierPiau Okay, then I still think my point stands: since the OP's quote was "... it is sometimes possible to choose a canonical version of $E(X|\mathcal{Q})$.", I am merely giving an example of one of these times. "Sometimes" means "under certain conditions", and the certain conditions in this case are that the underlying space is topological. - Quinn Culver
Sure... :-) Said certain conditions which you carefully avoid mentioning since the word topological does not even appear in your answer. Do as you wish but please spare me the if-something-is-continuous-then-that-means-there-must-be-a-t‌​opological-space part, I know and this is not the point. - Did
@DidierPiau Then what is the point? - Quinn Culver
Who knows... $ $ - Did
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