What 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 versionof $E(X| \mathcal{Q})$.

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!

[+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**

1

[+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-topological-space* part, I know and this is not the point. - ** Did**

@DidierPiau Then what is the point? - ** Quinn Culver**

Who knows... $ $ - ** Did**

2

Quinn CulverTimcontinuousrandom 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 CulverTimQuinn CulverTimsometimespossible to choose a canonical version..." (emphasis mine). -Quinn CulverGeorge LowtherTim