Someone knows of some definition or reference of how to define conditional expectation for a measure space with $\sigma$-finite measure.

I think it should be as follows:

Let $(X,\mathcal{B},\nu)$ be a measure space and let $\mathcal{F}\subset\mathcal{B}$ a sub$-\sigma-$algebra, such that $\nu$ is $\sigma-$finite in $\mathcal{F}$. Then for all $f\in L^1(X,\mathcal{B},\nu)$ there exists $g\in L^1(X,\mathcal{F},\nu|_{\mathcal{F}})$ such that $$\int_{E}fd\nu=\int_Egd\nu|_{\mathcal{F}},\qquad\forall E\in\mathcal{F};$$ then $g=:\mathbb{E}_{\nu}[f|\mathcal{F}]$ is called the conditional expectation of $f$ given $\mathcal{F}$.

Is this the correct way to define conditional expectation? There is another way to define it without requiring the hypothesis that $\nu$ be $\sigma$-finite in $\mathcal{F}$?


2 Answers 2


$\newcommand{\N}{\mathbb N} \newcommand{\R}{\mathbb R} \newcommand{\B}{\mathcal B} \newcommand{\F}{\mathcal F} \newcommand{\la}{\lambda} \newcommand{\si}{\sigma} \newcommand{\Si}{\Sigma} \renewcommand{\c}{\circ} \newcommand{\tr}{\operatorname{tr}}$

The definition you quoted is correct.

However, there can be no reasonable notion of the conditional expectation without the sigma-finiteness condition, even in the discrete setting. E.g., let $X=\N$, $\B=2^\N$, and let $\F$ be any sigma-algebra over $\N$ containing an infinite atom $A\subseteq\N$; for instance, one may take $\F=\{\emptyset,\N\}$, with $A=\N$. Let $\nu$ be the counting measure on $\B=2^\N$, and let $f(x)=1/2^x$ for $x\in\N$. Then $E_\nu f=1\in\R$.

However, on the atom $A$ one cannot reasonably ascribe any value to the conditional expectation $E_\nu(f|\F)$, because such a value (say $v$) could reasonably be only the $\nu$-average of $f$ on $A$. Indeed, if you take $v=0$, this would imply $\int_A f\,d\nu=0$, which is false; if you take $v\ne0$, this would imply $|\int_A f\,d\nu|=|v|\nu(A)=\infty$, which is also false.

The problem here is that, while the measure $\nu$ is sigma-finite, its restriction $\nu|_\F$ to $\F$ is not.


One can define a reasonable notion of conditional expectation for arbitrary localizable measurable spaces, not necessarily σ-finite. This is explained in great detail in the answer to Is there an introduction to probability theory from a structuralist/categorical perspective?

The “pushforward for L_1-spaces” mentioned there is precisely the conditional expectation. Let me offer a few comments and expand this a little bit.

First, one doesn't really need a measure μ to talk about conditional expectations, only a measure class [μ], or equivalently, a σ-ideal N of negligible sets (alias sets of measure 0). Given a set X with a σ-algebra M of measurable subsets and a σ-ideal N of negligible subsets, one can define the set of (finite complex-valued) measures on (X,M,N) as the set of additive functions M→C that vanish on N. Infinite measures can be defined using [0,∞] instead of C. Furthermore, given a faithful finite measure μ on (X,M,N), one can identify the set of μ-integrable functions f with the set of finite measures ν via the isomorphism f↦fν supplied by the Radon-Nikodym theorem.

Given a morphism (X,M,N)→(X',M',N'), one can pushforward a finite measure on (X,M,N) and get a finite measure on (X',M',N'), by taking the preimage of a measurable subset of X' and computing its measure as a subset of X. This is the conditional expectation. In particular, in the notation of the original post, one uses the morphism (X,B,N)→(X,F,N), where N is the σ-ideal of sets with ν-measure zero. Although both the domain and codomain have the same underlying set X, it's best to think of them as different spaces, and the pushforward can then be thought of as the fiberwise integration. (For instance, take X=[0,1]×[0,1] with the Borel σ-algebra and the σ-algebra of “vertical” Borel sets. The resulting morphism will be isomorphic to the projection [0,1]×[0,1]→[0,1], and the pushforward will be the fiberwise integration map.)

Let's illustrate the above construction on the example by Iosif Pinelis. The product of the infinite measure ν and the function f is the finite measure fν. Its pushforward along the map (X,B,∅)→(X,F,∅) can be computed as follows. The codomain is isomorphic to the measurable space consisting of a single point (which is how one should think about it geometrically). By definition, measures on (X,F,∅) can be identified with complex numbers, and the pushforward of a finite measure on (X,B,∅) simply computes the measure of X. Thus, in this example the conditional expectation is the measure on (X,F,∅) that assigns the number 2 (the sum of 1/2^x) to X.

Of course, most analysts are more comfortable with functions rather than measures and prefer to use the Radon-Nikodym theorem with respect to the pushforward of ν to convert the pushforward of fν to an integrable function. However, this is only possible if the pushforward of ν is a faithful semifinite measure, whereas in this example it is a purely infinite, nonsemifinite measure.

However, the failure of the Radon-Nikodym theorem doesn't mean that the conditional expectation doesn't exist, but rather that it exists only as a finite measure that cannot be converted to an integrable function.

  • 1
    $\begingroup$ It appears that you suggest to define the conditional expectation just as the push-forward measure, with some additional care taken of the null sets. This way, you can of course avoid division by $\infty$ (and any other division!) and thus avoid the need for the sigma-finiteness condition. However, I don't think many people in probability will agree that you may call push-forward measures conditional expectations, especially because the people already have and use the separate notion of the push-forward measure, and that use is quite different from the actual use of conditional expectations. $\endgroup$ Mar 11, 2018 at 4:02
  • 1
    $\begingroup$ Of course, one can apply the Radon--Nikodym (RN) theorem, with null sets taken care of. In particular, one can apply the RN theorem to the restriction of the measure to a sub-sigma-algebra. So then, what is your suggestion? To expropriate the term "conditional expectation" from its current usage and give it to the push-forward measure? But why do that? We already have the established notion of a push-forward measure, and we know that one does not need sigma-finiteness to have such a measure. $\endgroup$ Mar 11, 2018 at 4:56
  • 1
    $\begingroup$ I cannot see why what you proposed, "pushforward a finite measure on (X,M,N) and get a finite measure on (X',M',N') [...]. This is the conditional expectation", is an extension of the accepted notion of the conditional expectation to non-finite-measures. Indeed, in the sigma-finite case, what you proposed is not at all the same as the accepted notion of the conditional expectation. As you wrote, your "conditional expectation" is a push-forward measure. So, how can I see this as anything but a suggestion about terminology, to assign the name of a known object to another known object? $\endgroup$ Mar 11, 2018 at 5:52
  • 1
    $\begingroup$ Previous comment, continued: I am wondering if your notion of "conditional expectation" (which is actually a restriction or, more generally, a push-forward of a measure) has ever been used (under this name, "conditional expectation") in any publication. If not, do you know anyone except for yourself who believes that a push-forward of a measure is a useful extension of the accepted notion of conditional expectation? $\endgroup$ Mar 12, 2018 at 13:52
  • 2
    $\begingroup$ I don't know what you mean by my claims. I don't think I have made any claims; once again, I have only questioned the relevance and usefulness of your suggestion. I also don't know what you mean by "specific examples". You asked me for an example of what I consider impressive regarding conditional expectation. To that, I answered: "I think what is deep and impressive about the accepted notion of conditional expectation is its existence, which is the main content of the Radon--Nikodym (RN) theorem." I agree to terminate this discussion now, as I have no further questions to you on the matter. $\endgroup$ Mar 13, 2018 at 2:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.