• $E_1$ be a normed $\mathbb R$-vector space
  • $E_2$ be a separable $\mathbb R$-Banach space
  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $\mathcal F\subseteq\mathcal A$ be a $\sigma$-algebra of $\Omega$
  • $X$ be an $\mathfrak L(E_1,E_2)$-valued random variable on $(\Omega,\mathcal A,\operatorname P)$
  • $Y$ be an $E_1$-valued random variable on $(\Omega,\mathcal A,\operatorname P)$

I want to show, that if $X\in\mathcal L^1(\operatorname P,\mathfrak L(E_1,E_2))$, $XY\in\mathcal L^1(\operatorname P,E_2)$ and $Y$ is $\mathcal F$-measurable, then $$\operatorname E\left[XY\mid\mathcal F\right]=\operatorname E\left[X\mid\mathcal F\right]\:Y\;.\tag 1$$

If $E_1=E_2=\mathbb R$, then $(1)$ is an elementary result and I know how to prove it. How can we prove it in the more general case described here?

Besides the proof of $(1)$, I wonder if all the basic properties (like the "tower property", etc.) of the conditional expectation in the real-valued case generalize to the Banach space case. Unfortunately, I couldn't find any textbook which gives more than a simple existence/uniqueness result of the conditional expectation in Banach spaces. So, I would be happy if someone could give me a good reference.

  • $\begingroup$ Are you willing to assume $E_2$ is a dual space? $\endgroup$ – Uri Bader Oct 7 '16 at 11:54
  • $\begingroup$ @UriBader No, but I'm willing to assume that $E_1,E_2$ are separable Hilbert spaces or that $E_2$ admits a Schauder basis. $\endgroup$ – 0xbadf00d Oct 7 '16 at 12:02
  • $\begingroup$ You need an assumption on the integrability of $X$, don't you? $\endgroup$ – Nate Eldredge Oct 7 '16 at 12:20
  • $\begingroup$ @NateEldredge Of course! Thanks for pointing that out. $\endgroup$ – 0xbadf00d Oct 7 '16 at 12:27
  • 1
    $\begingroup$ You don't want to assume that $E_2$ is a dual space but a nevertheless Hilbert??? $\endgroup$ – Jochen Wengenroth Oct 7 '16 at 12:59

The answer is yes in full generality. You don't need the separability assumption on $E_2$ and the integrability assumption on $XY$ (though clearly you do have to assume that $X\in L^1(\Omega,L(E_1,E_2))$). In comments to the question I was concerned about $E_2$ being a dual space, but that was based on my own confusion and it is irrlevant.

Let me first note that if $\mathcal{F}$ is trivial then $Y$ is simply a vector in $E_1$ and your equation reads $$ (*)\quad E[X]Y=E[XY]$$ and is straight forward to prove.

For the genral case, I prefer to think of $\mathcal{F}$ as a factor. That is, I consider a measurable map $\pi:\Omega \to \Omega'$ where $(\Omega',\mathcal{A}',P')$ is another probability space in which the points are separated, $\mathcal{F}=\pi^*\mathcal{A}$ and $P'=\pi_*P$. It is known that such a space exists (and it is unique in an apropriate sense) and for every $\mathcal{F}$-measurable map $Z:\Omega\to E$ there exists an $\mathcal{A}'$-measurable map $Z':\Omega'\to E_1$ such that $Z=Z'\circ\pi$. We say that $Z$ is the pull-back of $Z'$.

Now we can disintegrate the measure $P$ over $P'$. That is, for a.e $\omega'\in\Omega'$ there exists a probability measure $P_{\omega'}$ on $\Omega_{\omega'}:=\pi^{-1}(\{\omega'\})$ (varying measurably wrt $\mathcal{A}$) such that $P=\int_{\Omega'} P_{\omega'}dP(\omega')$. For every Banach space $E$ and $\mathcal{A}$-measurable map $Z:\Omega\to E$, given $\omega'\in\Omega'$ we may consider the restriction $Z_{\omega'}:=Z|_{\Omega_{\omega'}}$ and its expectation wrt $P_{\omega'}$. This defines a function $Z':\Omega'\to E$. It is a basic fact that the conditional expectation $E(Z\mid \mathcal{F})$ is the pull back $Z'$.

The equation $E[XY\mid \mathcal{F}]=E[X\mid\mathcal{F}]Y$ now becomes simply the equation $(*)$, when reduced to the fibers of $\pi$. Indeed, fixing $\omega'\in\Omega'$ and applying $(*)$ for $X_{\omega'}$ and $Y'(\omega')$ (instead of $X$ and $Y$) we get $E[X_{\omega'}]Y(\omega')=E[X_{\omega'}Y(\omega')]$ and pulling back we get the required equation.

Let me conclude by the personal remark that I (a.s) prefer to think geometrically, replacing subalgebras by factors and conditional expectation by integration over fibers. In particular, I prefer "integral" over "expextation". Nevertheless, I am well aware of some advantages of the "probabilistic" mind set over the "geometric" one.

| cite | improve this answer | |
  • $\begingroup$ There is a remaining problem discussed in the comments below the question: Does $\operatorname E\left[X\mid\mathcal F\right]$ exist, even when $X$ takes values in a non-separable space? $\endgroup$ – 0xbadf00d Oct 8 '16 at 9:47
  • $\begingroup$ @0xbadf00d In my answer I didn't discuss the definition of what $L^1(\Omega,E)$ is. I assume (as is standard) that the functions there are strongly measurable, hence have (essentially) separable images. Are you using a different convention? $\endgroup$ – Uri Bader Oct 8 '16 at 10:09
  • $\begingroup$ Also I am assuming throughout that $\Omega$ is standard Borel space, eg $[0,1]$. $\endgroup$ – Uri Bader Oct 8 '16 at 10:22
  • $\begingroup$ What is strongly measurable? Why do you assume that $\Omega$ is standar Borel space? Please, could you give me a reference where to find it. $\endgroup$ – Daniel Camarena Perez Jan 19 '19 at 15:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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