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Let

  • $E_1$ 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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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