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Assume that we have probability spaces $\left(\Omega_i,\mathcal{A}_i,\mathcal{P}_i\right)$ for $i=1,2$. We form the probability space $\left(\Omega_1\times\Omega_2,\mathcal{A}_1\otimes\mathcal{A}_2,\mathcal{P}_1\times\mathcal{P}_2\right)$. What can we say about the conditional expectation $$ E_{\mathcal{P}_1\times\mathcal{P}_2}[X|\mathcal{A}_1'\otimes\mathcal{A}_2'] $$ for $\mathcal{A}_i'\subset\mathcal{A}_i$ and $X:\left((\Omega_1\times\Omega_2),(\mathcal{A}_1\otimes\mathcal{A}_2))\rightarrow(\mathbb{R},\mathcal{B}(\mathbb{R}))\right)$.

Is it possible to reduce it in some cases to the conditional expectation on the underlying spaces?

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    What is $X$ here? – 2017-02-03
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    @zhoraster I edited the question. – 2017-02-03
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    Ok, in this generality I think nothing can be said. If $\mathcal A$ itself is a product $\mathcal A_1'\otimes \mathcal A_2'$, then (I guess) the conditional expectation may be taken as repeated expectation: conditionally of $ \mathcal A_1'$ w.r.t. $\omega_1$ and then conditionally on $\mathcal A_2'$ w.r.t. $\omega_2$. – 2017-02-03
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    @zhoraster I think i am interested in this particular case. – 2017-02-03

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I claim that $$ \mathbf{E}_{\mathcal{P}_1\times\mathcal{P}_2}[X\mid\mathcal{A}_1'\otimes\mathcal{A}_2'] = \mathbf{E}_{\mathcal{P}_2}[\mathbf{E}_{\mathcal{P}_1}[X\mid\mathcal{A}_1']\mid\mathcal{A}_2']=:\eta, $$ where in the rhs the first expectation is taken w.r.t. $\omega_1$, the second, w.r.t. $\omega_2$. To this end, we need to show that the rhs is $\mathcal{A}_1'\otimes\mathcal{A}_2'$-measurable (I'll leave this for you) and that $$ \mathbf{E}_{\mathcal{P}_1\times\mathcal{P}_2}[X\mathbf{1}_A] = \mathbf{E}_{\mathcal{P}_1\times\mathcal{P}_2}[\eta\mathbf{1}_A] $$ for any $A\in\mathcal{A}_1'\otimes\mathcal{A}_2'$. By a monotone class argument, it is enough to show this for $A = A_1\times A_2$, where $A_1\in \mathcal{A}_1'$, $A_2\in\mathcal{A}_2'$. Using the Fubini theorem, $$ \mathbf{E}_{\mathcal{P}_1\times\mathcal{P}_2}[\eta\mathbf{1}_{A_1\times A_2}] = \mathbf{E}_{\mathcal{P}_1}[\mathbf{E}_{\mathcal{P}_2}[\eta\mathbf{1}_{\omega_1\in A_1}\mathbf{1}_{\omega_2\in A_2}]]\\ = \mathbf{E}_{\mathcal{P}_1}\big[\mathbf{E}_{\mathcal{P}_2}[\mathbf{E}_{\mathcal{P}_2}\big[\mathbf{E}_{\mathcal{P}_1}[X\mid\mathcal{A}_1']\mid\mathcal{A}_2']\mathbf{1}_{\omega_1\in A_1}\mathbf{1}_{\omega_2\in A_2}\big]\big] \\ = \mathbf{E}_{\mathcal{P}_1}\big[\mathbf{E}_{\mathcal{P}_2}[\mathbf{E}_{\mathcal{P}_2}\big[\mathbf{E}_{\mathcal{P}_1}[X\mid\mathcal{A}_1']\mid\mathcal{A}_2']\mathbf{1}_{\omega_2\in A_2}\big]\mathbf{1}_{\omega_1\in A_1}\big]\\ = \mathbf{E}_{\mathcal{P}_1}\big[\mathbf{E}_{\mathcal{P}_2}[\mathbf{E}_{\mathcal{P}_1}[X\mid\mathcal{A}_1']\mathbf{1}_{\omega_2\in A_2}\big]\mathbf{1}_{\omega_1\in A_1}\big]\\ = \mathbf{E}_{\mathcal{P}_2}\big[\mathbf{E}_{\mathcal{P}_1}[\mathbf{E}_{\mathcal{P}_1}[X\mid\mathcal{A}_1']\mathbf{1}_{\omega_1\in A_1}\big]\mathbf{1}_{\omega_2\in A_2}\big]\\ = \mathbf{E}_{\mathcal{P}_2}\big[\mathbf{E}_{\mathcal{P}_1}[X\mathbf{1}_{\omega_1\in A_1}\big]\mathbf{1}_{\omega_2\in A_2}\big] = \mathbf{E}_{\mathcal{P}_1\times \mathcal{P}_2}\big[X\mathbf{1}_{A_1\times A_2}\big], $$ as required.

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    Isn't your notation a bit confusing? Since $X$ is $\mathbb{R}$ valued, we have $\mathbf{E}_{\mathcal{P}_1\times\mathcal{P}_2}[X(\omega_1,\omega_2)\mid\mathcal{A}_1'\otimes\mathcal{A}_2']=X(\omega_1,\omega_2)$. – 2017-02-04
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    @peer, it is not my notation, but rather yours. However, I don't find it confusing. Why do you think the equality you've written is true? What this has to do with $\mathbb{R}$-valuedness of $X$? – 2017-02-04
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    Well, $\mathbf{E}_{\mathcal{P}_1\times\mathcal{P}_2}[x\mid\mathcal{A}_1'\otimes\mathcal{A}_2']=x$ for $x\in\mathbb{R}$ and $X(\omega_1,\omega_2)\in\mathbb{R}$ for $(\omega_1,\omega_2)\in\Omega_1\times\Omega_2$. Isn't it better to write $\mathbf{E}_{\mathcal{P}_1\times\mathcal{P}_2}[X\mid\mathcal{A}_1'\otimes\mathcal{A}_2'](\omega_1,\omega_2)$? – 2017-02-04
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    @peer, I see your point. Is it more clear now? – 2017-02-04
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    Yes, i get the point. Thank you. But still, from a notational point of view, it seems not correct to me to write down $\omega_1$ and $\omega_2$ in your calculation. – 2017-02-05
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    There are still a few points in your proof/claim, which i find annoying: First of all, $X$ is a mapping from $\Omega_1\times\Omega_2$ into $\mathbb{R}$, so it is problematic to write $E_{\mathcal{P}_1}[X\mid \mathcal{A}_1^{'}]=:E^{\mathcal{A}_1^{'},\mathcal{P}_1}(X)$, since $E^{\mathcal{A}_1^{'},\mathcal{P}_1}$ is a mapping from $L^1(\mathcal{A}_1,\mathcal{P}_1)$ into $L^1(\mathcal{A}_1^{'},\mathcal{P}_1)$, but $X\notin L^1(\mathcal{A}_1,\mathcal{P}_1)$. This can be fixed probably if we consider the function $X(.,\omega_2)$ for $\omega_2\in\Omega_2$. – 2017-07-17
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    And why do we have that $E_{\mathcal{P}_1}[X\mid \mathcal{A}_1^{'}]\in L^1(\mathcal{A}_2,\mathcal{P}_2)$ in order to be able to write $E^{\mathcal{A}_2^{'},\mathcal{P}_2}(E_{\mathcal{P}_1}[X\mid\mathcal{A}_1^{'}])$? – 2017-07-17
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    1. There were $\omega_1$ and $\omega_2$ initially in my post, but I removed them after you found them confusing. 2. Fubini theorem. – 2017-07-18
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    If i use the "basic" notation, then $$ E_{\mathcal{P}_1\times\mathcal{P}_2}[E_{\mathcal{P}_1\times\mathcal{P}_2}[X\mid\mathcal{A}_1^{'}\otimes\mathcal{A}_2^{'}];A_1\times A_2] =\int_{A_1}\int_{A_2}X(\omega_1,\omega_2)d\mathcal{P}_2(\omega_2)d\mathcal{P}_1(\omega_1) =\int_{A_1}\int_{A_2}E_{\mathcal{P}_2}[X(\omega_1,.)\mid\mathcal{A}_2^{'}](\omega_2)d\mathcal{P}_2(\omega_2)d\mathcal{P}_1(\omega_1) =\int_{A_2}\int_{A_1}E_{\mathcal{P}_2}[X(\omega_1,.)\mid\mathcal{A}_2^{'}](\omega_2)d\mathcal{P}_1(\omega_1)d\mathcal{P}_2(\omega_2)$$ – 2017-07-18
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    $$=\int_{A_2}\int_{A_1}E_{\mathcal{P}_1}[Y(.)(\omega_2)\mid\mathcal{A}_1^{'}](\omega_1)d\mathcal{P}_1(\omega_1)d\mathcal{P}_2(\omega_2) =\int_{A_1\times A_2}E_{\mathcal{P}_1}[Y(.)(\omega_2)\mid\mathcal{A}_1^{'}](\omega_1)d\mathcal{P}_1\times\mathcal{P}_2(\omega_2,\omega_1)) $$ where $Y:\Omega_1\rightarrow L^1(\mathcal{P}_1,\mathcal{A}_1^{'})$ is defined by $Y(\omega_1)=E_{\mathcal{P}_2}[\tilde{X}(\omega_1)\mid\mathcal{A}_2^{'}](.)$ for $\omega_1\in\Omega_1$ and $\tilde{X}(\omega_1)=X(\omega_1,.)$. What do you think of that? – 2017-07-18
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    This seems to say exactly what I wrote. If you find these equalities clearer, feel free to edit my post. – 2017-07-19