Conditional expectations on 2 independent sigma-algebras I saw the next property of Conditional Expectation:
If $\mathcal{G}_1$ and $\mathcal{G}_2$ are two independent $\sigma$-algebras, then:
$\mathbb{E}(X|\sigma(\mathcal{G}_1 \cup \mathcal{G}_1)) = \mathbb{E}(X|\mathcal{G}_1) + \mathbb{E}(X|\mathcal{G}_2)-\mathbb{E}(X)$.
I got an interesting answer stating that it is not always true. So the new question would be, what is the answer to $\mathbb{E}(X|\sigma(\mathcal{G}_1 \cup \mathcal{G}_2))$? Here is what I have done until now:
We know that:
$$\sigma(\mathcal{G}_1 \cup \mathcal{G}_2)=\sigma(S), \quad S=\{A_{1}\cap A_{2}, A_{1}\in \mathcal{G}_{1}, A_{2} \in \mathcal{G}_2\}$$
So, we have that:
$$\int_{C}\mathbb{E}(X|\sigma(\mathcal{G}_1 \cup \mathcal{G}_2))dP = \int_{C}XdP, \quad for \; any\;\, C\in \sigma(\mathcal{G}_1 \cup \mathcal{G}_2) $$
If we only focus on $S$, then we will have that $C=A_{1}\cap A_{2}$ then, the random variable $\mathbb{I}_{A_{1} \cap A_{2}} = \mathbb{I}_{A_{1} }+\mathbb{I}_{ A_{2}}-\mathbb{I}_{A_{1} \cup A_{2}}$ so:
$$\int_{{A_{1} \cap A_{2}}}\mathbb{E}(X|\sigma(\mathcal{G}_1 \cup \mathcal{G}_2))dP = \int_{{A_{1} \cap A_{2}}}XdP=\int_{\Omega}X\,\mathbb{I}_{A_{1} \cap A_{2}}dP$$
$$\int_{\Omega}X\,\mathbb{I}_{A_{1} \cap A_{2}}dP=\int_{\Omega}X\,(\mathbb{I}_{A_{1}}+\mathbb{I}_{A_{2}}-\mathbb{I}_{A_{1} \cup A_{2}})dP$$
$$\int_{\Omega}X\,(\mathbb{I}_{A_{1}}+\mathbb{I}_{A_{2}}-\mathbb{I}_{A_{1} \cup A_{2}})dP = \int_{\Omega}X\,\mathbb{I}_{A_{1}}dP+\int_{\Omega}X\,\mathbb{I}_{A_{2}}dP-\int_{\Omega}X\,\mathbb{I}_{A_{1} \cup A_{2}}dP$$
$$\int_{\Omega}X\,\mathbb{I}_{A_{1}}dP+\int_{\Omega}X\,\mathbb{I}_{A_{2}}dP-\int_{\Omega}X\,\mathbb{I}_{A_{1} \cup A_{2}}dP = \int_{A_{1}}XdP+\int_{A_{2}}XdP-\int_{A_{1} \cup A_{2}}XdP$$
So, what I've gotten is
$$\int_{C}\mathbb{E}(X|\sigma(\mathcal{G}_1 \cup \mathcal{G}_2))dP = \int_{A_{1}}XdP+\int_{A_{2}}XdP-\int_{A_{1} \cup A_{2}}XdP$$
$$\int_{C}\mathbb{E}(X|\sigma(\mathcal{G}_1 \cup \mathcal{G}_2))dP = \int_{A_{1}}\mathbb{E}(X|\mathcal{G}_1)dP + \int_{A_{2}}\mathbb{E}(X|\mathcal{G}_2)dP - \int_{A_{1} \cup A_{2}} \mathbb{E}(X|\mathcal{G}_1 \cup \mathcal{G}_2)dP$$
And $\mathbb{E}(X|\mathcal{G}_1)$, $\mathbb{E}(X|\mathcal{G}_2)$ and $\mathbb{E}(X|\mathcal{G}_1 \cup \mathcal{G}_2)$ are $\sigma(\mathcal{G}_1 \cup \mathcal{G}_2)$-measurables, so:
$$\mathbb{E}(X|\sigma(\mathcal{G}_1 \cup \mathcal{G}_2)) = \mathbb{E}(X|\mathcal{G}_1) + \mathbb{E}(X|\mathcal{G}_2) - \mathbb{E}(X|\mathcal{G}_1 \cup \mathcal{G}_2)$$
but, for any $C \in S$. The problem is that I don't see where to use the fact that the $\sigma$-algebras are independents.
Any comments?
 A: Where did you see this  property of Conditional Expectation?
It does not hold in general.
Suppose that $Y,Z$ are independent variables of mean zero taking values $\pm 1$, and that $X=YZ$. Let $\mathcal G_1=\sigma(Y)$ and $\mathcal G_2=\sigma(Z)$. Then $X$ is measurable under $\sigma(\mathcal G_1 \cup \mathcal G_2)$, but
$$\mathbb{E}(X)=\mathbb{E}(X|\mathcal{G}_1)=\mathbb{E}(X|\mathcal{G}_2)=0 \,.
$$
Therefore,
$$ \mathbb E(X|\sigma(\mathcal G_1 \cup \mathcal G_2)) =X \ne 0=
\mathbb{E}(X|\mathcal{G}_1) + \mathbb{E}(X|\mathcal{G}_2)-\mathbb{E}(X) \,.
$$
A: comment
If true, it would be an amazing way to avoid multiple integrals!
Let $\Omega = [0,1]\times[0,1]$ and $\mathbb P$ be $2$-dimensional Lebesgue measure.  Let $X : [0,1]^2 \to \mathbb R$ be Borel measurable.  Recall
$$
\mathbb{E}\left[X\right] = \int_0^1\int_0^1 X(x,y)\;dx\;dy .
$$
Let $\mathcal G_1$ be sets that depend only on the first coordinate and $\mathcal G_2$ the second coordinate, so
$$
\mathbb{E}\left[X\mid\mathcal G_1\right] = \int_0^1 X(x,y)\;dy,\qquad
\mathbb{E}\left[X\mid\mathcal G_2\right] = \int_0^1 X(x,y)\;dx
$$
Next, noting that $\sigma(\mathcal G_1\cup\mathcal G_2)$ is all Borel sets in $[0,1]^2$, we have
$$
\mathbb{E}\left[X\mid\sigma(\mathcal G_1\cup\mathcal G_2)\right] = X .
$$
Your "rule" lets us evaluate double integrals with great ease!
$$
\int_0^1\int_0^1 X(x,y)\;dx\;dy =
\int_0^1 X(x,y)\;dy + \int_0^1 X(x,y)\;dx - X(x,y) .
$$
Unfortunately, it is very easy to find counterexamples to this.
