Let $(\Omega, \mathcal{A}, P)$ be a probability space and let $X : \Omega \rightarrow \mathbb{R}$ be a random variable such that $ X\in \mathcal{L}^1(\mathcal{A})$.
Assume that $\mathcal{B_1}$ and $\mathcal{B_2}$ are two $\sigma$-sub-algebras of $\mathcal{A}$ such that $E[X | \mathcal{B}_i]=0$ for $i=1,2$.
I would expect that it now holds that $E[X | \sigma(\mathcal{B_1},\mathcal{B_2})]=0$, where $\sigma(\mathcal{B_1},\mathcal{B_2})$ is the smallest $\sigma$-sub-algebra of $\mathcal{A}$ containing $\mathcal{B}_1$ and $\mathcal{B}_2$, but I cannot seem to prove it. Any ideas if this is true, and if so how to prove it?
If we add the condition that $X$ is non-negative, then it seems doable. For example if $B_i \in \mathcal{B}_i$ for $i=1,2$, then
$$ \int_{B_1 \cup B_2} X dP = \int_{B_1}XdP + \int_{B_2}XdP – \int_{B_1 \cap B_2}XdP.
$$
Since $E[X | \mathcal{B}_i]=0$, we have $\int_{B_i}XdP = 0$, and since $X$ is non-negative we have $\int_{A} X dP \geq 0$ for any $A \in \mathcal{A}$. Hence, $$\int_{B_1 \cup B_2} X dP = 0.$$
Using this idea one can show that $$\int_{B} X dP = 0$$ for every $B \in \sigma(\mathcal{B_1},\mathcal{B_2})$ and hence that $E[X | \sigma(\mathcal{B_1},\mathcal{B_2})] = 0$.
Best Answer
The is false. There exist events $A,B,C$ such that they are pairwise independent but $P(A\cap B\cap C)\neq P(A)P(B)P(C)$. Take $X=I_A-P(A),\mathcal B_1 =\{\emptyset, B, B^{c},\Omega \},\mathcal B_2 =\{\emptyset, C, C^{c},\Omega \}$. Note that $EXI_{B\cap C} \neq 0$.