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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 $E[X | \sigma(\mathcal{B_1},\mathcal{B_2})] = 0$.

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  • $\begingroup$ Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: your background and motivation for asking this question, relevant definitions, the source of the question, possible strategies you've considered, your current progress in terms in trying to answer the question, why the question is interesting or important, etc. $\endgroup$
    – amWhy
    Dec 29, 2018 at 22:32
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    $\begingroup$ Hint: use the tower property of the conditional expectation $\endgroup$
    – Michh
    Dec 29, 2018 at 22:37
  • $\begingroup$ @Michh This does not work. See my answer. $\endgroup$ Dec 29, 2018 at 23:31
  • $\begingroup$ @KaviRamaMurthy My bad. Thank you for correcting me. $\endgroup$
    – Michh
    Jan 1, 2019 at 10:51
  • $\begingroup$ Given that this thread is relatively old I don't think OP is going to see this comment, but the main problem here is that $\mathcal{B}_1 \cup \mathcal{B}_2$ may not be a $\pi$ - system. If it were, then the claim would be true. $\endgroup$ Jan 7, 2023 at 14:26

1 Answer 1

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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 $EXI_{B\cap C} \neq 0$.

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