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In Adam Bobrowski's book "Functional Analysis for Probability and Stochastic Processes. An Introduction", the author introduces an interesting property for the conditional expectation:

Let $X$ be a mean-zero random variable on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ and let $\mathcal{G}, \mathcal{H}$ be two independent sub-sigma algebras. Then $$\mathbb{E}(X \mid \sigma(\mathcal{G}\cup \mathcal{H})) = \mathbb{E}(X \mid \mathcal{G}) + \mathbb{E}(X \mid \mathcal{H})$$ almost surely.

Now consider a iid sequence $X=(X_j)_{j \in \mathbb{N}}$ of real valued-random variables, consider a measurable function $g: \mathbb{R}^{\mathbb{N}} \to \mathbb{R}$ and let $Y = g(X)$ . Further let $I \subset \mathbb{N}$ be a finite subset and $I^c := \mathbb{N} \setminus I$. I think the result above doesn't imply the following $$\tag{*}Y = \mathbb{E}(Y \mid (X_j)_{j \in \mathbb{N}}) = \mathbb{E}(Y \mid (X_j)_{j \in I}) + \mathbb{E}(Y \mid (X_j)_{j \in I^c}),$$

which then would allow to compute $$Y - \mathbb{E}(Y \mid (X_j)_{j \in I}) = \mathbb{E}(Y \mid (X_j)_{j \in I^c}).$$

I'm interested in an expression for $Y - \mathbb{E}(Y \mid (X_j)_{j \in I})$ and hence I was wondering, if we could add an error term to the righthandside of $(*)$, so that the equation holds true, which then may help in finding the desired expression.

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    $\begingroup$ Is it $E[X \mid \mathcal{F}]$ or $E[X \mid \mathcal{H}]$ on the right-hand side? $\endgroup$
    – angryavian
    Jun 3, 2022 at 22:19
  • $\begingroup$ @angryavian I meant to write $\mathbb{E}(X\mid\mathcal{H})$ $\endgroup$
    – student7481
    Jun 5, 2022 at 12:47
  • $\begingroup$ Actually, the property described in the quote is not true. Here is a counter-example: Let $X_1$ and $X_2$ be independent random variables such that $\mathbf{E}[X_i^4]<\infty$ and $\mathbf{E}[X_i]=0$. If we write $\mathcal{F}=\sigma(X_1)$ and $\mathcal{G}=\sigma(X_2)$, then $X=X_1 X_2$ satisfies $\mathbf{E}[X]=\mathbf{E}[X_1]\mathbf{E}[X_2]=0$ and $$X=\mathbf{E}[X\mid\sigma(\mathcal{F}\cup\mathcal{G})]\neq\mathbf{E}[X\mid\mathcal{F}]+\mathbf{E}[X\mid\mathcal{G}]=X_1\mathbf{E}[X_2]+\mathbf{E}[X_1]X_2 = 0.$$ $\endgroup$
    – Sangchul Lee
    Jul 17, 2022 at 10:24
  • $\begingroup$ I checked the textbook, and the relevant theorem redirects reader to an exercise which essentially asks to show that $$X=\mathbf{E}[X\mid\mathcal{F}]+\mathbf{E}[X\mid\mathcal{G}]-\mathbf{E}[X]$$ for $X$ in the closed subspace $$H_0 = L^2(\Omega,\mathcal{F},\mathbf{P}) + L^2(\Omega,\mathcal{G},\mathbf{P})$$ of $L^2(\Omega,\sigma(\mathcal{F}\cup\mathcal{G}),\mathbf{P})$. The issue is that $H_0$ need not be the same as the ambient $L^2$-space. In short, the indicated equality only holds when $X$ is already of that form. $\endgroup$
    – Sangchul Lee
    Jul 17, 2022 at 10:28

1 Answer 1

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Let $\mathcal{G}=\sigma(\{X_i:i\in I\})$ and $\mathcal{H}=\sigma(\{X_i:i\in I^c\})$. Then, since $\mathcal{G}$ and $\mathcal{H}$ are independent, and $Y=\mathsf{E}[Y\mid \mathcal{G}\vee \mathcal{H}]$ a.s., $$ Y-\mathsf{E}[Y\mid \mathcal{G}]=\mathsf{E}[Y\mid \mathcal{H}]-\mathsf{E}[Y]\quad\text{a.s.} $$

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  • $\begingroup$ So indeed $Y - \mathbb{E}(Y \mid \mathcal{G})$ is independent from $\mathcal{H}$? $\endgroup$
    – student7481
    Jun 5, 2022 at 13:54
  • $\begingroup$ @student7481 Why is that? It is true, however, that $\mathsf{E}[Y\mid \mathcal{G}]$ is independent of $\mathsf{E}[Y\mid \mathcal{H}]$. $\endgroup$
    – user140541
    Jun 5, 2022 at 14:00
  • $\begingroup$ I guess I found my mistake. My thought was the following: $Y - \mathbb{E}(Y\mid\mathcal{G}) = \mathbb{E}(Y\mid \mathcal{H})$ and by the Doob-Dynkin lemma we find a measurable function $g$ such that $ \mathbb{E}(Y\mid \mathcal{H}) = g((X_j)_{j \in I})$. But $g$ may depend on the $(X_j)_{j \in I^c}$, right? $\endgroup$
    – student7481
    Jun 5, 2022 at 14:09
  • $\begingroup$ Wait. $g$ is a function of $(X_i)_{i\in I^c}$. $\endgroup$
    – user140541
    Jun 5, 2022 at 14:17
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    $\begingroup$ @student7481 $Y-\mathsf{E}[Y\mid\mathcal{G}]$ is indeed independent of $\mathcal{G}$ because the former equals to an $\mathcal{H}$-measurable random variable a.s. $\endgroup$
    – user140541
    Jun 5, 2022 at 14:44

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