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.
Best Answer
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., $$ Y-\mathsf{E}[Y\mid \mathcal{G}]=\mathsf{E}[Y\mid \mathcal{H}]-\mathsf{E}[Y]\quad\text{a.s.} $$