Assume we have an iid sequence $(\varepsilon_i)_{i \in \mathbb{Z}}$ of real valued random variables on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$. Let $\xi_n = (\dots,\varepsilon_{n-1},\varepsilon_n)$ be the one sided Bernoulli-Shift. Let $(\mathcal{F}_n)_{n \in \mathbb{Z}} = (\sigma(\xi_n))_{n \in \mathbb{Z}}$ the filtration generated by $\xi_n$. Let further $G: \mathbb{R} \to \mathbb{R}$ measurable and let $X_n = G(\xi_n)$ and let $l \in \mathbb{N}$.
Now let $$X_n' = \mathbb{E}(X_n \mid \varepsilon_{n-l},\dots,\varepsilon_{n}).$$
Then the sequence $(X_n')_{n \in \mathbb{Z}}$ is $l$-dependent ($X_i', X_{i+k}'$ are independent whenever $k > l$) as a consequence of the $\varepsilon_i$ being independent and the Doob-Dynkin lemma.
I'm currently try to handle $\mathbb{E}(X_n - X_n' \mid \mathcal{F}_j)$:
If $n \leq j$, then both $X_n$ and $X_n'$ are $\mathcal{F}_{j}$ measurable and hence $\mathbb{E}(X_n - X_n' \mid \mathcal{F}_j) = X_n - X_n'$.
If on the other hand $ j \leq n-l$, then (at least if I use the Doob-Dynkin lemma correctly) $X_n'$ is independent from $\mathcal{F}_j$ and hence $\mathbb{E}(X_n - X_n' \mid \mathcal{F}_j) = \mathbb{E}(X_n\mid \mathcal{F}_j) - \mathbb{E}(X_n')$.
Does anyone have an idea how an expression could look like for the case $ n-l > j > n$?
For a subset $I$ of $\mathbb Z$, let $\mathcal G_I$ be the $\sigma$-algebra generated by the random variables $\varepsilon_i, i\in I$. Then for each integrable random variable $Y$ and $I,J\subset\mathbb Z$, $$ \mathbb E\left[ \mathbb E\left[ Y\mid\mathcal G_I\right]\mid\mathcal G_J\right]=\mathbb E\left[Y \mid\mathcal G_{I\cap J}\right]. $$ To see this, let $J'=J\setminus I$ and let $Z=\mathbb E\left[ Y\mid\mathcal G_I\right]$. Then $$ \mathbb E\left[ \mathbb E\left[ Y\mid\mathcal G_I\right]\mid\mathcal G_J\right]=\mathbb E\left[Z\mid\mathcal G_{I\cap J}\vee \mathcal G_{J'}\right]. $$ Since $\mathcal G_{J'}$ is independent of $\sigma(Z)\vee \mathcal G_{J'}$, we get that $$ \mathbb E\left[Z\mid\mathcal G_{I\cap J}\vee \mathcal G_{J'}\right]=\mathbb E\left[Z\mid\mathcal G_{I\cap J}\right] $$ and going back to the expression of $Z$ gives the result.