Lately I worked with stationary processes of the following kind. Let $(\varepsilon_i)_{i\in\mathbb{N}_0}$ be iid. RVs. Let $\xi_n := (…,\varepsilon_{n-1}, \varepsilon_n)$ and further let $g:\mathbb{R}^{\mathbb{N}_0} \to \mathbb{R}$ measurable such that $X_n := g(\xi_n)$ describes a sequence of random variables. Let now $F_n := \sigma(\xi_n)$ be the sequence of sigma algebras generated by $\xi_n$, i.e. a filtration, since $F_{n-1} \subset F_n$.
By a result in probability theory – we called it the shifting lemma – one can notice that $X_n$ is stationary and ergodic.
In the case $j \geq i$, I'm wondering whether the following statements holds for the conditional expectation of this sequence of stationary processes:
$\mathbb{E}(X_i \mid F_j) = \mathbb{E}(X_0\mid F_{j-i})$
My idea to proof this, was rather simple and I'm wondering, if I'm missing something.
Let $F \in F_{j-i} \subset F_j$. Now
$\int_F \mathbb{E}(X_i \mid F_j) d\mathbb{P} \overset{(a)}{=} \int_F X_i d\mathbb{P} \overset{(b)}{=} \int_F X_0 d\mathbb{P} $
where (a) holds by definition of the conditional expectation and (b) due to stationarity. Since $F$ was arbitrary, this should, again by definition of the conditional expectation prove the statement.
Edit: (I messed up the indices)
What I wanted to show, is that
$\mathbb{E}[X_k\mid\mathcal{F}_{j}] = \mathbb{E}[X_{k+l}\mid\mathcal{F}_{j+l}]$
for any $j < k$ and $l \geq 0$.
Also I want to use a result, which says if $h(X)$ is a measurable solution of
$\int_F h(X) d\mathbb{P} = \int_F X d\mathbb{P}$ $\forall F \in \mathcal{F}$, then $h(X) = \mathbb{E}[X \mid \mathcal{F}]$ $\mathbb{P}-a.s.$
Since for any $F \in \mathcal{F}_j$ we already have $F \in \mathcal{F}_{j+l}$ and thus
$\int_F \mathbb{E}[X_{k+l}\mid\mathcal{F}_{j+l}] d\mathbb{P} \overset{F \in \mathcal{F}_{j+l}}{=} \int_F X_{k+l} d\mathbb{P} \overset{stationary}{=} \int_F X_{k} d\mathbb{P}$
holds true for any $F \in \mathcal{F}_j$, which yields $\mathbb{E}[X_{k+l}\mid\mathcal{F}_{j+l}] = \mathbb{E}[X_k \mid \mathcal{F}_j]$ $\mathbb{P}-a.s.$.
Does this now seem correct or am I still missing something?
Best Answer
Observe first that if $j\geqslant i$, then $X_i$ is $\mathcal F_j$-measurable because $X_i$ is $\mathcal F_i$-measurable. Moreover, by a similar reasoning, $X_0$ is $\mathcal F_{j-i}$-measurable. So your equality would be $X_j=X_0$, which cannot be true.
Instead, we can assume that $j<i$. Suppose first that $X_n=\sum_{k\geqslant 0}a_k\varepsilon_{n-k}$, where $\varepsilon_i$ is centered and has a finite moment of order two. Then $$ \mathbb E\left[X_i\mid\mathcal F_{j}\right]=\sum_{k\geqslant 0}a_k\mathbb E\left[\varepsilon_{i-k}\mid\mathcal F_{j}\right] $$ and $\mathbb E\left[\varepsilon_{i-k}\mid\mathcal F_{j}\right]=0$ if $i-k\geqslant j$ and $\mathbb E\left[\varepsilon_{i-k}\mid\mathcal F_{j}\right]=\varepsilon_{i-k} $ otherwise hence $$ \mathbb E\left[X_i\mid\mathcal F_{j}\right]=\sum_{k\geqslant i-j}a_k\ \varepsilon_{i-k} $$ and $$ \mathbb E\left[X_0\mid\mathcal F_{j-i}\right]=\sum_{k\geqslant i-j}a_k\ \varepsilon_{-k} . $$ Now let us go back to the general case. We can write $\mathbb E\left[X_i\mid\mathcal F_{j}\right]$ as $g_{i,j}\left(\xi_j\right)$ (by Doob-Dynkin's's theorem) and it turns out that $\mathbb E\left[X_0\mid\mathcal F_{j-i}\right]=g_{i,j}\left(\xi_ {j-i}\right)$, that is, the same representing function works. Indeed, let $A\in\mathcal F_j$. We can write this as $A=\{\xi_j\in B\}$ for some Borel subset $B$ of $\mathbb R^{\mathbb N_0}$. Then $$ \mathbb E\left[X_0\mathbf{1}_{\xi_{j-i}\in B}\right] =\mathbb E\left[g\left(\xi_0\right)\mathbf{1}_{\xi_{j-i}\in B}\right] =\mathbb E\left[g\left(\xi_i\right)\mathbf{1}_{\xi_{j}\in B}\right] $$ where we used stationarity in the last equality. By definition of conditional expectation, we thus get $$ \mathbb E\left[X_0\mathbf{1}_{\xi_{j-i}\in B}\right]=\mathbb E\left[g_{i,j}\left(\xi_j\right)\mathbf{1}_{\xi_{j}\in B}\right]. $$ Using again stationarity gives $$ \mathbb E\left[X_0\mathbf{1}_{\xi_{j-i}\in B}\right]=\mathbb E\left[g_{i,j}\left(\xi_{j-i}\right)\mathbf{1}_{\xi_{j-i}\in B}\right]. $$