I am working with an article, where it says: 



> 
> "that the discrete time stationary sequence $\{Y\_j\}\_{j\in Z}$ is
>  mixing and hence ergodic."
> 
> 
> 


where $Y\_t$ is defined as
$Y\_t = \int\_{-\infty}^{t} h\_k(t-s)dL\_s$
and $L$ is a Lévy process. 


I have been unable to find a definition of mixing and ergodic process. I can only find a definitions in (opens a PDF): 


<http://math.byu.edu/~tfisher/documents/classes/2008/winter/635/Lecture12.pdf>


Page 1 & 2. But this is not for processes and I am not sure if this relates in any way to processes. 


In this link I have found a theorem that says mixing implies ergodic. The proof of this is also not for process, but can you use this theorem anyway?



Ergodic and mixing processes
Yes, the results you quote are general statements on mixing and ergodicity, which can be translated to stochastic process as follows.


In the source you mention (and many other sources), mixing is defined as 


$\lim\_{n \rightarrow \infty} \mu(T^{-n}(A) \cap B) = \mu(A) \mu(B), $


where $T$ is a transformation of interest on a measure space $(X, \mathcal B, \mu)$ that preserves $\mu$, i.e. for every $A \in \mathcal B$, $\mu(T^{-1}A) = \mu(A)$.


In the context of stochastic processes, usually this is understood in the following way:


* $X$ is the path space of the stochastic process of interest, in your case $\mathbb R^{\mathbb Z}$, and $\mathcal B$ the $\sigma$-algebra generated by cylinder sets.
* $T$ is the shift operation, i.e. for $x \in X$, $x = (\dots, x\_{-1}, x\_0, x\_1, \dots)$, $(Tx)\_i := x\_{i+1}$.
* $\mu$ is a stationary measure on $\mathbb R^{\mathbb Z}$ corresponding to a stationary stochastic process $Y\_t$.


To illustrate this for your example, first one needs to check that the discrete time process $Y\_i$ is stationary. By a change of variables


$Y\_{i+1} = \int\_{-\infty}^{i+1} h(i+1-s) \ d L\_s = \int\_{-\infty}^t h(i -r) \ d \tilde L\_r$, 


where
$r = s-1$ and $\tilde L\_t = L\_{t+1}$. By stationarity of the Lévy process, it follows that $Y\_{i+1}$ has the same distribution as $Y\_i$. In order to fully show stationarity you have to check that the finite dimensional distributions are stationary, i.e. $(Y\_{i\_1},\dots,Y\_{i\_k})$ has identical distribution to $(Y\_{i\_1+1}, \dots, Y\_{i\_k+1})$ for any choice of indices $(i\_1,\dots, i\_k)$.


As for the mixing, you would need to check, for starters, that, as $n \rightarrow \infty$,


$ \mathbb P\left( \int\_{-\infty}^{i+n} h(i+n-s) \ d L\_s \in A \ \mbox{and} \ \int\_{-\infty}^j h(j-s) \ d L\_s \in B \right) \rightarrow \mathbb P\left(\int\_{-\infty}^i h(i-s) \ d L\_s \in A \right) \mathbb P\left(\int\_{-\infty}^j h(j-s) \ d L\_s \in B \right),$


which I guess will depend on the choice of $h$.
Again, you would need to check this for general finite dimensional distributions.


Once mixing is established, ergodicity follows as you mention. By the definition of ergodicity, any set $A \in \mathcal B$ which is "essentially" $T$-invariant (here: shift-invariant), is trivial (i.e. has probability measure 0 or 1). Translated back to stochastic processes, it follows that the infinite time average 


$\lim\_{n \rightarrow \infty} \frac 1 n \sum\_{i=1}^n Y\_i$


which exists by the Birkhoff ergodic theorem, and which is shift-invariant, is constant (except perhaps on a $\mu$-null-set): If it were not, this would split $X$ into non-trivial invariant sets.


Details can be found in the stochastic literature in various places, e.g.


* Durrett, *Probability: Theory and Examples*, 1996
* Kallenberg, *Foundations of Modern Probability*, 2002
* Da Prato & Zabczyk, *Ergodicity for infinite-dimensional systems*, 1996


