In Section 4.2.4 of [1], the authors write



> 
> In this section we consider a causal linear process
> $$
> X\_t = \sum\_{j = 0}^\infty a\_j \varepsilon\_{t - j}, \quad t \in \mathbb{N},
> $$
> where, without loss of generality, $\sum\_{j = 0}^\infty a\_j^2 = 1$ and $\varepsilon\_t$ $(t \in \mathbb{Z})$ are i.i.d. zero mean random variables with Var$(\varepsilon\_1) = \sigma^2\_\varepsilon < \infty$. Thus, Var$(X\_1) = \sigma^2\_X = \sigma^2\_\varepsilon$. Note that
> Gaussian processes are included in this definition, but the class is much more general.
> 
> 
> 


I read the last sentence as "all Gaussian processes are causal linear processes". And [2] seems to support that. It argues that *"...the Mallows MA closure is exhausted by three types of processes. The first type is the set of stationary Gaussian processes with mean zero..."*.


However, a counter-example can be constructed by a Gaussian process which is defined as $X(t) = Y$ for all $t \in \mathbb{N}$ where $Y$ is a zero-mean Gaussian random variable. This is an (admittedly degenerate) Gaussian process and I fail to see how it can be expressed as a linear process.


So, is my counter-example wrong? Or do we need additional assumptions for a Gaussian process to be a (causal) linear process? [1] often works with regularly varying auto-covariance functions resp. spectral densities. Has this maybe been assumed implicitly? If so, how do these assumptions help to prove the claim?


References
==========


[1] J. Beran, Y. Feng, S. Ghosh, and R. Kulik. Long-memory processes. Springer, 2016.


[2] <https://www.pnas.org/doi/pdf/10.1073/pnas.93.22.12128>



Why is every Gaussian process a linear process?
$\newcommand\ep\varepsilon\newcommand\si\sigma\newcommand\N{\mathbb N}$Your counterexample is correct. Indeed, if
$$X\_t=\sum\_{j=0}^\infty a\_j \ep\_{t-j} \tag{1}\label{1}$$
for $t\in\N$, $\sum\_{j=0}^\infty a\_j^2=1$, and the $\ep\_t$'s are iid zero-mean random variables with variance $\si^2\in(0,\infty)$, then for $u\in\N$
$$EX\_tX\_{t+u}=\si^2\sum\_{j=0}^\infty a\_ja\_{j+u}\to0$$
as $u\to\infty$, by the Cauchy--Schwarz inequality.


This contradicts the condition that $X\_t=Y$ for all $t$, because then we would have $EX\_tX\_{t+u}=EX\_1^2=\si^2>0$ for all $u\in\N$.




---


Generally (see e.g. [Theorem 2, p. 263](https://link.springer.com/content/pdf/bfm%3A978-3-642-61943-4%2F1.pdf)), a Gaussian process $(X\_t)$ is of the form \eqref{1} if and only if $(X\_t)$ is a zero-mean stationary process with $EX\_t^2<\infty$ which has an absolutely continuous spectral measure and its spectral
density $f$ satisfies the condition
$$\int\_{-\pi}^\pi\ln f(t)\,dt>-\infty$$
(so that $f$ does not get too close to $0$ on any set of significant enough Lebesgue measure).


(In your counterexample, the spectral measure is degenerate, a multiple of the Dirac measure supported on the singleton set $\{0\}$, and hence not all absolutely continuous.)


