3
$\begingroup$

A linear process $(X_t)_{t \in \mathbb{Z}}$ is usually written as a moving-average process with infinity order: \begin{equation}\label{linear_process}\tag{Eq. 1.1} X_{t} = \sum_{j =0 }^\infty \psi_{j} \varepsilon_{t-j}, \forall t \in \mathbb{Z}. \end{equation} where $\varepsilon_{t}$ is a i.i.d. white noise ($E(\varepsilon_{t})=0,\,\, E |\varepsilon_{t}|^2< \infty$) and $\sum_{j =0 }^\infty \psi_{j}^2 < \infty$.

According to this paper, page 12130, the author says:

Mallows (12) argues that a linear process such as in (\ref{linear_process}) is close to a Gaussian process if $\max_{j\geq 0}|\psi_j|$ is small.

I would like to know if you have any relatively simple examples for this statement. I tried to think of a simple example, but I couldn't. I don't want to go to the Mallows paper before I go through here.

$\endgroup$
2
  • 3
    $\begingroup$ $\max_j |\psi_j|$ small is just a rewriting of the usual assumption for the CLT that no one term contains a constant proportion of the overall variance: a necessary condition for the CLT of $\sum_i\sigma_i \epsilon_i$ is $\frac{\max_k \sigma_k}{(\sum_i \sigma_i^2)^{1/2}} \to 0$. $\endgroup$
    – jlewk
    Jun 7, 2022 at 13:38
  • $\begingroup$ Interesting comment and a good approach to try to find an example. Thanks. $\endgroup$
    – PSE
    Jun 7, 2022 at 14:34

2 Answers 2

3
$\begingroup$

$\newcommand{\R}{\mathbb R}\newcommand{\Z}{\mathbb Z}\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}$Let $\psi_j:=0$ for $j=-1,-2,\dots$. Then \begin{equation*} X_t=\sum_{j\in\Z}X_{t,j} \end{equation*} for $t\in\Z$, where \begin{equation*} X_{t,j}:=\psi_{t-j}\ep_j. \end{equation*} Let \begin{equation*} B:=\sqrt{\sum_{j\in\Z} \psi_i^2},\quad m:=\max_{j\ge0}|\psi_j|=\max_{j\in\Z}|\psi_j|. \end{equation*} Suppose that $B>0$ and $m$ vary in any manner such that \begin{equation*} m/B\to0. \tag{1}\label{1} \end{equation*} Let us show that then $X_t/B$ converges in distribution to a standard normal random variable, for each $t\in\Z$.


For each real $\de>0$, \begin{equation*} \begin{aligned} L&:=\frac1{B^2}\sum_{j\in\Z}EX_{t,j}^2\,1(|X_{t,j}|\ge\de B) \\ &=\frac1{B^2}\sum_{j\in\Z}E(\psi_{t-j}\ep_j)^2\,1(|\psi_{t-j}\ep_j|\ge\de B) \\ &\le\frac1{B^2}\sum_{j\in\Z}\psi_{t-j}^2E\ep_j^2\,1(|\ep_j|\ge\de B/m) \\ &=\frac1{B^2}\sum_{j\in\Z}\psi_{t-j}^2E\ep_0^2\,1(|\ep_0|\ge\de B/m) \\ &=E\ep_0^2\,1(|\ep_0|\ge\de B/m)\to0. \end{aligned} \end{equation*} Hence, \begin{equation*} \frac1{B^2}\sum_{j\in\Z}EX_{t,j}^2\,1(|X_{t,j}|<\de B)=1-L\to1, \end{equation*} \begin{equation*} \begin{aligned} &\frac1{B^2}\sum_{j\in\Z}(EX_{t,j}\,1(|X_{t,j}|<\de B))^2 \\ &=\frac1{B^2}\sum_{j\in\Z}(EX_{t,j}\,1(|X_{t,j}|\ge\de B))^2 \le L\to0, \end{aligned} \end{equation*} \begin{equation*} \begin{aligned} &\Big|\frac1B\sum_{j\in\Z}EX_{t,j}\,1(|X_{t,j}|<\de B)\Big| \\ &=\Big|\frac1B\sum_{j\in\Z}EX_{t,j}\,1(|X_{t,j}|\ge\de B)\Big| \\ &\le\frac1B\sum_{j\in\Z}E|X_{t,j}|\,1(|X_{t,j}|\ge\de B)\ \le \frac L\de\to0, \end{aligned} \end{equation*} \begin{equation*} \sum_{j\in\Z}P(|X_{t,j}|\ge\de B)\le L\to0. \end{equation*} So, by Theorem 18 in Chapter IV, $X_t/B$ converges in distribution to a standard normal random variable, for each $t\in\Z$.

Thus, under condition \eqref{1}, all the one-dimensional distributions of the process $(X_t)$ are asymptotically normal.


Similarly considered are all the finite-dimensional distributions of the process $(X_t)$ -- that is, all the joint distributions of $(X_{t_1},\dots,X_{t_p})$ for integers $t_1<\cdots<t_p$. This is done by writing \begin{equation*} \sum_{i=1}^p c_i X_{t_i}=\sum_{j\in\Z}Y_j \end{equation*} for any real $c_1,\dots,c_p$, where \begin{equation*} Y_j:=\phi_j\ep_j,\quad\phi_j:=\sum_{i=1}^p c_i \psi_{t_i-j}, \end{equation*} so that $\sum_{j\in\Z}\phi_j^2<\infty$ and $\max_{j\in\Z}|\phi_j|\le m\sum_{i=1}^p |c_i|$.

$\endgroup$
3
$\begingroup$

The definition of a Gaussian process is as follows. A stochastic process $X(t), t \in T $ is a Gaussian process if for all $n \in \mathbb{N}$, $a_i \in \mathbb{R}$, $t_i \in T$, $\sum_{i=1}^n a_i X(t_i)$ is normally distributed. You can use the definition and verify when it is not fulfilled.

This is Definition 2.1.8 in the Giné-Nickl book (full citation given below).

Giné, Evarist; Nickl, Richard, Mathematical foundations of infinite-dimensional statistical models, Cambridge Series in Statistical and Probabilistic Mathematics 40. Cambridge: Cambridge University Press (ISBN 978-1-108-99413-2/pbk; 978-1-00-902281-1/ebook). xiv, 690 p. (2021). ZBL1460.62007.

$\endgroup$
2
  • $\begingroup$ It is a beatiful book, thanks for the reference. I will try to think on an example. Thanks. $\endgroup$
    – PSE
    Jun 7, 2022 at 14:23
  • 2
    $\begingroup$ The question, though, is, not about the definition of a Gaussian process, but about the convergence of a linear process to a Gaussian one. $\endgroup$ Jun 7, 2022 at 15:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.