Show that the set of strictly stationary, mean zero and finite variance stochastic processes is closed (or not)

Question

Let $\mathcal{P}$ be the set of real-valued and strictly stationary processes with expectation zero and finite variance, i.e.: \begin{equation} \mathcal{P}:=\left\{ X = (X_t)_{t \in \mathbb{Z}} \, : \, X \hbox{ is strictly stationary, } \mathbb{E} X_t = 0 \hbox{ and } \mathbb{E}[X_t^2]< \infty, \, \forall\, t \in \mathbb{Z} \right\} \end{equation} Remark: for any stochastic process $X$, we consider $Q$ the Law of a stochastic process according this. We denote $X \sim Q$.

I'm trying to show whether or not $\mathcal{P}$ is closed according to the Mallows metric:

Let $X = (X_t)_{t \in \mathbb{Z}} \sim P$ and $Y = (Y_t)_{t \in \mathbb{Z}}\sim Q$ be two stochastic processes. In order to define the Mallows metric, for all $m\in \mathbb{N}$, let $\mathcal{M}_m$ be the random vectors $(\tilde{X},\tilde{Y})$ having marginals $P\circ\pi_{1,...,m}^{-1}$ and $Q\circ\pi_{1,...,m}^{-1}$, where $\pi_{1,...,m}( (X_t)_{t \in \mathbb{Z}} )= (X_{1},..., X_{m})$ . So: $$d( (X_t)_{t \in \mathbb{Z}},(Y_t)_{t \in \mathbb{Z}})= \sum_{m=1}^\infty d^{(m)}(P\circ\pi_{1,...,m}^{-1}, Q\circ\pi_{1,...,m}^{-1})2^{-m}$$ where $$d^{(m)}(P\circ\pi_{1,...,m}^{-1}, Q\circ\pi_{1,...,m}^{-1}) = \inf_{(\tilde{X},\tilde{Y})\in \mathcal{M}_m}{(E||\tilde{X}-\tilde{Y}||^2)^{\tfrac{1}{2}}}.$$

Some hint?

Answer 1

$\newcommand{\Z}{\mathbb Z}\newcommand{\PP}{\mathcal D}\newcommand{\R}{\mathbb R}$Your function $d$ is not a metric, for two reasons: (i) there may be many processes $(X_t)_{t\in\Z}$ with the same distribution $P$ and (ii) your function $d$ does not take into account the values of $X_t$ for negative $t\in\Z$. So, your $d$ is, not a metric, but a pseudometric, which does not allow one to identify limits uniquely.

We can fix these deficiencies as follows: Let $\PP$ denote the set of the distributions of the processes in $\mathcal P$.

Given $P$ and $Q$ in $\PP$, for any natural $m$ let \begin{equation} P_m:=P\circ\pi_{-m,\dots,m}^{-1},\quad Q_m:=Q\circ\pi_{-m,\dots,m}^{-1}, \end{equation} where $\pi_{r,\dots,s}((x_t)_{t\in\Z}):=(x_r,\dots,x_s)$ for any given integers $r,s$ such that $r\le s$. Let \begin{equation} d(P,Q):=\sum_{m=1}^\infty d^{(m)}(P_m,Q_m)2^{-m}, \end{equation}
where $d^{(m)}$ is the Wasserstein metric of order $2$.

We want then to show that $\PP$ is closed with respect to the metric $d$.

Suppose now that we have a sequence $(P^{(n)})$ in $\PP$ such that $d(P^{(n)},Q)\to0$ (as $n\to\infty$) for some probability measure $Q$ (on the cylindrical $\sigma$-algebra) over $\R^\Z$. Then for each natural $m$ we have $d^{(m)}(P^{(n)}_m,Q_m)\to0$. So, by the well-known characterization of the convergence in the Wasserstein metric, $P^{(n)}_m\to Q_m$ weakly, $\int_{\R^{\Z_m}} x_t^2\,Q_m(dx)=\lim_n\int_{\R^{\Z_m}} x_t^2\,P^{(n)}_m(dx)<\infty$, and $\int_{\R^{\Z_m}} x_t\,Q_m(dx)=\lim_n\int_{\R^{\Z_m}} x_t\,P^{(n)}_m(dx)=\lim_n0=0$ for $t\in\Z_m:=\{-m,\dots,m\}$.

So, $\int_{\R^\Z} x_t\,Q(dx)=0$ and $\int_{\R^\Z} x_t^2\,Q(dx)<\infty$ for all $t\in\Z$, and $P^{(n)}_{r,s}\to Q_{r,s}$ weakly for any given integers $r,s$ such that $r\le s$, where $P^{(n)}_{r,s}:=P^{(n)}\circ\pi_{r,\dots,s}^{-1}$ and $Q_{r,s}:=Q\circ\pi_{r,\dots,s}^{-1}$.

By the stationarity, $P^{(n)}_{r+1,s+1}=P^{(n)}_{r,s}$ for all suitable $r,s,n$. Letting now $n\to\infty$, we conclude that $Q_{r+1,s+1}=Q_{r,s}$, so that $Q$ is the distribution of a stationary process. Also, as we saw, $\int_{\R^\Z} x_t\,Q(dx)=0$ and $\int_{\R^\Z} x_t^2\,Q(dx)<\infty$ for all $t\in\Z$. So, $Q\in\PP$.

We conclude that $\PP$ is closed, as desired.

Answer 2

Yes, $\mathcal{P}$ is closed in the spaces \begin{equation} \mathcal{P}_1:=\left\{ X = (X_t)_{t \in \mathbb{Z}} \, : \, \mathbb{E} X_t = 0 \hbox{ and } \mathbb{E}[X_t^2]< \infty, \, \forall\, t \in \mathbb{Z} \right\} \end{equation} and \begin{equation} \mathcal{P}_2:=\left\{ X = (X_t)_{t \in \mathbb{Z}} \, : \, \mathbb{E}[X_t^2]< \infty, \, \forall\, t \in \mathbb{Z} \right\} \end{equation} on which the Mallows metric is defined. Note that the metric is not defined on all processes, a second moment of the marginals is required to be finite. Also note that this is a metric on laws: two processes could be different mappings from the same probability space to the space of sequences, yet if they have the same law the Mallows distance between them is zero.

That $\mathcal{P}_1$ is closed in $\mathcal{P}_2$ is a standard consequence of the Cauchy-Schwarz inequality, so we will focus on checking that $\mathcal{P}$ is closed in $\mathcal{P}_1$.

Suppose that the law $P$ of the process $X = (X_t)_{t \in \mathbb{Z}}$ is in the closure of $\mathcal{P}$ in $\mathcal{P}_1$. Let $S(P)$ be the law of the shifted process $S(X) = (X_{t+1})_{t \in \mathbb{Z}} \sim S(P)$.

Observe that a process $X$ in $\mathcal{P}_1$ is strictly stationary iff $d(X,S(X))=0$.

Given $\epsilon>0$, we can find a process $Y = (Y_t)_{t \in \mathbb{Z}}\sim Q$ in $\mathcal{P}$ such that $d(X,Y)<\epsilon$. The definition of the Mallows metric then yields that $d(S(X),S(Y)) <2\epsilon$. Therefore, $$d(X,S(X)) \le d(X,Y)+ d(Y,S(Y))+d(S(Y),S(X)) < \epsilon+0+2\epsilon=3\epsilon \,,$$ because $Y$ is strictly stationary. Finally, since $\epsilon>0$ is arbitrary and the Mallows metric is indeed a metric on laws in $\mathcal{P}_1$, we conclude that $X$ and $S(X)$ have the same law, so the law of $X$ is in $\mathcal{P}$, whence $\mathcal{P}$ is closed in in $\mathcal{P}_1$.