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

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

• What are the $t_j$'s? Aug 17, 2022 at 4:20
• They represent the indices of the marginal distributions of the stochastic process, more specifically, the distribution of an m-dimensional vector. Stationarity, I believe, is fundamental for the metric to be well defined.
– Fam
Aug 17, 2022 at 4:26
• I think I was confused such $t_j$'s indices. Now I've fixed it.
– Fam
Aug 17, 2022 at 4:38

$$\newcommand{\Z}{\mathbb Z}\newcommand{\PP}{\mathcal D}\newcommand{\R}{\mathbb R}$$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 $$$$P_m:=P\circ\pi_{-m,\dots,m}^{-1},\quad Q_m:=Q\circ\pi_{-m,\dots,m}^{-1},$$$$ where $$\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$$. Let $$$$d(P,Q):=\sum_{m=1}^\infty d^{(m)}(P_m,Q_m)2^{-m},$$$$
where $$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.

• @Fam : The weak convergence follows by the well-known characterization of the convergence in the Wasserstein metric. I have added a reference to this fact. Aug 18, 2022 at 17:19
• @fam : Also, you cannot actually take $\mathcal P_1$ for the set of probability measures $Q$ -- because $\mathcal P_1$ is a set of processes, not a set of measures. However, just as I wrote, $Q$ belongs to the set of all probability measures on the cylindrical $\sigma$-algebra. Aug 18, 2022 at 17:25
• @Fam : I can barely understand your latter comment. (i) You again are talking about "$\mathcal P_1$ (set of measures )". As I said, $\mathcal P_1$ is, not a set of measures, but a set of processes. Maybe, this is an issue of you expressing your thoughts. (ii) For any probability measures $P$ and $Q$ on the cylindrical $\sigma$-algebra, the probability measures $P_m$ and $Q_m$ are well defined, by the definition of the cylindrical $\sigma$-algebra (as is done in my answer). So, $d(P,Q)$ is well defined. Aug 18, 2022 at 22:23
• @Fam : Perhaps, you are concerned that for some probability measures $P$ and $Q$ on the cylindrical $\sigma$-algebra we may have $d(P,Q)=\infty$. But that is all right, because we assume that $d(P^{(n)},Q)\to0$ and hence $d(P^{(n)},Q)<\infty$ for all large enough $n$. So, the set $\mathcal D$ is closed, not only in the set of all probability measures $P$ on the cylindrical $\sigma$-algebra with $\int x_t^2\,P(dx)<\infty$ for all $t$, but even in the set of (absolutely) all probability measures $P$ on the cylindrical $\sigma$-algebra (even without finite second moments). Aug 18, 2022 at 22:43
• Sorry, I do not engage in such chat. I have answered all your questions, have I not? If you need more details, please let me know right here. Aug 19, 2022 at 11:11

Yes, $$\mathcal{P}$$ is closed in the spaces $$$$\mathcal{P}_1:=\left\{ X = (X_t)_{t \in \mathbb{Z}} \, : \, \mathbb{E} X_t = 0 \hbox{ and } \mathbb{E}[X_t^2]< \infty, \, \forall\, t \in \mathbb{Z} \right\}$$$$ and $$$$\mathcal{P}_2:=\left\{ X = (X_t)_{t \in \mathbb{Z}} \, : \, \mathbb{E}[X_t^2]< \infty, \, \forall\, t \in \mathbb{Z} \right\}$$$$ 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 $$\mathcal{P}_1$$.