Given a strictly stationary random process $(\xi_t)_{t \in \mathbb Z}$. Define $\mu:= E[\xi_t]$, for all $t$. Suppose $(\xi_t)_{t \in \mathbb Z}$ ergodic:

\begin{equation}\label{a}\tag{E} \frac{1}{n}\sum_{j=1}^n \xi_j \overset{a.s.}{\longrightarrow} \mu \quad (n \to \infty) \end{equation} (We will use the index $j$ instead of $t$ when talking about realizations of a stochastic process). Now, consider copies of $(\xi_t)_{t \in \mathbb Z}$. By copies of $(\xi_t)_{t \in \mathbb Z}$, we mean as sequence of random process $(\xi_{t;1})_{t \in \mathbb Z}, (\xi_{t;2})_{t \in \mathbb Z}, (\xi_{t;3})_{t \in \mathbb Z},... \overset{iid}{\sim} (\xi_{t})_{t \in \mathbb Z}$.

Let $(Y_t)_{t \in \mathbb Z}$ be a process compounding the copies of $(\xi_t)_{t \in \mathbb Z}$: \begin{equation}\label{CP}\tag{CP} Y_t = \sum_{\iota=1}^N \xi_{t;\iota}\quad \forall\, t\in \mathbb Z, \quad N \sim \hbox{Poisson}(\lambda) \end{equation} where the sequence of copies and $N$ are independent. To better understand the definition of copies and the process $(Y_t)_{t \in \mathbb Z}$, consider the environment of this question. Besides this, it says that we can approximate $(Y_t)_{t \in \mathbb Z}$, using the mallows metric, by the following sequence of linear processes $(X^{(n)} , n \geq 1)$: \begin{equation}\label{b}\tag{I} X^{(n)}_t = \sum_{j =1 }^n \bar{\xi}_j U_{t - j ;n}, \quad \, \forall t \in \mathbb Z. \end{equation} where the coeficients $(\bar{\xi}_j)_{j\in \mathbb Z}$ are a fixed realization of $(\xi_t)_{t \in \mathbb Z}$ and $(U_{t;n})_{t \in \mathbb Z} \overset{iid}{\sim} \hbox{Bernoulli}(\lambda/n)$ independent of $(\xi_t)_{t \in \mathbb Z}$. To verify convergence with respecto to the Mallows mectric, we have to prove two things:

$(1)$ $X_{t_1,...,t_m}^{(n)}\implies Y_{t_1,...,t_m}\, (n \to \infty)$ for all $t_1,...,t_m \in \mathbb{Z}$ and all $m \in \mathbb{N}$. This is a convergence in distributions, and the method of characteristic functions can be used.

$(2)$ $E[|X^{(n)}_{t}|^2] \to E[|Y_{t}|^2], (n \to \infty)$ for any $t$.

Sketch of proof:

The demo is a bit similar to this case, except that instead of using iid, we're going to use ergodicity (\ref{a}). Denote by $\varphi_n$ the ch. f. of $X^{(n)}_{t}$. Thus, for $m=1$, we have: $$\varphi_n(s)= E\left[\exp\left( isX^{(n)}_{t}\right)\right]= \prod_{j=1}^n\varphi_U(\bar{\xi}_j s) =\prod_{j=1}^n \left(1+ \frac{\lambda}{n}[\exp(is \bar{\xi}_j ) -1 ] \right) $$ By ergodicty (\ref{a}), we have that: $$\varphi_n(s) \overset{a.s.}{\longrightarrow} \exp\left( \lambda E[\exp(is \bar{\xi}_j ) -1 ] \right)= E\left[ \exp\left( is Y_t\right)\right]$$ For any $m \in \mathbb N$, the proof is similar. So $(1)$ is ok!

For $(2)$, we have again by (\ref{a}): $$ \begin{aligned} E\left|X_{t}^{(n)}\right|^2=& E\left|U_{1 ; n}\right|^2 \sum_{j=1}^n \bar{\xi}_{j}^2=\frac{\lambda}{n} \sum_{j=1}^n \bar{\xi}_{j}^2 \\ & \overset{a.s.}{\longrightarrow} \lambda E\left|\bar{\xi}_{1}\right|^2=E\left|Y_{t}\right|^2\quad (n \rightarrow \infty) \end{aligned} $$


Following the same strategy given by $(1)$ and $(2)$, I have been trying for a long time to "guess" where the following sequence of stochastic processes converges: \begin{equation}\label{c}\tag{II} X^{(n)}_t = \sum_{j =1 }^n \bar{\xi}_j U_{t - j ;n}Z_{t-j;n} \end{equation} where $(\bar{\xi}_j)_{j\in \mathbb Z}$ is a fixed realization of the (strictly stationary and ergodic) autoregressive model: \begin{equation}\label{ar}\tag{ar} \xi_t = \rho \xi_{t-1} + u_t, \quad (u_t)\overset{iid}{\sim} N(0,1), 0< |\rho|<1 \end{equation} with $(U_{t;n})_{t \in \mathbb Z} \overset{iid}{\sim} \hbox{Bernoulli}(\lambda/n)$ independent of $(\xi_t)_{t \in \mathbb Z}$ and $(Z_{t})_{t \in \mathbb Z} \overset{iid}{\sim}t_5$, a Student's-t distribution with 5 degrees of freedom also independent of $(\xi_t)_{t \in \mathbb Z}$ and $(U_{t;n})$. Note that, unlike Bernoulis, $Z_{t;n}$ does not depend on $n$. Note also that (\ref{c}) is almost the same case as (\ref{b}), only without the Student's-t distribution.

I have a sneaking suspicion that we can turn (\ref{c}) into (\ref{b}) and show that $(X^{(n)}, n \geq 1)$ given by (\ref{c}) still converges to the compound Poisson $(Y_t)_{t \in \mathbb Z}$, given by (\ref{CP}), compounding the autoregressive process given by (\ref{ar}).

Is this true or does the sequence given in (\ref{c}) converge to another process?

  • 1
    $\begingroup$ Your notations and phrasing are confusing. It looks like the $\xi_j$'s (not quite defined, as functions from what to what, as they should be) mean something different from the $\xi_t$'s, "defined" as some undefined "compounding copies", and then you also have undefined $\xi_{t;\iota}$'s. The post seems to requite quite a bit of further work. $\endgroup$ Jan 9 at 14:16
  • $\begingroup$ The indices $\iota$ just indicate that they are copies of the stochastic process $(\xi_t)_{t \in \mathbb Z}$ For every $\iota$, I have a stochastic process with the same distribution of $(\xi_t )_{t \in \mathbb Z}$ . By the other hand, I use the index $j$ to indicate that I am dealing with a realization of the stochastic process $(\xi_t)_{t \in \mathbb Z}$. I do this because I'm going to use this fixed realization as the coefficients of my linear processes. I tried to better the question. $\endgroup$
    – Fam
    Jan 9 at 15:41


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