# Finding the limit of a specific sequence of linear processes

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}

Question

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?

• 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. Jan 9 at 14:16
• 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.
– Fam
Jan 9 at 15:41