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Consider the sequence of stochastic processes $(X_n, n \geq 1)$, where $X_n = (X_{t;n})_{t\in \mathbb Z}$ and: \begin{equation}\label{I}\tag{SP} X_{t;n} = \sum_{j=0}^\infty \theta_{jn} \varepsilon_{t-j;n} \end{equation} with $\sum_{j=0}^\infty \theta_{jn}^2 < \infty$ and $(\varepsilon_{t;n})_{t\in \mathbb Z} \overset{\text{iid}}{\sim} \mu_n(dx)$ with zero mean and variance $\sigma_n =1$, for all $n$. Suppose: \begin{equation}\label{uan}\tag{Uan} \quad\quad \max_{0\leq j } |\theta_{jn}| \longrightarrow 0\quad (n \to \infty). \end{equation}

Note that each $X_n$ is strictly stationary. Suppose $(X_{t})_{t\in \mathbb Z}$ is another strictly stationary satisfying the following two conditions:

  1. $E[X_{t;n}^2]\longrightarrow E[X_{t}^2]< \infty$, as $n \to \infty$, for all $t$;
  2. $(X_{t_1n}, X_{t_2n},\dotsc, X_{t_pn}) \Longrightarrow (X_{t_1}, X_{t_2},\dotsc, X_{t_p})$ as $n \to \infty$ for all $t_1 <t_2<\cdots < t_p$ (weak convergence of finite dimensional vectors).

Now, fix $(t_1 <t_2<\cdots < t_p)$. We can show, starting from condition 2 and using (\ref{I}) and (\ref{uan}), that: $$\sum_{j=0}^n \left(\theta_{jn} , \theta_{(j+ t_2 - t_1)n }, \dotsc, \theta_{(j+ t_p - t_1)n } \right) \varepsilon_{t_1-j;n} \Longrightarrow (X_{t_1}, X_{t_2},\dotsc, X_{t_p})\quad (n \to \infty).$$ For simplicity, denote $X_{jn}:=\left(\theta_{jn} , \theta_{(j+ t_2 - t_1)n }, \dotsc, \theta_{(j+ t_p - t_1)n } \right) \varepsilon_{t_1-j;n}$ and $X:=(X_{t_1}, X_{t_2},\dotsc, X_{t_p})$ (note that $X_{jn}$ and $X$ depends on $t_1, t_2,\dotsc, t_p$). So, the last equation means that $$S_n := \sum_{j=0}^n X_{jn} \Longrightarrow X\quad (n \to \infty).$$ Finally, we can show that there exist a non-negative definite matrix $\Sigma$ and a Levy measure $\nu$ such that the characteristic function of $X$ is: $$\varphi_X(u)= \exp\left\{ \frac{-u' \Sigma u}{2} +\int_{\mathbb R^p} \left[e^{iu'x} - 1- i u'x \right] d\nu(x) \right\}.$$ We adopt the following notation: \begin{equation} X_{jn}\sim \nu_{jn}(dx), \,\, \nu_n(dx):= \sum_{j=0}^n\nu_{jn}(dx). \end{equation} Notice that $\nu_{jn}$ is a probability measure in $\mathbb R^p$, since it depends on $\mu_{n}(dx)$ — the probability measure of the iid $(\varepsilon_{t;n})_{t \in \mathbb Z}$ defined on borelians of $\mathbb R$ — but also depends on the vector $\left(\theta_{jn} , \theta_{(j+ t_2 - t_1)n }, \dotsc, \theta_{(j+ t_p - t_1)n } \right)\in \mathbb R^p$.

Moreover, the measure $\nu$ can be characterized as follows: let $\mathcal C_\#$ be the class of continuous and bounded functions vanishing on a neighborhood of $0$. Then: \begin{equation}\label{M}\tag{M} \int f \, \nu_n(dx) \to \int f \, \nu(dx),\quad \forall f \in \mathcal C_\# \quad (n \to \infty). \end{equation} or equivalently (See Barczy and Pap - Portmanteau theorem for unbounded measures): $$\nu_n(E) \longrightarrow \nu(E), \quad (E\,\,\ \nu\hbox{-contunity set}, 0 \notin \overline{E},\,\, n \to \infty )\label{MI}\tag{M'}$$

Question

Notice that $E[S_n]= \int_{\mathbb R^p} x \nu_n(dx) =0$, for all $n$. I want to show that: $$\int_{\mathbb R^p} x \nu(dx) =0\label{q1}\tag{I}$$ Can we show (\ref{q1}) or can we give a counterexample?

Attempt

First, I can show that condition 1 and (\ref{uan}) imply: \begin{equation}\label{ui}\tag{UI} \int_{\mathbb R^p} |x|^2 \nu_{n}(dx) = \sum_{j=0}^n \int_{\mathbb R^p} |x|^2 \nu_{jn}(dx) \longrightarrow \int_{\mathbb R^p} |x|^2 \nu(dx)< \infty,\quad ( n\to \infty) \end{equation}

It is worth noting that we can define $$m_n(B):= \int_{B} |x|^2 \nu_n(dx)< \infty\quad\hbox{ and }\quad m(B):= \int_{B} |x|^2 \nu(dx)< \infty$$ for all borelian $B$. Since $E[S_n]=0$, we have $$\int_{\mathbb R^p} \frac{x}{|x|^2} m_n(dx)=0$$ and (\ref{q1}) is equivalent to: $$\int_{\mathbb R^p} \frac{x}{|x|^2} m(dx)$$ I would venture to say that this is true due to (\ref{ui}), using an argument similar to uniform integrability.

Sorry if I contextualized the issue too much, but I needed to avoid counterexamples like this answer. In this same question, we can also find more technical details about the context given here.

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  • $\begingroup$ Please interpret $(\epsilon_{t:n})_{t\in\mathbb{Z}}\overset{\text{iid}}{\sim}\nu_n(dx)$(a compound Poisson Distribution?) $\endgroup$
    – JGWang
    Oct 10, 2023 at 9:58
  • $\begingroup$ No. It shouldn't be a CP distribution. It is just that, for each $n$, a sequence iid according to a distribution $\nu_n$. For example $(\epsilon _{t,n})_{t \in \mathbb Z}\sim N(0, \sigma_n)$ or $(\epsilon _{t,n})_{t \in \mathbb Z }\sim Bernoulli (1/n)$. $\endgroup$
    – PSE
    Oct 10, 2023 at 17:57
  • $\begingroup$ What the expression (Uan) means? $\max\limits_{1\le j\le n}|\theta_{jn}|\to0$ or $\max\limits_{j\in\mathbb{Z}_+}|\theta_{jn}|\to0$. $\endgroup$
    – JGWang
    Oct 11, 2023 at 2:48
  • $\begingroup$ Okay, you can assume the second option. I'll fix this now. Some minutes later: fixed. $\endgroup$
    – PSE
    Oct 11, 2023 at 3:32
  • $\begingroup$ @JGWang now we have $(\varepsilon_{t;n})_{t \in \mathbb Z} \overset{iid}{\sim} \mu_n(dx)$. $\endgroup$
    – PSE
    Nov 11, 2023 at 9:56

1 Answer 1

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At first, we consider an example.

Let \begin{gather*} f(x)=\frac{I_{\{x>0\}}(x)}{2x^2(1\vee x^2)} =\frac{I_{\{(0,1)\}}(x)}{2x^2} + \frac{I_{\{[1,\infty)\}}(x)}{2x^4},\\ \nu(\mathrm{d}x)=f(x)\,\mathrm{d}x, \end{gather*} where $\nu$ is a $\sigma-$finite measure on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$ and \begin{equation*} \int_0^\infty (1\wedge x^2)\nu(\mathrm{d}x)<\infty \tag{1} \end{equation*} Also, let \begin{align*} \psi_f(u)&=\int_{0}^{\infty}(e^{iux}-1-iux)\,\nu(\mathrm{d}x)\\ &=\int_{0}^{\infty}(e^{iux}-1-iux)f(x)\,\mathrm{d}x,\\ \bar{\phi}_f(u)&=\exp(\psi_f(u))=\exp\Big[\int_{0}^{\infty} (e^{iux}-1-iux)f(x)\,\mathrm{d}x\Big]. \tag{2} \end{align*} Then, from (1), $\bar{\phi}_f(u)$ is a (infinitely divisible) characteristic function and \begin{equation*} \bar{\phi}_f^\prime(0)=\psi_f^\prime(0)=0, \quad -\psi_f^{\prime\prime}(0)=\int_{0}^{\infty}x^2f(x)\,\mathrm{d}x=1. \tag{3} \end{equation*}

Now we are ready to construct a special $(\epsilon_{t;n})$ and $\theta_{jn}$. Let $(\epsilon_{t;n})_{t\in\mathbb{Z}}\stackrel{\mathrm{iid}}{\sim}\mu_n(dx)$, where the characteristic function of $\mu_n$ is \begin{equation*} \phi_{\epsilon;n}(u)\stackrel{\text{def}}{=}\int_{\mathbb{R}}\exp(iux)\mu_n(dx) =\exp\Big[\frac1n \psi_f(\sqrt{n}u)\Big], \end{equation*} then \begin{equation*} \mathsf{E}[\epsilon_{t;n}]=0, \qquad \mathsf{var}[\epsilon_{t;n}]=1,\qquad \forall n\ge 1. \tag{4} \end{equation*} Meanwhile, let \begin{equation*} \theta_{jn}=\frac{1}{\sqrt{n}}, \qquad 0\le j\le n-1,\quad \forall n\ge 1. \tag{5} \end{equation*} Hence, \begin{align*} X(t;n)&\stackrel{\text{def}}{=}X_{t;n}=\sum_{j=0}^{n-1}\theta_{jn}\epsilon_{t-j;n}=\frac{1}{\sqrt{n}} \sum_{j=0}^{n-1}\epsilon_{t-j;n},\qquad \forall n\ge 1,\\ \phi_{X(t;n)}&\stackrel{\text{def}}{=}\mathsf{E}[\exp(iX(t;n)u)] =[\phi_{\epsilon;n}(u/\sqrt{n})]^n \\ &=\exp[\psi_f(u)]=\bar{\phi}(u), \qquad \forall n\ge 1.\tag{6} \end{align*} From (4)(5), (Uan) and conditions 1,2 hold. If $\phi_{X(t_1)}(u) $ is the characteristic function of one dimensional distribution of limit process $X=\{X_t\} $, than, from (6), \begin{equation*} \phi_{X(t_1)}(u)=\phi_{X(t;n)}=\exp[\psi_f(u)]=\bar{\phi}(u). \end{equation*}

Now from above example we could discuss your questions.

Question 1 In above example, \begin{equation*} \int_{\mathbb{R}}|x|\nu(\mathrm{d}x)=+\infty. \end{equation*}

Question 2 In fact, if \begin{equation*} \int_{\mathbb{R}^p}|x|^2 \nu(\mathrm{d}x)<\infty, \tag{7} \end{equation*} $\bar{\phi}_f(u) $, defined by (2), may be $\phi_{X(t_1)}(u) $, the characteristic function of one dimensional distribution of limit process $X=\{X_t\} $. Hence your question transfer to ``if $\nu $ satisfy (7), under what conditions we have $\nu(\mathbb{R}^p)<\infty$ or $\nu(\mathbb{R}^p)=\infty$''.

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  • $\begingroup$ Thank you so much for your answer. I have already reviewed your answer and it seems mathematically valid, although I expected that the distribution of sequences $X_n$ would not be the same as the limit. But mathematically, your counterexample is valid. Thanks. I just didn't understand very well regarding the possibility of $\nu(\mathbb R)= \infty $ or $\nu(\mathbb R)<\infty $. Could you comment on this? $\endgroup$
    – PSE
    Nov 26, 2023 at 21:24
  • $\begingroup$ @PSE Thank you for your replication. Regarding the Question II(in the early version of this post), I could not find a reasonable answer in term $\nu_n$ right now. If you still interesting in it, I agree the Thomas Kojar's suggestion. $\endgroup$
    – JGWang
    Nov 27, 2023 at 3:19

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