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: $$\varphi_X(u)= \exp\left\{ \frac{-u' \Sigma u}{2} +\int_{\mathbb R^p} \left[e^{iu'x} - 1- i u'x \right] d\nu(x) \right\}.$$ Adopting the following notation: \begin{equation} X_{jn}\sim \nu_{jn}(dx), \,\, \nu_n(dx):= \sum_{j=0}^n\nu_{jn}(dx). \end{equation} 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} 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 I

Notice that $E[S_n]=0$. I want to show that $$\int_{\mathbb R^p} x \nu(dx)\overset{*}{=}\int_{\mathbb R^p} \frac{x}{|x|^2} m(dx) =0,\\ \hbox{where}\quad m(B):= \int_{B} |x|^2 \nu(dx)< \infty, \,\,\, \forall\, B \in \mathcal{B}(\mathbb R^p)\label{q1}\tag{I}$$ $(*)$ is by definition of change of measure. Can we show (\ref{q1}) or can we give a counterexample? I would venture to say that this is true due to (\ref{ui}), using an argument similar to uniform integrability.

Question II

I'm trying to investigate under what conditions we have $\nu(\mathbb R^p)< \infty$ or $\nu(\mathbb R^p)=\infty$.

For that, let $A= [-a,a]^p = [-a,a] \times \cdots \times [-a,a]$, with $a >0$. Note that $\nu_n(\mathbb R^p)= \sum_{j=0}^n \nu_{jn}(\mathbb R^p)=n+1$. So $$n+1= \nu_n(A) + \nu_n (A^c) \label{II}\tag{II}$$

It is easy to show the following:

Statement: If $\limsup \nu_n(A) < \infty$ (or $\liminf \nu_n(A) < \infty$) for som $a>0$, then $\nu(A) = \infty$ (Consequently, $\nu(\mathbb R^p) = \infty$ ).

The proof is by contradiction: suppose $\limsup \nu_n(A) < \infty$ and $\nu(A)< \infty$. Taking the $\limsup$ in (\ref{II}) and using (\ref{MI}), we have: \begin{equation} \infty = \limsup \nu_n(A) + \nu(A^c) \end{equation} Since $\nu(A^c)< \infty$ (by the definition of a Levy measure) we have a contradiction.

If this is correct, it would remain to analyze under what hypotheses we have $\nu(A)<\infty$?

Do you have any ideas?

Some facts that may be useful

• For $A= [-a,a]^p = [-a,a] \times \cdots \times [-a,a]$, with $a >0$, we have that \begin{align*} \nu_{jn}(A) &= \mathbb{P}(|\theta_{jn}\varepsilon_{t_1-j;n}|\leq a, |\theta_{(j+ t_2 - t_1)n }\varepsilon_{t_1-j;n}|\leq a , \ldots, |\theta_{(j+ t_p - t_1)n }\varepsilon_{t_1-j;n}|\leq a) \\ &=\mathbb{P}\bigg(|\varepsilon_{t_1-j;n}|\leq \frac{a}{\max(|\theta_{jn}|, |\theta_{(j+ t_2 - t_1)n }| \ldots,|\theta_{(j+ t_p - t_1)n }|)}\bigg)\\ &\overset{(**)}{=}\mathbb{P}\bigg(|\varepsilon_{t_1-j;n}|\leq \frac{a}{|\theta_{jn}|}\bigg). \end{align*} $(**)$ If we assume that $(\theta_{jn})_{j=0}^{\infty}$ is decreasing in $j$.

$\quad\,\,\,$Note that $\nu_{jn}(A)$ depends on $t_1, t_2,\dotsc, t_p$. Moreover \begin{equation}\label{N}\tag{N} \nu_n(A)= \sum_{j=0}^n \mathbb{P}\bigg(|\varepsilon_{t_1-j;n}|\leq \frac{a}{\max(|\theta_{jn}|, |\theta_{(j+ t_2 - t_1)n }| \ldots,|\theta_{(j+ t_p - t_1)n }|)}\bigg) \end{equation}

• It is possible to show that if $\nu(\mathbb R^p)=\infty$, then the law of $X$ is difusse (non-atomic). See Proposition 7.16 from Foundations of Modern Probability, by Olav Kallenberg:

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.

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{I} 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 $(\varepsilon_{t;n})_{t\in \mathbb Z} \overset{\text{iid}}{\sim} \nu_n(dx)$ 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}| \quad (n \to \infty). \end{equation}

Note that each $X_n$ is strictly stationary. Suppose $X^* = (X_{t})_{t\in \mathbb Z}$ is another strictly stationary process such that:

1. $E[X_{tn}^2]\longrightarrow E[X_{t}^2]< \infty$, 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 (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 $. Moreover, I think I managed to demonstrate that (1) and (\ref{uan}) imply: \begin{equation} \sum_{j=0}^n \int_{\mathbb R^p} |x|^2 \nu_{jn}(dx) = \int_{\mathbb R^p} |x|^2 \nu_{n}(dx)\longrightarrow \int_{\mathbb R^p} |x|^2 \nu(dx),\quad ( n\to \infty) \\ \hbox{where }\,\,\,\,\ X_{jn}\sim \nu_{jn}(dx), \,\, \nu_n(dx):= \sum_{j=0}^n\nu_{jn}(dx),\,\, X \sim \nu(dx). \end{equation} Notice that $\nu_{jn}$ is a measure on borelians of $\mathbb R^p$, since it depends on $\nu_{n}(dx)$ — a measure 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:

• For $A= [-a,a]^p = [-a,a] \times \cdots \times [-a,a]$, with $a >0$, we have that \begin{align*} \nu_{jn}(A) &= \mathbb{P}(|\theta_{jn}\varepsilon_{t_1-j;n}|\leq a, |\theta_{(j+ t_2 - t_1)n }\varepsilon_{t_1-j;n}|\leq a , \ldots, |\theta_{(j+ t_p - t_1)n }\varepsilon_{t_1-j;n}|\leq a) \\ &=\mathbb{P}\bigg(|\varepsilon_{t_1-j;n}|\leq \frac{a}{\max(|\theta_{jn}|, |\theta_{(j+ t_2 - t_1)n }| \ldots,|\theta_{(j+ t_p - t_1)n }|)}\bigg). \end{align*} Note that $\nu_{jn}(A)$ depends on $t_1, t_2,\dotsc, t_p$. If $G_n$ is the CDF of $\varepsilon_{t_1-j;n}$, we have: \begin{equation}\label{N}\tag{N} \nu_n(A)= \sum_{j=0}^n \mathbb{P}\bigg(|\varepsilon_{t_1-j;n}|\leq \frac{a}{\max(|\theta_{jn}|, |\theta_{(j+ t_2 - t_1)n }| \ldots,|\theta_{(j+ t_p - t_1)n }|)}\bigg)\\ = \sum_{j=0}^n \left[G_n(\kappa_{jn}) - G_n(-\kappa_{jn})\right] \end{equation} where $\kappa_{jn}= \frac{a}{\max(|\theta_{jn}|, |\theta_{(j+ t_2 - t_1)n }| \dotsc,|\theta_{(j+ t_p - t_1)n }|)}$. Similarly, we can show that: $$\nu(A^c)= \sum_{j=0}^n \mathbb P (|\varepsilon_{t_1-j;n}|\geq \kappa_{jn} ) $$

• The measure $\nu$ is a Lévy measure and 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 \, d\nu_n \to \int f \, d\nu \quad (n \to \infty),\quad \forall f \in \mathcal C_\#. \end{equation} Check out How to show that $\int x \,d\nu = 0$ using a pseudo-weak convergence of measures? for a little more detail on why this characterization. It is worth mentioning that (\ref{uan}) implies $\max_{0\leq j \leq < n}P(|X_{jn}|> \epsilon) \to 0$, as $n \to \infty$ for all $\epsilon >0$.

Question 1

For any $a>0$ and $A=[-a,a]^p$ defined above. I want to show: \begin{equation}\label{C2}\tag{E} \int_{A^c} x \nu(dx)=0. \end{equation}

Attempt

I endeavored to discover a counterexample involving a sequence of processes, denoted as $(X_n, n \geq 1)$, with $X_{t;n}$ conforming to the conditions outlined in (\ref{I}) while simultaneously satisfying (\ref{uan}), along with a process $X^{*}=(X_t)_{t \in \mathbb Z}$ that adheres to both criteria 1 and 2. This pursuit was motivated by the insights provided in response to the same question previously referenced above. However, despite my efforts, I was unable to identify such a counterexample.

Another attempt is to note that $0=\int x \nu_n(dx)= \int_{A} x \nu_n(dx) + \int_{A^c} x \nu_n(dx)$. Note that we can not conclude that $\int_{A^c} x \nu_n(dx) \to \int_{A^c} x \nu(dx)$ using Barczy and Pap (See Theorem 2.1), because $f(x)=x$ is not bounded in $A^c$. However, one attempt is to use the fact that $\int_{\mathbb R^p}|x|^2 \nu_n(dx)$ is uniformly bounded and try to use some uniform integrability argument. If this allows us to conclude that $\int_{A^c} x \nu_n(dx) \to \int_{A^c} x \nu(dx)$, we can conlude that $$\int_{A^c} x \nu(dx) = - \lim_{n \to \infty}\int_{A} x \nu_n(dx)$$ So this can help to construct a counterexample for (\ref{C2}).

Question 2

We can show that (\ref{M}) is equivalent to $\nu_n(E) \longrightarrow \nu(E)$, as $n \to \infty$, for all $\nu-$conntinuity set $E$ with $0 \notin \bar{E}$ (closure of $E$) (See Barczy and Pap - Portmanteau theorem for unbounded measures). So, we have that $$\nu_n(A^c) \longrightarrow \nu(A^c), \quad (n \to \infty)\label{M'}\tag{M'}$$

However, my objective is to investigate whether $\nu(\mathbb R^p)< \infty$ or $\nu(\mathbb R^p)=\infty$.

Attempt

Note that $\nu_n(\mathbb R^p)= \sum_{j=0}^n \nu_{jn}(\mathbb R^p)=n+1$. So

$$n+1= \nu_n(A) + \nu_n (A^c) = \sum_{j=0}^n \nu_{jn}(A) + \nu_n (A^c)\label{II}\tag{II}$$

Statement 1: If $\limsup \nu_n(A) < \infty$ (or $\liminf \nu_n(A) < \infty$) for som $a>0$, then $\nu(A) = \infty$ (Consequently, $\nu(\mathbb R^p) = \infty$ ).

The proof is by contradiction: suppose $\nu(A)< \infty$. So if $\limsup \nu_n(A) < \infty$, taking the $\limsup$ in (\ref{II}) and using (\ref{M'}), we have: \begin{equation} \infty = \limsup \nu_n(A) + \nu(A^c) \end{equation} Since $\nu(A^c)< \infty$ (by the definition of a Levy measure) we have a contradiction.

If this is correct, it would remain to analyze under what hypotheses we have $\nu(A)<\infty$.

This suggests analyzing how $\mathbb{P}\bigg(|\varepsilon_{t_1-j;n}|\leq \frac{a}{\max(|\theta_{jn}|, |\theta_{(j+ t_2 - t_1)n }| \ldots,|\theta_{(j+ t_p - t_1)n }|)}\bigg)$ approaches 1 in order to conclude something.

Remarks

Here I will make some corrections to some wrong statements I was making and some updates or advances of the question. This can help to better understand the progress.

• I was stating, very wrongly, that $\nu_{n}(A)$ approaches to $n+1$ as $n$ grows. In fact, this is not true;
• Regarding Question 2, I think I managed to show that if $\lim_{n \to \infty} \nu_{n}(A)< \infty$ for some $a>0$, then necessarily $\nu(\mathbb R^p)=\infty$ (See Statement 2). But it remains to show under what conditions I have $\nu(\mathbb R^p)<\infty$;
• Question 1 is important to me, and I have a certain priority on it.

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: $$\varphi_X(u)= \exp\left\{ \frac{-u' \Sigma u}{2} +\int_{\mathbb R^p} \left[e^{iu'x} - 1- i u'x \right] d\nu(x) \right\}.$$ Adopting the following notation: \begin{equation} X_{jn}\sim \nu_{jn}(dx), \,\, \nu_n(dx):= \sum_{j=0}^n\nu_{jn}(dx). \end{equation} 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} 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 I

Notice that $E[S_n]=0$. I want to show that $$\int_{\mathbb R^p} x \nu(dx)\overset{*}{=}\int_{\mathbb R^p} \frac{x}{|x|^2} m(dx) =0,\\ \hbox{where}\quad m(B):= \int_{B} |x|^2 \nu(dx)< \infty, \,\,\, \forall\, B \in \mathcal{B}(\mathbb R^p)\label{q1}\tag{I}$$ $(*)$ is by definition of change of measure. Can we show (\ref{q1}) or can we give a counterexample? I would venture to say that this is true due to (\ref{ui}), using an argument similar to uniform integrability.

Question II

I'm trying to investigate under what conditions we have $\nu(\mathbb R^p)< \infty$ or $\nu(\mathbb R^p)=\infty$.

For that, let $A= [-a,a]^p = [-a,a] \times \cdots \times [-a,a]$, with $a >0$. Note that $\nu_n(\mathbb R^p)= \sum_{j=0}^n \nu_{jn}(\mathbb R^p)=n+1$. So $$n+1= \nu_n(A) + \nu_n (A^c) \label{II}\tag{II}$$

It is easy to show the following:

Statement: If $\limsup \nu_n(A) < \infty$ (or $\liminf \nu_n(A) < \infty$) for som $a>0$, then $\nu(A) = \infty$ (Consequently, $\nu(\mathbb R^p) = \infty$ ).

The proof is by contradiction: suppose $\limsup \nu_n(A) < \infty$ and $\nu(A)< \infty$. Taking the $\limsup$ in (\ref{II}) and using (\ref{MI}), we have: \begin{equation} \infty = \limsup \nu_n(A) + \nu(A^c) \end{equation} Since $\nu(A^c)< \infty$ (by the definition of a Levy measure) we have a contradiction.

If this is correct, it would remain to analyze under what hypotheses we have $\nu(A)<\infty$?

Do you have any ideas?

Some facts that may be useful

• For $A= [-a,a]^p = [-a,a] \times \cdots \times [-a,a]$, with $a >0$, we have that \begin{align*} \nu_{jn}(A) &= \mathbb{P}(|\theta_{jn}\varepsilon_{t_1-j;n}|\leq a, |\theta_{(j+ t_2 - t_1)n }\varepsilon_{t_1-j;n}|\leq a , \ldots, |\theta_{(j+ t_p - t_1)n }\varepsilon_{t_1-j;n}|\leq a) \\ &=\mathbb{P}\bigg(|\varepsilon_{t_1-j;n}|\leq \frac{a}{\max(|\theta_{jn}|, |\theta_{(j+ t_2 - t_1)n }| \ldots,|\theta_{(j+ t_p - t_1)n }|)}\bigg)\\ &\overset{(**)}{=}\mathbb{P}\bigg(|\varepsilon_{t_1-j;n}|\leq \frac{a}{|\theta_{jn}|}\bigg). \end{align*} $(**)$ If we assume that $(\theta_{jn})_{j=0}^{\infty}$ is decreasing in $j$.

$\quad\,\,\,$Note that $\nu_{jn}(A)$ depends on $t_1, t_2,\dotsc, t_p$. Moreover \begin{equation}\label{N}\tag{N} \nu_n(A)= \sum_{j=0}^n \mathbb{P}\bigg(|\varepsilon_{t_1-j;n}|\leq \frac{a}{\max(|\theta_{jn}|, |\theta_{(j+ t_2 - t_1)n }| \ldots,|\theta_{(j+ t_p - t_1)n }|)}\bigg) \end{equation}

• It is possible to show that if $\nu(\mathbb R^p)=\infty$, then the law of $X$ is difusse (non-atomic). See Proposition 7.16 from Foundations of Modern Probability, by Olav Kallenberg:

Statement 1: If $\limsup \nu_n(A) < \infty$ (or $\liminf \nu_n(A) < \infty$) for som $a>0$, then $\nu(A) = \infty$.

• I was stating, very wrongly, that $\nu_{n}(A)$ approaches to $n+1$ as $n$ grows. In fact, this is not true;
• Regarding Question 2, I think I managed to show that if $\lim_{n \to \infty} \nu_{n}(A)< \infty$ for some $a>0$, then necessarily $\nu(A )=\infty$ (See Statement 2). But it remains to show under what conditions I have $\nu(A)<\infty$;
• Question 1 is important to me, and I have a certain priority on it.

Statement 1: If $\limsup \nu_n(A) < \infty$ (or $\liminf \nu_n(A) < \infty$) for som $a>0$, then $\nu(A) = \infty$ (Consequently, $\nu(\mathbb R^p) = \infty$ ).

• I was stating, very wrongly, that $\nu_{n}(A)$ approaches to $n+1$ as $n$ grows. In fact, this is not true;
• Regarding Question 2, I think I managed to show that if $\lim_{n \to \infty} \nu_{n}(A)< \infty$ for some $a>0$, then necessarily $\nu(\mathbb R^p)=\infty$ (See Statement 2). But it remains to show under what conditions I have $\nu(\mathbb R^p)<\infty$;
• Question 1 is important to me, and I have a certain priority on it.