We have a classic theorem about convergence of infinitely divisible (ID) distributions. Before, remember that we say that the ID random vector $X$ has a Lévy-Khintchine representation, for a suitable integrable function $c:\mathbb R^p \to \mathbb R$, if
\begin{equation}
\varphi_X(u) = \exp \left\{ i a_c^\intercal u – \frac{u ^\intercal \sigma u}{2} + \int_{\mathbb R^p} \left[e^{i u^\intercal x } – 1 – i u^\intercal x c(x) \right] d\nu(x) \right\},\quad u \in \mathbb R^p,
\end{equation}
We denote this by $X\sim(a, \sigma, \nu)_c$.
The following classical theorem is given in Sato, Theorem 8.7:
Let $c: \mathbb R^p \to \mathbb R$ be a function such that:
\begin{equation}\label{I}\tag{c} c(x)= 1+ o(|x|)\quad \hbox{ as }\,\,
x \to 0\\ c(x)=O(1/|x|)\quad \hbox{ as }\,\, x \to \infty.
\end{equation}Suppose $X\sim(a_{n}, \sigma_n, \nu_n)_c$ be a sequence of $ID$
random variables and $X$ some random variable. Then $X_n
\overset{d}{\to} X$ iff $X$ is ID and has a Lévy-Khintchine
representation triplet $(a, \sigma, \nu)_c$ satisfying:
- For any continuity set $E$ (i.e. $\nu(\partial E)=0$) with $0\notin \bar{E}$, $$\nu_n(E) \to \nu(E), \quad (n \to \infty)$$
- For any $\epsilon > 0$, defining the symmetric nonnegative-definite matrices $\sigma_{n, \varepsilon}$ by
\begin{equation}
\left\langle u, \sigma_{n, \varepsilon} u\right\rangle :=\left\langle u, \sigma_n u\right\rangle+\int_{|x| \leq
> \varepsilon}\langle u, x\rangle^2 \nu_n(dx) . \end{equation} we have
\begin{equation}\label{II}\tag{II} \lim_{\varepsilon \to 0} \limsup_{n
> \rightarrow \infty}\left|\left\langle u, \sigma_{n, \varepsilon}
u\right\rangle-\langle u, \sigma u\rangle\right|=0 \quad \text { for }
u \in \mathbb{R}^n. \end{equation}- $a_{n} \to a$, as $n \to \infty$.
I want to use this Theorem to show another version of the same result (please, see the fact A from the firts two page of this paper). For convenience, let's print the paper's statement specifying items 1, 2 and 3, analogous to Theorem 8.7. Item 2 is marked with a red rectangle to indicate the difference between versions of the same result.
- First, note that the paper uses $c(x)= \frac{1}{1+|x|^2}$. It's easy to show that $c(x)= \frac{1}{1+|x|^2}$ satisfies (\ref{I}).
- So, Items 1 and 3 of the paper (analogous to Theorem 8.7 in Sato's book) are obtained directly.
- Thus, it remaims to show that the second item on the paper marked with the red rectangle is equivalent to (\ref{II}). For convenience, we rewrite it.
\begin{equation}\label{2}\tag{2}
\sigma_n + \int_{\mathbb R^p} \frac{xx'}{1+ |x|^2}d \nu_n \to \sigma + \int_{\mathbb R^p} \frac{xx'}{1+ |x|^2} d \nu
\end{equation}
Problem 1
How to show that (\ref{II}) and (\ref{2}) are equivalent?
Considerations
Sato says that equation (8.20) (page number 43 from his book) given by:
\begin{equation}\label{II'}\tag{II'}
\lim_{\epsilon \to 0} \limsup_{n \to \infty} \left[\left\langle u, \sigma_n u\right\rangle+\int_{|x| \leq \varepsilon}\langle u, x\rangle^2 \nu_n(dx)\right] = \langle u, \sigma u \rangle, \quad \forall u \in \mathbb R^p
\end{equation}
is equivalent to (\ref{II}). Note that (\ref{II'}) is also true for $\liminf$ instead of $\limsup$.
If we are willing to assume that (\ref{2}) is equivalent to
\begin{equation}\label{2'}\tag{2'}
\langle u , \sigma_n u \rangle + \int_{\mathbb R^p} \frac{\langle u, x\rangle^2}{1+ |x|^2}d \nu_n \to \langle u , \sigma u \rangle + \int_{\mathbb R^p} \frac{\langle u, x\rangle^2}{1+ |x|^2} d \nu, \quad \forall u \in \mathbb R^p
\end{equation}
So maybe it's easier to show the equivalence between (\ref{II'}) and (\ref{2'}). In this case, Problem 1 would be rewritten as:
Problem 1'
How to show that (\ref{II'}) and (\ref{2'}) are equivalent?
A Third Version
Another approach might be to use Theorem 7.7 from Kallenberg's Book (see JGWang's comment). First, Corollary 7.6 below shows that an ID random vector has a canonical representation – i.e. the case $c(x)= \mathbf{1}_{D}$, where $D=\{x: |x|\leq 1\}$. To show this, Kallenberg shows and uses a third version of the same Theorem 8.7 with some truncated version. Below is the print. I got by omitting statements that are not of interest to us. (Just be careful noting that the parameters are different. Just to situate the reader, the parameter $a$ and $\sigma$, in the version of Sato's book, correspond respectively to the parameter $b$ and $a$ of Kallemberg's book. The Levy measures in both cases are denoted by $\nu$ and the convergence is the same in the three versions.)
But this version for me is more challenging, because it involves $h$ and I don't know how to deal with it. So a third question that is also interesting (but not as urgent for me) is:
Problem 2 (not as urgent but might be helpful)
How are the three versions related?
Help!
Best Answer
Let \begin{gather*} \text{(I): For any continuity set $E$(i.e. $\nu(\partial E)=0$) with } 0\notin \bar{E},\\ \nu_n(E) \to \nu(E), \quad \text{as } n\to\infty. \tag{I}\\ \lim_{\epsilon\to0}\varlimsup_{n\to\infty}\Big[\langle u,\sigma_n u \rangle +\int_{|x|\le\epsilon}\langle u,x\rangle^2 \nu_n(\mathrm{d}x)\Big] =\langle u,\sigma u\rangle. \tag{II'} \\ \lim_{n\to\infty}\Big[\langle u,\sigma_n u \rangle+ \int_{\mathbb{R}^p} \frac{\langle u,x\rangle^2}{1+|x|^2}\nu_n(\mathrm{d}x)\Big] = \langle u,\sigma u \rangle+ \int_{\mathbb{R}^p} \frac{\langle u,x\rangle^2}{1+|x|^2}\nu(\mathrm{d}x) \tag{2'}\\ \lim_{n\to\infty}\Big[\langle u,\sigma_n u \rangle+ \int_{\mathbb{R}^p} \frac{\langle u,x\rangle^2}{1+|x|^2}\nu_n(\mathrm{d}x) -\int_{\mathbb{R}^p}\frac{\langle u,x\rangle^2}{1+|x|^2}\nu(\mathrm{d}x) \Big] = \langle u,\sigma u \rangle. \tag{2"} \end{gather*}
The following will prove that \begin{equation*} (I) + (II^\prime) \Longleftrightarrow (I) +(2') \quad (\text{i.e., } (I) + (II^\prime) \Longleftrightarrow (I) +(2")). \tag{3} \end{equation*} From (I), \begin{equation} \sup_n \nu_n(|x|\ge \epsilon)\le C_1(\epsilon),\quad \forall \epsilon>0. \tag{4} \end{equation} From (II'), \begin{equation*} \sup_n\int_{|x|\le\epsilon} \frac{\langle u,x\rangle^2}{1+|x|^2} \nu_n(\mathrm{d}x)\le C_2(u, \epsilon),\quad \forall \epsilon>0, u\in\mathbb{R}^p. \end{equation*} Combining above and (4), \begin{equation*} \sup_n\int_{\mathbb{R}^p} \frac{\langle u,x\rangle^2}{1+|x|^2}\nu_n(\mathrm{d}x)\le C_3(u),\quad \forall u\in\mathbb{R}^p. \tag{5} \end{equation*} (5) holds too from (2').
Now, we are ready to prove (3). In the following proof, suppose the $(2')$ or $(II')$ holds, therefore (5) holds too. Subtracting the left side of (II') and (2"), \begin{align*} &\Big[\langle u,\sigma_n u \rangle+ \int_{\mathbb{R}^p} \frac{\langle u,x\rangle^2}{1+|x|^2}\nu_n(\mathrm{d}x) -\int_{\mathbb{R}^p}\frac{\langle u,x\rangle^2}{1+|x|^2}\nu(\mathrm{d}x) \Big]\\ &\quad -\Big[\langle u,\sigma_n u \rangle +\int_{|x|\le\epsilon}\langle u,x\rangle^2 \nu_n(\mathrm{d}x)\Big] = E_1+E_2+E_3, \tag{6} \end{align*} where \begin{align*} E_1 &= \int_{|x|>\epsilon} \frac{\langle u,x\rangle^2}{1+|x|^2}\nu_n(\mathrm{d}x) -\int_{|x|>\epsilon}\frac{\langle u,x\rangle^2}{1+|x|^2}\nu(\mathrm{d}x) \\ E_2 &=-\int_{|x|\le\epsilon}\frac{\langle u,x\rangle^2}{1+|x|^2}\nu(\mathrm{d}x) \\ E_3 &=\int_{|x|\le\epsilon}\frac{\langle u,x\rangle^2}{1+|x|^2}\nu_n(\mathrm{d}x) - \int_{|x|\le\epsilon}\langle u,x\rangle^2 \nu_n(\mathrm{d}x)\\ &\quad \epsilon \in\{a>0:\nu(\{x:|x|=a\})=0\} \end{align*}
Now from (1), \begin{align*} \lim_{n\to\infty}|E_1| &=\lim_{n\to\infty}\Big|\int_{|x|>\epsilon} \frac{\langle u,x\rangle^2}{1+|x|^2}\nu_n(\mathrm{d}x) -\int_{|x|>\epsilon} \frac{\langle u,x\rangle^2}{1+|x|^2}\nu(\mathrm{d}x) \Big| \\ &=0, \quad \forall u\in\mathbb{R}^p. \tag{7} \end{align*} Since $\int_{\mathbb{R}^p}(|x|^2\wedge 1)\nu(\mathrm{d}x)<\infty $ and $\nu(\{0\})=0$, then \begin{gather*} |E_2|=\int_{|x|\le\epsilon} \frac{\langle u,x\rangle^2}{1+|x|^2}\nu(\mathrm{d}x) \le |u|^2 \int_{|x|\le\epsilon}|x|^2\nu(\mathrm{d}x),\\ \lim_{\epsilon\to 0}|E_2|=0. \tag{8} \end{gather*} \begin{align*} |E_3| &=\Big|\int_{|x|\le\epsilon}\langle u,x\rangle^2 \nu_n(\mathrm{d}x) -\int_{|x|\le\epsilon} \frac{\langle u,x\rangle^2}{1+|x|^2}\nu_n(\mathrm{d}x)\Big|\\ & =\int_{|x|\le\epsilon} \frac{\langle u,x\rangle^2|x|^2 }{1+|x|^2}\nu_n(\mathrm{d}x) \le \epsilon^2\sup_n\int_{\mathbb{R}^p}\frac{\langle u,x\rangle^2}{1+|x|^2}\nu_n(\mathrm{d}x)\\ &\le \epsilon^2C_3(u), \end{align*} by (5) and \begin{equation*} \lim_{\epsilon\to0}\varlimsup_{n\to\infty}|E_3|=0 \tag{9} \end{equation*}
Using (7)-(9) and (6), got \begin{align*} &\lim_{\epsilon\to0}\varlimsup_{n\to\infty}\biggl|\Big[\langle u,\sigma_n u \rangle+ \int_{\mathbb{R}^p} \frac{\langle u,x\rangle^2}{1+|x|^2}\nu_n(\mathrm{d}x) -\int_{\mathbb{R}^p}\frac{\langle u,x\rangle^2}{1+|x|^2}\nu(\mathrm{d}x) \Big]\\ &\qquad\quad -\Big[\langle u,\sigma_n u \rangle +\int_{|x|\le\epsilon}\langle u,x\rangle^2 \nu_n(\mathrm{d}x)\Big]\biggr|=0, \end{align*} and \begin{equation*} (I) + (II^\prime) \Longleftrightarrow (I) +(2') \quad (\text{i.e., } (I) + (II^\prime) \Longleftrightarrow (I) +(2")). \end{equation*}