Let $(X\_{jn})\_{1\leq j \leq n}$, $n\in \mathbb N$, be a triangular array of random vectors in $\mathbb R^d$ (the $X\_{jn}$ are understood to be independent in $j$ for fixed $n$.). We say that the triangular array is *null* if $X\_{jn} \overset{p}{\to} 0$, as $n \to \infty$, uniformly in $j$, i.e.,:
\begin{equation}
\lim\_{n \to \infty} \max\_{1\leq j\leq n} P( \, |X\_{jn}|> \epsilon \, )=0, \quad (\forall \, \epsilon>0).
\end{equation}
For any random vector $X$ with distribution $\mu$, we introduce an associated compound Poisson random vector $X^{\tilde{}}$ (Note the tilde, please) with characteristic measure $\mu$, i.e.:
\begin{equation}
\log \varphi\_{X^{\tilde{}}}(u) = \int\_{\mathbb R^d} (e^{iux}-1)\mu (dx), \quad u \in \mathbb R^d.
\end{equation}
For a triangular array $(X\_{jn})\_{1\leq j \leq n}$, the corresponding compound Poisson vectors $(X\_{jn}^{\tilde{}})\_{1\leq j \leq n}$ are again assumed to have row-wise independent entries.


By $X\_n \overset{d}{\sim} Y\_n$ we mean that, if either side converges in distribution along a sub-sequence, then so does the other along the same sequence.


We can show the proposition about compound Poisson approximation: Let $(X\_{jn})\_{1\leq j \leq n}$, $n\in \mathbb N$, be a null array of random vectors in $\mathbb R^d$. Fix $h > 0$, and define:
\begin{equation}\label{k1}\tag{K1}
b\_{jn}= E(X\_{jn}\,;\,|X\_{jn}|\leq h).
\end{equation}
Then
\begin{equation}\label{k2}\tag{K2}
\sum\_j X\_{jn} \overset{d}{\sim} \sum\_j \left\{ (X\_{jn} - b\_{jn})^{\tilde{}} +b\_{jn} \right\}
\end{equation}
(This the Proposition 7.11 from [Foundations of Modern Probability](https://link.springer.com/book/10.1007/978-3-030-61871-1), Third edition)


**Update**


Allow me to rephrase my question. We say that a Infinite Divisible r. vector has the levy kintchine representations if its characteristic function is given by:
$$\varphi\_X(u)= \exp\left\{i u'b + \frac{1}{2}u'au + \int\_{\mathbb R^d} \left[e^{iu'x}-1 - i u'x c(x)\right] d\nu(x) \right\} $$
where $c(x)$ is a integrable function. We denote this as $X \sim (b\_c, a, \nu)\_c$. In the case above, we have that $X^{\tilde{}} \sim (0,0,\mu)\_0$.


Now, we can change the truncation function $c(x)$ by other $h(x)$. If $X \sim (b\_c, a, \nu)\_c$, we have that
$$X \sim (b\_h, a, \nu)\_h, \quad b\_h = b\_c + \int\_{\mathbb R^d}x [ h(x)- c(x)]d\nu(x)$$


**Question:**


Given a **Null** triangular array $(X\_{jn})\_{1\leq j \leq n}$ with $X\_{jn} \sim \mu\_{jn}$. Suppose $E[X\_{jn}]=0$ and we also have that:
\begin{equation}\label{I}\tag{I}
\sum\_{j=1}^n \mathbf{v}(X\_{jn})\leq C< \infty,\quad \forall \, n
\end{equation}
where $X\_{jn}=(X\_{jn\_{1}},..., X\_{jn\_{d}})$ and:
$$\mathbf{v}(X\_{jn}):=\hbox{trace} \left(E\left[X\_{jn}X\_{jn}^{\,\,'}\right]\right)= \sum\_{k=1}^d \hbox{var}(X\_{jn\_{k}})$$
Note that $X\_{jn\_{k}}$ is unidemsional, $E[X\_{jn\_{k}}]=0$ and $\hbox{var}(X\_{jn\_{k}})= E[(X\_{jn\_{k}})^2]$.


Define $S\_n' \sim (0,0,\mu\_n)\_h$ with $h(x)\equiv 1$ and $$\mu\_n := \sum\_{j=1}^n \mu\_{jn}$$
So, how to show, using (\ref{k1}) and (\ref{k2}), that:
$$S\_n := \sum\_j X\_{jn} \overset{d}{\sim} S\_n' $$



Convergence of the row sums in a triangular null array with zero mean
$\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}$Let us rephrase the question a bit. For each natural $n$, let $X\_{1,n},\dots,X\_{j\_n,n}$ be independent zero-mean random vectors in $\R^d$ such that (i) for each real $\ep>0$
\begin{equation\*}
 \max\_{j\in J\_n} P(\|X\_{j,n}\|>\ep)\to0 \tag{0}\label{0}
\end{equation\*}
(as $n\to\infty$) and (ii)
for some real $C>0$ and all $n$
\begin{equation\*}
 \sum\_{j\in J\_n} E\|X\_{j,n}\|^2\le C, \tag{10}\label{10}
\end{equation\*}
where $j\_n$ is a positive integer, $J\_n:=\{1,\dots,j\_n\}$, and $\|\cdot\|$ denotes the Euclidean norm. Let
\begin{equation\*}
 S\_n:=\sum\_{j\in J\_n} X\_{j,n}. 
\end{equation\*}


For any random vector $Z$ in $\R^d$, let $f\_Z$ denote the characteristic function (ch. f.) of $Z$, so that $f\_Z(t)=Ee^{it\cdot Z}$ for $t\in\R^d$, where $\cdot$ denotes the dot product.


For each natural $n$, let $Y\_{1,n},\dots,Y\_{j\_n,n}$ be independent random vectors in $\R^d$ such that for all $j\in J\_n$ and all $t\in\R^d$
\begin{equation\*}
 f\_{Y\_{j,n}}(t)=\exp(f\_{X\_{j,n}}(t)-1). 
\end{equation\*}
Let
\begin{equation\*}
 T\_n:=\sum\_{j\in J\_n} Y\_{j,n}.
\end{equation\*}


The problem is then to show that
\begin{equation\*}
 \text{$S\_n$ converges in distribution iff $T\_n$ converges in distribution.} \tag{20}\label{20}
\end{equation\*}


Here is a proof. Take any $t\in\R^d$. Note that $|e^{iu}-1|\le\min(2,|u|)$ for all real $u$. So, for each real $\ep>0$,
\begin{equation\*}
 |f\_{X\_{j,n}}(t)-1|\le E|e^{it\cdot X\_{j,n}}-1|\,1(\|X\_{j,n}\|\le\ep)
 +E|e^{it\cdot X\_{j,n}}-1|\,1(\|X\_{j,n}\|>\ep)
 \le\|t\|\ep+2P(\|X\_{j,n}\|>\ep)
\end{equation\*}
and hence, by \eqref{0},
\begin{equation\*}
 \limsup\_n\max\_{j\in J\_n}|f\_{X\_{j,n}}(t)-1|\le\|t\|\ep. 
\end{equation\*}
So,
\begin{equation\*}
\max\_{j\in J\_n}|f\_{X\_{j,n}}(t)-1|\to0\text{ uniformly in $t$ in any bounded set}. \tag{30}\label{30}
\end{equation\*}


Note also that $|e^{iu}-1-iu|\le u^2/2\le u^2$ for all real $u$. So,
\begin{equation\*}
 |f\_{X\_{j,n}}(t)-1|=|E(e^{it\cdot X\_{j,n}}-1-it\cdot X\_{j,n})|\le E(t\cdot X\_{j,n})^2
 \le\|t\|^2 E\|X\_{j,n}\|^2. 
\end{equation\*}
So, by \eqref{10},
\begin{equation\*}
\sum\_{j\in J\_n}|f\_{X\_{j,n}}(t)-1|\le C\|t\|^2. \tag{40}\label{40}
\end{equation\*}


It follows from \eqref{30} that for all large enough $n$ (depending on $t$) and all $j\in J\_n$ the value of $\ln f\_{X\_{j,n}}(t)$ is defined and
\begin{equation}
 |\ln f\_{X\_{j,n}}(t)-(f\_{X\_{j,n}}(t)-1)|\le|f\_{X\_{j,n}}(t)-1|^2. 
\end{equation}
So, by \eqref{40} and \eqref{30},
\begin{equation}
 \Big|\sum\_{j\in J\_n}\ln f\_{X\_{j,n}}(t)-\sum\_{j\in J\_n}(f\_{X\_{j,n}}(t)-1)\Big|
 \le\sum\_{j\in J\_n}|f\_{X\_{j,n}}(t)-1|^2 \\ 
 \le\max\_{j\in J\_n}|f\_{X\_{j,n}}(t)-1|\,\sum\_{j\in J\_n}|f\_{X\_{j,n}}(t)-1|
 \le C\|t\|^2\max\_{j\in J\_n}|f\_{X\_{j,n}}(t)-1|\to0. 
 \tag{50}\label{50}
\end{equation}


Thus, in view of \eqref{50} and \eqref{30},


$S\_n$ converges in distribution


iff $\prod\_{j\in J\_n}f\_{X\_{j,n}}$ converges pointwise to a function continuous at $0$


iff $\sum\_{j\in J\_n}\ln f\_{X\_{j,n}}$ converges pointwise to a function continuous at $0$


iff $\sum\_{j\in J\_n}(f\_{X\_{j,n}}-1)$ converges pointwise to a function continuous at $0$


iff $T\_n$ converges in distribution.


So, we have \eqref{20}. $\quad\Box$


