We know that a Poisson distribution can be approximated by a binomial distribution. More exactly, let $(X\_{jn})\_{1\leq j \leq n}$ be a i.i.d. triangular array such that
$$P[X\_{jn}= 1 ] = p\_n = 1- P[X\_{jn}=0]$$
and:


1. $p\_n \to 0$ as $n \to \infty$;
2. $np\_n \to \lambda$ as $n \to \infty$


So we have the following convergence in distribution:
$$S\_n = \sum\_{j=1}^n X\_{jn} \overset{d}{\to} N= \sum\_{j=1}^N 1, \quad N \sim \hbox{Poisson}(\lambda)$$
Thus, if we want approximate $N\sim \hbox{Poisson}(\lambda)$ by a binomial distribution, we can set $X\_{jn} \sim \hbox{Bernoulli}(\lambda/n)$ or $S\_n \sim \hbox{Binomial}(\lambda/n , n)$


Now, given $(\xi\_j)\_{j=1}^\infty$ a i.i.d. sequence of random variables independent of $N$. Consider $Y = \sum\_{j=1}^N \xi\_j$. Something tells me that I can find a sum $S\_n$ that converges in distribution to $Y$:
$$S\_n \overset{d}{\to} Y = \sum\_{j=1}^N \xi\_j, \quad N \sim \hbox{Poisson}(\lambda)$$
Since $Y$ is infinitely divisible, so it there is a triangular array $(X\_{ij})$ such that $S\_n$ converges in distribution to $Y$, But I think I should adjust some weights in the summations: $S\_n= \sum\_{j=1}^n w\_j X\_{jn}$, where $X\_{jn} \sim \hbox{Bernoulli}(\lambda/n)$.


Is there any constructive way to express this sum $S\_n$?



Approximation of a random sum of random variables (infinitely divisible distribution) by a triangular array
$\newcommand\la\lambda$Let
$$S\_n:=\sum\_{j=1}^n\xi\_j X\_{j,n},$$
where the $\xi\_j$'s are iid random variables (r.v.'s) and, for each $n$, the $X\_{j,n}$'s are iid r.v.'s independent of $\xi\_j$'s and such that each $X\_{j,n}$ has the Bernoulli distribution with parameter $p\_n$.
Suppose that $n\to\infty$ and $np\_n\to\la$ for some real $\la>0$.


Then
\begin{equation\*}
 S\_n\to Y:=\sum\_{j=1}^N\xi\_j \tag{1}\label{1}
\end{equation\*}
in distribution, where
$N$ is a Poisson r.v. with parameter $\la$ independent of the $\xi\_j$'s.


This follows easily by the method of characteristic functions: If $f(t):=Ee^{it\xi\_1}$ for real $t$, then
\begin{equation\*}
 Ee^{itS\_n}=(Ee^{it\xi\_1 X\_{1,n}})^n
=(1-p\_n+p\_n f(t))^n\to e^{\la(f(t)-1)} \tag{2}\label{2}
\end{equation\*}
and
\begin{equation\*}
 Ee^{itY}=\sum\_{n=0}^\infty P(N=n)f(t)^n
=\sum\_{n=0}^\infty \frac{\la^n}{n!}\,e^{-\la}f(t)^n
=e^{\la(f(t)-1)}, \tag{3}\label{3}
\end{equation\*}
so that $Ee^{itS\_n}\to Ee^{itY}$.


The result you quoted in the beginning of your post is the special case of \eqref{1} with $\xi\_j=1$ for all $j$.




---


**Details on \eqref{2}:**
\begin{equation\*}
\begin{aligned}
 Ee^{it\xi\_1 X\_{1,n}}&=Ee^{it\xi\_1 X\_{1,n}}\,1(X\_{1,n}=0)+Ee^{it\xi\_1 X\_{1,n}}\,1(X\_{1,n}=1) \\ 
 &=E1(X\_{1,n}=0)+Ee^{it\xi\_1}\,1(X\_{1,n}=1) \\ 
 &=1-p\_n+Ee^{it\xi\_1}\,E1(X\_{1,n}=1) \\ 
 &=1-p\_n+f(t) p\_n. 
\end{aligned}
\end{equation\*}
Here we used the equality $Ee^{it\xi\_1}\,1(X\_{1,n}=1)=Ee^{it\xi\_1}\,E1(X\_{1,n}=1)$, which holds because $\xi\_1$ and $X\_{1,n}$ are independent.


**Details on \eqref{3}:**
\begin{equation\*}
\begin{aligned}
 Ee^{itY}&=\sum\_{n=0}^\infty E1(N=n)e^{itY} \\ 
 &=\sum\_{n=0}^\infty E1(N=n)\exp\Big(it\sum\_{j=1}^n\xi\_j\Big) \\ 
 &=\sum\_{n=0}^\infty E1(N=n)\,E\exp\Big(it\sum\_{j=1}^n\xi\_j\Big) \\ 
&=\sum\_{n=0}^\infty P(N=n)f(t)^n. 
\end{aligned}
\end{equation\*}
Here we used the equality $E1(N=n)\,\exp\big(it\sum\_{j=1}^n\xi\_j\big)=E1(N=n)\,E\exp\big(it\sum\_{j=1}^n\xi\_j\big)$, which holds because the $\xi\_j$'s and $N$ are independent.


