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:

- $p_n \to 0$ as $n \to \infty$;
- $np_n \to \lambda$ 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 $S_n$?