For some $\lambda$, suppose for each $n > \lambda$, there are independent random variables $(Z_{j,n})_{j = 1}^n$ such that
$$
P(Z_{j,n} = 1) = 1 – P(Z_{j,n} = 0) = \frac\lambda n
$$
I am trying to calculate
$$
\lim_{n\to\infty} P(\sum_{j=1}^n Z_{j,n} = k).
$$
Since for each $n$, $E(\sum_{j=1}^n Z_{j,n})= Var(\sum_{j=1}^n Z_{j,n}) = \lambda$, this reminds me of Poisson distribution. I found on Wikipedia that this is called free Poisson distribution. But I did not find any helpful material elsewhere. Can someone point to how to approach this? Thanks in advance!
Best Answer
For any given $n$ we have $$ \mathbb P\left(\sum_{j=1}^n Z_{j,n}=k\right) = \binom nk \left(\frac\lambda n\right)^k\left(1-\frac\lambda n\right)^{n-k}. $$ As $n\to\infty$ the above limit is $e^{-\lambda}\frac{\lambda^k}{k!}$, and hence $Z_{j,n}\stackrel{n\to\infty}\longrightarrow \mathsf{Pois}(\lambda)$.
More generally it holds for a sequence $p_n$ of real numbers in $[0,1]$ such that $np_n\to \lambda$, $$ \lim_{n\to\infty} \binom nk p_n^k (1-p_n)^{n-k} = e^{-\lambda}\frac{\lambda^k}{k!}$$ (this is commonly known as the Poisson limit theorem).