# (PTE) 3.6. Poisson convergence

$\newcommand{\argmin}{\mathop{\mathrm{argmin}}\limits}$ $\newcommand{\argmax}{\mathop{\mathrm{argmax}}\limits}$

I would like to finish reviewing Probability theory I by briefly mentioning the Poisson convergence (section 3.6) and limit theorems in $\mathbb{R}^d$ (3.10).

Poisson convergence is about limiting laws of a sequence of independent discrete Bernoulli-like random variables with $ES_n$ converges to a constant $\lambda.$ I used the term “Bernoulli-like” because for large enough $n,$ such $X_n$ should behave similar to Bernoulli random variables. For those sequences, the limiting law is not normal, but actually Poisson.

## Basic Poisson convergence

Let $(X_{nk})_{k=1}^n$ be a rowwise independent triangular array with $P(X_{nk}=1)=p_{nk}$ and $P(X_{nk}=0)=1-p_{nk}.$ As $n\to\infty,$ if
(i) $\sum_{k=1}^n p_{nk} \to \lambda \in (0,\infty).$
(ii) $\max_{1\le k\le n} p_{nk} \to 0.$
Then $S_n=\sum_{k=1}^n X_{nk} \overset{w}{\to} \mathcal{P}(\lambda).$
expand proof

As in the proof of the CLT, it suffices to show $\varphi_{S_n}(t) \to \exp(\lambda(e^{it}-1)).$ \begin{aligned} \varphi_{S_n}(t) &= \prod_{k=1}^n \varphi_{nk}(t) \\ &= \prod_{k=1}^n \left( p_{nk}(e^{it}-1) + 1 \right). \end{aligned} Thus \begin{aligned} &\left| \varphi_{S_n} - e^{\lambda(e^{it}-1)} \right| \\ &= \left| \prod_{k=1}^n \left( p_{nk}(e^{it}-1) + 1 \right) - \prod_{k=1}^n e^{p_{nk}(e^{it}-1)} \right| \\ &= \sum_{k=1}^n \left| p_{nk}(e^{it}-1)+1 - e^{p_{nk}(e^{it}-1)} \right| \\ &\;\;\;\;(b = p_{nk}(e^{it}-1),~ |b| \le 2p_{nk} \le 1 \text{ for large n.}) \\ &\le \sum_{k=1}^n p_{nk}^2 |e^{it}-1|^2 \le 4 \sum_{k=1}^n p_{nk}^2 \\ &\le 4 \max_{1\le k\le n} p_{nk} \sum_{k=1}^n p_{nk} \to 0. \end{aligned} The first inequality is from the lemmas we covered before \begin{aligned} &|p_{nk}(e^{it}-1)+1| \le 1-p_{nk}+p_{nk}|e^{it}| \le 1.\\ &|e^{p_{nk}(e^{it}-1)}| \le e^{p_{nk} |e^{it}-1|} \le e^{2p_{nk}} \le 1 \text{ for large } n. \end{aligned}

For a special case, consider $X_k \overset{\text{iid}}{\sim} \mathcal{B}(p_n)$ for $k=1,\cdots,n$ where $np_n\to\mu.$ Then $S_n \overset{w}{\to} \mathcal{P}(\mu).$

## General Poisson convergence

Let $(X_{nk})_{k=1}^n$ be rowwise independent, non-negative, integer-valued triangular array with $P(X_{nk}=1) = p_{nk}$ and $P(X_{nk} \ge 2) = \epsilon_{nk}.$ As $n\to\infty$ if
(i) $\sum_{k=1}^n p_{nk} \to \lambda \in (0,\infty).$
(ii) $\max_{1\le k\le n} p_{nk} \to 0.$
(iii) $\sum_{k=1}^n \epsilon_{nk} \to 0.$
then $S_n = \sum_{k=1}^n X_{nk} \overset{w}{\to} \mathcal{P}(\lambda).$

Let $X_{nk}' = X_{nk}$ if $X_{nk}=1$ and $0$ otherwise. Then \begin{aligned} P(S_n \ne S_n') \le \sum_{k=1}^n P(X_{nk} \ge 2) = \sum_{k=1}^n \epsilon_{nk} \to 0. \end{aligned} By basic Poisson convergence theorem, $S_n' \overset{w}{\to} \mathcal{P}(\lambda).$ By the converging together lemma, we get the desired result.

## Total variation and weak convergence

Lastly, I will introduce total variation, a metric on a space of discrete probability measures. Like Levy metric, convergence in total variation distance is equivalent to weak convergence on such space.

Let $\mu,\nu$ be measures on countable space $\mathcal{S}.$
\begin{aligned} \|\mu-\nu\| &:= \frac{1}{2} \sum_{z\in\mathcal{S}} |\mu(z) - \nu(z)| \\ &= \sup_{A\subset\mathcal{S}} |\mu(A) - \nu(A)| \end{aligned} is the total variation distance between $\mu$ and $\nu.$

While the first formula is the definition, the second one is a derived property. Note that

\begin{aligned} \sum_{z\in\mathcal{S}}|\mu(z)-\nu(z)| &\ge |\mu(A) - \nu(A)| + |\mu(A^c)-\nu(A^c)| \\ &= 2|\mu(A)-\nu(A)|. \end{aligned}

Let $A=\{z:~ \mu(z) \ge \nu(z)\}$ then the equality holds.

(i) $d(\mu, \nu) = \|\mu-\nu\|$ is a metric on the space of probability measures on $\mathbb{Z}.$
(ii) $\|\mu_n-\mu\|\to0$ if and only if $\mu_n(z)\to\mu(z)$ for all $z\in\mathbb{Z}.$ (i.e. $\mu_n \overset{w}{\to} \mu.$)

(ii)
($\Rightarrow$) Given $z\in\mathbb{Z},$ \begin{aligned} |\mu_n(z)-\mu(z)| \le \sup_{A\subset\mathbb{Z}}|\mu_n(A)-\mu(A)| \to 0. \end{aligned} ($\Leftarrow$) $\mu_n \overset{w}{\to} \mu$ then \begin{aligned} \frac{1}{2}\sum_{z\in\mathbb{Z}} |\mu_n(z)-\mu(z)| = \sum_{z\in\mathbb{Z}} (\mu_(z)-\mu(z))^+ \to 0 \end{aligned} by DCT.

Acknowledgement

This post series is based on the textbook Probability: Theory and Examples, 5th edition (Durrett, 2019) and the lecture at Seoul National University, Republic of Korea (instructor: Prof. Johan Lim).