# 3.9. Infinitely divisible distributions

A certain kind of well behaving distributions has characteristic functions that can be represented in canonical form. In this section we cover conditions that such distributions have and its canonical representation.

## Infinitely divisible distributions

(i) There exists a squence of distributions $(F_n)$ such that $F=F_n*\cdots* F_n$ for all $n\in\mathbb{N}.$

(ii) There exists random variables $X,X_{nk}$ in a probability space $(\Omega, \mathcal{F}, P)$ such that $X\overset{d}{=}X_{n1}+\cdots+X_{nn}$ for all $n,$ where $X\sim F$, $X_{nk}\sim F_n$ for all $k$ and $X_{nk}$'s are rowwise independent.

(iii) There exists a sequence of characteristic functions $(\varphi_n)$ such that $\varphi = (\varphi_n)^n.$

Here $*$ denotes convolution. In fact all three conditions are equivalent. As an example, we can easily check that a normal distribution $X\sim\mathcal{N}(\mu,\sigma^2)$ is infinitely divisible since $X\overset{d}{=}X_{n1}+\cdots+X_{nn}$ for rowwise independent $X_{nk}\sim\mathcal{N}(\frac{\mu}{n},\frac{\sigma^2}{n}).$

First important property is that characteristic functions of ID distributions always have non-zero values. For this, we need a lemma that applies to all characteristic functions.

Proof is simple using elementary trigonometrics. Notice that $|\varphi|^2$ is a real-valued ch.f. so it suffices to show that $\text{Re}(1-\varphi(2t)) \le 4\text{Re}(1-\varphi(t))$ for a given $\varphi.$ $$\begin{aligned} \text{Re}(1-\varphi(t)) &= \int(1-\cos tx)dF(x) \\ &= \int2\sin^2 \frac{t}{2}dF(x) \\ &= \int\frac{\sin^2 tx}{2\cos^2\frac{t}{2}x}dF(x) \\ &= \int\frac{1}{2}\sin^2 tx dF(x) \\ &= \int\frac{1}{4}(1-\cos 2tx) dF(x) \\ &= \frac{1}{4}\text{Re}(1-\varphi(2t)). \end{aligned}$$

Proof is by induction. Since $\varphi$ is ID, there exists $\varphi_n$ such that $\varphi=(\varphi_n)^n.$ We know that $\varphi\to1$ as $t\to0,$ so there exists $a>0$ such that $|\varphi(t)|>0$ for all $|t|\le a.$

Given $t$ that $|t|\le a,$ $$|\varphi_n(t)| = |\varphi(t)^\frac{1}{n}| \ge \left(\inf_{|t|\le a}|\varphi(t)|\right)^\frac{1}{n} \to_n 1.$$ so for all $0\le\epsilon\le1,$ there exists $N>0$ such that $|\varphi_n(t)| > 1-\epsilon$ for all $n\ge N.$ By the lemma, for $n\ge N$ and $|t|\le a,$ $$1-|\varphi_n(2t)|^2 \le 4(1-(1-\epsilon)^2) \le 8\epsilon.$$ Thus for large n, $|\varphi_n(2t)|^2 \ge 1-8\epsilon > 0,$ for all $0 < \epsilon < 1/8.$ This gives $\varphi(2t) \ne 0$ for all $|t|\le a.$ Repeatedly apply the process to get $\varphi \ne 0$ for all $t.$

## Canonical representation

Characteristic functions of Infinitely divisible distributions can be uniquely represented in a certain form. Furthermore, if a characteristic function can be written in such form, then it is infinitly divisible. We call it a *canonical form*. While there are several equivalent canonical representations, I would like to cover the one by Kolmogorov. The first theorem is about sufficiency of ID distribution.

### Sufficiency

## expand proof

(case 1. $\mu$ has a mass only at 0.)

Let $\sigma^2 = \mu(\mathbb{R}) = \mu\{0\}>0$ then $\varphi(t) = e^\frac{t^2\sigma^2}{2}$ which is the ch.f. of $\mathcal{N}(0,\sigma^2)$ so it is ID.

(case 2. $\mu$ has a mass only at $x\ne0.$)

Let $\mu\{x\}=\lambda x^2$ for some $\lambda>0.$ Then $\varphi(t)=e^{\lambda(e^{itx}-1-itx)}$ which is a ch.f. of $x(Z_\lambda-\lambda)$ where $Z\sim\mathcal{P}(\lambda).$ Let $X_{nk}\overset{\text{iid}}{\sim} \mathcal{P}(\frac{\lambda}{n})$ for $1\le k\le n.$ $x(Z_\lambda-\lambda) \overset{d}{=} x\sum_{k=1}^n(X_{nk}-\frac{\lambda}{n})$ so it is ID.

(case 3. $\mu$ has masses at $x_1,\cdots,x_k.$)

Let $\mu\{x_i\}=\delta_i>0$ and $\varphi_i(t) = \exp\{ \int_{-\infty}^\infty \frac{e^{itx}-1-itx}{x^2} d\mu_i(x) \}$ where $\mu_i(\mathbb{R})=\delta_i.$ By case 2, $\varphi_i$ is ID with mean 0 and variance $\delta_i.$ Thus for all $n,$ there exists ch.f.s $\varphi_{jn}$ such that $\varphi_n = (\varphi_{jn})^n.$ It follows that $\varphi = \prod_{j=1}^k \varphi_j = \left(\prod_{j=1}^k \varphi_{jn}\right)^n$ thus $\varphi$ is ID. Let $X\sim\varphi$ and $X_i\overset{\text{indep}}{\sim} \varphi_i$ then $X\overset{d}{=}X_1+\cdots+X_k$ so $EX=0,\text{Var}(X)=\mu\{x_1,\cdots,x_k\}.$

(case 4. general finite $\mu$.)

Let $\mu_k\{j\cdot2^{-k}\} = \mu(j\cdot2^{-k}, (j+1)2^{-k}],$ $j\in J_k = \{0,\pm1,\pm2,\cdots,\pm2^{2k}\}.$ Then $\mu_k$ has masses on $\{j\cdot2^{-k}:~ j \in J_k\}.$ Since $\mu_k(\mathbb{R}) \to \mu(\mathbb{R}) > 0$ as $k\to\infty,$ $\mu_k(\mathbb{R})>0$ for all large $k.$

Now assume that $f:\mathbb{R}\to\mathbb{R}$ is continuous and vanishes at infinity (i.e. $\lim_{|x|\to\infty}f(x)=0$). Let $$f_k = \begin{cases} f(j\cdot2^{-k}) &,~ x\in (j\cdot2^{-k}, (j+1)2^{-k}] \\ 0 &,~ \text{otherwise} \end{cases}$$ be a step function, then $\int{f}d\mu_k = \int{f_k}d\mu.$ As $k\to\infty,$ $f_k\to f.$ Since $|f_k| \le |f| \le \sup_x |f(x)| < \infty,$ apply BCT and we get $\int{f_k}d\mu \to \int{f}d\mu.$

By the case 3, $\varphi_k(t) = \exp\{ \int_{-\infty}^\infty \frac{e^{itx}-1-itx}{x^2} d\mu_k(x) \}$ is ID. since the integrand is continuous and vanishes at infinity, $\varphi_k \to \varphi$ as $k\to\infty.$ Since $\varphi(0)=1$ and $\varphi$ is continuous at 0, by continuity theorem $\varphi$ is a ch.f. for some random variable.

In addition, $EX^2 \le \liminf_k EX_k^2 < \infty$ for $X\sim\varphi$ and $X_k\sim\varphi_k.$ By moment generating property of ch.f., $iEX = \varphi'(0) = 0$ and $-\text{Var}(X) = -\mu(\mathbb{R}).$ Let another $\psi_n(t) = \exp\{ \int_{-\infty}^\infty \frac{e^{itx}-1-itx}{x^2} d\frac{\mu}{n}(x) \}$ then it is a ch.f. Observe that $\varphi = (\psi_n)^n$ so $\varphi$ is ID.

In other words,
\(\log \varphi(t) = \int_{-\infty}^\infty \frac{e^{itx}-1-itx}{x^2} d\mu(x).\)
We call the right hand side the *canonical representation of $\varphi$* and $\mu$ the *canonical measure*. Note that $\frac{|e^{itx}-1-itx|}{x^2} \le t^2$ so the integral is well-defined. For $x=0,$ we define $\frac{e^{itx}-1-itx}{x^2} \big\vert_{x=0} = -\frac{t^2}{2}$ by continuity. Also note that
\(\frac{|e^{itx}-1-itx|}{x^2} \le t^2 \wedge \frac{2|t|}{|x|} \to 0 \text{ as } |x|\to\infty.\)
This follows from error estimation of the second-order Taylor series.

### Necessity

To show the necessity part for more general class of characteristic functions, we define the condition R.

(i) $EX_{nk}=0,$ $\sigma_{nk}^2 = EX_{nk}^2 < \infty,$ $s_n^2 = \sum_{k=1}^{r_n} \sigma_{nk}^2 > 0.$

(ii) $\sup_n s_n^2 < \infty.$

(iii) $\max_{1\le k\le r_n} \sigma_{nk}^2 \to 0$ as $n\to\infty.$

For the proof of the next theorem, we need the following lemma.

## expand proof

$$\begin{aligned} \bigg| \underbrace{\prod_{k=1}^{r_n} \varphi_{nk}(t)}_\text{(i)} - \underbrace{\prod_{k=1}^{r_n} e^{\varphi_{nk}(t)-1}}_\text{(ii)} \bigg| &\le \sum_{k=1}^{r_n} |\varphi_{nk}(t) - e^{\varphi_{nk}(t)-1}| \\ &\le \sum_{k=1}^{r_n} |\varphi_{nk}(t) - 1|^2 \\ &\le \sum_{k=1}^{r_n} (t^2\sigma_{nk}^2)^2 \\ &\le t^4 \max_{1\le k\le r_n} \sigma_{nk}^2 \cdot s_n^2 \to 0. \end{aligned}$$ The first inequality is from 3.4.3, the second is from 3.4.4, the third is from 3.3.19, and $\to0$ is by condition R. In addition, $$\text{(i)} \to \varphi \text{ as } n\to\infty$$ also by condition R. $$\begin{aligned} \text{(ii)} &= \sum_{k=1}^{r_n} \int(e^{itx}-1) dF_{nk}(x) \\ &= \sum_{k=1}^{r_n} \int\frac{e^{itx}-1-itx}{x^2} x^2 dF_{nk}(x) \\ &= \int\frac{e^{itx}-1-itx}{x^2} d\left(\sum_{k=1}^{r_n}x^2 F_{nk}(x)\right) \end{aligned}$$ Let $\mu_n(-\infty,x] = \sum_{k=1}^{r_n} \int_{-\infty}^xy^2 dF_{nk}(y),$ then $$\text{(ii)} = \int\frac{e^{itx}-1-itx}{x^2} d\mu_n(x)$$ and $\mu_n(\mathbb{R}) = s_n^2.$ So $\sup_n \mu_n(\mathbb{R}) < \infty$ and there exists $(\mu_{nj}),\mu$ such that $\mu_{nj} \overset{w}{\to} \mu$ and $\int{h}d\mu_{nj} \to \int{h}d\mu$ for all $h$ that is continuous and vanishes at infinity. By the above mentioned fact, $$\int\frac{e^{itx}-1-itx}{x^2} d\mu_{nj}(x) \to \int\frac{e^{itx}-1-itx}{x^2} d\mu(x).$$ By convergence of (i) and (ii), the existence part of the proof is done.

For the uniqueness part, we only need to show that such $\mu$ is unique. Suppose $$\int\frac{e^{itx}-1-itx}{x^2} d\mu(x) = \int\frac{e^{itx}-1-itx}{x^2} d\nu(x),~ \forall t.$$ This implies $\int e^{itx}d\mu(x) = \int e^{itx}d\nu(x).$ Put $t=0$ to both sides and we get $c:=\mu(\mathbb{R})=\nu(\mathbb{R}).$ Dividing both sides with $c,$ $\mu/c$ and $\nu/c$ becomes probabilty measures with identical ch.f.s and the proof is done.

*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. Sangyeol Lee).