# Basic definitions

We will review definition and some basic properties of Levy process.

The characteristic function of r.v. ${X\sim \mu}$ is the mapping ${\Phi: \mathbb{R}^d \rightarrow \mathbb{C}}$ defined by

$\displaystyle \Phi(u) = \mathbb{E} (e^{iu\cdot X}) = \int_{\mathbb{R}^d} e^{iu\cdot y} \mu(dy).$

A random variable ${X}$ is infinitely divisible if its law ${p_x}$ is infinitely divisible, e.g. ${X =^d Y_1^{(n)}+\cdots + Y_n^{(n)}}$, where ${Y_1^{(n)}, \ldots Y_n^{(n)}}$ are i.i.d., for each ${n\in \mathbb{N}}$. Note that, the characteristic function of ${X}$ can be written as ${\Phi_X(u) = (\Phi_{Y_1^{(n)}}(u))^n}$ in the above.

Definition 1 Levy measure is a measure ${\nu}$ on ${\mathbb{R}^d \setminus \{0\}}$ such that

$\displaystyle \int (|y|^2 \wedge 1) \nu (dy) < \infty. \ \ \ \ \ (1)$

The characteristic functions of infinitely divisible probability measures were completely characterized by Levy and Khintchine in the 1930s. This is a fundamental result.

Theorem 2 (Levy-Khintchine) A Borel probability measure ${\mu}$ on ${\mathbb{R}^d}$ is infinitely divisible if there exists a vector ${b\in \mathbb{R}^d}$, a non-negative symmetric ${d\times d}$ matrix ${A}$ and a Levy measure ${\nu}$ on ${\mathbb{R}^d\setminus \{0\}}$ such that for all ${u\in \mathbb{R}^d}$,

$\displaystyle \phi_\nu(u) = \exp \Big\{i b\cdot u - \frac 1 2 u \cdot A u + \int_{\mathbb{R}^d \setminus \{0\}} \Big(e^{iu\cdot y} - 1 - iu\cdot y 1_{B_1(0)}(y)\Big) \nu (dy) \Big\}. \ \ \ \ \ (2)$

Conversely, any mapping of the form (2) is the characteristic function of an infinitely divisible probability measure on ${\mathbb{R}^d}$.

The triple ${(b, A, \nu)}$ is called the characteristics of the infinitely divisible random variable ${X}$, and ${\eta := \log \phi_{\mu}}$ is called the Levy symbol or Characteristic exponent.

Definition 3 A Levy process ${X = (X(t), t\ge 0)}$ is a stochastic process satisfying the following:

1. ${X(0) = 0}$ w.p.1
2. ${X}$ has independent and stationary increments,
3. ${X}$ is stochastically continuous, i.e. for all ${a>0}$ and for all ${s\ge 0}$, ${\lim_{t\rightarrow s} P(|X(t) - X(s)|>a) = 0}$.

It follows from the Definition 3 (2) that each Levy process ${X(t)}$ is infinitely divisible. Therefore,

Theorem 4 If ${X}$ is a Levy process, then ${X(t)}$ is infinitely divisible for each ${t\ge 0}$. Furthermore,

$\displaystyle \phi_{X(t)} (u) = e^{t\eta(u)}, \forall u\in \mathbb{R}^d, t\ge 0,$

where ${\eta}$ is the Levy symbol of ${X(1)}$.

Here are examples of Levy processes.

Example 1 (Brownian motion with drift) Let ${B(t)}$ be a standard Brownian motion in ${\mathbb{R}^d}$. Then, ${C(t) = bt + \sigma B(t)}$ is a Levy process with its characteristic function given by

$\displaystyle \Phi_{B(t)} (u) = \exp \{ t \eta_C(u)\},$

where ${\eta_C(u)}$ is Levy symbol of ${C(1)}$ of the form

$\displaystyle \eta_C(u) = i(b,u) - \frac 1 2 (u, \sigma \sigma^T u).$

In fact a Levy process has a continuous sample paths if and only if it is of the form ${C(t)}$.

Example 2 (The compounded Poisson process) Let ${\{Z(n): n\in \mathbb{N}\}}$ be a sequence of i.i.d r.v. in ${\mathbb{R}^d}$ with law ${\mu_Z}$. Let ${N}$ be a Poisson process of intensity ${\lambda}$, which is independent of ${Z(n)}$. The compound Poisson process ${Y}$ is defined as follows:

$\displaystyle Y(t) = Z(1) + \cdots + Z(N(t)).$

Then, ${Y}$ has Levy symbol of

$\displaystyle \eta_Y(u) = \int (e^{i(u,y)} - 1) \lambda \mu_Z(dy).$

If ${\mu_Z = \delta_{1}}$, then ${Y}$ is said to be Poisson process. The characteristics of ${Y(1)}$ is

$\displaystyle b = \lambda \mathbb E[Z(1) I_{B_{1}(0)}(Z(1))], \ A = 0, \ \nu = \lambda \mu_{Z}.$

Example 3 (Subordinators) A subordinator is a one-dimensional Levy process which is increasing a.s. Such processes can be thought of as a random model of time eveolution. A Levy process ${T = \{T(t): t\ge 0\}}$ be a subordinator if and only if its Levy symbol takes the form

$\displaystyle \eta (u) = i b u + \int_{(0,\infty)} (e^{iuy} - 1) \lambda(dy),$

for some ${b\ge 0}$ and a Levy measure ${\lambda}$ satisfying

$\displaystyle \lambda(-\infty, 0) = 0, \quad \int_{(0,\infty)} (y\wedge 1) \lambda(dy) <\infty.$