# Levy processes

(Difference between revisions)
 Revision as of 17:24, 25 April 2012 (view source)Luis (Talk | contribs)m (→Stochastic control and fully non-linear integro-differential operators)← Older edit Latest revision as of 21:02, 25 January 2016 (view source)Luis (Talk | contribs) (→Connection with linear integro-differential operators) Line 35: Line 35: == Connection with linear integro-differential operators == == Connection with linear integro-differential operators == - Any Lévy process $X(t)$ such that $X(0)=0$ almost surely defines a linear semigroup $\{U_t\}_{t\geq0}$ on the space  of continuous functions $f:\mathbb{R}^d\to\mathbb{R}^d$ as follows + Any Lévy process $X(t)$ such that $X(0)=0$ defines a linear semigroup $\{U_t\}_{t\geq0}$ on the space  of continuous functions $f:\mathbb{R}^d\to\mathbb{R}^d$ as follows $(U_tf)(x)= \mathbb{E}\left [ f(x+X(t)) \right ]$ $(U_tf)(x)= \mathbb{E}\left [ f(x+X(t)) \right ]$

## Latest revision as of 21:02, 25 January 2016

A Lévy process is an important type of stochastic process (namely, a family of $\mathbb{R}^d$ valued random variables each indexed by a positive number $t\geq 0$). In the context of parabolic integro-differential equations they play the same role that Brownian motion and more general diffusions play in the theory of second order parabolic equations.

Informally speaking, a Lévy process is a random trajectory, generalizing the concept of Brownian motion, which may contain jump discontinuities. A prototypical example would be $X(t)=B(t)+N(t)$ where $B(t)$ is the standard Brownian motion and $N(t)$ is a Compound Poisson process, the trajectory described by typical sample path of this process would look like the union of several disconnected Brownian motion paths.

## Definition

A stochastic process $X=\{X(t)\}_{t \geq 0}$ with values in $\mathbb{R}^d$ is said to be a Lévy process if

1.For any sequence $0 \leq t_1 < t_2 <...<t_n$ the random variables $X(t_0),X(t_1)-X(t_0),...,X(t_n)-X(t_{n-1})$ are independent.

2.For any positive times $s\leq t$ the random variables $X(t-s)$ and $X(t)-X(s)$ have the same probability law.

3.Almost surely, the trajectory of $X(t)$ is continuous from the right, with limit from the left also known as "càdlàg" for its acronym in french.

## Lévy-Khintchine Formula

It follows from the first two properties above that if $X$ is a Lévy process and we further assume $X(0)=0$ a.s. then for each fixed positive $t$ the random variable $X(t)$ is infinitely divisible, that is, it can be written as the sum of $n$ independent and identically distribued random variables, for all $n\in\mathbb{N}$. Indeed, let $h=\tfrac{t}{n}$, then

$X(t) = \left( X(h)-X(0)\right)+\left( X(2h)-X(h)\right)+...+\left( X(t)-X((n-1)h)\right)$

and by the above definition the differences $X(kh)-X((k-1)h)$ are independent and distributed the same as $X(h)$. From the infinite divisibility of $X(t)$ it follows by a theorem of Lévy and Khintchine that for any $\xi \in \mathbb{R}^d$ we have

$\mathbb{E} \left [ e^{i\xi\cdot X_t}\right ] = e^{t\eta(\xi)}$

the function $\eta(\xi)$ given by

$\eta(\xi)=i y\cdot b -\tfrac{1}{2}(A\xi,\xi)+\int_{\mathbb{R}^d} \left ( e^{i \xi\cdot y}-1-i\xi\cdot y \chi_{B_1}(y) \right ) d\mu(y)$

where $b$ is a vector, $A$ is a positive matrix, $B_1$ is the unit ball and $\mu$ is a Lévy measure, that is, a Borel measure in $\mathbb{R}^d$ such that

$\int_{\mathbb{R}^d}\frac{|y|^2}{1+|y|^2}d\mu(y) <+\infty.$

The interpretation of this measure $\mu$ is that jumps from some point $x$ to $x+y$ with $y$ in some set $A$ occur as a Poisson process with intensity $\mu(A)$.

## Connection with linear integro-differential operators

Any Lévy process $X(t)$ such that $X(0)=0$ defines a linear semigroup $\{U_t\}_{t\geq0}$ on the space of continuous functions $f:\mathbb{R}^d\to\mathbb{R}^d$ as follows

$(U_tf)(x)= \mathbb{E}\left [ f(x+X(t)) \right ]$

Given the initial assumption on $X(0)$ it is clear that $U_0$ is the identity, and given that $X(t)-X(s)$ is distributed as $X(t-s)$ it follows that $U_t \circ U_s = U_{t+s}$.

As a semigroup, $U_t$ has an infinitesimal generator which turns out to be a Linear integro-differential operator. More precisely, if we let $f(x,t):=(U_tf)(x)$, then, assuming that $f(x,t)$ has enough regularity it can be checked that

$\partial_t f = Lf \;\;\;\mbox{ for all } (x,t)\in\mathbb{R}^d\times \mathbb{R}_+$

where for any smooth function $\phi$, we have

$L\phi(x) = b \cdot \nabla \phi(x) +\mathrm{tr} \,( A\cdot D^2 \phi )+ \int_{\R^d} (\phi(x+y) - \phi(x) - y \cdot \nabla \phi(x) \chi_{B_1}(y)) \, \mathrm{d} \mu(y)$

## Stochastic control and fully non-linear integro-differential operators

A similar connection holds via the Isaacs-Bellman equation arising in stochastic control problems (or more generally in stochastic games). In this case the corresponding semigroup is not linear, and instead one must work in terms of viscosity solutions to build it.