5
$\begingroup$

I have just started learning about stochastic processes and I am confused with the notion of Brownian motion. The text defines (linear) Brownian motion under measure $\mathbb{P}$ as $B=(B_t; t\geq 0)$ where each $B_t$ is some random variable such that:

$\bullet$ $t\to B_t$ is a continuous function

$\bullet$ $B_t$ is distributed as $N(0,t)$.

$\bullet$ $B_{s+t}-B_{s}$ is distributed as $N(0,t)$

I have some questions regarding this definition. What is the sample space (space of events) on which each random variable $B_t$ is defined? What does it mean that $t\to B_t$ is continuous? Is it that for every $\omega$ in the sample space $t\to B_t(\omega)$ is continuous? What kind of values is $B_t$ taking?

Hopefully I formulated my questions clearly. Any help is appreciated.

  • 2
    @ dmm : I think you missed an important hypothesis which is independence of increments. Best regards.2012-11-07

1 Answers 1

5

The Brownian motion $B$ is a mapping $I \times \Omega \ni (t,w) \mapsto B(t,w) \in \mathbb{R}$ where $I \subseteq \mathbb{R}$ (usually $I=[0,\infty)$ or $I=[0,1]$) and $(\Omega,\mathcal{A},\mathbb{P})$ is an arbitrary probability space. For fixed $w \in \Omega$ the mapping

$I \ni t \mapsto B(t,w) \in \mathbb{R}$

is called sample path. And you are correct: "$t \mapsto B_t$ continuous" means that the sample paths $t \mapsto B_t(w)$ are continuous for all $w \in \Omega$.

(As @TheBridge already observed: You forgot to mention the independence of the increments.)