0
$\begingroup$

Let triple $(\Omega, \Sigma, P)$ be a probability space. A (temporal) stochastic process $X(t, \omega): (0,\infty]\times\Omega\rightarrow R$ can be viewed as:

  1. for each $t$, $X(t, \cdot)$ is a random variable
  2. for each fixed $\omega\in\Omega$, $X(\cdot, \omega)$ is a sample path (measurable function).

(1) is easy to understand. I'm confused with (2). What does it mean by fixing $\omega$?

For example, let's say ($\mathbb{R}$, $\mathcal{B}(\mathbb{R})$, $P$), by fixing $\omega=a\in\mathbb{R}$ with a number, then the sample path is

$$ X(1, \omega=a), X(2, \omega=a), X(3, \omega=a)\,\cdots, X(t, \omega=a) $$

Is this correct by choosing an $\omega\in\Omega$?

Could someone give an explicit example of $(\Omega, \Sigma, P)$, and one fixed $\omega\in\Omega$ of Brownian motion? What are the $\Omega, \Sigma, \omega$ here


Similar question:

  1. Stochastic Processes and Trajectories He says $\omega$ should be a sequence. But doesn't this violate $\omega\in\Omega$? Does it mean for every time step $t$, we have to "enlarge" the sample space?

  2. An example of stochastic process

$\endgroup$
1
  • $\begingroup$ For $\omega\in\Omega$, you can simply define a mapping $X':(0,\infty]\rightarrow R$ by $X'(t):=X(t,\omega)$. $\endgroup$
    – peer
    Jan 10, 2020 at 15:40

1 Answer 1

5
$\begingroup$

Let us start with a simple example. Assume that $( \Omega, \mathcal{F}, P)$ is some probability space and let $X : \Omega \rightarrow [0, 10]$ be a random variable which is uniformly distributed on $[0,10]$.

Let us define a stochastic process $X( t, \omega)$ in the following way: $$ X ( t, \omega ) := t X(\omega), \quad ( t, \omega) \in [0, \infty) \times \Omega. $$ Now let us first fix $t=100$. Then $X ( 100 , \omega) = 100X(\omega)$. In other words, we get $100X$, which is again a random variable, since a random variable multiplied by a constant is again a random variable. This argument can be used to conclude that $X(t, \omega) = t X(\omega)$ is a random variable for every fixed $t \geq 0$.

Let us now fix some $\omega_0 \in \Omega$ and assume that $X(\omega_0) = 3$. This then means that for all $t \geq 0$ $$ X( t, \omega_0 ) =tX(\omega_0) = 3t. $$ And if we let $t$ vary over $[0, \infty)$ we will simply get a linear function of the form $$ y(t)=t X( \omega_0) = 3t, \quad t \in [ 0, \infty) $$ which will be the sample path of the stochastic process $X(t, \omega)$ for the fixed $\omega_0 \in \Omega$. Extending this argument, we see that all the sample paths of the stochastic process $X(t,\omega) = t X( \omega )$ are from the following class of linear functions: $$ \{ y : [0, \infty ) \rightarrow \mathbb{R} : y(t) = bt, \ b \in [ 0, 10 ]\}. $$ Let us now turn to a Brownian $B(t, \omega)$ defined on (some other) probability space $( \Omega, \mathcal{F}, P )$. The Brownian motions starts almost surely at $0$, which means that $B(0, \omega)$, which is just a random variable for the fixed $t = 0$, is equal to $0$ with a probability of $1$, i.e. $P( B(0, \omega ) = 0 ) = 1$. Moreover, we know that a Brownian motion has continuous sample paths, which means that for every fixed $\omega_0 \in \Omega$, $y(t) = B(t, \omega_0)$ is a continuous function of $t$. Therefore, with probability $1$, the sample paths of the Brownian motion will be from the following class of functions $$ \{ y : [0, \infty ) \rightarrow \mathbb{R} : y(t) \ \text{is continuous}, \ y(0) = 0 \}. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy