Sample path of stochastic process: What is "by fixing $\omega$"?

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

• For $\omega\in\Omega$, you can simply define a mapping $X':(0,\infty]\rightarrow R$ by $X'(t):=X(t,\omega)$.
– peer
Jan 10 '20 at 15:40

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 $$\{ y : [0, \infty ) \rightarrow \mathbb{R} : y(t) \ \text{is continuous}, \ y(0) = 0 \}.$$