I try to understand the nature of the sample space $\Omega$ for a stochastic process $(X_t)_{t \in I}$. Similar to the question here, I wonder what a fixed $\omega \in \Omega$ means, when we talk about a trajectory or path given by the function on $I$ , $t \rightarrow X_t(\omega)$. I can follow the explanation here that $\omega$ encodes a whole sequence and in the book from Oksendal "Stochastic Differential Equations" it is written that $\Omega$ should be regognized as a subset of the space $\tilde{\Omega}=(\mathbb{R}^n)^I$ of all function from $I$ into $\mathbb{R}^n$, if the stochastic process maps to $\mathbb{R}^n$. However, checking the consistency of these statements (and my poor imagination) with examples of stochastic processes, I stumbled over the following example for a modification of a stochastic process see also here (which is quite standard as far as I get it):
$\textbf{Example:}$ Let $\Omega = [0,\infty), \mathcal{A} = \mathcal{B}([0,\infty))$ and $P$ be a probability measure on $\mathcal{A}$ which has a density. Define two stochastic processes $(X(t): t \ge 0)$ and $(Y(t): t \ge 0)$ by
\begin{align}
X(t)(\omega) =
\begin{cases}
1, \text{ if $t = \omega$},\\
0, \text{ otherwise}
\end{cases}
\quad Y(t)(\omega) = 0 \quad \text{for all $t \ge 0$ and all $\omega \in \Omega$.}
\end{align}
Then $X$ and $Y$ are modifications of each other.
I wonder how the idea that $\omega$ should encode the whole sequence relates to the specific sample space $\Omega = [0,\infty)$ given by the example. In this case $\omega \in \Omega$ is just a real number, isn't it? If this statement is true, then dimensionality of $\omega$ does not match my expectations.
On the one hand, $I$ seems to be the non-negative real number line $\mathbb{R}_{\geq 0}$. On the other hand, we obtain a single non-negative real number $\omega \in \Omega \, \forall t \in I$? Where is the randomness (in time) in this case?
Best Answer
Well, this is a nice question, but I would recommend you not to care too much about the structure of $\Omega$, as this is quite a delicate matter.
Say, it is possible to define the process "canonically", i.e. on $\Omega = \mathbb R^{[0,\infty)}$: define $\xi:[0,\infty) \to \mathbb R^{[0,\infty)}$ by $\xi: x\mapsto \mathbf{1}_{\{x\}}(\cdot)$ and let $$ \mathrm P(A) = P(\xi^{-1}(A)), A\in \mathcal B(\mathbb R^{[0,\infty)}), $$ where $P$ is some probability measure on $\mathcal{B}([0,\infty))$. Then, $X_t(\omega) = \omega(t)$ is as in your example. However, the $\sigma$-algebra $B(\mathbb R^{[0,\infty)})$ is quite poor, for instance, the event $\sup_{t\in[0,1]} X_t>0$ will not be measurable unless $P|_{[0,1]}$ is discrete (supported by at most countable set). The problem appears since the whole space $\mathbb R^{[0,\infty)}$ is "too large". To resolve the issue, one can work in a smaller space. In this situation, the space of functions having left-hand and right-hand limits at each points may be suitable, but one has also to consider appropriate topology. So perhaps the best choice is to work, as proposed in the original post, on $\Omega=[0,\infty)$, which we are best acquainted with.
Long story short, there is no problem with a "small" $\Omega$. Vice versa, too large sample space may cause a problem. Actually, any sample space is fine as long as it allows you to pose questions and answer them.