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In his book Stochastic Differential Equations - An Introduction with Applications, Øksendal gives the following definition of a stochastic process:

A stochastic process is a parametrized colletcion of random variables $$\{ X_t\}_{t\in T} $$ defined on a probability space $(\Omega, \mathcal{F}, P)$ and assuming values in $\mathbb{R}^n$.

He then notes that it may be useful to think of $t$ as time and each $\omega \in \Omega$ as an individual experiment, such that $X_t(\omega)$ would represent the result at time $t$ of the experiment $\omega$. He also notes that a path of a stochastic process is obtained by the mapping $t \mapsto X_t(\omega)$ for a fixed $\omega \in \Omega$.

This seems to indicate that the outcome space $\Omega$ does not vary with time, and that the set of possible outcomes for each experiment, parametrized by $t$, is not dependent on $t$. However, it is not clear to me how this view would represent such experiments in this context. Take for instance the example of a random walk. At each time $t \in \mathbb{N}^+$ a coin is flipped. If the outcome is $H$, a step is taken vertically upwards, if $T$ a step downwards.

If each $X_t$ would represent the step taken at time $t$, would not the outcome of the experiment (the coin toss at time $t$) be $\omega \in \{ H, T, \emptyset \} = \Omega$? But then, fixing $\omega' \in \Omega$, for each time $t$ the variable $X_t(t)$ would have the same outcome, so this cannot be the correct interpretation.

The question becomes:

  1. What would each experiment be in this context, and is it then true that $\Omega = \{H, T, \emptyset\}$?

  2. In this context, how would the accumulated position of the random walk at time $t$ be formulated?

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$\Omega$ often takes the form of the so-called canonical space: $\Omega=(\mathbb{R}^n)^T,$ rather than the $\mathbb{R}^n$ you seem to propose. In this case, we can take a slightly different approach. A natural choice would be $\Omega=\{1,-1\}^{\mathbb{N}^+}$. You might then define $X_t(\omega)=\sum_{i=1}^{\lfloor t\rfloor}\omega_i$.

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  • $\begingroup$ Thanks, this was clarifying. I think what confuses me still is how one views this. I like to think each coin flip is done independently and without information about the prior and subsequent flips. But in this view, with $\Omega = \{-1, 1\}^{\mathbb{N}^+}$, we already "decided" to flip the coin countably many times. However, if we only wanted to flip a coin once, should we still take the same $\Omega$, and evaluate all the possible outcomes in $\mathcal{F}_1 = \{-1, 1, \emptyset\}$? $\endgroup$ Jul 7 '20 at 13:26

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