Let $(\Omega, \mathcal{F}, P)$ be a probability space. Let's define a stochastic process as a function

$$ S: \Omega \times \mathbb{R} \rightarrow \mathbb{R} \\ \omega \times t \mapsto S(\omega, t) \, . $$

It makes sense that at fixed $t$, $$X_t(\cdot) \equiv S(\cdot, t)$$ is by definition a random variable. However I don't understand what is the meaning of $$X_\omega(\cdot) \equiv S(\omega, \cdot) \, .$$ An example could help. How do we fix $\omega$ in, for instance, a random walk or in the Gaussian noise? If we fix $\omega$, doesn't it imply that we will always get the same value for $X_\omega(\cdot)$?


If we treat $\omega$ as a constant, then $S(\omega, \cdot): \mathbb{R} \to \mathbb{R}$ is a function. For example, any set of time series data such as a set of stock price data at hand is a "function of time", which is mathematically viewed as a realization of a stochastic process. To see this, recall what the horizontal axis measures and what the vertical axis measures in the stock price data set.

Note that the set $\Omega$ is simply an abstract space, whose elements need not be numbers. A typical example is the coin-tossing one. Tossing a fair coin can give us either the result "head" or the result "tail". So here we may take $\Omega := \{ \text{''head", "tail"} \}$. But can math speak something directly from $\Omega$? I am afraid not so. But with the help of the concept of random variable, which is a "nice" function on $\Omega$ in $\mathbb{R}^{n}$, math starts working.

A phrase such as "we fix $\omega$" is a mathematical one, which does not mean that any one of us did manually somehow "determine" a value of $\omega$ in whatever sense you probably are thinking of :).

  • $\begingroup$ Thanks for answering @GudsonChou. What would be $\omega$ in your example? $\endgroup$
    – dapias
    Aug 4 '16 at 3:30
  • $\begingroup$ No problem. The symbol '$\omega$' is your notation :). $\endgroup$
    – Megadeth
    Aug 4 '16 at 3:31
  • $\begingroup$ I mean, the problem I am having is that I cannot figure out what $\omega$ would be in a particular example. Consider the most simple Bernoulli trial with possible outcomes ${{H}, {T}}$. Let's fix $\omega = {H}$. Is this choice the initial condition of the random process? What does it mean to fix $\omega$ in your example of stock price? $\endgroup$
    – dapias
    Aug 4 '16 at 3:36
  • $\begingroup$ That simply means that the function ate an element of the set $\Omega$. $\endgroup$
    – Megadeth
    Aug 4 '16 at 3:41
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    $\begingroup$ One may note that the sample space $\Omega := \{ \text{''head", "tail"} \}$ can be used to model one coin toss. For an infinite sequence of independent coin tosses (the archetypal situation the OP is interested in), one would need to move to a sample space similar to $\Omega := \{h,t\}^\mathbb N$. $\endgroup$
    – Did
    Aug 4 '16 at 9:31

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