In the Wikipedia article about stochastic processes, a sample function is defined by specifying some $\omega\in\Omega$, as $X(\cdot,\omega):T\to S$, which then is a non-random function.

My (conceptual) question is: with $\{X(t,\cdot)\}_{t\in T}$ being a given family of random variables (i.e. a stochastic process), how would we get more than $|\Omega|$ different sample functions (which is something I intuitively expect to be the case)?

Let me illustrate this question with a Bernoulli process, e.g. a sequence of coin tosses. Let $X_i$ be the random variable that describes the outcome of the $i$-th toss. The $X_i$ are all independent, $\Omega =\{H, T\}$, and I may (may I?) specify that $X_i(H) = 1$ for all $i$, and $X_i(T) = 0$. In this case I would have exactly two choices for sample functions: Either $\omega = H$, giving me the sequence $[1,1,1,\ldots]$, or $\omega = T$, giving me $[0,0,\ldots]$. Intuitively, however, I would expect any binary sequence, e.g. $[1,0,0,1,0,\ldots]$, to be a possible sample function.

How can this be explained by picking a single $\omega$, and not a sequence of $\omega_i$? What am I missing in the definition?


1 Answer 1


There are at most $|\Omega|$ sample functions (a.k.a. sample paths) for $\{X_t\}_{t \geq 0}$.

The source of your confusion in your example is that you have taken a probability space $\Omega$ that simply doesn't have a sequence of independent Bernoullis defined on it, as is evidenced by your observation that either all of them are $1$ or all of them are $0$ (so they definitely aren't independent).

As a result, you do not have that $\Omega = \{H,T\}$. This is only a suitable state space for a single Bernoulli random variable and not for the whole sequence of independent Bernoullis in your example.

When you take the proper state space to allow the $X_i$ to be independent then we will have that $X_1(\omega) = 1$ does not imply that $X_i(\omega) = 1$ for all $i$ in your example (indeed, subsequent coin tosses are independent). That means you can have that e.g. $$(X_1(\omega), X_2(\omega), X_3(\omega), X_4(\omega), \dots) = (1,0,0,1,\dots) $$ for some fixed $\omega \in \Omega$.

  • $\begingroup$ I see, my example violates the independence! So a 'correct' $\Omega$ would be $\{H,T\}^{\mathbb N}$, i.e. the product space? $\endgroup$
    – DominikS
    Jun 5, 2018 at 12:07
  • 2
    $\begingroup$ @DominikS Yes and then you want to equip it with the product $\sigma$-algebras and measure. $\endgroup$ Jun 5, 2018 at 12:09
  • $\begingroup$ Thank you so much.I think now I properly understood definition of random variable and stochastic process $\endgroup$
    – manifold
    Aug 6, 2019 at 13:47

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