5
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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 $\omega \in \Omega$.

$\endgroup$
3
  • $\begingroup$ 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 '18 at 12:07
  • 2
    $\begingroup$ @DominikS Yes and then you want to equip it with the product $\sigma$-algebras and measure. $\endgroup$ Jun 5 '18 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 '19 at 13:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.