I use the following definition for a stochastic process.

Let $(\Omega, \mathcal F, P)$ be a probability space, $(E, \mathcal E)$ be a measurable space, and $T$ be a non-empty set. A collection $X:=(X_t)_{t\in T}$ of random variables is called a stochastic process with state space $(E, \mathcal E)$ indexed by $T$ and defined on $\Omega$ if for all $t \in T$, $X_t: \Omega \to E$ is $\mathcal F/\mathcal E$ measurable. In addition, for $\omega \in \Omega$ the function $t \mapsto X_t(\omega)$ is called the sample path of $X$.

My question is how to interpret the $\omega$ in $X_t(\omega)$ in the above definition. Consider the infinite fair coin tossing experiment. In this case, it is clear that one sample path can be $$X_t(\omega) = 0101011101010 \cdots, {t \in \{1, 2, 3, \cdots\}}.$$ What exactly is the $\omega$ in this sample path, please? How do I understand it? Thank you!

  • 3
    $\begingroup$ The best option is often to leave ω and Ω with no precise definition. On the one hand general theorems ensure that in every situation you will encounter such spaces Ω do exist. On the other hand one gains nothing in specifying Ω. What must be specified are the distributions of the events and/or random variables the problem entails, that is, in your setting, the distribution of each $X_t$ and of each family $(X_s)_{s\in S}$ for every finite $S\subseteq T$. $\endgroup$
    – Did
    Aug 2, 2014 at 7:46
  • $\begingroup$ @Did Do you mean the probability measure $P$ on $\mathcal F$ when you say distribution, please? $\endgroup$
    – LaTeXFan
    Aug 2, 2014 at 7:49
  • $\begingroup$ No, rather the images of $P$ by $X_t$ and every $(X_s)$, thus, probability measures on $\mathcal E$ and on products of $\mathcal E$. $\endgroup$
    – Did
    Aug 2, 2014 at 7:56
  • $\begingroup$ @Did OK, but all those things you mentioned are derived from $P$. Right? $\endgroup$
    – LaTeXFan
    Aug 2, 2014 at 8:01
  • $\begingroup$ No--this is the point. Actually, probabilists rarely get interested in/even mention the space $\Omega$ or the measure $P$. $\endgroup$
    – Did
    Aug 2, 2014 at 8:32

1 Answer 1


Let's say we want to model the fivefold throw of a fair coin. Then we define the corresponding process via the outcome of the throws, i.e. we set

$$X_t := \begin{cases} 1 & \text{t-th throw is head} \\ 0 & \text{otherwise} \end{cases}$$

for $t \in \{1,\ldots,5\}$. Now, since we have a fair coin, the probability that $X_t$ equals $1$ is 0.5 for each $t$. In probability theory, this is translated in the following abstract way: For a probability space $(\Omega,\mathcal{A},\mathbb{P})$, the random variables $X_t$ have to satisfy $\mathbb{P}(X_t=1)=\tfrac{1}{2}$. This means in particular that we do not care how the probability space looks like; only the distribution of the random variables is of importance.

Moreover, for any $\omega \in \Omega$ the mapping $t \mapsto X_t(\omega)$ is a realization of our process. If we throw the coin five times and observe e.g. $$0 \, \, 1 \, \, 0 \, \, 0 \, \, 1,$$ then there exists $\omega \in \Omega$ which "symbolizes" this outcome, i.e.


Usually, we are interested in questions like "What is the probability that we throw head 3 out of 5 times?"; this probability equals $\mathbb{P}(\sum_{t=1}^5 X_t = 3)$. This question can be answered if we know the (finite dimensional) distributions of the stochastic process $(X_t)_t$.

So, basically a stochastic process (on a given probability space) is an abstract way to model actions or events we observe in the real world; for each $\omega \in \Omega$ the mapping $t \mapsto X_t(\omega)$ is a realization we might observe. The likeliness of the realization is characterized by the (finite dimensional) distributions of the process.


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