A sample path of a stochastic process $x:\Omega\times T\to\mathbb{R}^n$ is defined as taking $x(\omega,t)$ with fixed $\omega\in\Omega$, thus $x(\omega,\cdot)$.

However I have the following issue with this definition: the process is supposed to be a sequence of random variables $X_t:\Omega\to\mathbb{R}^n$, so a sample path, or realization would be a realization for each of the random variables in the sequence, leading to (maybe) different values of $\omega\in\Omega$, one for each $t\in T$.

For example if $\Omega = \{H,T\}$ and $T=\{0,1,\dots\}$ and all random variables $X_t$ are independent with $P(X_t = H) = p\ \ \forall t\in T$, then a realization would be something like $\{H,T,T,H,\dots\}$.

But, what am supposed to interpret by "fixing $\omega\in\Omega$", since one would only get sequences like $\{H,H,H,\dots\}$.

I've seen other answers here, and some say that you should not fix $\omega$ for all random variables in the sequence, or that $\omega$ is fixed on sequences in $\Omega^T$ (then $\omega\in\Omega^T$?). Of course that seems reasonable, but it doesn't follow the definition.

Am I missing something?

(Im using "Introduction to Stochastic Control" by Åström, Karl Johan and "Stochastic Differential Equations" by Øksendal, Bernt)


1 Answer 1


you are mixing up a couple of things.

Assume we are considering a sequence of independent random variables $X_t$ with $\Bbb{P}(X_t = H) = p$ for all $t \in T$. Then the corresponding state space has to be much larger than just $\Omega = \{H, T\}$. Otherwise the $X_t$ cannot be independent.

Why is that? Assume that $\Omega$ only has two elements and let's call them $\omega_H$ and $\omega_T$. We define $X_1(\omega_H) = H$ and $X_1(\omega_T) = T$. The probability measure $\Bbb{P}$ is then chosen such that $\Bbb{P}( \{\omega_H\}) = p$ such that $\Bbb{P}(X_1 = H) = p$. So far so good. But how do you define $X_2$ now? There are two possibilities:

  • If we choose $X_2(\omega_H) = H$ and $X_2(\omega_T) = T$ then $X_1$ and $X_2$ will not be independent. The reason is that $X_1$ and $X_2$ will always be equal.
  • If we choose $X_2(\omega_H) = T$ and $X_2(\omega_T) = H$ then $X_1$ and $X_2$ will not be independent either: every time $X_1$ is heads $X_2$ will be tails and vice versa.

So you see that you need a larger sample space to define two independent Bernoulli random variables. A quick checks shows that you will need at last four elements in your state space to define two independent random variables, e.g. $\Omega = \{\omega_{HH}, \omega_{HT}, \omega_{TH}, \omega_{TT}\}$. Then you can set:

\begin{align} X_1(\omega_{HH}) = X_1(\omega_{HT}) &= H, \\ X_1(\omega_{TH}) = X_1(\omega_{TT}) &= T, \\ X_2(\omega_{HH}) = X_1(\omega_{TH}) &= H, \\ X_2(\omega_{HT}) = X_2(\omega_{TT}) &= T. \end{align}

This way knowing the outcome of $X_1$ does not tell you anything about the outcome of $X_2$. Hence $X_1$ and $X_2$ are independent.

A sample path of this stochastic process can now be obtained by fixing one the four elements of the state space. Say we fix $\omega_{HT}$. Then the path of this stochastic process would be $HT$.

Hope this helps.

  • $\begingroup$ Thanks! Just to clarify: if $\hat{\Omega} = \{\omega_{H},\omega_{T}\}$ then the stochastic process is a mapping $X_t$ which takes values on $\Omega = \hat{\Omega}^T$ instead of $\Omega$?. Or in other words, to define the sample space $\Omega$ from which the stochastic process act on, the set $T$ is needed before hand? $\endgroup$ Nov 8, 2019 at 9:31
  • $\begingroup$ Not necessarily. You can do it like you described (i.e. product spaces) but you can also start with a (large enough) $\Omega$ without having to go through product spaces. $\endgroup$
    – Cettt
    Nov 8, 2019 at 9:33
  • $\begingroup$ Thanks! I will, I was just thinking about this in depth. This was really helpful. $\endgroup$ Nov 8, 2019 at 11:04

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