# Definition of sample path of stochastic process

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)

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.

• 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? Nov 8, 2019 at 9:31
• 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. Nov 8, 2019 at 9:33
• Thanks! I will, I was just thinking about this in depth. This was really helpful. Nov 8, 2019 at 11:04