Is sample path of Brownian Motion deterministic?

Question

Consider a standard continuous Brownian Motion $B_t(\omega): \Omega \times [0,T] \to \mathbb{R}$, where: $B_0(\omega) = 0$; $B_t(\omega)$ has stationary independent increments; $B_t-B_s \sim N(0,t-s)$. Then what can we say about the sample path of a Brownian Motion? In my point of view, a sample path is such that we fix $\omega = \omega_0 \in \Omega$ and get a function $B(\omega_0,t)$, and we plot the graph of that function with respect to $t$. If so, can we say that $\textbf{a sample path is deterministic}$? Thanks in advance for any help.

Answer 1

What you have said is literally true, but is misleading, and shows confusion about what is meant by the term "sample path".

A "sample path" is not, in itself, a mathematical object; it is not a path or a function or a number or anything like that. Rather, any sentence involving the phrase "sample path" is just a slightly informal way of expressing a more formal statement about a certain event having probability 1.

For instance, when we say "A Brownian motion sample path is 1/4-Hölder continuous", what we really mean is the following. Let $A$ be the set of all outcomes $\omega \in \Omega$ such that $t \mapsto B_t(\omega)$ is 1/4-Hölder continuous; that is $$A = \left\{ \omega \in \Omega : \sup_{s,t \in [0,T]} \frac{|B_t(\omega) - B_s(\omega)|}{|t-s|^{1/4}} < \infty\right\}.$$ Then "A Brownian motion sample paths is 1/4-Hölder continuous" should be understood as the assertion that $\mathbb{P}(A) = 1$; i.e. $A$ happens almost surely.

It may help to rephrase "sample path" as "generic path" or "typical path". It does not refer to any specific path; it makes a statement about the set of paths as a whole, and how prevalent among them are paths with a certain property. For example, a statement like "the typical American eats between 50 and 100 hamburgers per year" does not give us any information about any particular American, but it tells us something about the set of Americans.

And describing a sample path as deterministic is rather nonsensical on another level. To use an analogy from lower-level mathematics, think of a function like $f(x)=x^2$. Your saying that "$B_t(\omega_0, t)$ is deterministic" is like saying "$f(5)$ is constant". Okay, well, $f(5)$ is the number $25$, which is not a function of $x$, so in that sense $f(5)$ is a constant. But when people use the word "constant", they mean that some object that might potentially depend on $x$, does not actually depend on it. So it doesn't really make much sense to say "$f(5)$ is constant" because there is no possibility for $f(5)$ to depend on $x$; substituting in $5$ got rid of $x$. Such a statement is literally true, but also totally pointless to say.

A statement like "$B(\omega_0, t)$ is deterministic" is similarly pointless. "Deterministic" means that something which potentially might depend on $\omega$, does not actually depend on it. But obviously $B_t(\omega_0, t)$ does not depend on $\omega$, because you just replaced $\omega$ by $\omega_0$. So what you said has no useful content.