Stochastic Processes – Problem 41: Prove that, with probability one, the set of local extrema of a standard Brownian path is…

Prove that, with probability one, the set of local extrema of a standard Brownian path is dense. (Note: A point t is a local extremum if and only if there exists an open neighborhood containing that point such that the maximum of the Brownian path in that neighborhood is attained at t.)

Step 1. Let $D$ denote the collection of all open intervals in $[0, \infty)$ with rational endpoints. Since the rationals are countable, $D$ is also countable; write $D = \{I_n\}_{n=1}^{\infty}$, where $I_n = (a_n, b_n)$ with $0 \le a_n < b_n$. For any fixed interval $[a, b]$, because the sample paths of standard Brownian motion $B_t$ are almost surely continuous on $[0, \infty)$, by the Weierstrass theorem a continuous function on the closed interval $[a, b]$ must attain its maximum. Denote the time at which this maximum is attained by $T_{max}$, i.e., $B_{T_{max}} = \sup_{t \in [a, b]} B_t$.

Step 2. Use properties of Brownian motion to show that the maximum is almost surely not attained at an endpoint. For any fixed time $t \ge 0$, by the law of the iterated logarithm or local properties of Brownian motion, for any $\epsilon > 0$ we almost surely have $\sup_{s \in (t, t+\epsilon)} B_s > B_t$. This means that for the fixed left endpoint $a$, there almost surely exists $s \in (a, b)$ with $B_s > B_a$, so $T_{max} \neq a$. Similarly, for the fixed right endpoint $b$, there almost surely exists $s \in (a, b)$ with $B_s > B_b$, so $T_{max} \neq b$. In summary, for any given $n$, the maximum point $\tau_n$ on $[a_n, b_n]$ almost surely lies in $(a_n, b_n)$.

Step 3. Since $\tau_n \in (a_n, b_n)$ and $\tau_n$ is the point at which $B_t$ attains its maximum on $[a_n, b_n]$, there exists an open neighborhood $(a_n, b_n)$ in which the maximum of $B_t$ is attained at $\tau_n$. By the definition given in the problem, $\tau_n$ is a local extremum (local maximum). Define the event $E_n$ as "the Brownian path has a local extremum in the interval $I_n$." By the above analysis, $P(E_n) = 1$.

Step 4. Let $E$ denote the event "the set of local extrema is dense in $[0, \infty)$." This event is equivalent to every open interval $I_n$ containing at least one local extremum, i.e., $E = \bigcap_{n=1}^{\infty} E_n$. Since $P(E_n) = 1$ and $\{E_n\}$ is a countable collection, by properties of probability measures the intersection of countably many probability-one events still has probability one. Therefore $P(E) = P(\bigcap_{n=1}^{\infty} E_n) = 1$. That is, with probability 1, the set of local extrema of a standard Brownian path is dense.

QED.