Local maxima and minima of Brownian motion

Question

I have trouble understanding why the Brownian motion is nowhere differentiable, and I found somewhere that after showing the total variation of the Brownian motion is $+\infty$, the author claims that "as a result (of the infinite total variation), the path of the Brownian motion $B$ has an infinite number of local minima and maxima on any interval $[0,T]$, where $T>0$. "

I don't get how to show the part that there exists an infinite number of local minima and maxima on the interval $[0,T]$.

Any help will be extremely appreciated!

Best and regards

Thank you saz! but how the result from that question applies to this one? I mean, that one showed that the B.M attains local max at some times in the interval..but does it imply we have inifite number of them? s — amazonfire, Sep 19 '13 at 14:50

score 1 · Accepted Answer · edited Apr 13 '17 at 12:20

1

The first step is to prove that for arbitrary $0<a<b$, the Brownian motion $B$ attains almost surely a local maximum (minimum) in $(a,b)$ (see this question for a proof).

Now choose a sequence $(t_n)_{n \in \mathbb{N}}$ such that $t_n < t_{n+1}$ and $t_n \to T$ as $n \to \infty$, e.g. $t_n := T- \frac{1}{n}$. By the first part, $(B_t)_{t \geq 0}$ attains almost surely a local maximum at some time $s_n \in (t_n,t_{n+1})$. Consequently, the set $\{s_n; n \in \mathbb{N}\}$ consists of local maxima in the interval $[0,T]$.

edited Apr 13 '17 at 12:20

Community

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answered Sep 19 '13 at 15:55

saz

120,083

Brilliant! Thank you so much, i think this also implies that the set of local maxima (minima) for any given interval [0,T] is dense?! – amazonfire Sep 19 '13 at 18:21
@sobaya Exactly. (By the way, you can upvote/accept answers which satisfy you.) – saz Sep 19 '13 at 18:38
sure =) thank you! – amazonfire Sep 19 '13 at 18:48

Local maxima and minima of Brownian motion

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