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Let $B=(B_t)_{t\geq 0}$ be the standard Brownian motion. I want to show that

for every $t_0 \geq 0$

$\mathbb{P}$($B$ has a local maximum in $t_0$)=0.

I've already shown that for every $0<a<b<\infty$ $B$ is $\mathbb{P}$-a.s. not monotone on the interval [$a,b$].

My ideas were the following:

Suppose $B_t$ attains a local maximum in $t_0$. Can I assume that since $B$ has a.s. continuous paths, $t_0$ is preceded by an interval ($t_0-\epsilon, t_0$) where $B_t$ increases and is followed by an interval ($t_0,t_0+\epsilon$) where $B_t$ decreases? ($\epsilon > 0$) Then I would have two intervals where $B$ is monotone and since $B$ is not monotone on these intervals $\mathbb{P}$-a.s., I get

$\mathbb{P}$($B$ has a local maximum in $t_0$) = $\mathbb{P}$($B$ is monotone on ($t_0-\epsilon,t_0$) and $(t_0,t_0+\epsilon)$)=0.

Is that correct or do I have to argue in a different way? Can I always find those non-empty increasing and decreasing intervals "before" the next extremum?

Thanks in advance.

Max93
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    No, your argument doesn't work. There are, in fact, many functions which have a local maximum at a point but are not monotone on any one-sided neighborhood of that point. Think about the continuous function $f$, defined to be $f(x)=-|x\sin \frac{1}{x}|$ when $x \neq 0$, and $f(0)=0$. Look at behavior near zero. One can modify this example to get a strict maximum at $0$, if desired. – shalin May 15 '15 at 02:30
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    Moreover, your $\epsilon$ depends on the realization of the process $(B_t)$, and so from your monotonicity statement, it would not immediately follow that $(B_t)$ is necessarily non-monotone on $(t_0,t_0 + \epsilon)$. You've shown that for a fixed interval $[a,b]$, the process $(B_t)$ is a.s. not monotone on $[a,b]$. What you would need to show, though, is that, almost surely, for any interval $[a,b]$, the process $(B_t)$ is not monotone on $[a,b]$. Do you see the difference between the two statements? This is an easier problem to fix, however, than the other one... – shalin May 15 '15 at 02:47
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    Thank you for your answer! Then I have to use a different argument...I have already shown that $B_t$ is a.s. not monotone on [a,b] for a fixed interval [a,b]; if I take the countable union over all the intervals [a,b] where a and b are rational, it follows that $B_t$ is not monotone on any interval with rational endpoints and since every other interval contains a sub-interval with rational endpoints, $B_t$ is not monotone on [a,b] for any interval. But how can I proof now that for every $t_0 \geq 0$ $\mathbb{P}$($B$ has a local maximum in $t_0$)=0? – Max93 May 15 '15 at 02:57
  • Yes, this is a good proof to fix the second problem that I stated. The first one, however... I'll let you know – shalin May 15 '15 at 03:04

1 Answers1

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Here is one argument that would work, assuming that one already knows the following facts:

  1. For any $s>0$, the process $(B_{s+t}-B_s)_{t \geq 0}$ is a Brownian motion.

  2. The time inversion $(tB_{\frac{1}{t}})$ of a Brownian motion is a Brownian motion (defined to start at $0$ at $t=0$).

  3. $\limsup_{t \to \infty} B_t >0$ (in fact, the lim sup is $+\infty$).

There exist very succinct proofs of each of these statements, each of which can be derived directly from the basic axioms of Brownian motion. See theorem 1.9, proposition 1.23, and theorem 2.3 in the book by Morters and Peres, for example.

We can use these three things to get a short proof of your statement:

Fix $t_0 \geq 0$.

By $(1)$, we know that the process $Y_t = B_{t+t_0}-B_{t_0}$ is a Brownian motion.

By $(2)$, we know that the process $(X_t)$, defined by $X_t = tY_{\frac{1}{t}}$ when $t>0$ (and $X_0=0$), is a Brownian motion.

Now convince yourself that the following inclusion of events is true: $$\{ t_0 \text{ is a local maximum of } (B_t)\} \subseteq \{ \limsup_{t\to \infty} X_t \leq 0\}$$ The reason that this is true, is that if $t_0$ is a local maximum of $(B_t)$, then $Y_t \leq 0$ for small enough $t$, and thus $X_t \leq 0$ for large enough $t$.

By statement $(3)$, we see that the event on the right has probability $0$, and so the event on the left has probability $0$.

(I do concede, however, that there may be a shorter proof which relies on less statements. But I don't know one off the top of my head...)

shalin
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    Indeed one can shorten this slightly, noting that $t_0$ is a local maximum if $Y_t\leqslant 0$ for every $t$ in some $(0,\theta)$ with $\theta>0$, thus, one asks that the maximum process $M$ is such that $M_\theta=0$, but $M_\theta$ is distributed as $|B_\theta|$ hence $M_\theta>0$ almost surely, QED. – Did May 15 '15 at 10:40