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I don't quite understand a property of the Wiener process. Such process has the property that
$W(t) - W(s) \sim \mathcal{N}(0, t-s)$
where $t > s > 0$. What I don't understand is this. As the distance between $t$ and $s$ increases you have a gaussian variable with a variance that increases with it, so you would expect "big variations". However, if $t-s$ is large enough, w.h.p. somewhere between $t$ and $s$, say at $t'$, the process crosses the zero, i.e., $W(t')=0$.

So my question is this: if $s=0$ you can say that at time $t>>1$, the process will be likely to be very far from zero, because the variance is $t$. However, there is a $t' >>1$, with $t'<t$ where $W(t')=0$, a point close to $t$ where the process is equal to zero. With respect to that point you expect to see $W(t)$ not to far from $W(t')=0$ which looks a bit contradictory, but I may be missing something.

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1 Answer 1

up vote 4 down vote accepted

Your trouble might come from a confusion between the behaviour of $W(t)$ for a fixed $t$ and the behaviour of the path $(W(t'))$ for $t'$ in the interval $(0,t)$.

For a given large $t$, the probability that $W(t)$ is close to zero is small in the sense that, for every given $K$, $P(|W(t)|\le K)$ goes to zero when $t$ goes to infinity. But the probability that there exists a time $t'$ in the interval $(\frac45t,t)$, say, such that $W(t')=0$, does not go to zero. However, for every given $K$, the probability that there exists a time $t'$ in the interval $(t-K,t)$, say, such that $W(t')=0$, does go to zero (and it seems this last fact contradicts something you write).

More precisely, the time of the last zero before $t$ scales like $t$, in the following sense. Let $T_t=\sup\{t'<t;W(t')=0\}$, then $T_t$ is distributed like $tT_1$ and the distribution of $T_1$ is absolutely continuous with a positive density on $(0,1)$ (called the arcsine law, see here for example).

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+1: The first sentence seems to hit the nail on the head. –  Tim van Beek Jun 11 '11 at 9:49
    
@Tim: Thanks. $ $ –  Did Jun 11 '11 at 11:03
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