# Does Brownian Motion return to the origin infinitely soon? [closed]

Consider a standard unidimensional Brownian Motion $B_t$ (Wiener process).

Fact: This process returns to the origin infinite number of times with probability one.

Consider a stopping time $\tau = \inf \{t>0: B_t=0 \}$ (time of the first return).

Question: what can we say about the distribution of $\tau$? In particular, is it true that $\forall t >0$: $\Pr(\tau < t)=1$, i.e. BM almost surely returns to the origin infinitely soon?

## closed as off-topic by Did, Christopher, graydad, kjetil b halvorsen, TravisJMay 28 '15 at 19:28

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• What can "we" say, meaning posters on this board, as opposed to what the person assigned the homework can say without someone else doing it for them? – Mark L. Stone May 27 '15 at 2:40
• Yes; the set of zeros of a Wiener trajectory forms a perfect set almost surely. – Ian May 27 '15 at 2:47
• 2Mark: Not a homework; independent research, if that's crucial (which it shouldn't be since I've seen plenty of much easier homework questions answered here). I always write "we" in formal texts. – nevvy May 27 '15 at 2:52
• @nevvy Please describe context of said "independent research" (since, as it is, your question provides NO personal input). – Did May 27 '15 at 9:10
• 2Did: matching with experimentation and private types. Fine, next time I ask a question here I'll make it less formal "haha lol". – nevvy May 27 '15 at 13:57

## 1 Answer

Let $C_t = \dfrac{B_{at}}{\sqrt{a}}$. Then $C_t$ has the same distribution as $B_t$ and for $0<t_1<\cdots<t_n$, $(C_{t_1},\ldots,C_{t_n})$ has the same distribution as $(B_{t_1},\ldots,B_{t_n})$.

$C_t=0$ if and only if $B_{at}=0$. Thus rescaling the set of all returns to $0$ results in a set of all returns to $0$ that has the same probability distribution.

• Is this supposed to prove that $\tau=0$ almost surely? – Did May 27 '15 at 9:11
• Yes, it just lacks the last step. This argument implies in particular that the distribution of $\tau$ is described by cdf $F_\tau (t)=const$ for $t>0$. Since $\lim _{t\to\infty} F_\tau (t)=1$, we get $F_\tau (t)=1$ $\forall t>0$. – nevvy May 27 '15 at 13:50