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As I was trying to exhibit new interesting(?) path transformations of Brownian motion, I became interested in the (random) set of times $t$ such that $B(t)=B(t+1)=0$, where $B(t)$ denotes a standard Brownian motion. Not surprisingly, I found that this set is almost surely empty. By scaling invariance of Brownian motion, the same result holds if $1$ is replaced by any positive real $s$. Precisely, for all $s>0$, $$P[\exists t, B(t)=B(t+s)=0]=0.$$ Let me rephrase this a bit differently. Let $Z$ be the set of Brownian motion zeros and $D(Z)$ be the associated algebraic self-difference set $$D(Z)=Z-Z=\{u-v : (u,v)\in Z^2\}.$$ In this setting, the preceding result says that for all $s\not=0$, $$P[s\in D(Z)]=0.$$ Integrating this relation over $s$ with respect to Lebesgue measure $\lambda$ and exchanging the order of integration gives $$E[\lambda(D(Z))]=0.$$ In other words, the set $D(Z)$ is almost-surely negligible.

I've never seen any property of this set in the literature, thus my question is:

Is this a well-known result and, if so, what other properties are known, regarding its Hausdorff dimension for example ?

[EDIT] It seems that some of you doubt that the first probability is zero, so let me sketch a proof. (I apologize if there is a stupid mistake I've not been able to see). First, it is clear that, $$P[\cup_{n\geq 0}\{B_n=B_{n+1}=0\}]=0$$ since this is a countable union of zero probability events. Therefore, $$P[\exists t\geq 0, B_t=B_{t+1}=0]\leq\sum_{n=0}^{\infty}P[\exists t\in (n,n+1),B_t=B_{t+1}=0]=\sum_{n=0}^\infty p_n.$$ I asserts that each term $p_n$ inside the sum is equal to zero. For simplicity, I will do it only for $n=0$. (For other values of $n$, it suffices to condition first on the value of $B_n$ and then apply the same argument). Set $X_t=B_t$ and $Y_t=B_{t+1}-B_{1}$. Then, $$p_0=P[\exists t\in(0,1),(X_t,X_1+Y_t)=(0,0)],$$ and it suffices to show that $$P[\exists t\in(0,1),(X_t,x+Y_t)=(0,0)\vert X_1=x]=0$$ for (almost) all $x\in\mathbb{R}$. Now, $X=(X_t,0\leq t\leq 1)$ and $Y=(Y_t,0\leq t\leq 1)$ are two independent Brownian motions, hence conditionally to $X_1=x$ the two-dimensional process $(X,x+Y)$ is a Brownian ``bridge'' (starting from $(0,x)$ and ending on the line $\{(x,r):r\in\mathbb{R}\}$). This process inherits from the standard two-dimensional Brownian motion the property that it does not hit $(0,0)$. Hence $p_0=0$.

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    $\begingroup$ Is it really clear that your first probability is zero? That definitely works for fixed $t$, but then you have uncountably many $t$ to consider. Or, viewed from another angle, $D(Z)$ is a set of differences of numbers taken from a $\dim$ $1/2$ Cantor set, so it's not clear there can be a quick argument showing this is small ($C-C$ contains an interval). $\endgroup$ – Christian Remling May 18 '14 at 23:46
  • $\begingroup$ Well, I'm definitely not saying that this is straightforward, but I think I have a proof that the first probability is zero. I will try to post it in a few hours. $\endgroup$ – MassiveJack May 19 '14 at 6:53
  • $\begingroup$ This looks fine to me. Thanks for the clarification. $\endgroup$ – Christian Remling May 20 '14 at 0:22
  • $\begingroup$ "Not surprisingly, I found that this set is almost surely empty." This isn't true: clearly there are times when $B(t) > B(t+1)$ and times when $B(t)<B(t+1)$, and so the existence of times when $B(t)=B(t+1)$ follows from the intermediate value theorem applied to $B(t)-B(t+1)$. Sorry if I have misinterpreted your original statement. $\endgroup$ – tmh May 23 '14 at 0:18
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    $\begingroup$ @Tom Hutchcroft : You're right, you have misunderstood the statement. I claim that there is no time $t$ such that both $t$ and $t+1$ are zeros of $B$. $\endgroup$ – MassiveJack May 23 '14 at 5:30
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Regarding Hausdorff dimension $\dim_H$, the following holds:

$$ \dim_H(Z-Z)=1 \quad\text{a.s.} $$

This is essentially a consequence of Marstrand's projection theorem, the (stochastic) self-similarity of $Z$, and the well known fact that $\dim_H Z=1/2$ (and the box counting dimension is also $1/2$).

Recall that Marstrand's projection theorem says that for a Borel set $E\subset\mathbb{R}^2$, the orthogonal projection onto almost every line has Hausdorff dimension $\min(\dim_H E,1)$. In particular, taking $E=Z_1\times Z_2$, where $Z_1, Z_2$ are independent realizations of the zero set of Brownian bridge (see below for def.), then $$ \dim_H(Z_1-\lambda Z_2)=1 \quad\text{for almost all }\lambda. $$ (For "nice" sets such as $Z_1, Z_2$, we have $\dim_H(Z_1\times Z_2)=\dim_H(Z_1)+\dim_H(Z_2)$).

Now let us look at $Z-Z$. By scale invariance of Brownian motion, without loss of generality we can look at the zeros of the Brownian bridge instead, i.e. Brownian motion started at $0$ conditioned on $B(1)=0$. In this case, $Z$ can be constructed by replacing $[0,1]$ by the random union $[0,\tau_1]\cup [\tau_2,1]$, where $\tau_1=\max\{ t\le 1/2: B_t=0\}$ and $\tau_2=\min\{ t\ge 1/2: B_t=0\}$, and continuing inductively on these intervals. See the Chapter 14.3 of Lyons and Peres' Probability on trees and networks for the details.

Conditioned on $\tau_1,\tau_2$, $Z_1$ has the law of $\tau_1 Z$, $Z_2$ the law of $(1-\tau_2)Z$ up to a translation, and $Z_1, Z_2$ are independent. On the other hand, the distribution of $(1-\tau_2)/\tau_1$ is absolutely continuous (it has some explicitly computable density). This means that $Z-Z\supset Z_1 - Z_2$ which, conditioned on the abs. continuous random variable $\lambda=(1-\tau_2)/\tau_1$ is homothetic to $Z'-\lambda Z''$, where $Z',Z''$ are two independent realizations of the zeros of the Brownian bridge. But we have seen before that for a.e. realization of $(Z', Z'')$, we have $\dim_H(Z'-\lambda Z'') = 1$ for almost all $\lambda$, so we are done (after Fubini).

In fact, using newer (and deeper) techniques such as those in M. Hochman's paper "Dynamics on fractals and fractal distributions" and my own paper with Hochman, one can show that a.s. $$ \dim_H(Z-\lambda Z) = 1 \quad\text{for all }\lambda\in\mathbb{R}\setminus\{0\}. $$

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