I am curious about the following simple problem but I couldn't do any progress on it. I would like to know whether it is possible to prove (with probabilistic proof) that a brownian trajectory contains the vertices of a equilateral triangle.

Mathematical Formulation

Let $(W_t)_{t\in[0,1]}$ a standard Wiener process on the interval $[0,1]$. Does there exist $\ell>0$ and $(t_1,t_2,t_3)\in [0,1]^3$ such that: $$ \begin{cases} \|(t_1,W_{t_1})-(t_2,W_{t_2})\| = \ell\\ \|(t_3,W_{t_3})-(t_2,W_{t_2})\| = \ell\\ \|(t_1,W_{t_1})-(t_3,W_{t_3})\| = \ell\\ \end{cases} ~~~~\mathrm{?} $$


The problem I mention is then to find a equilateral triangle of length $\ell$ as in the drawing. enter image description here

  • $\begingroup$ Hi @NateEldredge, to me one dimensional was a standard name. It just means that the trajectory is a function $W:t\mapsto W_t \in \mathbb{R}$ $\endgroup$
    – NancyBoy
    Commented May 17 at 14:01
  • $\begingroup$ I am really sorry @NateEldredge but I can't get what your are pointing out. I don't see why it is absurd. I have added a drawing to make it more clear. $\endgroup$
    – NancyBoy
    Commented May 17 at 14:49
  • $\begingroup$ Your formula and picture are not consistent. You should replace $W_t$ in the formula to a point $(t,W_t)\in \mathbb{R}^2$ $\endgroup$ Commented May 17 at 14:54
  • $\begingroup$ It is now correct I think, I was too deeply focused that I couldn't figure out the error. Thank you! $\endgroup$
    – NancyBoy
    Commented May 17 at 14:58
  • $\begingroup$ @NancyBoy Nate's problem is that your picture doesn't actually correspond to your question... Anyway, if we interpret your question by the picture, then the answer is yes (and one vertex can be $(0,0)$) by the intermediate value theorem and the scaling properties of the Wiener process. $\endgroup$ Commented May 17 at 14:58

1 Answer 1


The answer is yes, here is a sketch of proof. Take a decreasing sequence $\{t_n\}_{n \ge 0}$ such that $W_{t_n} = 0$ and such that $\lim_{n \to \infty}t_n = 0$. Consider the triangles formed by $(0,0)$, $(t_n,0)$, $(t_n/2, \sqrt 3 t_n/2)$. Then, for large values of $n$, the probability of lying below the graph of $W$ approaches $1/2$ and these events are approximately independent for values such that the ratio of the corresponding times is large enough. Pick now two values $n$ and $m$ such that the third vertex lies above the graph of $W$ at $t_n$ and below at $t_m$. The claim then follows from the intermediate value theorem by having the second vertex tracing out the graph of $W$ for $t \in [t_n,t_m]$.

  • $\begingroup$ Thank you very much @Martin Hairer for this answer. Two quick question: (i) how do you link this proof on $\mathbb{R}$ with the time interval $[0,1]$. Do you do a variable change in time ? (ii) How do you get that starting with the probability of lying below with probability equal $1/2$ for vey negative $n$ that a.s., it exists $n\in\mathbb{Z}$ such that it is below ? $\endgroup$
    – NancyBoy
    Commented May 17 at 17:14
  • $\begingroup$ Moreover, how do you apply the itermediate value theorem ? Assuming that you are below for small $n$ and above for large $n$, your "set" of triangle to test (with $W_{t_n} = 0$) is discrete. Could you explain me where am I misunderstanding ? $\endgroup$
    – NancyBoy
    Commented May 17 at 17:37
  • $\begingroup$ @NancyBoy I hadn't realised you were restricting to $[0,1]$ but it makes no difference, see updated version. I also clarified the use of the intermediate value theorem. $\endgroup$ Commented May 17 at 17:50
  • $\begingroup$ Thanks @martinHairer, but how can you be sure that when the third vertex hit the line $t\mapsto \frac{3}{2} t$ at $t^*\in(t_n,t_m)$, the second vertex will be a zero of the brownian motion ($W_{t^*}=0$)? $\endgroup$
    – NancyBoy
    Commented May 17 at 18:10
  • $\begingroup$ Why would you want to have $W_{t^*} = 0$? You don't impose the orientation of the triangle. $\endgroup$ Commented May 18 at 9:00

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