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 it exist $\ell>0$ and $(t_1,t_2,t_3)\in [0,1]^3$ such that:
$$
\begin{cases}
||(t_1,W_{t_1})-(t_2,W_{t_2})|| = \ell\\
||(t_3,W_{t_3})-(t_2,W_{t_2})|| = \ell\\
||(t_1,W_{t_1})-(t_3,W_{t_3})|| = \ell\\
\end{cases}
~~~~\mathrm{?}
$$
Illustration
The problem I mention is then to find a equilateral triangle of length $\ell$ as in the drawing.
Best Answer
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 $t \in [t_n,t_m]$.