Let $B_t$ be a standard Brownian motion. Let $J(j, n) = [j/n, (j+1)/n]$. We will call $J(j, n)$ an increase interval if$$B_s \le B_t,\text{ }0 \le s \le {j\over{n}},\text{ }{{j+1}\over{n}} \le t \le 3.$$I am wondering, do there exist constants $0 < c_1$, $c_2 < \infty$ such that for $j = 1, 2, \dots, 2n$,$${{c_1}\over{\sqrt{jn}}} \le \textbf{P}\{J(j, n) \text{ is an increase interval}\} \le {{c_2}\over{\sqrt{jn}}}?$$Thanks!


1 Answer 1


Rescaling, splitting, reversing time, etc., we arrive at the following reformulation:

Let $B_1(t)$ and $B_2(t)$ be two independent Brownian motions and $A,B\ge 2$. Estimate the probability that $\min_{1\le t\le A}B_1(t)\ge \max_{1\le t\le B}B_2(t)$.

The lower bound is trivial: $B_1(1)\ge 1$ with fixed positive probability, after which the probability that we never go down to $0$ is about $A^{-1/2}$. Similarly, $B_2$ stays non-positive with probability about $B^{-1/2}$.

The upper bound is similar. Condition upon $B_1(1)=x$, $B_2(1)=y$. Obviously, we must have that $x>y$ and that $B_2\le x$, $B_1\ge y$. This immediately gives the bound like $$ \iint_{x>y}e^{-x^2/2}e^{-y^2/2} (x-y)^2A^{-1/2}B^{-1/2} \,dx\,dy $$ (the probability to never go above $r$ before time $T$ is at most $r/\sqrt T$), which is enough to yield the desired estimate.


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