MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have what should be a very simple questions for Brownian motion experts... Let $[a,b]$ be a given time interval. Let $f(x)$ be the probability that a linear Brownian motion with initial value $x$ at time $t=0$ has a zero in the interval $[a,b]$. I want to argue that $f(x)$ is maximal for $x=0$. This seems intuitively clear but I cannot figure out a simple proof of this (other than writing the exact expression for $f$ which is a double integral and analyzing its variations via long calculations). I would be interested, in order of preference, by such a simple proof or by a reference where this lemma is proven.


share|cite|improve this question
up vote 8 down vote accepted

For any other initial x, construct a coupling between BMs started at x and 0, where the processes move in opposite directions until they meet (if they do), then they stick together afterwards. Then the answer is apparent.

share|cite|improve this answer

There is an easy way to do it via calculation. Start by supposing that $x\geq 0$ Then we're asking for the probability that $x+W_t$ reaches $0$ in the interval $[a,b]$ where $W_t$ is a standard Brownian motion. This the same as $P(\min_{a\leq t\leq b}x+W_t<0)=P(\min_{a\leq t\leq b}W_t<-x)$ Then $$\begin{eqnarray} P(\min_{a\leq t\leq b}W_t<-x)&=& E[1\{\min_{a\leq t\leq b}W_t<-x \}| W_a]\\ &=& E[1\{\min_{a\leq t\leq b}W_{t-a}<-x-W_a \}| W_a]\\ &=& E[1\{\min_{0\leq t\leq b-a}W_{t}<-x-W_a \}| W_a]\\ &=& E[1\{-|W_{b-a}|<-x-W_a \}| W_a]\\ \end{eqnarray}$$ Where we used the fact that $W_t-W_a$ has the same law as $W_{t-a}$ and $\min_{0\leq t\leq b-a}W_{t}$ has the same law as $-|W_{b-a}|$ then we have

$$P(\min_{a\leq t\leq b}W_t<-x)=2\int_R \frac{\exp(-y^2/2a)}{\sqrt{2\pi a}}N(-y-x)dy $$

Where $N$ is the Gaussian cumulative. Now it is easy to see that the right hand side is decreasing as function of $x$. We treat then $x\leq 0$ in the same way by symmetry of the Brownian motion.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.