# Probability that a Brownian motion takes value 0 in an interval.

Given $$W(t)$$ a standard brownian motion. I.e. $$W(0) = 0$$. Find the probability that $$W(t) = 0$$ for $$3 \le t \le 4$$

The book I am using has an example where:

$$\displaystyle P(W(s) = 0, 1 \le s \le t) = 1 - \frac{2}{\pi}tan^{-1}\frac{1}{\sqrt{t - 1}}$$

I thought initally I could use this result to calculate this but it is not immediately clear. I know that $$W(4) - W(3)$$ will be standard normal. I don't see how to relate this back to the previous result. Of course it being $$0$$ for a particular value of $$t$$ has probability $$0$$ but that isn't the same as calculating the probability it is $$0$$ for some value of $$t$$.

In order for it to take value 0 somewhere in the interval it would need to take values greater than or equal to 0 and less than or equal to 0. But I would need to do some condition on what happens between 0 and 3 I believe.

$$\mathbb P(\exists 1\le s\le\frac43,W_s=0)=\mathbb P(\exists 3\le s\le 4,W_{s/3}=0)=\mathbb P(\exists 3\le s\le4,\sqrt 3W_{s/3}=0)$$.

$$(\sqrt 3W_{t/3})_{t\ge0}$$ is a standard brownian motion.

So $$\mathbb P(\exists 3\le s\le4,\sqrt 3W_{s/3}=0)=P(\exists 3\le s\le4,W_s=0)=1-\frac2\pi\tan^{-1}\sqrt{3}=\frac13$$