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


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