for a fixed $\epsilon$ $> 0$, I want to show that almost surely (ie with probability 1), a standard brownian motion $W_t$ would change sign over [o,$\epsilon$ ].
I thought about defining a random variable $U_t = sign (W_t)$ which would satisfy :
$(1+U_t)/2$ follows symmetric Bernoulli distribution
But I can't conclude with this.
Am I on a good track?
Best Answer
The reflection principle says that for any $\epsilon>0$ and $x>0$, we have $$\mathbb{P}_0(\sup_{0\leq t\leq \epsilon} W_t \geq x)=2\mathbb{P}_0(W_\epsilon\geq x)= 2\mathbb{P}\left(Z\geq x/\sqrt{\epsilon}\right). $$ Letting $x\downarrow 0$ we conclude that $$\mathbb{P}_0\left(\sup_{0\leq t\leq \epsilon } W_t >0 \right)=2\mathbb{P}(Z> 0)=1.$$ In any time interval, no matter how short, the Brownian motion $(W_t)$ must take a strictly positive value. Applying this to the Brownian motion $(-W_t)$, we see that $(W_t)$ must also take a strictly negative value. So the process $(W_t)$ changes sign with probability one.