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Let $(B_t)_{t\geq 0}$ a standard Brownien motion. I have to show that $\dim Z=\frac{1}{2}$ where $Z=\{t\in [0,1]\mid B_t=0\}$. Why to do this, I have to show that $$\mathbb P\{\dim Z=1/2\}=1\ \ ?$$ I don't see the correlation between $ \mathbb P\{\dim Z=1/2\}=1$ and $\dim Z=1/2$.

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    Brownian motion is a collection of random variables. So $\dim(Z)$ is a random variable. It's impossible to show that $\dim(Z)$ is $1/2$, because it might not be! The most you can expect any time you're showing that any random variable has any property is to show that the probability it has that property is $1$. – David C. Ullrich Apr 19 '16 at 14:21

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