Probability distribution of sign changes in Brownian motion Let us consider a 1d Brownian motion. Displacements in space will be positive or negative and this is a random variable $U(t)$ that characterizes a random process and that can take just the values $\pm 1$. The question I am looking for an answer is: What is the probability distribution for $U$? 
Of course, the problem can be generalized to $\mathbb{R}^d$ and, in this case, one should look for a distribution of directions. It would be nice to get an answer also in this case.
Thanks.
 A: Let $U_t = \operatorname{sign}(W_t)$, where $\operatorname{sign}(x) = +1$ for $x \geqslant 0$ and $\operatorname{sign}(x) = -1$ for $x<0$. Since the joint distribution of $W_t$ is known to be multinormal, we can compute joint probabilities for $U_t$ as well.
$$
   \mathbb{P}( U_t = 1) = \mathbb{P}(W_t \geqslant 0) = \frac{1}{2} \qquad
   \mathbb{P}( U_t = -1) = \mathbb{P}(W_t < 0) = \frac{1}{2}
$$
which means that $(1+U_t)/2$ follows symmetric Bernoulli distribution.
For the joint probabilities of $U_{t_1}$ and $U_{t_2}$ for $t_2 > t_1 > 0$ are multinormal orthant probabilities:
$$ \begin{eqnarray}
  \mathbb{P}(U_{t_1} \geqslant 0 \land U_{t_2} \geqslant0) &=& \frac{1}{4} + \frac{1}{2 \pi} \arcsin(\sqrt{\frac{t_1}{t_2}}) \\ 
   \mathbb{P}(U_{t_1} \geqslant 0 \land U_{t_2} < 0) &=& \frac{1}{4} - \frac{1}{2 \pi} \arcsin(\sqrt{\frac{t_1}{t_2}}) \\
    \mathbb{P}(U_{t_1} < 0 \land U_{t_2} \geqslant 0) &=& \frac{1}{4} - \frac{1}{2 \pi} \arcsin(\sqrt{\frac{t_1}{t_2}}) \\
    \mathbb{P}(U_{t_1} < 0 \land U_{t_2} < 0) &=& \frac{1}{4} + \frac{1}{2 \pi} \arcsin(\sqrt{\frac{t_1}{t_2}}) 
\end{eqnarray}
$$
These can be computed by representing $(W_{t_1}, W_{t_2}) \stackrel{d}{=}  ( \sqrt{t_1}  Z_1, \sqrt{t_1}  Z_1 + \sqrt{t_2-t_1} Z_2)$, where $Z_i$ are independent standard normals.
