Probability that the product of Brownian motion at two dates is positive Let $(B_t)_{t \ge 0}$ be a standard Brownian motion, $B_0 = 0$.
$0 \le t_1 \le t_2$
As the title says, I am trying to calculate the following probability : 
$\text{Pr}(B_{t_1}B_{t_2} \ge 0)$
Any leads ?
 A: Fix $s < t$. By symmetry of Brownian motion, we have $$\mathbb{P}(B_s B_t \geq 0) = \mathbb{P}(B_s \geq 0, B_t \geq 0) + \mathbb{P}(-B_s \geq 0, -B_t \geq 0) = 2 \mathbb{P}(B_s \geq 0, B_t \geq 0).$$ Consequently, it suffices to compute the probability on the right-hand side. Conditioning on $\mathcal{F}_s := \sigma(B_r; r \leq s)$, we find from the Markov property that
$$\mathbb{P}(B_s \geq 0, B_t \geq 0)= \mathbb{E}\left( 1_{\{B_s \geq 0\}} \mathbb{P}(B_{t-s}+x \geq 0) \bigg|_{x=B_s}\right). \tag{1}$$
Denote by $\Phi$ the cdf of the standard Gaussian distribution. Since $B_{t-s}$ is Gaussian with mean $0$ and variance $t-s$, we have
$$\mathbb{P}(B_{t-s}  +x \geq 0) = \mathbb{P}(B_{t-s} \leq x) = \Phi(x/\sqrt{t-s}).$$
Hence, by $(1)$,
$$\mathbb{P}(B_s \geq 0,B_t \geq 0) = \mathbb{E}(1_{\{B_s \geq 0\}} \Phi(B_s/\sqrt{t-s})).$$
If we denote by $\varphi$ the density of the Gaussian distribution, then we can write this equivalently as
$$\mathbb{P}(B_s \geq 0,B_t \geq 0) = \int_0^{\infty} \Phi(\sqrt{s} x/\sqrt{t-s}) \varphi(x) \, dx.$$
The latter integral can be calculated explicitly, see e.g. this answer, and we get
$$\mathbb{P}(B_s \geq 0,B_t \geq 0) = \frac{1}{4} + \frac{1}{2\pi} \arctan \sqrt{\frac{s}{t-s}}.$$
Hence,
$$\mathbb{P}(B_s B_t \geq 0)= \frac{1}{2} + \frac{1}{\pi} \arctan \sqrt{\frac{s}{t-s}}.$$
A short sanity check: For $s=t$ both sides equal $1$ - as they should.
