As stated, this is an interview question.
Given Brownian motion $B_t,B_s$ and $t>s$, how to calculate $P(B_t>0,B_s<0)$?
As stated, this is an interview question.
Given Brownian motion $B_t,B_s$ and $t>s$, how to calculate $P(B_t>0,B_s<0)$?
Set $X_t=B_t-B_s$ and $Y_t=-B_t$. $X_t\sim N(0,t-s)$ and $X_t$ , $Y_s$ are independent. $$I=P(B_t>0, B_s<0)=P(B_t-B_s>-B_s\,,\, -B_s>0)=P(X_t>Y_s\,, Y_s>0)$$ $$I=\frac{1}{2\pi\sqrt{s(t-s)}}\int_{0}^{\infty}\int_{y}^{\infty}\exp\left(-\frac{y^2}{2s}-\frac{x^2}{2(t-s)}\right)dxdy$$ Set $$y={\sqrt{s}}\,\,r\sin \theta$$ $$\quad x={\sqrt{t-s}}\,\,r\cos \theta$$ we have $$dx\,dy=\sqrt{s(t-s)}\,r \,dr d\theta$$ $y>0$ and $x>y$ in other words $${\sqrt{s}}\,\,r\sin \theta<{\sqrt{t-s}}\,\,r\cos \theta$$ i.e $$\tan \theta <\sqrt{\frac{t-s}{s}}$$ or $$\theta<{\tan^{-1}\left({\sqrt{\frac{t-s}{s}}}\right)}=\cos^{-1}{\left({\sqrt{\frac{s}{t}}}\right)}$$ therefore $$I=\frac{1}{2\pi}\int_{0}^{{\cos^{-1}{\left({\sqrt{\frac{s}{t}}}\right)}}}\int_0^{\infty}r\exp\left(-\frac{r^2}{2}\right)drd\theta$$ $$I=\frac{1}{2\pi}\int_{0}^{{\cos^{-1}{\left({\sqrt{\frac{s}{t}}}\right)}}}-\exp\left(-\frac{r^2}{2}\right)\Big{|}_{0}^{\infty}d\theta$$ $$I=\frac{1}{2\pi}\int_{0}^{{\cos^{-1}{\left({\sqrt{\frac{s}{t}}}\right)}}}d\theta=\frac{1}{2\pi}\cos^{-1}\left(\sqrt{\frac{s}{t}}\right)$$