# 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 $P(B_t>0,B_s<0)$?

• Could you please relate your question to quantitative finance, it looks like it is off topic – lehalle Jul 25 '16 at 20:01
• @lehalle, come on, this is a quant interview question and a Brownian motion. – PythonNewHand Jul 25 '16 at 23:31
• I am not sure all quant interview questions can be asked here. I ask usually questions about the differences between PCA and ICA, and I would close a question here like that, except if it is related to quant finance (say about yield curve modeling)... – lehalle Jul 26 '16 at 5:41

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 $$I=\frac{1}{2\pi}\int_{0}^{{\cos^{-1}{\left({\sqrt{\frac{s}{t}}}\right)}}}\int_0^{\infty}r\exp\left(-\frac{r^2}{2}\right)drd\theta$$ $$I=\frac{1}{2\pi}\int_{0}^{{\cos^{-1}{\left({\sqrt{\frac{s}{t}}}\right)}}}-\exp\left(-\frac{r^2}{2}\right)\Big{|}_{0}^{\infty}d\theta$$ $$I=\frac{1}{2\pi}\int_{0}^{{\cos^{-1}{\left({\sqrt{\frac{s}{t}}}\right)}}}d\theta=\frac{1}{2\pi}\cos^{-1}\left(\sqrt{\frac{s}{t}}\right)$$
• can you specify more details about why dydx = $\sqrt{t(t-s)}rdrd\theta$ – PythonNewHand Jul 24 '16 at 18:39
• \begin{align} & dydx=Jdrd\theta \\ & dydx=\left| \begin{matrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta } \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta } \\ \end{matrix} \right|drd\theta \\ & dydx=\left| \begin{matrix} \sqrt{t}\cos \theta & -\sqrt{t}r\sin \theta \\ \sqrt{t-s}\sin \theta & \sqrt{t-s}r\cos \theta \\ \end{matrix} \right|drd\theta \\ & dydx=\left( \sqrt{t(t-s)}\,r{{\cos }^{2}}\theta +\sqrt{t(t-s)}\,r{{\sin }^{2}}\theta \right)drd\theta =\sqrt{t(t-s)}\,rdrd\theta \\ \end{align} – user16651 Jul 24 '16 at 18:46