Calculating Brownian Motion Sign-Crossing Probabilities

Summary

The document derives the probability that Brownian motion is positive at a later time and negative at an earlier time. It rewrites the event using the increment from the earlier to the later time and the negative of the earlier value. These variables are independent centered normal variables, with variances determined by the two time intervals.

After expressing the joint probability as an integral over a region, the derivation rescales the variables and uses polar coordinates. The event becomes an angular sector, yielding a closed-form probability in terms of the inverse cosine of the square root of the ratio of the earlier time to the later time. This is a mathematical derivation rather than a trading strategy; its direct value is in probability calculations involving Brownian paths. The result assumes standard Brownian motion started at zero and times with the earlier time positive.

Key ideas

• The Brownian increment after the earlier time is independent of the earlier Brownian value.
• The joint sign event can be written as an inequality between independent normal variables.
• Rescaling the normal variables turns the probability region into an angular sector.
• The resulting probability depends on the ratio of the two observation times.

Tags

# Given Brownian motion $B_t,B_s$ and $t>s$, how to calculate $P(B_t>0,B_s<0)$?


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

## Answer by user16651 (score 4, accepted)

https://quant.stackexchange.com/a/28250

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 $$I=\frac{1}{2\pi}\int_{0}^{{\cos^{-1}{\left({\sqrt{\frac{s}{t}}}\right)}}}\int_0^{\infty}r\exp\left(-\frac{r^2}{2}\right)drd\theta$$ $$I=\frac{1}{2\pi}\int_{0}^{{\cos^{-1}{\left({\sqrt{\frac{s}{t}}}\right)}}}-\exp\left(-\frac{r^2}{2}\right)\Big{|}_{0}^{\infty}d\theta$$ $$I=\frac{1}{2\pi}\int_{0}^{{\cos^{-1}{\left({\sqrt{\frac{s}{t}}}\right)}}}d\theta=\frac{1}{2\pi}\cos^{-1}\left(\sqrt{\frac{s}{t}}\right)$$

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.