Probability That Brownian Motion Values at Two Times Have Opposite Signs
Summary
The problem asks for the probability that standard Brownian motion at time t and at time 2t have opposite signs. The key step is to express the later value as the sum of the earlier value and an independent increment. Both components have centered normal distributions with the same variance, so the question reduces to the signs of one normal variable and its sum with an independent normal variable.
The solution separates the two possible sign patterns, which have equal probability by symmetry. For one pattern, conditioning on the earlier value gives a normal cumulative distribution probability; integrating over negative earlier values yields one eighth. Doubling this contribution gives a total probability of one quarter. A geometric argument using the rotational symmetry of the pair of independent normal variables reaches the same result. The calculation relies on Brownian independent increments and normality, and applies to the stated two-time comparison rather than arbitrary dependent processes.
Key ideas
- The Brownian value at the later time is the earlier value plus an independent increment.
- The earlier value and the increment are independent centered normal variables with equal variance.
- The two opposite-sign cases have equal probability by symmetry.
- Conditioning on the earlier value reduces one case to an integral involving the normal density and cumulative distribution.
- The total probability of opposite signs is one quarter.
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Full text
# probability question about brownian motion
# probability question about brownian motion
Assume $W_{t}$ is a standard Brownian Motion, calculate the the probability that $W_{t}*W_{2t}$ is negative, i.e., $P(W_{t}*W_{2t}<0)$. I find it tricky to calculate the probability.Thank you.
## Answer by Gordon (score 4, accepted)
https://quant.stackexchange.com/a/17647
Your decomposition is correct. I will show here the computation for one term: \begin{align*} P(W_t < 0, W_{2t} >0) &= P(W_t < 0, W_{2t}-W_t > -W_t)\\ &= E\Big(E\big(\mathbb{1}_{\{W_t < 0\}}\mathbb{1}_{\{W_{2t}-W_t > -W_t\}}\mid W_t\big)\Big)\\ &= E\Big(\mathbb{1}_{\{W_t < 0\}} \Phi\big(W_t/\sqrt{t}\big)\Big) \\ &=E\Big(\mathbb{1}_{\{W_t/\sqrt{t} < 0\}} \Phi\big(W_t/\sqrt{t}\big)\Big) \\ &=\int_{-\infty}^0 \phi(x) \Phi(x) dx\\ &=\frac{1}{2}\Phi(x)^2\mid_{-\infty}^0\\ &=\frac{1}{8}, \end{align*} where \begin{align*} \phi(x)=\frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} \end{align*} is the density of a standard normal random variable, and $\Phi(x)$ is the cumulative distribution function.
## Answer by vanna (score 6)
https://quant.stackexchange.com/a/17651
Since $W_{2t}-W_{t}$ is independent of $W_t$ and has the same law as $W_{2t-t}=W_t$ we only have to compute $$P(X(X+Y)<0)$$ where $(X,Y)$ follows a bivariate normal distribution (with zero correlation). From there you can split the probability in two cases : either $X<0$ and $X+Y>0$ or the opposite. The two events have the same probability since $(-X,-Y)\sim (X,Y)$. You are left with the computation of $$ P(X<0,X+Y>0)$$ since the distribution of $(X,Y)$ is invariant by rotation around the z-axis) this probability can be computed geometrically (think cutting a cake, the cake being the bivariate density) : it is equal to $1/8$. The final result is thus $1/4$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.