Calculating the Probability of a Brownian Motion Crossing Zero Between Times
Summary
The document calculates the probability that standard Brownian motion is positive at the first observation time and negative at the second. It represents the first value as one standard normal variable and the later value as that variable plus an independent standard normal increment. The event therefore becomes a pair of inequalities involving two independent Gaussian variables.
A geometric symmetry argument identifies the relevant region as half of a quadrant, yielding a probability of one eighth. Another answer describes the two observations as a bivariate normal pair with known covariance and gives an approximate value. A third derives the same result by conditioning and integrating the normal density. Together these approaches illustrate independent increments, Gaussian dependence across time, and symmetry-based probability calculations. The result applies to standard Brownian motion at the specified times; changing the time points or process assumptions changes the covariance structure and may change the probability.
Key ideas
- Brownian motion’s later value is the earlier value plus an independent Gaussian increment.
- The two observations are jointly Gaussian but correlated because they share the earlier increment.
- The sign event can be transformed into inequalities for two independent standard normal variables.
- Symmetry gives a probability of one eighth for the stated pair of sign conditions.
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Full text
# For $B_t$ a Brownian motion what is the probability that $B_1>0$ and $B_2<0$?
# For $B_t$ a Brownian motion what is the probability that $B_1>0$ and $B_2<0$?
Let $B_t$ be a Brownian Motion. What's the probability that $B_1>0$ and $B_2<0$?
## Answer by Mark Joshi (score 7)
https://quant.stackexchange.com/a/17818
The problem is equivalent to given to 2 independent standard normals $W$ and $Z$ the probability of $$ W > 0, \text{ and } W+Z<0. $$ or $$ W > 0, \text{ and } Z<-W. $$ Plotting this set we see it is the bottom half of the lower right quadrant. The probability of being in the lower right quadrant is clearly $0.25$ by symmetry. The probability of being in the bottom half is half again by symmetry so the answer is $0.125.$
## Answer by Kiwiakos (score 4)
https://quant.stackexchange.com/a/17808
B1~N(0,1) and B2=B1+Z, for Z~N(0,1). From that E(B1*B1)=E(B1*B2)=1, E(B2*B2)=2. Therefore they are bivariate Gaussian with covariance matrix (1,1;1,2) therefore probability is around 12%, which is the volume over the bottom-right quadrant.
## Answer by Yuri Kulchitsky (score 2)
https://quant.stackexchange.com/a/17813
Let $Z_1,Z_2\sim N(0,1), B_1=Z_1,B_2=Z_1+Z_2.$ Construct a random variable $Y$ as following: $$\left\{ \begin{array}{cc} Y=1 & B_1 > 0, B_2 < 0\\ Y=0 & otherwise \end{array} \right. $$ Note that $\mathbb{P}(Z_1+Z_2 < 0\mid Z_1 > 0)=\mathbb{P}(Z_2<-Z_1\mid Z_1>0)=\mathbb{E}Y$.
Use that to construct the integral. Everything that is not relevant adds up to zero as we obtain $$\mathbb{P}(Z_2>-Z_1\mid Z_1<0)=\int_0^\infty f(-x)\cdot (1-F(x))dx,$$ where $f(x)=\dfrac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$, a probability distribution function of a normal distribution. As in our case it is standard normal distribution, we have $f(x)=f(-x)$, so $$ \int_0^\infty f(-x)\cdot (1-F(x))dx=\int_0^\infty f(x)\cdot (1-F(x))dx=\int_0^\infty f(x)dx-\int_0^\infty F(x)\cdot f(x)dx $$
First part obviously is equal to $F(0)=\dfrac{1}{2}$.
Now consider the second part. As $F(x)=\int_{-\infty}^x f(y)dy$, and $f(x)$ is continuous, so we have $F'(x)=f(x)$.
Then $f(x)dx=dF(x)$, giving us $$ \int_0^\infty F(x)\cdot f(x)dx=\int_0^\infty F(x)dF(x)=\dfrac{1}{2}\int_0^\infty dF^2(x)=\dfrac{1}{2}\left(F(\infty)^2-F(0)^2\right)=\dfrac{1}{2}(1-0.25)=\dfrac{3}{8} $$
After all, we get an answer $$ \mathbb{P}(Z_1+Z_2>0\mid Z_1<0)=\int_0^\infty f(x)\cdot (1-F(x))dx=\dfrac{1}{2}-\dfrac{3}{8}=\dfrac{1}{8}=0.125 $$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.