# 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 $B_2<0$?

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 $0.125.$

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.

• Please can you elaborate why? – Antonius Gavin May 15 '15 at 12:13
• Which part you didn't understand? Basically @Kiwiakos transformed your problem into a multi-normal, then he figured out the distribution. His derivation based on the fact that the expectation of a standard-normal is 0 and variance is 1 etc. – SmallChess May 16 '15 at 12:38

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 $$\int_0^\infty F(x)\cdot f(x)dx=\int_0^\infty F(x)dF(x)=\dfrac{1}{2}\int_0^\infty dF^2(x)=\dfrac{1}{2}\left(F(\infty)-F(0)\right)=\dfrac{1}{2}(1-0.25)=\dfrac{3}{8}$$

After all, we get an answer $$\mathbb{P}(Z_1+Z_2>0\mid Z_1<0)=\int_0^\infty f(x)\cdot (1-F(x))dx=\dfrac{1}{2}-\dfrac{3}{8}=\dfrac{1}{8}=0.125$$