4
$\begingroup$

Let $B_t$ be a Brownian Motion. What's the probability that $B_1>0$ and $B_2<0$?

$\endgroup$
6
$\begingroup$

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 $0.125.$

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Please can you elaborate why? $\endgroup$ – Antonius Gavin May 15 '15 at 12:13
  • $\begingroup$ 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. $\endgroup$ – SmallChess May 16 '15 at 12:38
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.