Positive Then Negative

Let $B_t$ be a standard Brownian motion. What is the probability that $B_1 > 0$ and $B_2 < 0$ ?

Reveal solutionHide solution

#The joint is bivariate normal

The pair $(B_1, B_2)$ is jointly normal with means $0$ , variances $\Var(B_1) = 1$ and $\Var(B_2) = 2$ , and covariance $\Cov(B_1, B_2) = \min(1,2) = 1$ . So their correlation is $\rho = \tfrac{1}{\sqrt{2}}$ .

#An orthant probability

For a bivariate normal with correlation $\rho$ , the chance of landing in a same-sign quadrant is $\tfrac{1}{4} + \tfrac{\arcsin\rho}{2\pi}$ . The opposite-sign quadrant takes the rest of a half,

\PP(B_1>0,\,B_2<0) = \tfrac{1}{2} - \left(\tfrac{1}{4} + \tfrac{\arcsin\rho}{2\pi}\right) = \tfrac{1}{4} - \tfrac{\arcsin\rho}{2\pi}. \tag{1}

With $\arcsin\tfrac{1}{\sqrt{2}} = \tfrac{\pi}{4}$ ,

\PP(B_1>0,\,B_2<0) = \tfrac{1}{4} - \tfrac{\pi/4}{2\pi} = \tfrac{1}{4} - \tfrac{1}{8} = \tfrac{1}{8}. \tag{2}

The two times are positively correlated, so the contour leans along the diagonal and the mass piles into the like-sign corners. The opposite-sign quadrant, first positive and second negative, holds only one eighth.

#Read it off

The answer is $\tfrac{1}{8}$ . The positive correlation drags the two values onto the same side, so a sign flip between time $1$ and time $2$ is only half as likely as the naive independent guess of $\tfrac{1}{4}$ .