Conditional Probability that a Brownian Motion is positive given that it is positive at a previous time

Question

If we let $X_t$ be a standard Brownian motion, then what is $P(X_2 > 0 | X_1 > 0)$? I've tried to apply Baye's theorem and got the above to be equal to $2P(X_2 > 0, X_1 > 0) = 2P((X_2 - X_1) > -X_1, X_1 > 0)$ but now I'm not sure how to set up the integrals of the normal distribution. Is there a more elegant way?

Answer 1

$EX_tX_s=\min \{t,s\}$ so $EX_1X_2=1$. Thus $(X_1,X_2)$ is jointly normal with variance-covariance matrix $$M=\left[\begin{array}{llll}1 & 1 \\ 1 & 2 \end{array}\right].$$ Now integrate the density function $\frac 1 {2\pi} e^{-(x,y)M(x,y)^{T}/2}dxdy$ over $(0,\infty)\times (0,\infty)$ to find $P(X_1>0,X_2>0)$.