Brownian motion probability P(B2>B1>B3)

Question

For standard Brownian motion, I'm trying to find $P(B_2>B_1>B_3)$, which I thought I could split into:

$=P(B_2>B_1)*P(B_1>B_3)$
$=P(B_2-B_1>0)*P(B_1-B_3>0)$
$=(1-P(B_2-B_1\leq0))*(1-P(B_1-B_3\leq0))$

which in R I calculated using:

 (1-pnorm(0,0,1))*(1-pnorm(0,0,sqrt(2)))

But I'm not sure if this is the right answer! Is what I'm doing sound?

Answer 1

Hint

$Z_1= B_2-B_1$ and $Z_2= B_3-B_2$ are independent $N(0,1)$ and your event is equivalent to $\{Z_1>0,Z_2+Z_1<0\}.$ (And no, you can’t split this up by independence either, but it’s a pretty easy to understand region of $Z_1-Z_2$ space.)