In one of the question on forum $Pr (Z(1) < 0, Z(2) < 0)$ is calculated. I have a slightly similar question $Pr ( Z(1) < 0, Z(2) < 0, Z(3) < 0)$ ?
$Z(1), Z(2) \; and \; Z(3)$ are brownian motion points at $t=1,2,3$
Is there a formula simpler than the one involving integrals of Gaussian distributions?
The first one is $$p:=P(Z_1<0,Z_2<0)=P(Z_1<0,Z_2-Z_1<-Z_1)$$ But $Z_2-Z_1$ is independent of $Z_1$ so eventually we are just computing $$p=P(X<0,X+Y<0)$$ where $X,Y \sim \mathcal{N}(0,1)$. Now you can see this as an integral w.r.t. the bivariate Gaussian density of course. Or you can just say that it all boils down to cutting a piece of cake. If the bivariate density is a cake then the value $p$ is a $3/8$ slice (the two lines delimitates $3/8$ of the plane). Exact same reasoning in higher dimension. $$p=P(X<0,X+Y<0,X+Y+Z<0)$$ Imagine a unit sphere in 3d and cut it according to the 3 lines.