Brownian motion: find the backward conditional probability

Question

Let $W_t$ be a standard Brownian motion.

I want to compute $P[W_1+W_2>3|W_3=4]$.

Any help? Thanks.

Answer 1

Let $U=W_1+W_2,V=W_3$, then $(U,V)^{'}$ obeys bivariate normal distribution with mean $(0,0)^{'}$ and

$Var[U]=Var[W_1]+Var[W_2]+2Cov[W_1,W_2]=5$

$Var[V]=3$

$Cov[U,V]=Cov[W_1,W_3]+Cov[W_2,W_3]=3$

Thus, we obtain the conditional distribution $U|V=4$~$N(4,2)$.

$P[U>3|V=4]=\Phi(-\frac{\sqrt{2}}{2})$