# Browian motion: $P(B_1<4 | B_2 =1)$

I want to calculate $P(B_1<4 | B_2 =1)$ for the B.M.

What I tried: $P(B_1<4 | B_2 =1)=P(B_1 - B_2 < 3- B_2 | B_2 =1)$

but I cant use any independence to calculate further.

• Isn't this the same as $P(B_2<4|B_1=1)$? . Think of running paths backwards through the known point. – dm63 Apr 22 '18 at 12:09
• @dm63 How is it the same? – izimath Apr 22 '18 at 12:48
• Actually, I think I'm wrong if it is assumed that $B_0=0$. – dm63 Apr 22 '18 at 16:05
• Using Cholesky decomposition, you can express $B_1$ as a linear combination of $B_2$ and another normal random variable that is independent of $B_2$. – Gordon Apr 22 '18 at 17:29

Consider the multivariate normally distributed vector $(B_1,B_2)$, in particular
$\begin{pmatrix} B_1 \\ B_2 \\ \end{pmatrix}$ ~$N $$\left(\begin{pmatrix} 0 \\ 0 \\ \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 1 & 2 \\ \end{pmatrix} \right) . Then it holds that B_1 |B_2=1 ~ N(1,1/2) (wikipedia). • Isn't it N(½,½) ? – Appliqué Apr 19 at 12:32 This is related to the concept of the Brownian Bridge. If B_t is Brownian motion then W_t defined as$$ W_t = (B_t|B_T = 0) $$is the Brownian bridge with end point 0. There is version with a and b too. You know where it starts and where it ends (that's why is the bridge) and there is uncertainty inbetween. One could read-up on the topic. If I just quote wikipedia then for B_(t_1) = a and B_(t_2) = b the Brownian bridge is normal with mean$$ a + \frac{t-t_1}{t_2-t_1}(b-a) $$and covariance between W(s) and W(t)$$ \frac{(t_2-t)(s-t_1)}{t_2 - t_1}.$$Thus in your case we tie the Brownian bridge at$B_0 = 0$then$t_1 = 0$and$a=0$,$t_2 = 2$and$b = 1$and you have to find the probability that the Brownian bridge at time$1$is less than$4$. Here is what I found:$P(B_1<4 |B_2 =1) =P(2B_1 -B_2 < 7 |B_2 =1) =P(2B_1 -B_2<7 )= P(\sqrt{2} Z<7 ) = N(7/ \sqrt{2})$where$Z$is the standard normal r.v.. I have used the fact$2B_1 -B_2 $is indep of$B_2\$ since they are multivariate normal and their covariance is 0.