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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.

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

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  • $\begingroup$ Isn't it N(½,½) ? $\endgroup$ – Appliqué Apr 19 at 12:32
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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 $4$.

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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.

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