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Here is my solution, which I am hoping can either be verified or corrected.

$B(t)$ is normal with mean 0 and variance $t$.

$P(B(4)>1\;|\;B(8)=0) = P(B(4)+B(8)>1+0\;|\;B(8)=0) = P(B(12)>1) = P(Z>\frac{1}{\sqrt{12}}) = 1-P(Z<0.29) = 1-0.6141 = 0.3859$.

D Clark
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  • $B(8)=0$ is null and you are conditioning on a null event... – user10354138 Apr 11 '21 at 16:35
  • Not sure where you are going with that. Could you elaborate? – D Clark Apr 11 '21 at 16:37
  • You cannot condition on a null event: if $\mathbb{P}(F)=0$, then $\mathbb{P}(-\mid F)$ is not well-defined. – user10354138 Apr 11 '21 at 16:40
  • Interesting. This was a practice question given by my professor. – D Clark Apr 11 '21 at 16:46
  • By the way, do you have any idea for how I can get more practice with elementary Brownian motion? – D Clark Apr 11 '21 at 16:47
  • The only equation defining $\mathbb{P}(-\mid F)$ is $\mathbb{P}(E\cap F)=\mathbb{P}(E\mid F)\mathbb{P}(F)$, but if $\mathbb{P}(F)=0$ then $\mathbb{P}(E\cap F)=0$ also, and hence $\mathbb{P}(E\mid F)$ gives you a $0/0$. If your professor wants a Brownian bridge, he/she should start with $B(t)=W(t)-tW(8)/8$ instead. – user10354138 Apr 11 '21 at 16:49
  • Interesting. I have no idea what a Brownian bridge is, but, your explanation of why the problem is not solvable makes perfect sense. Thank you. – D Clark Apr 11 '21 at 16:51
  • The problem is solvable, but the notation is just some matter of convention. As @user10354138 mentioned, ${B(8)=0}$ is a null set, so it certainly isn't "standard" definition of conditional probability, no. It has to do with conditional expectation. Namely, you are to find random variable $\mathbb E[ 1_{{B(4) > 1}} | B(8) ] =: \mathbb P(B(4) > 1 | B(8))$. From general theory, the result you'll get will be in form $F(B(8))$ for some borel function $F$. Notation $\mathbb P(B(4) > 1 | B(8) = 0)$ means then value $F(0)$. – Presage Apr 11 '21 at 17:57

1 Answers1

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Because the pair $(B(4),B(8))$ has a joint normal distribution, the (regular) conditional distribution of $B(4)$, given that $B(8) = z$ is normal with mean $z/2$ and variance $2$.

Alternatively, again because $(B(4),B(8))$ has a joint normal distribution, the fact that $B(4)-.5B(8)$ and $B(8)$ are uncorrelated implies that they are independent, and the former has a normal distribution with mean $0$ and variance $2$. Thus, $B(4) = [B(4)-.5B(8)]+.5B(8)$, conditional on $B(8) = z$ is normal with mean $z/2$ and variance that of $B(4)-.5B(8)$, namely $2$.

John Dawkins
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