Let $B(t) : t ≥ 0$ be a standard Brownian motion process. Find the following:
(i) $Pr(−1 ≤ B(2) ≤ 3|B(1) = 1)$
(ii) $Pr(−2 ≤ B(3) − B(4) ≤ 1|B(1) = 3)$
(iii) $Pr(0 ≤ B(3) ≤ 4|B(5) = 3)$
For (i), the distribution of $B(2)-B(1)$ is normal with mean $0$, variance $1$ so it is equal to $Pr(-1-1 ≤ B(2)-B(1) ≤ 3-1)=Pr(-2 ≤ Z ≤ 2)$. (Note independence).
I'm having trouble with (ii), mostly because I can't express it in terms that make sense to me. I'm very new to Brownian Motion. Thank you!
$Pr(−2 ≤ B(3) − B(4) ≤ 1|B(1) = 3)=Pr(−1 ≤ B(4) − B(3) ≤ 2|B(1) = 3)$
Is it true that $B(4) − B(3)|B(1)=3$ is normally distributed with mean $3$, variance $2$?
Making it $Pr(-2*\sqrt2+3 ≤ Z ≤ 2*\sqrt2+3)$
For (iii), the the distribution of $B(3)|B(5)$ is normal with mean $\frac35*3=1.8$, variance $\frac35(5-3)=1.2$ so it is equal to $Pr(0*\sqrt{1.2}+1.8 ≤ Z ≤ 4\sqrt{1.2}+1.8)$
Best Answer
I will explain in a simple context, the increments of brownian is independent for each time step and is normal distributed to mean 0 and variance corresponding to the size of time step.
Let's say $t_1<t_2<t_3<t_4$. $B(t_4)-B(t_3)$ is normal distributed to mean 0 and variance $t_4-t_3$ and is independent from $B(t_2)-B(t_1)$
For ii), $B(1)=3$ is just an event from the $\sigma$-algebra generated by $B(1)$. However, the independence of increments tell us that $P(B(t_j)-B(t_i)|B(t_k))=P(B(t_j)-B(t_i))$ for all $t_j>t_i \geq t_k$
For iii), you are interested in increments between $B(3)$ and $B(5)$. $P(B(5)-B(3))$ is also normal distributed with mean 0 and variance 2. You can do almost the same as in i).
Only note, you are calculating the joint probability. Conditional probability is defined as(for example ):$P(B(2)=2|B(1)=1)=\frac{P[B(2)=2,B(1)=1]}{P[B(1)]=1)}=\frac{P[(B(2)-B(1)]=1]}{P[B(1)=1]}$.