# Brownian Motion, Conditional Probability Problem

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 $Pr(0*\sqrt{1.2}+1.8 ≤ Z ≤ 4\sqrt{1.2}+1.8)$

• What you calculated in i) is just the joint probability. Dividing with P(B(1)=1) you will have the conditional probability. For ii) or iii), jus imagine the immediate step B(2) or B(4) as a fixed value x. Then integrate over all possible probabilities. – quallenjäger Mar 30 '17 at 20:02
• @quallenjäger I'm not sure what you mean – Yeah.. Mar 30 '17 at 20:05
• ii) is just independent from B(1)=3, because you are interested in increments. The iii) just rewrite it as increment between time 3 and 5. Only in this case your distribution have variance2, instead of 1. This is equivalent to take an immediate step B(4) and integrate over all possibilities. – quallenjäger Mar 30 '17 at 20:15
• I still can't wrap my head around it; a string of equalities would be great so I could follow the process – Yeah.. Mar 30 '17 at 20:18

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 ):$P(B(2)=2|B(1)=1)=\frac{P[B(2)=2,B(1)=1]}{P[B(1)]=1)}=\frac{P[(B(2)-B(1)]=1]}{P[B(1)=1]}$.

• In your final note, why does the numerator and denominator of the conditional probability didn't cancel like this? $P(B(2)=2|B(1)=1)=\frac{P[B(2)=2,B(1)=1]}{P[B(1)=1]}=\frac{ P[B(1)]=1 P[(B(2)-B(1)]=1]}{P[B(1)=1]}=P[B(2)-B(1)=1]$. – Ying Jan 18 '18 at 18:52
• Also, does the following derivation in (iii) correct? $Pr(0 \le B(3) \le 4|B(5) = 3)=\frac{Pr( B(5)=3 ~\cup~ 0 \le B(3) \le 4 )}{Pr(B(5)=3)}=\frac{ Pr( 0 \le B(3) \le 4 ~\cup~ -1 \le B(5)-B(3) \le 3) }{Pr(B(5)=3)}=\frac{ Pr( 0 \le B(3) \le 4 )Pr( -1 \le B(5)-B(3) \le 3) }{Pr(B(5)=3)}$. Then, in this case, I cannot cancel the denominator and numerator. – Ying Jan 18 '18 at 19:07