How to show that $P(B_1>0, B_2<0)= P(B_1>0)P(B_2-B_1 <0) P(|B_2 - B_1| > |B_1|)$?

Question

The following is an interview question from Zhou Xinfeng's book entitled 'A Practical Guide to Quantitative Finance Interview'.

Question: Let $B_t$ be a Brownian motion. What is the probability that $B_1 > 0$ and $B_2 < 0$?

The author provided two methods, first uses double integral and second uses independence and conditional probability. The following is the second approach.

$$P(B_1>0, B_2<0) = P(B_1>0)P(B_2-B_1 <0) P(|B_2 - B_1| > |B_1|) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}.$$

I do not understand how the author decompose $P(B_1>0, B_2<0)$ into $P(B_1>0)P(B_2-B_1 <0) P(|B_2 - B_1| > |B_1|).$

Answer 1

There are two aspects to this; I don't know which aspect you have trouble with.

Logic:

$$B_1 > 0 \cap B_2 < 0 \iff B_1 > 0 \cap B_2 - B_1 < 0 \cap |B_2 - B_1| > |B_1|$$

The latter two RHS terms literally say, between $t=1$ to $2$, you have to have a net leftward movement, and by an absolute distance bigger than the rightward movement between $t=0$ and $1$ (in order to end up $B_2 < 0$). I am not sure I can explain better than this.

Probability: From the logic formula above, we have:

$$P(B_1 > 0 \cap B_2 < 0) = P( B_1 > 0 \cap B_2 - B_1 < 0 \cap |B_2 - B_1| > |B_1|)$$

So next we need to argue that the three RHS terms are mutually independent. This comes from:

• $X_1 = B_1$ and $X_2 = (B_2 - B_1)$ are independent because they don't overlap in time

• $Sign(X_j)$ and $|X_j|$ are independent, because the Brownian motion is unbiased (symmetric about $0$).

• Therefore all four of $Sign(B_1), |B_1|, Sign(B_2 - B_1), |B_2 - B_1|$ are mutually independent.

Now realizing that $B_1 > 0 \iff Sign(B_1) = +1$ etc allows us to break the three terms of the RHS apart.

$$P( B_1 > 0 \cap B_2 - B_1 < 0 \cap |B_2 - B_1| > |B_1|) \\ = P( B_1 > 0) ~~P(B_2 - B_1 < 0) ~~P(|B_2 - B_1| > |B_1|)$$ The first two terms each equals $1/2$ because of the Brownian motion being unbiased (symmetric about $0$), and the last term equals $1/2$ because $|B_1|$ and $|B_2 - B_1|$ are i.i.d. variables (and they're equal with prob $0$).

Answer 2

Let $\omega \in \{B_1>0 \}\ \cap \{ B_2<0 \}.$ Let's think about the path $t \rightarrow B_t(\omega)$ as a continuous function such that $B_0(\omega)=0.$ We have the time in the x-axis and $B_t(\omega)$ in the y-axis. If $B_t(\omega)$ traveled $B_1(\omega)>0$ units from $t=0$ to $t=1$, then for $B_2(\omega)$ to be negative, the distance $|B_2(\omega)-B_1(\omega) |$ traveled over the y-axis from $t=1$ to $t=2$ must have been greater than the distance $|B_1(\omega)|$ traveled from $t=0$ to $t=1.$ There are 2 possibilities equally provable$_{(*)}$ for the distance $|B_2(\omega)-B_1(\omega) |$ to be greater than the distance $|B_1(\omega)|$. One possibility is that $B_2(\omega)$ ended above $B_1(\omega),$ (i.e $B_2(\omega)>B_1(\omega)$), the other one is that it ended below (i.e $B_2(\omega)<B_1(\omega)$). Obviously, if it ended above, it can't be $B_2(\omega)<0,$ (since $B_1(\omega)>0$) so it has to be $B_2(\omega)<B_1(\omega)$. So $\omega \in \{B_1>0 \} \cap \{ |B_2-B_1|>|B_1| \} \cap \{ B_2<B_1 \}.$ The reverse can be proven similarly.

(*) the two possibilities are equally probable because the Brownian motion is a Markov process.

Since $B_2-B_1$ is independent from $B_1$, these last three sets are independent, so you can multiply the probabilities.

Answer 3

Use the independence of $B_1$ and $B_2-B_1$ along with the identity \begin{align} \{B_1>0,B_2<0\}&=\{B_1>0,B_2-B_1<0,B_1-B_2>B_1\}\\ &=\{B_1>0,B_2-B_1<0,|B_2-B_1|>|B_1|\} \end{align}

Answer 4

$P(B_1 > 0 \land B_2 < 0 ) = P(B_1 > 0)P(B_2<0 | B_1 > 0)$

$B_1-B_0 = B_1 \thicksim \mathcal N(0, 1)$

$B_2-B_1=N \thicksim \mathcal N(0, 1)$

$P(B_2<0 | B_1 > 0) = P(B_2-B_1<B_1 | B_1 > 0) = P(N<B_1 | B_1 > 0)$ because we need to jump from positive value to negative value so if in step 1 we got value $x>0$ now in step 2 we need to make jump by value smaller than $-x$ (negative and bigger in terms of absolute values)

$P(N<B_1 | B_1 > 0) = P(N<0)P(|N| > |B_1|) = P(B_2-B_1 < 0)P(|B_2-B_1| > |B_1|)$

*answer probably needs a bit of polishing