standard Brownian motion. Calculate the probability that W(3) > W(2) > W(1).

Question

Assume that W(t) is the standard Brownian motion. Calculate the probability that W(3) > W(2) > W(1).

Hi I am really bad with BM so can anyone please help me here?

Answer 1

Equivalently you want to calculate $W(3)-W(2)>0$ and $W(2)-W(1)>0$.

Recall that $W(3)-W(2)\sim N(0,1)$ and $W(2)-W(1) \sim N(0,1)$ are independent.

Therefore, $$\mathbb{P}(W(3)>W(2)>W(1)) =\mathbb{P}(W(3)-W(2)>0)\cdot\mathbb{P}(W(2)-W(1)>0)=\frac{1}{4}$$