Finding the probability that $P \{W(2)> |W(1)|\}$, where $W(t)$ is the standard Wiener process. Brownian motion.

Question

I know that the increments of the Brownian motion are independent, and it looks like that might be of use in calculating this probability.

I thought that this might be the case:

$$P\{W(2)> |W(1)|\}= P\{W(2)>W(1);W(2)>-W(1) \} = \\ P\{W(2)-W(1)>0 , W(2)+W(1)>0\}$$

$$W(2)-W(1): N(0,1) -\text{normal distribution parameters 0,1}$$ $$W(2)+W(1): N(0,3) -\text{normal distribution parameters 0,3}$$

Then I have two independent variables, allowing me to multiply the density functions and integrating over the first quadrant with limits $0, +\infty$

Answer 1

Let $X=W(2)-W(1)$ and $Y=W(2)+W(1)$. We can see that the vector $(X,Y)$ is Gaussian hence its density can be computed. However, it seems that $X$ and $Y$ are correlated hence the expression of the density is unfortunately not the product of densities.