While studying for my exam I came across the following question:
Let $W(t)$ be a standard Brownian motion. Find
$$\mathbb{P}(0<W(1)+W(2)<2, 3W(1)-2W(2)>0)$$
Attempt:
(1) Attempt 1: Finding the distribution of W(1)+W(2) which I found to be $\mathcal{N}(0,5)$ and same for $3W(1)-2W(2) $ which is $\mathcal{N}(0,5)$ my attempt is to prove that the two things are independent so I could separate both terms of the probability and multiply two quantities as
$$\mathbb{P}(0<W(1)+W(2)<2). \mathbb{P}(3W(1)-2W(2)>0)$$
but this didn't work out yet.
(2) Attempt 2: Working with the inequalities to find common bounds i.e I found $$\dfrac{2}{3} Y <X <2-Y$$ but this isn't so useful.
$$\mathbb{P}(0<W(1)+W(2)<2, 3W(1)-2W(2)>0)$$
$$=\mathbb{P}(0<2W(1)+\Delta W(2)<2, W(1)-2\Delta W(2)>0)$$
where $\Delta W(2)=W(2)-W(1)$. Given the properties of a Wiener process $[W(1),\Delta W(2)]$ is jointly normal with mean (0, 0) and covariance $diag(1, 1)$.
Now
$$\mathbb{P}(0<W(1)+W(2)<2, 3W(1)-2W(2)>0)$$
$$=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}I(0<2x+y<2, x-2y>0)f(x)f(y)dxdy.$$
Where $f$ is the pdf of a standard normal variable. This can be evaluated numerically.