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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.

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$$\mathbb{P}(0<W(1)+W(2)<2, 3W(1)-2W(2)>0)$$

$$=\mathbb{P}(0<2W(1)+\Delta W(2)<2, W(1)-2\Delta W(2)>0)$$

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.

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