Compute $P\{W_{1} > 0, W_{4} < 0\}$, where $W_{t}$ is a standard Brownian Motion. Options are 1/6, 1/4, 1/12, 1/3.
Bit stuck on this, here is my working out, not sure how to proceed.
$W_{1} \sim N(0,1)$ and $W_{4} \sim N(0,4)$. We also know $W_{4} - W_{1} \sim N(0,3)$ since $W_{t-s} \sim N(0,t-s)$.
$P\{W_{1} > 0, W_{4} < 0\} = P\{W_{1} > 0, (W_{4} - W_{1}) + W_{1} < 0\} = P\{W_{1} > 0, (W_{4} - W_{1}) < - W_{1}\}$.
Let $X=W_{1}$ and $Y = W_{4} - W_{1}$, where $X, Y$ are now normally distributed random variables. Now, I'm not sure how to proceed. Normally, I'd just find the overlapping region for the two equations $X>0$ and $Y<-X$, but that gives me 1/8, which isn't an option. I then realised that maybe I need to transform Y to be a standard normal random variable, and so the correct equations would be $X>0$ and $Y<\frac{-X}{\sqrt{3}}$, which I think gives an overlapping region of 1/6. Is that correct? Is there a more efficient way I can compute these? Would like to maybe utilise a double integral and the joint PDF, which I think looks like this:
$\int_{0}^{\infty} \int_{-\infty}^{-x} \frac{1}{2\pi \sqrt{3}} e^{\frac{-(3x^2 + y^2)}{6}} dydx$
but how do I then evaluate this double integral? Bit out of touch of my maths. Think we can use polar co-ordinates, but I don't understand similar solutions online, including how and why we use polar co-ordinates.
Set $U = W_1, V = (W_4-W_1)/\sqrt{3}$. Then we know that $U,V$ are independent standard Gaussians, and you're interested in $$ P(U > 0, U + \sqrt{3} V < 0).$$ Now, set $Z = \begin{pmatrix} U \\ V\end{pmatrix}$. Note that since $Z$ is standard Gaussian in $\mathbb{R}^2,$ it is isotropic, and its norm and angle are independent, i.e., in polar coordinates, $Z = (R,\Theta)$ for $R\perp \Theta$, and $\Theta$ distributed uniformly on $[-\pi,\pi)$ [prove this fact, if you're not familiar with it]. But note that $R > 0$ a.s., so $$ P(U > 0, U +\sqrt{3}V < 0) = P(\cos\Theta > 0, \cos\Theta + \sqrt{3}\sin \Theta < 0)\\ = P(\cos\Theta > 0, \sin(\Theta + \pi/6) < 0),$$ which should be straightforward to work out.