$X$ and $Y$ have a joint normal distribution with unity variance, zero mean, and correlation = $0.5$ -- what is $P(X > 2Y | X > 0)$?

Question

I know how to find $P(X > 2Y | X > 0)$ for the case where $X$ and $Y$ are independent. I would use a graphical approach. It would come out to be $\frac{\frac{\pi}{2} - \arctan(0.5)}{\pi}$.

With a non-zero correlation, the bivariate distribution is no longer radially symmetric, and I believe for a correlation of $0.5$, it would stretch out the distribution along the $y = x$ line. How can I apply the graphical approach here?

Answer 1

I will assume that the joint mean is $(\mu_X, \mu_Y) = (0,0)$, otherwise the question becomes much more difficult. Then let

$$X = U, \quad Y = \frac{U + \sqrt{3} V}{2},$$

or equivalently

$$U = X, \quad V = \frac{2Y - X}{\sqrt{3}}.$$

Under this transformation, $(U, V)$ will be bivariate standard normal. Then the desired probability becomes

$$\Pr[X > 2Y \mid X > 0] = \Pr[\sqrt{3} V < 0 \mid U > 0] = \frac{1}{2}$$ trivially. The main computation is showing that $(U,V)$ has identity covariance matrix.