Given
$$
f_{XY}(x,y) = \frac{1}{2\pi \sqrt{1-\rho^2}} \exp \left( -\frac{x^2 +y^2 – 2\rho xy}{2(1-\rho^2)} \right)
$$
$Y = Z\sqrt{1-\rho^2} + \rho X$
And
$$
f_{XZ}(x,z) = \frac{1}{2\pi } \exp \left( -\frac{x^2 +z^2}{2} \right)
$$
Show that $P(X>0,Y>0)= \frac{1}{4}+\frac{1}{2\pi}(\arcsin \rho) $
I'm supposed to use the fact that X and Z are independent standard normal random variables, but I don't quite understand how. Any help would be greatly appreciated.
Best Answer
Here's a solution that only uses linear algebra and geometry.
If $\pmatrix{X\\ Y}$ is bivariate normal with mean $\pmatrix{0\\0}$ and covariance matrix $\Sigma=\pmatrix{1&\rho\\\rho&1}$, then $\pmatrix{U\\V}=\Sigma^{-1/2} \pmatrix{X\\Y}$ is bivariate normal with mean $\pmatrix{0\\0}$ and covariance matrix $\pmatrix{1&0\\ 0&1}.$ That is, $U$ and $V$ are independent, standard normal random variables.
The illustration below shows that the probability that $\pmatrix{X\\Y}$ lies in the upper quadrant (in blue), is the same as the probability that $\pmatrix{U\\V}$ lies in the wedge (in orange). Since the distribution of $\pmatrix{U\\V}$ is rotationally invariant, simple geometry gives $\mathbb{P}(X>0,Y>0)={\theta\over 2\pi}.$
With $v=\Sigma^{-1/2}\pmatrix{0\\1}$ and $w=\Sigma^{-1/2}\pmatrix{1\\0},$ we have $\cos(\theta)=\langle v,w\rangle /\|v\| \|w\|.$ But \begin{eqnarray*} \langle v,w\rangle &=& (0\ 1)\,\Sigma^{-1}\pmatrix{1\\0}=-\rho/(1-\rho^2)\\[5pt] \|v\|^2&=&(0\ 1)\,\Sigma^{-1}\pmatrix{0\\1}=1/(1-\rho^2)\\[5pt] \|w\|^2&=&(1\ 0)\,\Sigma^{-1}\pmatrix{1\\0}=1/(1-\rho^2), \end{eqnarray*} so that $\cos(\theta)=-\rho.$ Putting it all together gives $$\mathbb{P}(X>0,Y>0)={\arccos(-\rho)\over 2\pi}.$$