I am in the middle of an exercise dealing with two random variables $(X,Y)$ having the bivariate normal density function:
$$f_{(X,Y)}(x,y)=\frac{1}{2\pi\sqrt{1-\rho^{2}}}\exp {\left[-\frac{1}{2(1-\rho^{2})}(x^{2}-2\rho xy + y^{2})\right]}$$
The exercise problem asks :
(1) Show that $X$ and $Z=(Y-\rho X)/\sqrt{1-\rho^{2}}$ are independent $N(0,1)$ variables, and
(2)Deduce that :
$$P(X>0,Y>0) =\frac{1}{4}+\frac{1}{2\pi}\arcsin\rho$$
I am not correctly formulate the solution to part (2) of the problem.
One clever step is to write down the event $\{X>0,Y>0\}$ in terms of $(W,Z)$, because they are independent gaussians, and easier to integrate.
$$\{X>0,Y>0\} = \{W>0,Z>-\frac{\rho W}{\sqrt{1-\rho^2}}\}$$
But, how to compute probabilities of such events of the type $\{X>a,Y>f(X)\}$.
For completeness, I reproduce my solution to part (1) of the problem.
My Solution.
Consider the map $T$ defined by:
\begin{align*} w & =x\\ z & =\frac{(y-\rho x)}{\sqrt{1-\rho^{2}}} \end{align*}
The inverse map $T^{-1}$ is given by:
\begin{align} x & =w\\ y & =\rho w+z\sqrt{1-\rho^{2}} \end{align}
The Jacobian $\frac{\partial(x,y)}{\partial(w,z)}$ is given by:
\begin{align} \frac{\partial(x,y)}{\partial(w,z)} & =\left|\begin{array}{cc} 1 & 0\\ \rho & \sqrt{1-\rho^{2}} \end{array}\right|\\ & =\sqrt{1-\rho^{2}} \end{align}
Hence, the joint density function of $(X,Z)$ is given by :
\begin{align} f_{(W,Z)} &=f_{(X,Y)}(w,\rho w+z\sqrt{1-\rho^{2}})\sqrt{1-\rho^{2}}\\ &=\frac{1}{2\pi}\exp\left( -\frac{1}{2(1-\rho^{2})}\left[w^{2}-2\rho w(\rho w+z\sqrt{1-\rho^{2}})+(\rho w+z\sqrt{1-\rho^{2}})^{2}\right]\right) \\ & =\frac{1}{2\pi}\exp\left( -\frac{1}{2(1-\rho^{2})}\left[w^{2}-2\rho^{2}w^{2}+\rho^{2}w^{2}+z^{2}(1-\rho^{2})\right]\right) \\ & =\frac{1}{2\pi}\exp\left( -\frac{1}{2(1-\rho^{2})}\left[w^{2}-\rho^{2}w^{2}+z^{2}(1-\rho^{2})\right]\right) \\ & =\frac{1}{2\pi}\exp\left( -\frac{1}{2}(w^{2}+z^{2})\right) \\ & =\frac{1}{\sqrt{2\pi}}\exp\left[-\frac{w^{2}}{2}\right]\cdot\frac{1}{\sqrt{2\pi}}\exp\left[-\frac{z^{2}}{2}\right] \end{align}
Consequently, $X$ and $Z$ are independent $N(0,1)$ random variables.
A common trick for bivariate independent standard normals $(X, Z)$ is to use polar transformation. This is because of the presence of the term $x^2 + z^2$ in the joint pdf which is "related to circles" so to speak.
So a natural approach is to use polar transformation $(x, z) \to (r, \theta)$ which gives the pdf of the random variables $R, \Theta$ with supports $\mathbb{R}_+, (-\pi, \pi)$: $$ f_{R, \Theta}(r, \theta) = \frac{r}{2\pi}e^{-r^2/2} \text{ for } r > 0, -\pi < \theta < \pi $$ and $0$ elsewhere.
Now in $\mathbb{R}^2$, the Cartesian region $\{x > 0, z > -(\rho/\sqrt{1 - \rho^2})x\}$ gets transformed to the polar region $\{r > 0, \arctan(-\rho/\sqrt{1 - \rho^2}) < \theta < \pi/2\}$ and notice that $\arctan(-\rho/\sqrt{1 - \rho^2}) = -\arcsin(\rho)$.
So \begin{align*} \int\int_{\{x > 0, z > -(\rho/\sqrt{1 - \rho^2})x\}} f_{X, Z}(x, z) \ dxdz &= \int\int_{\{r > 0, -\arcsin(\rho) < \theta < \pi/2\}} f_{R, \Theta}(r, \theta) \ drd\theta \\ &=\int_{-\arcsin(\rho)}^{\pi/2} \underbrace{ \int_0^\infty \frac{r}{2\pi}e^{-r^2/2} dr }_{\frac{1}{2\pi}} \ d\theta \\ &= \frac{1}{2\pi}\Big(\frac{\pi}{2} + \arcsin(\rho)\Big) = \frac{1}{4} + \frac{\arcsin(\rho)}{2\pi} \end{align*}