Csdrhrt

$X$ and $Y$ have a bivariate normal distribution with $rho$ as covariance. $X$ and $Y$ are standard normal variables.

I showed that $X$ and $Z= {frac{Y-rho X}{sqrt{(1-rho^2)}}}$ are independent standard normal variables.

Using this I need to show that

$$P(X >0,Y>0) = frac14 + frac{1}{2pi} cdotmathrm{arcsin}(rho).$$

Let $left( V, W right)$ be i.i.d standard normal, let $U$ be uniform on
$left[ - pi, pi right)$ independent of $R = sqrt{V^2 + W^2}$ and let
$varphi = arcsin rho$. Then the following vectors have the same
distributions (think about the Box-Muller transformation)
begin{eqnarray*}
left( X, Y right) & overset{d}{=} & sqrt{2} left( V cos varphi + W
sin varphi, W right)\
left( V, W right) & overset{d}{=} & R left( cos U, sin U right)
end{eqnarray*}
Implying
begin{eqnarray*}
P left[ X > 0, Y > 0 right] & = & P left[ cos U cos varphi + sin U
sin varphi > 0, sin U > 0 right]\
& = & P left[ U in left( varphi - frac{pi}{2}, varphi +
frac{pi}{2} right) cap left( 0, pi right) right]\
& = & frac{varphi}{2 pi} + frac{1}{4}
end{eqnarray*}

Here is an alternative proof:

Let $phi$ be the density of the standard normal distribution
begin{eqnarray*}
P left[ X > 0, Y > 0 right] & = & P left[ X < 0, Y < 0 right]\
& = & int_{- infty}^0 phi left( z right) int_{- infty}^0
frac{1}{sqrt{1 - rho^2}} phi left( frac{x - rho z}{sqrt{1 - rho^2}}
right) mathrm{d} x mathrm{d} z\
& = & int_{- infty}^0 phi left( z right) int_{- infty}^{- frac{rho
z}{sqrt{1 - rho^2}}} phi left( x right) mathrm{d} x mathrm{d} z
end{eqnarray*}
Let's call the above integral $h left( rho right)$, then after some
simplifications
$$frac{partial h left( rho right)}{partial rho} = frac{1}{2 pi
sqrt{1 - rho^2}} $$
By integrating back (or considering the problem as a first-order ordinary differential equation), $h left( rho right) = frac{1}{2 pi} arcsin rho +
K$ where $K$ is some constant. By the special case of independence, $h left( 0 right) =
frac{1}{4}$, you get the final solution
$$ P left[ X > 0, Y > 0 right] = frac{1}{2 pi} arcsin rho + frac{1}{4}$$

Using the OP's hint: The event ${X>0,Y>0}$ is the same as the event ${X>0,Z>frac{-rho}{sqrt{1-rho^2}}X}$, where now $X$ and $Z$ are independent standard normal variables. Writing $a:=frac{-rho}{sqrt{1-rho^2}}$ for brevity, the desired probability is expressible as a double integral involving the joint density of $(X,Z)$:
$$
P(X>0,Y>0)=P(X>0,Z>aX)=int_{x=0}^inftyint_{z=ax}^inftyfrac1{sqrt{2pi}}e^{-x^2/2}frac1{sqrt{2pi}}e^{-z^2/2},dz,dx.
$$
Switching to polar coordinates, this equals
$$
begin{align}
int_{theta=arctan(a)}^{pi/2}int_{r=0}^inftyfrac1{2pi}e^{-r^2/2},rdrdtheta&=int_{theta=arctan(a)}^{pi/2}frac1{2pi},dtheta\
&=frac1{2pi}left(fracpi2-arctan aright)\
&=frac14+frac1{2pi}arctanfracrho{sqrt{1-rho^2}};
end{align}
$$
for the last equality we substitute $a:=frac{-rho}{sqrt{1-rho^2}}$ and use the fact that the arctan function is odd. To finish off, remember that if $thetain[-pi/2,pi/2]$ then $$theta=arctanfrac{rho}{sqrt{1-rho^2}} Longleftrightarrow tantheta=frac{rho}{sqrt{1-rho^2}} Longleftrightarrow sintheta=rho Longleftrightarrow theta=arcsin rho.$$