# Getting $P(X>0,Y>0)$ for a bivariate distribution

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

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

Using this I had to show that

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

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As a hint for the first question, notice that $Z$ is the conditional of $Y$ on $X$. – Learner Dec 10 '12 at 11:03
it is easy to solve part. As X and Y are standard normal variables. But I am stuck at part two. – user669083 Dec 10 '12 at 11:10
If it is easy for part 1, why are you asking the question? – Learner Dec 10 '12 at 11:10
Because the second part is to use first part to solve second. – user669083 Dec 10 '12 at 11:12
@user669083 Learner is right. If you want an answer for part II, then you shouldnt say: "I had to show..." You need to say "using the fact that .....", I need to show : "part II" – Seyhmus Güngören Dec 10 '12 at 12:23

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 $$P \left[ X > 0, Y > 0 \right] = \frac{1}{2 \pi} \arcsin \rho + \frac{1}{4}$$

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is there a simpler solution ? – user669083 Dec 10 '12 at 17:58
@user669083 If you do not understand the above proof, you should point to the part where you have problems. Anyway, I am going to add an alternative proof (I am not sure it is simpler). – Learner Dec 11 '12 at 1:08
Let's call the above integral h(ρ), then after some simplifications. What is your simplification? Thanks in advance. – user52613 Dec 11 '12 at 9:38
@Fafsads Please check en.wikipedia.org/wiki/Leibniz_integral_rule for how to differentiate with respect to $\rho$. After that, all is needed is some simple manipulations of one-dimensional normal density integrals. Let me know after that if you are still having trouble. – Learner Dec 11 '12 at 10:34
I solved it by writing the equation as $(PX>0,Z> {\frac{-\rho X}{\sqrt{(1-\rho^2)}}})$. After that write in terms of integral, convert into polar co-ordinates and voila I got the answer – user669083 Dec 16 '12 at 9:52