# Correlation of signs of a jointly Gaussian RV

Given $$\left(\begin{array}{c} X_1 \\ X_2 \end{array}\right) \sim \mathcal{N} \left(\left(\begin{array}{c} 0 \\ 0 \end{array}\right), \left(\begin{array}{cc} 1 & \rho \\ \rho & 1 \end{array}\right)\right),$$

I want to show that $$\mathbb{E}\left[ \text{sign}(X_1) \text{sign}(X_2)\right] = \frac{2}{\pi}\sin^{-1}(\rho)$$

This seems to be the essence of the proof of Geomans-Williamson MAX-CUT SDP relaxation. Is there an easy way to see it?

For convenience let's call $\operatorname{sgn}(X_1),\operatorname{sgn}(X_2)$ as $S_1$ and $S_2$, respectively.

There are only $9$ possible combinations of $(S_1,S_2)$: $(\pm1,\pm1)$, and at least one of the $S$ being $0$. Now, since we are looking for $E[S_1S_2]$, ignoring the states of $S=0$ will not affect the result. Hence, \begin{align*}E[S_1S_2]&=1\times P(S_1=1,S_2=1)+1\times P(S_1=-1,S_2=-1)\\ &\quad+(-1)\times P(S_1=1,S_2=-1)+(-1)\times P(S_1=-1,S_2=1)\\ &=1\times P(X_1>0,X_2>0)+1\times P(X_1<0,X_2<0)\\ &\quad+(-1)\times P(X_1>0,X_2<0)+(-1)\times P(X_1<0,X_2>0).\end{align*}

Further, one can show that $$P(X_1>0,X_2>0)=P(X_1<0,X_2<0)=\frac{1}{4}+\frac{1}{2\pi}\sin^{-1}(\rho),$$ and $$P(X_1>0,X_2<0)=P(X_1<0,X_2>0)=\frac{1}{2\pi}\cos^{-1}(\rho).$$ So \begin{align*}E[S_1S_2]&=\frac{1}{2}+\frac{1}{\pi}\sin^{-1}(\rho)-\frac{1}{\pi}\left(\frac{\pi}{2}-\sin^{-1}(\rho)\right)\\ &=\frac{2}{\pi}\sin^{-1}(\rho).\end{align*}

• Is there an easy way to see the integral? That was where I was stuck before as well. All references I see seem to do it the messy way. – Devil Aug 3 '16 at 18:06
• @Devil: it's possible to turn integrations into trigonometric inequalities with the help of Box-Muller transformation, this thread may help. – Francis Aug 3 '16 at 20:47

$$\mathbb{E}[ \text{sign}(X_1) \text{sign}(X_2)] = 1 * (P(X_1 \ge 0,X_2 \ge 0) + P(X_1 \le 0,X_2 \le 0)) - (P(X_1 \ge 0,X_2 \le 0) + P(X_1 \le 0,X_2 \ge 0))$$ which in turn $$= 2P(X_1 \ge 0,X_2 \ge 0) - 2P(X_1 \ge 0,X_2 \le 0)$$ by symmetry.

Plugging in the Bivariate Normal density, this evaluates (integrates) to $\frac{2}{\pi} sin^{-1}(\rho)$. The details of performing the integration are left to you.

Edit: Have changed what was $\sigma^2$ to $\rho$ to match edit of question.

• +1 for that last comment since the implications of writing $\sigma^2$ are that the parameter cannot be negative and that it can exceed $1$. – Dilip Sarwate Aug 3 '16 at 15:39
• Ah. I did not realise that. Have fixed it. – Devil Aug 5 '16 at 3:05