# Sign of product of standard normal random variables

The question is mainly in the title:

Given two standard normal random variables with correlation $$\rho\rho$$, what is the distribution of sign of their product?

I understand that when $$\rho=0\rho=0$$, we have two iid standard normal random variables and therefore, they take positive and negative values independently with probability $$\frac12\frac12$$.

But I don’t know what to do if $$\rho\ne0\rho\ne0$$. We can take $$\rho>0\rho>0$$ without loss of generality (because if $$XX$$ and $$YY$$ are std normal with correlation $$\rho\rho$$, then $$-X-X$$ and $$YY$$ are std normal with correlation $$-\rho-\rho$$). But I could not proceed further.

As long as $$(X,Y)(X,Y)$$ is standard bivariate normal with correlation $$\rho\rho$$, the probability that $$XYXY$$ is positive or negative can be found using the well-known result for the positive quadrant probability $$P(X>0,Y>0)=\frac14+\frac1{2\pi}\sin^{-1}\rho \tag{1}P(X>0,Y>0)=\frac14+\frac1{2\pi}\sin^{-1}\rho \tag{1}$$

(This is likely discussed here before but I cannot quite find the question.)

You have

\begin{align} P(XY>0)&=P(X>0,Y>0)+P(X<0,Y<0) \\&=P(X>0,Y>0)+P(-X>0,-Y>0) \end{align}\begin{align} P(XY>0)&=P(X>0,Y>0)+P(X<0,Y<0) \\&=P(X>0,Y>0)+P(-X>0,-Y>0) \end{align}

Because $$(-X,-Y)(-X,-Y)$$ has the same distribution as $$(X,Y)(X,Y)$$, this probability is just $$P(XY>0)=2P(X>0,Y>0)P(XY>0)=2P(X>0,Y>0)$$

Similarly,

\begin{align} P(XY<0)&=P(X>0,Y<0)+P(X<0,Y>0) \\&=P(X>0,-Y>0)+P(-X>0,Y>0) \end{align}\begin{align} P(XY<0)&=P(X>0,Y<0)+P(X<0,Y>0) \\&=P(X>0,-Y>0)+P(-X>0,Y>0) \end{align}

Again, $$(X,-Y)(X,-Y)$$ and $$(-X,Y)(-X,Y)$$ have the same distribution, so

$$P(XY<0)=2P(X>0,-Y>0)P(XY<0)=2P(X>0,-Y>0)$$

And since $$(X,-Y)(X,-Y)$$ is bivariate normal with correlation $$-\rho-\rho$$, we have from $$(1)(1)$$ that

$$P(X>0,-Y>0)=\frac14-\frac1{2\pi}\sin^{-1}\rhoP(X>0,-Y>0)=\frac14-\frac1{2\pi}\sin^{-1}\rho$$