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The question is mainly in the title:

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

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

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

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3 Answers 3

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

(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}

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

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}

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

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

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

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

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    $\begingroup$ The initial formula follows immediately from the analysis and illustrations at stats.stackexchange.com/a/71303/919. The first quadrant corresponds to the sector $X^\prime \gt 0, Y^\prime \gt -\rho/\sqrt{1-\rho}^2$ where $(X^\prime,Y^\prime)$ has a standard Normal distribution. Because the latter is rotationally symmetric, the chance of this sector is simply its angle divided by $2\pi.$ For positive $\rho$ this sector consists of the entire first quadrant plus the wedge $X\gt 0, -\rho/\sqrt{1-\rho^2}\lt Y \le 0,$ whose angle (by elementary geometry) is $\sin^{-1}\rho.$ $\endgroup$
    – whuber
    Commented Oct 12, 2021 at 14:46
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    $\begingroup$ There are equivalent formulas, btw. One of them is that the expectation of $\operatorname{sgn}(XY)$ equals $2\arctan(\rho/\sqrt{1-\rho^2})/\pi.$ (Of interest is the appearance of the slope $\rho/\sqrt{1-\rho^2}$ of the boundary line, expressed as an angle.) I believe this form of the result has appeared elsewhere on CV. $\endgroup$
    – whuber
    Commented Oct 12, 2021 at 14:48
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    $\begingroup$ @whuber's first comment is also nicely illustrated in this answer on Math.SE. $\endgroup$ Commented Oct 12, 2021 at 15:21
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Consider $X,Y\sim N(0,1)$ with correlation $\rho$. Then (Nadarajah & Pogány, 2016; Gaunt, 2018) their product is variance-gamma distributed: $$ XY \sim \text{VG}(1,\rho,\sqrt{1-\rho^2},0). $$

Its PDF is $$ f_{XY}(x) = \frac{1}{\pi\sqrt{1-\rho^2}}\exp\left(\frac{\rho x}{1-\rho^2}\right) K_0\left(\frac{|x|}{1-\rho^2}\right), $$ where $K_0$ is the modified Bessel function of the second kind of order $0$.

Thus $$ P(\text{sign}(XY) = -1) = \int_{-\infty}^0 f_{XY}(x)\,dx. $$

You may be able to evaluate this using your favorite computer algebra system. Unfortunately, this exceeds the standard computation time for WolframAlpha. COOLSerdash notes that the integral evaluates to a nice round $\frac{\arccos\rho}{\pi}$.

Alternatively, you could map the above parameterization to the one employed in the VarianceGamma package for R and use the functions in there, if all you are interested in is numerical results.

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    $\begingroup$ The integral evaluates to $\arccos{(\rho)}/\pi$. $\endgroup$ Commented Oct 12, 2021 at 10:57
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    $\begingroup$ But you don't need the pdf of $XY$ to find this probability assuming $(X,Y)$ is bivariate normal. $\endgroup$ Commented Oct 12, 2021 at 11:16
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The correlation alone is not enough to be able to derive the probability distribution for the sign.

See the example below for a case where $\rho = 2/\pi$ and the sign of $XY$ is always positive.

example

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    $\begingroup$ Normality is given in the question. $\endgroup$
    – Martund
    Commented Oct 13, 2021 at 12:28
  • $\begingroup$ @Martund The variables $x$ and $y$ are following a standard normal distribution here. $\endgroup$ Commented Oct 13, 2021 at 12:29
  • $\begingroup$ It is impossible. You claim that product of two distinct standard normal random variables takes positive values with probability $1$???!!!!!!!!!!!!!!! $\endgroup$
    – Martund
    Commented Oct 13, 2021 at 12:32
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    $\begingroup$ @Martund I created the pairs of variables $X_i, Y_i$ from the answer in the following way. Let $X_i \sim N(0,1)$, let $Z_i \sim N(0,1)$ and let $Y_i = \text{abs}(Z_i) \times \text{sign}(X_i)$. Then $X_i$ and $Y_i$ each follow a standard normal distribution. Of course, they won't be independent and that is how you get that their product is always positive. $\endgroup$ Commented Oct 13, 2021 at 12:35
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    $\begingroup$ Martund, please see stats.stackexchange.com/a/30205/919. It will erase all the exclamation points in your comment. $\endgroup$
    – whuber
    Commented Oct 13, 2021 at 14:06

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