Let $$X=(X_1,X_2)$$ be a random vector in $$\mathbb{R}^2$$. Two realizations $$x$$ and $$x'$$ of $$X$$ form a concordant pair if $$(x_2'-x_2)$$ and $$(x_1'-x_1)$$ have the same sign. What’s the probability of sampling a concordant pair when $$X$$ is bivariate normal?

For example, suppose $$X_1$$ and $$X_2$$ have zero means, unit variances, and a correlation of $$\rho$$. The scatter plots below show 100 realizations of $$(X_1,X_2)$$ when $$\rho\in\{-0.5,0,0.5\}$$. These realizations contain $$\binom{100}{2}=4,\!950$$ pairs, of which 36% are concordant when $$\rho=-0.5$$. This percentage rises to 48% when $$\rho=0$$ and to 71% when $$\rho=0.5$$. Increasing $$\rho$$ makes concordance more likely because it makes $$(X_2-X_1)$$ larger and less noisy. Different samples give different concordance rates due to sampling variation. We can remove this variation by deriving the concordance rate analytically. To begin, suppose $$X$$ has mean $$\mathrm{E}[X]=(\mu_1,\mu_2)$$ and covariance matrix $$\mathrm{Var}(X)=\begin{bmatrix} \sigma_1^2 & \rho\sigma_1\sigma_2 \\ \rho\sigma_1\sigma_2 & \sigma_2^2 \end{bmatrix}.$$ Then $$X_2\mid X_1$$ is normal with mean $$\mathrm{E}[X_2\mid X_1]=\mu_2+\frac{\rho\sigma_2}{\sigma_1}(X_1-\mu_1)$$ and variance $$\mathrm{Var}(X_2\mid X_1)=(1-\rho^2)\sigma_2^2.$$ So for any two realizations $$x$$ and $$x'$$ of $$X$$ we can write $$\renewcommand{\epsilon}{\varepsilon} x'_2-x_2=\frac{\rho\sigma_2}{\sigma_1}\left(x'_1-x_1\right)+\epsilon$$ with $$\epsilon\sim N(0,2(1-\rho^2)\sigma_2^2)$$. Now $$x'_1-x_1\sim N(0,2\sigma_1^2)$$ is normal, and so $$z\equiv \frac{x'_1-x_1}{\sigma_1\sqrt{2}}$$ is standard normal and exceeds zero if and only if $$x'_1>x_1$$. Letting $$f$$ and $$\phi$$ be the density functions for $$\epsilon$$ and $$z$$ then gives \newcommand{\der}{\mathrm{d}} \begin{align} \Pr(x'_2>x_2\ \text{and}\ x'_1>x_1) &= \Pr(\sqrt{2}\rho\sigma_2 z+\epsilon>0\ \text{and}\ z>0) \\ &= \int_0^\infty\left(\int_{-\sqrt{2}\rho\sigma_2 z}^\infty f(\epsilon)\,\der \epsilon\right)\phi(z)\,\der z \\ &\overset{\star}{=} \int_0^\infty\left(\int_{\frac{-\rho z}{\sqrt{1-\rho^2}}}^\infty \phi(w)\,\der w\right)\phi(z)\,\der z \\ &= \int_0^\infty\left(1-\Phi\left(\frac{-\rho z}{\sqrt{1-\rho^2}}\right)\right)\phi(z)\,\der z \\ &\overset{\star\star}{=} \frac{1}{2}-\int_0^\infty\Phi\left(\frac{-\rho z}{\sqrt{1-\rho^2}}\right)\phi(z)\,\der z, \end{align} where $$\Phi$$ is the standard normal CDF, where $$\star$$ uses the change of variables $$w\equiv \frac{\epsilon}{\sigma_2\sqrt{2(1-\rho^2)}},$$ and where $$\star\star$$ uses the symmetry of $$\phi$$ about $$z=0$$. But $$f$$ is symmetric about $$\epsilon=0$$, which implies $$\Pr(x'_2>x_2\ \text{and}\ x'_1>x_1)=\Pr(x'_2<x_1\ \text{and}\ x'_1<x_1),$$ and therefore \begin{align} C(\rho) &\equiv \Pr(x\ \text{and}\ x'\ \text{are concordant}) \\ &= \Pr(x'_2>x_2\ \text{and}\ x'_1>x_1)+\Pr(x'_2<x_1\ \text{and}\ x'_1<x_1) \\ &= 1-2\int_0^\infty\Phi\left(\frac{-\rho z}{\sqrt{1-\rho^2}}\right)\phi(z)\,\der z. \end{align} The concordance rate $$C(\rho)$$ depends on the correlation $$\rho$$ of $$X_1$$ and $$X_2$$, but not their means or variances. It has value $$C(0)=0.5$$ when $$\rho=0$$ because $$\Phi(0)=0.5$$ is constant. Intuitively, if $$X_1$$ and $$X_2$$ are uncorrelated then we can’t use $$(x'_1-x_1)$$ to predict $$(x'_2-x_2)$$, which is equally likely to be positive or negative. Whereas if $$\lvert\rho\rvert=1$$ then $$(x'_1-x_1)$$ predicts $$(x'_2-x_2)$$ perfectly, and so $$\lim_{\rho\to1}C(\rho)=1$$ and $$\lim_{\rho\to-1}C(\rho)=0.$$ The chart below verifies that the concordance rate $$C(\rho)$$ grows with $$\rho$$. It also shows that $$C(\rho)+C(1-\rho)=1.$$ Thus, for example, we have $$C(-0.5)=1/3$$ and $$C(0.5)=2/3$$. These values remove the sampling error from the estimates 0.36 and 0.71 obtained using the 100 realizations above. 