Correlation and concordance

Let $X=(X_1,X_2)$ be a random vector in $R^2$. Two realizations $x$ and $x^{\prime }$ of $X$ form a concordant pair if $(x_2^{\prime }-x_2)$ and $(x_1^{\prime }-x_1)$ 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 $$(\frac{100}{2})=4,950$$ 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 $E\left[X\right]=({\mu }_1,{\mu }_2)$ and covariance matrix $$Var\left(X\right)=\left[\begin{array}{cc}{\sigma }_1^2& \rho {\sigma }_1{\sigma }_2\\ \rho {\sigma }_1{\sigma }_2& {\sigma }_2^2\end{array}\right].$$ Then $X_2\mid X_1$ is normal with mean $$E[X_2\mid X_1]={\mu }_2+\frac{\rho {\sigma }_2}{{\sigma }_1}(X_1-{\mu }_1)$$ and variance $$Var(X_2\mid X_1)=(1-{\rho }^2){\sigma }_2^2.$$ So for any two realizations $x$ and $x^{\prime }$ of $X$ we can write $$x_2^{\prime }-x_2=\frac{\rho {\sigma }_2}{{\sigma }_1}(x_1^{\prime }-x_1)+\epsilon$$ with $\epsilon \sim N(0,2(1-{\rho }^2\left){\sigma }_2^2\right)$. Now $x_1^{\prime }-x_1\sim N(0,2{\sigma }_1^2)$ is normal, and so $$z\equiv \frac{x_1^{\prime }-x_1}{{\sigma }_1\sqrt{2}}$$ is standard normal and exceeds zero if and only if $x_1^{\prime }>x_1$. Letting $f$ and $\varphi$ be the density functions for $\epsilon$ and $z$ then gives $$\begin{array}{cc}Pr(x_2^{\prime }>x_2\text{and}{x}_1^{\prime }>x_1)& =Pr(\sqrt{2}\rho {\sigma }_{2}z+\epsilon >0\text{and}z>0)\\ & ={\int }_0^{\infty }\left({\int }_{-\sqrt{2}\rho {\sigma }_{2}z}^{\infty }f\right(\epsilon \left)d\epsilon \right)\varphi \left(z\right)dz\\ & \stackrel{\star }{=}{\int }_0^{\infty }\left({\int }_{\frac{-\rho z}{\sqrt{1-{\rho }^2}}}^{\infty }\varphi \right(w\left)dw\right)\varphi \left(z\right)dz\\ & ={\int }_0^{\infty }(1-\Phi \left(\frac{-\rho z}{\sqrt{1-{\rho }^2}}\right))\varphi \left(z\right)dz\\ & \stackrel{\star \star }{=}\frac{1}{2}-{\int }_0^{\infty }\Phi \left(\frac{-\rho z}{\sqrt{1-{\rho }^2}}\right)\varphi \left(z\right)dz,\end{array}$$ 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 $\varphi$ about $z=0$. But $f$ is symmetric about $\epsilon =0$, which implies $$Pr(x_2^{\prime }>x_2\text{and}{x}_1^{\prime }>x_1)=Pr(x_2^{\prime }<x_1\text{and}{x}_1^{\prime }<x_1),$$ and therefore $$\begin{array}{cc}C\left(\rho \right)& \equiv Pr\left(x\text{and}{x}^{\prime }\text{are concordant}\right)\\ & =Pr(x_2^{\prime }>x_2\text{and}{x}_1^{\prime }>x_1)+Pr(x_2^{\prime }<x_1\text{and}{x}_1^{\prime }<x_1)\\ & =1-2{\int }_0^{\infty }\Phi \left(\frac{-\rho z}{\sqrt{1-{\rho }^2}}\right)\varphi \left(z\right)dz.\end{array}$$ The concordance rate $C\left(\rho \right)$ depends on the correlation $\rho$ of $X_1$ and $X_2$, but not their means or variances. It has value $C\left(0\right)=0.5$ when $\rho =0$ because $\Phi \left(0\right)=0.5$ is constant. Intuitively, if $X_1$ and $X_2$ are uncorrelated then we can’t use $(x_1^{\prime }-x_1)$ to predict $(x_2^{\prime }-x_2)$, which is equally likely to be positive or negative. Whereas if $\left|\rho \right|=1$ then $(x_1^{\prime }-x_1)$ predicts $(x_2^{\prime }-x_2)$ perfectly, and so $$\underset{\rho \to 1}{lim}C\left(\rho \right)=1$$ and $$\underset{\rho \to -1}{lim}C\left(\rho \right)=0.$$ The chart below verifies that the concordance rate $C\left(\rho \right)$ grows with $\rho$. It also shows that $$C\left(\rho \right)+C(1-\rho )=1.$$ Thus, for example, we have $C(-0.5)=1/3$ and $C\left(0.5\right)=2/3$. These values remove the sampling error from the estimates 0.36 and 0.71 obtained using the 100 realizations above.