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.
