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.