# Kendall's tau coefficent of Bivariate Normal [duplicate]

Let the joint distribution of (πΏ, π) be bivariate normal with mean vector $$\begin{pmatrix} 0 \\ 0\end{pmatrix}$$ and variance-covariance matrix $$\begin{pmatrix} 1 & π \\ π & 1 \end{pmatrix}$$, where $$βπ < π < π$$. Let $$π½_{π}(π, π) = π·(πΏ β€ π, π β€ π)$$. Then what will be Kendallβs π coefficient between πΏ and π equal to?

Since I am new to statistics I have no idea where to start?

• Kendall's $\tau$ is a measure of correlation that is non-parametric. It can be applied to any observed data $(X_1, Y_1), \ldots, (X_n, Y_n).$ Mar 14, 2022 at 18:29
• Mar 14, 2022 at 18:52

Let $$\Phi_\rho(x,y),\Phi(x),\Phi(y)$$ be the bivariate and the univariate CDFs of the standard normal distribution. Then the Gaussian Copula is defined as $$C_\rho(x,y)=\Phi_\rho(\Phi^{-1}(x),\Phi^{-1}(y))$$ Kendall's tau is then defined as \begin{align} \rho_\tau&=\mathbb E\Big[{\rm sign}[(X-\tilde{X})(Y-\tilde{Y})]\Big]\\ &=\mathbb P\Big[(X-\tilde{X})(Y-\tilde{Y})>0\Big]-P\Big[(X-\tilde{X})(Y-\tilde{Y})<0\Big]\,. \end{align} where $$(X,Y)$$ is bivariate standard normal, and $$(\tilde{X},\tilde{Y})$$ has the same distribution but is independent of $$(X,Y).$$ It can be shown (see [1] and duplicate) that $$\rho_\tau=4\int_0^1\int_0^1C_\rho(x,y)\,dC_\rho(x,y)-1=\frac{2}{\pi}\arcsin\rho\,.$$