# Kendall's Tau of bivariate normal

Prove Kendall's tau of a bivariate normal is given by

$$\rho_\tau (X_1,X_2)=\frac{2}{\pi}\arcsin\rho$$

I can derive the bivariate normal as

$$F(x_1,x_2)=\frac{1}{2\pi\sqrt{1-\rho^2}}e^{-\frac{1}{2}x_1\Sigma^{-1}x_2}$$

where, for simplicity, I am assuming $$X_1$$ and $$X_2$$ have mean $$0$$. I also am aware that the formula for Kendall's tau involves an integral over some domain involving $$X_1$$, $$X_2$$, and $$\rho$$, multiplied by $$4$$.

So it seems like all the pieces are there; the integral should result in the answer, if only the integral over the exponential term w.r.t. $$X_1$$ and $$X_2$$ were $$1$$. But this is not the case, so I am stuck.

• See here Jan 1, 2019 at 21:16
• I saw this but, as @Avraham writes, access is not given to the necessary papers for a rigorous proof. Jan 1, 2019 at 21:29
• Anyway just stating a task is not appropriate for MSE. Therefore please include some more details with an edit. Jan 1, 2019 at 21:41
• At least write down the expression for Kendall's tau and indicate where you are stuck. Jan 2, 2019 at 14:35

Kendall's $$\tau$$ is defined as $$\tau(X_1,X_2)=\mathsf{P}((X_1-Y_1)(X_2-Y_2)>0)-\mathsf{P}((X_1-Y_1)(X_2-Y_2)<0),$$ where $$Y\equiv(Y_1,Y_2)$$ is an independent copy of $$X\equiv(X_1,X_2)$$.
Let $$Z:=X-Y$$ and note that $$Z\sim N(0,2\Sigma)$$, where $$\Sigma=\operatorname {Var}(X)$$. In addition, $$Z\overset{d}{=}\sqrt{2}(\Sigma_{11}(V_1\cos(\varphi)+V_2\sin(\varphi)),\Sigma_{22}V_2),$$ where $$\varphi\equiv \arcsin\rho$$ and $$V\equiv(V_1,V_2)\sim N(0,I_2)$$. Then, by symmetry, \begin{align} \tau(X_1,X_2)&=2\mathsf{P}(Z_1Z_2>0)-1=4\mathsf{P}(Z_1>0,Z_2>0)-1 \\ &=4\mathsf{P}(V_1\cos(\varphi)+V_2\sin(\varphi)>0,V_2>0)-1. \end{align}
It is known that $$V\overset{d}{=}R(\cos(\Phi),\sin(\Phi))$$, where $$\Phi\sim U[-\pi,\pi]$$ is independent of $$R=\|V\|_2$$. Therefore, \begin{align} \tau(X_1,X_2)&=4\mathsf{P}(\cos(\Phi)\cos(\varphi)+\sin(\Phi)\sin(\varphi)>0,\sin(\Phi)>0)-1 \\ &=4\mathsf{P}(\Phi\in (\varphi-\pi/2,\varphi+\pi/2)\cap[0,\pi])-1 \\ &=\frac{2}{\pi}(\varphi+\pi/2)-1=\frac{2}{\pi}\arcsin \rho. \end{align}
• @DaeHyun $$\mathsf{P}(Z_1Z_2>0)=\mathsf{P}(Z_1>0,Z_2>0)+\mathsf{P}(Z_1<0,Z_2<0)=2\mathsf{P}(Z_1>0,Z_2>0).$$ Mar 18 at 15:54