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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.

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  • $\begingroup$ See here $\endgroup$ – mrtaurho Jan 1 '19 at 21:16
  • $\begingroup$ I saw this but, as @Avraham writes, access is not given to the necessary papers for a rigorous proof. $\endgroup$ – E Werner Jan 1 '19 at 21:29
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    $\begingroup$ Anyway just stating a task is not appropriate for MSE. Therefore please include some more details with an edit. $\endgroup$ – mrtaurho Jan 1 '19 at 21:41
  • $\begingroup$ At least write down the expression for Kendall's tau and indicate where you are stuck. $\endgroup$ – StubbornAtom Jan 2 '19 at 14:35
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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, $$ Z\overset{d}{\sim}\sqrt{2}(\Sigma_{11}(V_1\cos(\varphi)+V_2\sin(\varphi)),\Sigma_{22}V_2), $$ 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 $\Phi\sim U[-\pi,\pi]$ 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}

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