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


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.