Is Kendall's tau uniquely determined by Pearson rho?

Question

I know that two data sets can have the same Kendall's $\tau$ but different Pearson's $\rho$.

1. What about the opposite? Can two different data sets have the same Pearson's $\rho$, but different Kendall's $\tau$? Or rather, if we know the correlation matrix Pearson's $\rho_{ij}$, does it correspond to precisely one matrix of Kendall's $\tau_{ij}$, obtained by transforming the original matrix to Kendall's $\tau$ by $$\frac{2}{\pi}\arcsin{(\rho_{ij})}$$
2. If we sample from a multivariate distribution with a certain population correlation matrix $\rho_{ij}$, is this equivalent to sampling from a distribution with a population matrix $\tau_{ij}$ obtained by the previously mentioned transformation? I.e. do the samples, which have a sample Pearson correlation matrix $\Sigma$ and associated Kendall's correlation matrix, come from the population with this $\tau_{ij}$ obtained by transforming population $\rho_{ij}$?

Answer 1

Can two different data sets have the same Pearson's ρ, but different Kendall's τ?

Anscombe's quartet gives you four two-dimensional datasets with (almost) identical Pearson correlations, but their Kendall correlations are quite different. In R:

> with(anscombe,cor(x1,y1,method="pearson"))
[1] 0.8164205
> with(anscombe,cor(x2,y2,method="pearson"))
[1] 0.8162365
> with(anscombe,cor(x1,y1,method="kendall"))
[1] 0.6363636
> with(anscombe,cor(x2,y2,method="kendall"))
[1] 0.5636364

In general, there is no bijective transformation between the two correlations. You can also see this by the fact that small changes in data points will always change the Pearson correlation continuously - but the Kendall correlation will not change at first and suddenly change when data points change position, because the Kendall correlation only looks at whether two points are to the left or the right, respectively above or below each other.

Answer 2

An indirect approach:

If two $\rho_1 > \rho_2$ can relate to the same $\tau$, then imagine adjusting the dataset a little such that we increase all these three values.

We should be able to arrive to some values $\rho^\prime_1 > \rho_1$, $\rho^\prime_2 > \rho_2$ and $\tau^\prime > \tau$. This should be possible in a continuous way such that we get at some point $\rho^\prime_2 = \rho_1$.

That means that we have got the same Pearson's $\rho$, but with different $\tau^\prime$ and $\tau$.