# Can you use Kendall's Tau to compute covariance matrix?

I am working with multivariate archimedean copulas, and I am wondering how I can extract a covariance matrix out of them? I can get Kendall's Tau matrix of correlation so I was thinking that maybe I can swap the usual correlation matrix in the covariance calculation, as follows:

$$cov(X,Y) = \tau * \sigma_X * \sigma_Y$$ instead of $$cov(X,Y) = \rho * \sigma_X * \sigma_Y$$

Anyone knows if this would be mathematically correct? If not, is there a way to compute a covariance matrix from Copulas? I am using the HACopula package from J. Górecki on Matlab to fit the multivariate Clayton and Gumbel.

In general this will not be a suitable estimate of the covariance matrix. Indeed the copula alone cannot be used to get a reasonable estimate of covariance, since the copula is invariant to monotonic increasing transformation of the margins, but the covariance is certainly not.

However, in some particular situations it may be possible to get a good estimate of a covariance matrix.

For example, consider he situation where $$(X,Y)$$ is bivariate Gaussian but where the only information you have on the strength of the bivariate relationship is the value of the Kendall correlation.

Even in this case, it's not going to be suitable to use the Kendall correlation directly. Here's the result for 1000 simulated samples across the range of possible correlation values, each with sample size 100:

We can see that the sample Kendall correlation is closer to 0 than the population Pearson correlation in general, and the difference is fairly substantial when $$|\rho|$$ is between about 0.5 and 0.975 or so. The relationship is fairly linear near 0 but nonlinear further away.

You could use the relationship between the population Kendall and Pearson correlations for the bivariate normal case ($$\tau=\frac{2}{\pi}\arcsin({\rho})$$) to estimate the Pearson correlation from the sample Kendall correlation - and from there, obtain an estimate of the covariance.

Here's a simulation to show this approach in action. In this case I just used the naive estimator $$\hat{\rho}=\sin(\frac{\pi}{2}\hat{\tau})$$, and samples of size 100:

As we see, the naive estimator of $$\rho$$ based off a transformed sample Kendall correlation is excellent (bottom left); its variance is only a little larger than using the usual sample Pearson correlation (top left). Indeed the transformed sample Kendall correlation is very close to the corresponding value of the sample Pearson correlation (bottom right). There may well be a better estimator than the one I used, but this indicates that an approach based on the sample Kendall correlation may well be feasible.

If you do this at other typical sample sizes the spread around the relationships changes (the variance of the sample correlations will be proportional to $$\frac{1}{n}$$) but the basic pattern of results is similar.

The relationship between the Kendall and Pearson correlations will depend on the bivariate distribution (not just the copula) so I think you'll need a new analysis for each such case.

• Thank you for the detailed answer. I see that it indeed, Kendall's Tau is not a suitable replacement for a correlation matrix. Since I work with multivariate archimedean copulas, the relation you gave between Kendall's Tau and Spearman's Rho is not applicable for me. I am trying to find ressources about a relationship between Kendall's Tau and Spearman's Rho for non-normal multivariate copulas. – scp_34 Aug 16 at 14:22
• To clarify: nothing anywhere in my answer refers to Spearman's correlation coefficient; instead $\rho$ refers to the population value of the Pearson correlation throughout. That's the relevant correlation measure when discussing population covariance, which is what you're trying to estimate. In particular, when you write $\text{cov}(X,Y) = \rho \, \sigma_X\, \sigma_Y$, that $\rho$ is the population Pearson correlation. A relationship between Spearman and Kendall is (in general) useless for calculating covariance, because they both only depend on the copula but covariance doesn't. – Glen_b Aug 16 at 23:27
• +1 @Glen_b your answer is amazing. I learn new things. – Mary Aug 17 at 14:36