1
$\begingroup$

Is one of these statements correct for Gauss copula?

  • For normal marginals Pearson correlation equal to Spearman correlation
  • Pearson correlation is less or equal than Spearman correlation
$\endgroup$
  • $\begingroup$ for normal marginals Pearson correlation and Spearman correlation are almost equal exactly equal can not be said. Second statement is incorrect. $\endgroup$ – Prayalankar Feb 15 '17 at 14:07
2
$\begingroup$

A dependence measure is a parameter of a Gauss copula. Let denote by $\rho$ the Pearson correlation, $\rho_s$ the Spearman correlation, and $\tau_k$ the Kendell tau.

It's known $\rho \in [-1,1]$, and $$\rho_s = \frac{6}{\pi}\cdot arcsin(\frac{\rho}{2}) ,$$ $$\tau_k = \frac{2}{\pi}\cdot arcsin(\rho).$$

In the plot below one can see the dependencies between $\rho_s$, $\tau_k$ and $\rho$.

enter image description here

    rho <- seq(-1,1,0.01)     # Pearson correlation
    rho_s <- 6/pi*asin(rho/2) # Spearman correlation
    tau_k <- 2/pi*asin(rho)   # Kendell tau
    plot(rho, rho, type="n", xlab="rho", ylab="rho_s, tau_k")
    lines(rho, tau_k, col = "red")
    lines(rho, rho_s, col = "blue")
    legend("topleft", lty=c(1,1), col = c("red", "blue"), legend=c("tau_k", "rho_s"))
$\endgroup$
  • $\begingroup$ Put in text explaining.... $\endgroup$ – Carl Feb 22 '17 at 5:01

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.