2
$\begingroup$

Considering that θ in Frank copula is a function of the Kendall's tau as follows:

enter image description here

I would like to know the range of Kendall's tau that can be used in the Frank copula.

Thank you in advance for any helps.

$\endgroup$
  • $\begingroup$ Could any close voter please indicate what's unclear about this? $\endgroup$ – Glen_b -Reinstate Monica Nov 12 at 22:28
4
$\begingroup$

$\tau = 1+\frac{4}{\theta}[D_1(\theta)-1]$ with $D_1(\theta)= \frac{1}{\theta}\int_0^\theta \frac{t}{e^t-1} dt$,

where is the $D_n(x)$ is the Debye function. This is set as an exercise in Nelsen's book.

As far as I can see, the possible range of population $\tau$ values appears to be $(-1,1)\,\text{\\}\,\{0\}$. That is, between $-1$ and $1$, but not including the endpoints; you get as close to $1$ or $-1$ as you like by choosing sufficiently large parameter values with sign the same as $\tau$, and (strictly speaking) omitting $0$ (though again, you can get as close as you like by choosing the parameter close to $0$; however the independence copula is normally regarded as included).

$\endgroup$
  • $\begingroup$ Thank you for the response. I would like to know the range of τ as θ ∈ R/0. Can we determine a range for τ explicitly? $\endgroup$ – Saba Ghotbi Oct 30 at 1:32
  • $\begingroup$ Sorry, I made an edit. $\endgroup$ – Glen_b -Reinstate Monica Nov 1 at 9:42
1
$\begingroup$

The table at the bottom of this copula page gives Kendall's tau as a function of $\theta$. In general,

... Joe, H. (2014). Dependence Modeling with Copulas. Chapman & Hall / CRC.

contains many formulas and other results on specific copula families.

$\endgroup$
  • $\begingroup$ Thank you for the response. I would like to know the range of τ as θ ∈ R/0. Can we determine a range for τ explicitly? $\endgroup$ – Saba Ghotbi Oct 30 at 1:32

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.