BiCopPar2Tau(family, par, par2=0)0 = independence copula
1 = Gaussian copula
2 = Student t copula (t-copula)
3 = Clayton copula
4 = Gumbel copula
5 = Frank par2 = 0).
Note that the degrees of freedom parameter of the t-copula does not need to be set,
because the theoretical Kendall's tau value of the t-copul1, 2 $\frac{2}{\pi}\arcsin(\theta)$
3, 13 $\frac{\theta}{\theta+2}$
4, 14 $1-\frac{1}{\theta}$
5 $1-\frac{4}{\theta}+4\frac{D_1(\theta)}{\theta}$
with $D_1(\theta)=\int_0^\theta \frac{x/\theta}{\exp(x)-1}dx$ (Debye function)
6, 16 $1+\frac{4}{\theta^2}\int_0^1 x\log(x)(1-x)^{2(1-\theta)/\theta}dx$
7, 17 $1-\frac{2}{\delta(\theta+2)}$
8, 18 $1+4\int_0^1 -\log(-(1-t)^\theta+1)(1-t-(1-t)^{-\theta}+(1-t)^{-\theta}t)/(\delta\theta) dt$
9, 19 $1+4\int_0^1 ( (1-(1-t)^{\theta})^{-\delta} - )/( -\theta\delta(1-t)^{\theta-1}(1-(1-t)^{\theta})^{-\delta-1} ) dt$
10, 20 $1+4\int_0^1 -\log \left( ((1-t\delta)^\theta-1)/((1-\delta)^\theta-1) \right)$
$* (1-t\delta-(1-t\delta)^{-\theta}+(1-t\delta)^{-\theta}t\delta)/(\theta\delta) dt$
23, 33 $\frac{\theta}{2-\theta}$
24, 34 $-1-\frac{1}{\theta}$
26, 36 $-1-\frac{4}{\theta^2}\int_0^1 x\log(x)(1-x)^{-2(1+\theta)/\theta}dx$
27, 37 $-1-\frac{2}{\delta(2-\theta)}$
28, 38 $-1-4\int_0^1 -\log(-(1-t)^{-\theta}+1)(1-t-(1-t)^{\theta}+(1-t)^{\theta}t)/(\delta\theta) dt$
29, 39 $-1-4\int_0^1 ( (1-(1-t)^{-\theta})^{\delta} - )/( -\theta\delta(1-t)^{-\theta-1}(1-(1-t)^{-\theta})^{\delta-1} ) dt$
30, 40 $-1-4\int_0^1 -\log \left( ((1+t\delta)^{-\theta}-1)/((1+\delta)^{-\theta}-1) \right)$
$* (1+t\delta-(1+t\delta)^{\theta}-(1+t\delta)^{\theta}t\delta)/(\theta\delta) dt$
}CDVinePar2Tau, BiCopTau2Par## Example 1: Gaussian copula
tt1 = BiCopPar2Tau(1,0.7)
# transform back
BiCopTau2Par(1,tt1)
## Example 2: Clayton copula
BiCopPar2Tau(3,1.3)Run the code above in your browser using DataLab