Kendall's Tau and Copula Parameter Conversion

Why Convert Kendall’s Tau?

Kendall’s tau summarizes dependence by comparing pairs of observations: do larger values of one variable tend to occur with larger values of the other? Copula parameters, by contrast, describe dependence within a specified model. A tau-to-parameter relationship lets an analyst translate a rank-association target into a candidate parameter for a Gaussian, Student-t, Clayton, Gumbel, or Frank copula.

The conversion is model-specific. The same value of tau generally corresponds to different parameter values in different copula families. Tau is also not Pearson correlation: for Gaussian and Student-t copulas the conversion is between tau and a latent correlation parameter, not necessarily the Pearson correlation of the modeled losses.

Definition: For two independent, identically distributed draws $(X_1,Y_1)$ and $(X_2,Y_2)$ from a continuous bivariate distribution, Kendall’s tau is $$ \tau = P\big((X_1-X_2)(Y_1-Y_2)>0\big)-P\big((X_1-X_2)(Y_1-Y_2)<0\big). $$ The first event is a concordant pair and the second is a discordant pair. With continuous margins, ties have probability zero, so equivalently $\tau=2P(\text{concordance})-1$. Because strictly increasing transformations preserve pair order, tau depends on the copula rather than on the chosen continuous marginal distributions.

The formulas below concern population Kendall’s tau under the stated copula models. Converting a supplied tau does not by itself establish that the model is appropriate or that a sample estimate is reliable. Those are separate questions.

Gaussian and Student-t Copulas

For a Gaussian copula with latent standard normal variables having correlation parameter $\rho$, the population relationship is $\tau=(2/\pi)\arcsin(\rho)$. The same relationship holds for a Student-t copula with degrees of freedom $\nu>0$ and latent correlation parameter $\rho$. In the Student-t case, tau does not depend on $\nu$, although other dependence features, such as joint-tail behavior, do.

Theorem: For either a Gaussian copula or a Student-t copula with parameter $\rho$, $$ \tau = \frac{2}{\pi}\arcsin(\rho), \qquad \rho = \sin\left(\frac{\pi\tau}{2}\right). $$ Here $-1\leq \rho\leq 1$, and therefore $-1\leq \tau\leq 1$. For the Student-t copula, $\nu$ must also be specified, but it is not determined by tau.

A short derivation explains the formula and, importantly, why the Student-t result remains valid. In the Gaussian case, compare two independent draws. Their coordinate differences are jointly normal, and their correlation is $\rho$: each difference has variance 2 and their covariance is $2\rho$. Write $Q(\rho)$ for the probability that both differences are positive. For a standard bivariate normal pair,

Q(\rho)=\frac14+\frac{1}{2\pi}\arcsin(\rho).

One justification is that $Q(0)=1/4$, and differentiating the bivariate normal quadrant probability with respect to $\rho$ gives $Q'(\rho)=1/(2\pi\sqrt{1-\rho^2})$. Integrating this derivative from 0 to $\rho$ gives the displayed expression. By symmetry, the probability of both differences being negative is also $Q(\rho)$. Thus the concordance probability is $2Q(\rho)$, and

\tau=2\{2Q(\rho)\}-1 =4\left(\frac14+\frac{1}{2\pi}\arcsin(\rho)\right)-1 =\frac{2}{\pi}\arcsin(\rho).

For a Student-t copula, represent each latent draw as $(A_jZ_j,A_jW_j)$, where $A_j>0$ is its scale factor and $(Z_j,W_j)$ is a standard bivariate normal pair with correlation $\rho$. The two independent draws have independent scale factors $A_1,A_2$. Conditional on those factors, the differences are jointly normal:

\begin{aligned} \operatorname{Var}(A_1Z_1-A_2Z_2\mid A_1,A_2)&=A_1^2+A_2^2,\\ \operatorname{Var}(A_1W_1-A_2W_2\mid A_1,A_2)&=A_1^2+A_2^2,\\ \operatorname{Cov}(A_1Z_1-A_2Z_2,A_1W_1-A_2W_2\mid A_1,A_2) &=\rho(A_1^2+A_2^2). \end{aligned}

Their conditional correlation is therefore $\rho$, for every positive pair of scale factors. The conditional sign-agreement probability is the same bivariate normal quadrant probability as before. Averaging a probability that is constant in $A_1,A_2$ leaves it unchanged, proving the tau relationship for the Student-t copula as well. The scale factors must be conditioned on separately for the two draws; they are not generally a common multiplier on the differences.

Worked Example: Converting Tau for a Gaussian Copula

A hypothetical portfolio model uses a Gaussian copula, and its target population Kendall’s tau is $0.40$. Find the latent correlation parameter $\rho$. State what the result does and does not represent.

Method. Invert the Gaussian tau relationship: $$ \rho=\sin\left(\frac{\pi\tau}{2}\right) =\sin\left(\frac{\pi(0.40)}{2}\right) =\sin(0.20\pi). $$

Since $0.20\pi=36^\circ$, $\rho=\sin(36^\circ)=0.587785\ldots$. Rounded to four decimal places, the candidate latent correlation is $\rho=0.5878$.

Check. Substitution gives $(2/\pi)\arcsin(0.587785\ldots)=0.4000$. This is the Gaussian copula’s latent correlation, not a claim that the portfolio losses’ Pearson correlation is 0.5878. The common mistake is to set $\rho=\tau$, which would give 0.40 and fail the substitution check.

Worked Example: Student-t Tau Does Not Determine Degrees of Freedom

A hypothetical dependence specification uses a Student-t copula with $\nu=5$ degrees of freedom and target Kendall’s tau $0.60$. Determine $\rho$, and explain whether the tau conversion also determines $\nu$.

Method. The Student-t tau relationship is the same arcsine formula: $$ \rho=\sin\left(\frac{\pi(0.60)}{2}\right) =\sin(0.30\pi) =\sin(54^\circ) =0.809016\ldots. $$

Thus the candidate latent correlation, rounded to four decimal places, is $\rho=0.8090$. The specified degrees of freedom remain $\nu=5$; tau alone does not identify them.

Check and trap. $(2/\pi)\arcsin(0.809016\ldots)=0.6000$. It would be incorrect to argue that scale mixtures leave the signs of differences unchanged: independent draws use different scale factors, which can change those signs. The valid argument conditions on both factors; the resulting difference correlation is still $\rho$, as derived above.

Clayton and Gumbel Copulas

For the standard bivariate Clayton copula with positive parameter $\theta$, and the standard Gumbel copula with parameter $\theta\geq1$, tau has a simple one-to-one relationship with the parameter. These standard forms describe nonnegative association. If a target tau is negative, it is not valid to force it into either of these unrotated parameter ranges.

Key relationships: For the standard Clayton copula, $\theta>0$ and $\tau=\theta/(\theta+2)$, so $\theta=2\tau/(1-\tau)$. For the standard Gumbel copula, $\theta\geq1$ and $\tau=1-1/\theta$, so $\theta=1/(1-\tau)$. Independence is approached at the boundary $\theta\to0$ for Clayton and occurs at $\theta=1$ for Gumbel; in each case $\tau=0$.

Worked Example: Clayton Parameter From Target Tau

A hypothetical paired-loss model is to use a standard Clayton copula with target tau $0.60$. Find the parameter and verify both its range and its implied tau.

Method. Rearrange $\tau=\theta/(\theta+2)$: $$ \theta=\frac{2\tau}{1-\tau} =\frac{2(0.60)}{1-0.60} =\frac{1.20}{0.40} =3. $$

The candidate Clayton parameter is $\theta=3$, which satisfies the standard Clayton restriction $\theta>0$. Substitution gives $\tau=3/(3+2)=3/5=0.60$.

Common mistake. Using $\theta=\tau/(\tau+2)$ reverses the rearrangement and gives $0.60/2.60$, which does not reproduce the target tau. Always substitute the proposed parameter into the original relationship.

Worked Example: Gumbel Parameter and Boundary Check

A hypothetical aggregation model requires a standard Gumbel copula with target tau $0.40$. Convert the target and check whether the parameter is admissible.

Method. From $\tau=1-1/\theta$, $$ \theta=\frac{1}{1-\tau} =\frac{1}{1-0.40} =\frac{1}{0.60} =\frac53 =1.6667\ldots. $$

The candidate is $\theta=5/3$, which is at least 1 as required. Substitution gives $1-1/(5/3)=1-3/5=2/5=0.40$.

Common mistake. The standard Gumbel parameter cannot be below 1. For example, a negative target tau would produce a value below 1 using the algebraic inverse, but that value is not admissible for this standard copula. A different dependence construction would be needed.

Frank Copula: Numerical Conversion

Unlike the Clayton and Gumbel relationships, the Frank relationship is usually inverted numerically. For positive Frank parameter $\theta$, define the first Debye function by

D_1(\theta)=\frac{1}{\theta}\int_0^\theta \frac{t}{e^t-1}\,dt.

Key relationship: For the Frank copula with $\theta>0$, $$ \tau=1+\frac{4}{\theta}\{D_1(\theta)-1\}. $$ At $\theta=0$, the Frank copula is the independence copula and $\tau=0$. Negative parameters give negative tau, with $\tau(-\theta)=-\tau(\theta)$. The displayed Debye representation is for positive $\theta$; negative values can be handled using this symmetry.

For positive parameters, the relationship increases from 0 toward 1, so a positive target tau can be converted by solving the scalar equation numerically. A root finder or a numerical table can be used; the equation should then be checked by substituting the candidate parameter back into the integral relationship.

Worked Example: Numerically Inverting the Frank Relationship

A hypothetical dependence specification has target tau $0.50$ and uses a positive-parameter Frank copula. Find a numerical candidate for $\theta$ and verify it.

Method. Solve $$ 1+\frac{4}{\theta}\{D_1(\theta)-1\}=0.50. $$

Numerical root finding gives $\theta\approx5.736$. To check, use the equivalent integral form $$ \tau=1-\frac{4}{\theta} +\frac{4}{\theta^2}\int_0^\theta\frac{t}{e^t-1}\,dt. $$ At $\theta=5.736$, numerical integration gives approximately $1.62317$ for the integral. Also, $4/5.736\approx0.69735$ and $4/(5.736)^2\approx0.12157$. Therefore, $$ \tau\approx1-0.69735+(0.12157)(1.62317) \approx1-0.69735+0.19733 \approx0.5000. $$

The candidate is $\theta\approx5.736$. A bracket such as $5.7$ to $5.8$ is consistent with the target: the corresponding tau values are approximately $0.4979$ and $0.5035$. The common mistake is to omit the factor $1/\theta$ in the definition of $D_1$, or to use an incorrect power of $\theta$ when rewriting the integral form. Either error changes the parameter materially.

Using the Conversion Correctly

A converted parameter is a candidate within a selected copula family, not a universal measure of dependence. Model choice matters: for example, the Gaussian and Student-t copulas can have the same tau and latent correlation while differing in degrees of freedom and joint-tail behavior. Similarly, Clayton, Gumbel, and Frank parameters are on different scales and cannot be compared directly.

For continuous data, tau is invariant under strictly increasing transformations of either margin. With discrete or heavily rounded observations, ties can occur; the simple continuous-pair definition and formulas then require care, because the sample’s treatment of ties affects the reported association. Also distinguish a population relationship from a sample calculation: a sample estimate of tau has sampling uncertainty, while these formulas map a population tau to a model parameter.

Identify the copula family and parameter convention.
Use the formula for the specified standard or rotated form; do not transfer a parameter formula between families.

Check the target sign and parameter range.
For example, the standard Clayton and Gumbel forms above do not admit negative tau.

Invert the relationship.
Use the closed-form inverse where available; solve the Frank equation numerically.

Substitute back and interpret.
Confirm that the parameter reproduces the target tau and label it correctly, especially when it is a latent correlation.

Key takeaway: Kendall’s tau provides a common rank-association scale, but its conversion to a copula parameter is family-specific. State the model and parameter range, invert the correct relationship, and verify the result by substitution.

Check Your Understanding

For each problem, identify the relevant relationship, check admissibility, and verify the conversion where practical.

A Gaussian copula has target population tau $0.20$. Find its latent correlation parameter to four decimal places. (Answer: $\rho=\sin(0.10\pi)=0.3090$.)
A Student-t copula has $\nu=8$ and target tau $-0.50$. Find $\rho$, and state whether tau determines $\nu$. (Answer: $\rho=\sin(-\pi/4)=-0.7071$; tau does not determine $\nu$.)
A standard Clayton copula is intended to have tau $0.75$. Find $\theta$ and verify the implied tau. (Answer: $\theta=2(0.75)/(0.25)=6$; $6/(6+2)=0.75$.)
A standard Gumbel copula is proposed for target tau $0.20$. Find $\theta$ and check the range. (Answer: $\theta=1/(1-0.20)=1.25$, admissible since $\theta\geq1$.)
A positive-parameter Frank copula is calibrated to target tau $0.25$. Use numerical inversion of the Debye relationship to give a candidate parameter to three decimals. (Answer: $\theta\approx2.371$.)

What You'll Learn