Tail Dependence in Elliptical Distributions

Why Dependence in the Extremes Needs Its Own Measure

Two variables can be strongly associated overall yet behave differently when both take unusually large values. A correlation summarizes a broad feature of joint variation; it does not, by itself, say whether an extreme value of one variable makes an extreme value of the other persistently likely. Tail dependence isolates that question by examining joint exceedances at increasingly high thresholds.

For elliptical models, this distinction is especially clear when comparing Gaussian and Student t copulas. At any fixed correlation strictly between $-1$ and $1$, their upper-tail dependence differs: the Gaussian copula has coefficient zero, while the t copula has a positive coefficient for every finite number of degrees of freedom. The t coefficient also makes explicit how correlation and degrees of freedom jointly govern extreme co-movement.

Definition and Copula Interpretation

Definition: Let $X$ and $Y$ have continuous marginal distribution functions $F_X$ and $F_Y$. Their upper-tail dependence coefficient is $$ \lambda_U = \lim_{u\uparrow 1}P\!\left(Y>F_Y^{-1}(u)\mid X>F_X^{-1}(u)\right), $$ if this limit exists. For continuous margins, this is equivalently $$ \lambda_U = \lim_{u\uparrow 1} \frac{P(F_X(X)>u,\;F_Y(Y)>u)}{1-u}. $$ The coefficient lies in $[0,1]$. A positive value means that, conditional on one variable exceeding its $u$-quantile, the probability that the other exceeds its own $u$-quantile does not vanish as $u$ approaches 1.

The quantile thresholds are marginal-specific, so this definition compares equally rare events rather than equally sized raw values. If $U=F_X(X)$ and $V=F_Y(Y)$, then both variables are uniform on $(0,1)$, and the coefficient is $\lim_{u\uparrow1}P(V>u\mid U>u)$. Thus it depends on the copula, the part of the joint distribution that describes dependence, rather than on the choice of continuous marginal scales.

The coefficient is an asymptotic quantity. A value of zero does not mean that joint exceedances never occur at finite thresholds; it means their conditional probability tends to zero in the limit. Conversely, a positive coefficient is not the probability of joint exceedance at an arbitrary operational threshold.

The Gaussian Copula Has No Upper-Tail Dependence

Let $(Z_1,Z_2)$ be standard bivariate normal with correlation $\rho$, where $-1<\rho<1$. Its Gaussian copula has the same tail coefficient as this normal pair. By symmetry, it is enough to use a common standard normal threshold $q$ tending to infinity: $$ \lambda_U=\lim_{q\to\infty}\frac{P(Z_1>q,\;Z_2>q)}{P(Z_1>q)}. $$ If both components exceed $q$, their sum exceeds $2q$. Since $Z_1+Z_2$ is normal with variance $2(1+\rho)$, $$ P(Z_1>q,\;Z_2>q) \leq P(Z_1+Z_2>2q). $$ The normal tail has exponential rate $\exp(-x^2/(2\sigma^2))$ for variance $\sigma^2$. The numerator bound therefore has rate $\exp(-q^2/(1+\rho))$, whereas $P(Z_1>q)$ has rate $\exp(-q^2/2)$. Their ratio tends to zero because $1/(1+\rho)>1/2$ when $\rho<1$. Hence

Result: For a Gaussian copula with $-1<\rho<1$, $\lambda_U=0$. At the degenerate endpoint $\rho=1$, the two variables coincide and $\lambda_U=1$.

The bound proves the claim without requiring an exact bivariate normal tail approximation: its exponential decay is already faster than the marginal tail. In particular, a Gaussian copula can have a high correlation and still have zero asymptotic upper-tail dependence.

Deriving the Student t Copula Coefficient

Now let $(X,Y)$ have a standard bivariate Student t distribution with $\nu>0$ degrees of freedom and scatter matrix with unit diagonal and off-diagonal element $\rho$, where $-1<\rho<1$. Each margin is standard $t_\nu$. Write $T_\nu$ for the standard univariate $t_\nu$ distribution function and set $q=T_\nu^{-1}(u)$. The conditional distribution of $Y$ given $X=x$ is Student t with $\nu+1$ degrees of freedom, location $\rho x$, and scale $$ s(x)=\sqrt{\frac{\nu+x^2}{\nu+1}(1-\rho^2)}. $$ Thus, for $q>0$, $$ P(Y>q\mid X=x) = 1-T_{\nu+1}\!\left(\frac{q-\rho x}{s(x)}\right). $$ This conditional law, together with the heavy-tailed marginal, gives a direct route to the coefficient.

Let $q$ tend to infinity and condition on $X>q$. The ratio $S=X/q$, given $X>q$, converges in distribution to a variable on $[1,\infty)$ with density $\nu s^{-\nu-1}$. To see the source of this limit, the $t_\nu$ density is asymptotic to a positive constant times $x^{-\nu-1}$, so its survival probability is asymptotic to the same constant divided by $\nu$, times $x^{-\nu}$. Consequently, for fixed $s\geq1$, $$ P(X/q>s\mid X>q) = \frac{P(X>sq)}{P(X>q)} \longrightarrow s^{-\nu}. $$ Given $S=s$, the conditional exceedance probability has limit $$ P(Y>q\mid X=sq) \longrightarrow T_{\nu+1}\!\left( \sqrt{\frac{\nu+1}{1-\rho^2}} \left(\rho-\frac{1}{s}\right) \right). $$ The function on the right is bounded and continuous in $s$. The convergence of the conditional scale and location is uniform on each compact interval in $s$; the limiting conditional distribution of $S$ assigns arbitrarily small probability to $s>M$ by taking $M$ large. These facts justify passing to the limit in the conditional expectation. Therefore, $$ \lambda_U = \nu\int_1^\infty s^{-\nu-1} T_{\nu+1}\!\left( \sqrt{\frac{\nu+1}{1-\rho^2}} \left(\rho-\frac{1}{s}\right) \right)\,ds. $$

For completeness, this integral can be evaluated directly. Put $m=\nu+1$, $c=\sqrt{m/(1-\rho^2)}$, and $z(s)=c(\rho-1/s)$. Integration by parts, using $d(-s^{-\nu})=\nu s^{-\nu-1}ds$, gives $$ \lambda_U = T_m(z(1)) + \int_1^\infty s^{-\nu}t_m(z(s))\frac{c}{s^2}\,ds, $$ where $t_m$ is the standard $t_m$ density. Here $$ z(1)=-b,\qquad b=\sqrt{\frac{m(1-\rho)}{1+\rho}}. $$ In the remaining integral use $z=c(\rho-1/s)$, so that $dz=c\,ds/s^2$ and $s^{-\nu}=(\rho-z/c)^\nu$. Substitution of the $t_m$ density $$ t_m(z)= \frac{\Gamma((m+1)/2)}{\sqrt{m\pi}\,\Gamma(m/2)} \left(1+\frac{z^2}{m}\right)^{-(m+1)/2} $$ and simplification yield $$ \int_1^\infty s^{-\nu}t_m(z(s))\frac{c}{s^2}\,ds = T_m(-b). $$ The last equality follows by substituting $z=c(\rho-1/s)$ in the integral and using $m=\nu+1$; its limits are $-b$ and $c\rho$. Thus the two terms are equal, and the coefficient is

Theorem: For a bivariate Student t copula with $\nu>0$ degrees of freedom and correlation parameter $-1<\rho<1$, $$ \lambda_U = 2T_{\nu+1}\!\left( -\sqrt{\frac{(\nu+1)(1-\rho)}{1+\rho}} \right). $$

This derivation uses the conditional t distribution and the regularly varying t marginal tail; it does not require the t vector to have finite covariance. The copula and its coefficient are defined for every $\nu>0$. When $\nu>2$, the standard t covariance exists and is proportional to its scatter matrix, but that additional moment condition is not needed here.

Worked Calculations

Worked Example: A t Copula with Two Degrees of Freedom

Take $\nu=1$ and $\rho=-1/2$. The coefficient formula gives $$ \lambda_U = 2T_2\!\left( -\sqrt{\frac{(1+1)(1-(-1/2))}{1+(-1/2)}} \right) = 2T_2(-\sqrt{6}). $$ For a standard $t_2$ variable, direct integration of its density gives $T_2(x)=\tfrac12+x/(2\sqrt{x^2+2})$. Hence $$ 2T_2(-\sqrt6) = 2\left(\frac12-\frac{\sqrt6}{2\sqrt{6+2}}\right) = 1-\frac{\sqrt6}{\sqrt8} = 1-\frac{\sqrt3}{2} \approx 0.1340. $$ Even though the correlation parameter is negative, the coefficient is positive: joint upper extremes still occur asymptotically under a finite-degree-of-freedom t copula. The probability is small in this case, but it is not zero.

Worked Example: Zero Correlation Does Not Mean Tail Independence

For $\nu=1$ and $\rho=0$, the coefficient is $$ \lambda_U = 2T_2(-1). $$ Using the same $t_2$ CDF, $$ T_2(-1) = \frac12-\frac{1}{2\sqrt{3}}, \qquad \lambda_U = 1-\frac{1}{\sqrt3} \approx 0.4226. $$ Thus a t copula with zero correlation parameter can have substantial upper-tail dependence. Zero correlation describes linear association when the relevant moments exist; it does not imply independence. In this example $\nu=1$, so the marginal variance is not finite, and correlation as a Pearson moment is not defined. The parameter $\rho$ here is the t copula's scatter/correlation parameter.

Worked Example: Comparing Gaussian and t Copulas

Compare a Gaussian copula and a $t$ copula at $\rho=1/2$, taking $\nu=1$ for the t copula. The Gaussian coefficient is zero because $-1<1/2<1$. For the t copula, $$ \lambda_U = 2T_2\!\left(-\sqrt{\frac{1-1/2}{1+1/2}}\right) = 2T_2\!\left(-\frac{1}{\sqrt3}\right). $$ Since $T_2(x)=\tfrac12+x/(2\sqrt{x^2+2})$, $$ \lambda_U = 1-\frac{1/\sqrt3}{\sqrt{1/3+2}} = 1-\frac{1}{\sqrt7} \approx 0.6220. $$ The two copulas have the same dependence parameter but sharply different asymptotic joint-extreme behavior. This is the practical distinction the coefficient is designed to expose.

How Correlation and Degrees of Freedom Matter

For fixed $\nu$, increasing $\rho$ from $-1$ toward $1$ decreases the positive quantity $(1-\rho)/(1+\rho)$. The argument of $T_{\nu+1}$ therefore becomes less negative, and since a distribution function is increasing, $\lambda_U$ increases. Its limiting values are zero as $\rho\downarrow-1$ and one as $\rho\uparrow1$. At the endpoints the model is degenerate; the formula's limits describe the corresponding tail behavior.

For fixed $-1<\rho<1$, the t coefficient tends to zero as $\nu$ tends to infinity. This agrees with the Gaussian limit of the t copula. At finite $\nu$, the common random scale in the t construction can make both components large together; that mechanism persists even when $\rho$ is zero or negative. Smaller degrees of freedom generally correspond to stronger tail dependence, though the displayed formula, rather than a slogan about “heaviness,” should be used for numerical comparisons.

A common mistake is to equate “high correlation” with “tail dependent,” or to infer tail independence from a zero correlation parameter. The correction is to identify the dependence model and calculate or characterize its tail limit. A Gaussian copula with correlation $0.9$ still has $\lambda_U=0$; a finite-$\nu$ t copula with correlation zero has $\lambda_U>0$. Tail dependence is a property of the joint tail, not a synonym for correlation.

Key takeaway: For nondegenerate Gaussian copulas, upper-tail dependence is zero. For a t copula it equals $2T_{\nu+1}(-\sqrt{(\nu+1)(1-\rho)/(1+\rho)})$, so both the correlation parameter and degrees of freedom matter.

Check Your Understanding

Use the definitions and coefficient formula to answer these questions.

Express the upper-tail dependence coefficient as a limit of a conditional probability using marginal quantiles.
Why does the Gaussian copula coefficient equal zero for every $-1<\rho<1$, even when $\rho$ is close to one?
For a t copula with fixed $\nu$, how does the coefficient change as $\rho$ increases?
What does the t copula coefficient say about the relationship between zero correlation parameter and upper-tail independence?
Does the t copula coefficient formula require finite marginal variances? Explain.