Risk Modelling and Survival Analysis · Introduction to copulas

Tail Dependence Coefficients and Copulas Explained

Q: Which copula has lower tail dependence?

The Clayton copula has lower tail dependence, with λL = 2^(−1/θ) for θ > 0. The Student t copula also has lower tail dependence, and it has equal upper tail dependence. The Gaussian, Frank and Gumbel copulas have none in the lower tail.

Updated 11 October 2026 · Fact-checked

Tail dependence measures the chance that one variable is extreme given that the other is extreme. Upper λU = lim P(V > u | U > u) as u→1; lower λL = lim P(V ≤ u | U ≤ u) as u→0. Clayton has only lower tail dependence, Gumbel only upper, Gaussian and Frank neither, and the t copula has both.

Understand Tail Dependence

Correlation tells you about the average relationship between two risks. It does not tell you what happens when both risks are extreme at the same time. For insurers, joint extremes matter most: a flood hitting property and motor portfolios together, or two asset classes crashing together.

Tail dependence measures this. Take two random variables with uniform marginals U and V, joined by a copula C. The upper tail dependence coefficient is the limiting probability that V is large, given that U is large. The lower tail dependence coefficient is the limiting probability that V is small, given that U is small. Both are limits as you move further into the tail.

If the coefficient is greater than 0, the copula has tail dependence. Extremes tend to occur together even far into the tail. If the coefficient is 0, the variables are asymptotically independent in that tail. Joint extremes become rare relative to single extremes as you go further out. This can hold even when the overall dependence is strong.

The copula family decides the answer. The Clayton copula (θ > 0) has lower tail dependence only. The Gumbel copula (θ ≥ 1) has upper tail dependence only. The Gaussian copula has none in either tail when |ρ| < 1. The Frank copula has none. The Student t copula has both tails equal and positive, and it is larger when the degrees of freedom are smaller.

So copula choice matters for risk. Using a Gaussian copula for losses that really cluster in the tail will understate the probability of joint large losses. A Gumbel or t copula gives a more prudent view of joint large losses. Use Clayton when joint low values matter, for example joint falls in asset returns.

Key rules to remember

Upper tail dependence: λU = lim (u→1⁻) P(V > u | U > u) = lim (u→1⁻) [1 − 2u + C(u,u)] ÷ (1 − u); U and V are uniform(0,1). C is the copula. The limit must exist.
Lower tail dependence: λL = lim (u→0⁺) P(V ≤ u | U ≤ u) = lim (u→0⁺) C(u,u) ÷ u; Evaluate the copula on the diagonal, then divide by u.
Clayton copula: C(u,v) = (u^(−θ) + v^(−θ) − 1)^(−1/θ), θ > 0; λL = 2^(−1/θ) and λU = 0. Larger θ gives larger λL.
Gumbel copula: C(u,v) = exp{ −[(−ln u)^θ + (−ln v)^θ]^(1/θ) }, θ ≥ 1; λU = 2 − 2^(1/θ) and λL = 0. θ = 1 is independence.
Student t copula: λU = λL = 2 × t(ν+1)( −√[ (ν+1)(1 − ρ) ÷ (1 + ρ) ] ); t(ν+1) is the cdf of a t distribution with ν+1 degrees of freedom. ν is the degrees of freedom, ρ the correlation parameter. Positive for every finite ν and ρ > −1.
Gaussian and Frank copulas: λU = λL = 0; For the Gaussian copula this holds when |ρ| < 1.
Survival copula link: λU of C = λL of the survival copula of C, and the reverse; Reversing the direction of both variables swaps the two tails.

How to solve Tail Dependence questions

Use this method for any question on tail dependence, whether it asks you to calculate a coefficient, derive it, or choose a copula.

1Identify the copula and its parameter, and check the parameter is in its allowed range.
2Decide which tail is asked: upper (both large) or lower (both small).
3If the family is standard, recall the result from your formula list (for example λL = 2^(−1/θ) for Clayton).
4If you must derive it, write C(u,u) for the lower tail, or 1 − 2u + C(u,u) for the upper tail.
5Divide by u (lower) or by 1 − u (upper). Simplify, then take the limit. For Clayton, factor out u first.
6Substitute the parameter and compute a number. State it as a probability limit.
7Interpret the answer. If it is 0, say the copula is asymptotically independent in that tail. If it is positive, say joint extremes cluster.
8For choice questions, match the feature to the copula: joint large losses point to Gumbel or t, joint low values to Clayton or t.

Quickest way: Match the tail to the copula, then plug in

When to use it: Use in multiple-choice questions and short calculations where the copula family is named.

Memorise: Clayton = lower only, Gumbel = upper only, t = both, Gaussian and Frank = neither.
If the question asks for a tail the copula lacks, the answer is 0. Stop there.
Otherwise use λL = 2^(−1/θ) for Clayton or λU = 2 − 2^(1/θ) for Gumbel.
To find θ from a given λ, rearrange: Clayton θ = −ln 2 ÷ ln λL, and Gumbel θ = ln 2 ÷ ln(2 − λU).
Check the sense: Clayton λL rises with θ, and Gumbel λU rises with θ.

Common mistakes in Tail Dependence

Saying the Gaussian copula has tail dependence because it has high correlation.
Students mix up strong overall dependence with dependence in the extremes.
Fix: For |ρ| < 1 the Gaussian copula has λU = λL = 0, however large ρ is. Only the limit in the tail matters.
Giving Clayton an upper tail dependence value, or Gumbel a lower one.
Students remember the formulas but forget which tail each belongs to.
Fix: Clayton clusters at low values, so λL = 2^(−1/θ) and λU = 0. Gumbel clusters at high values, so λU = 2 − 2^(1/θ) and λL = 0.
Using the wrong denominator in the upper tail formula.
Students copy the lower tail form C(u,u) ÷ u for both tails.
Fix: The upper tail needs the joint survival probability 1 − 2u + C(u,u), divided by 1 − u, with u→1.
Taking θ = 0 or θ < 1 for Gumbel, or θ ≤ 0 for Clayton, in a calculation.
Students ignore the parameter ranges.
Fix: Clayton needs θ > 0 and Gumbel needs θ ≥ 1. Gumbel at θ = 1 gives λU = 2 − 2 = 0, which is independence.
Forgetting that the t copula is symmetric, so it cannot model tail dependence in only one tail.
Students assume any heavy-tailed copula can be one-sided.
Fix: The t copula has λU = λL. If the question needs an asymmetric tail, choose Clayton or Gumbel.
Evaluating the limit by substituting u = 0 directly and getting 0 ÷ 0.
Students skip the algebra for the indeterminate form.
Fix: Factor u out of C(u,u) first. For Clayton, C(u,u) = u(2 − u^θ)^(−1/θ), so the ratio tends to 2^(−1/θ).

Worked examples

Example 1

A Clayton copula with θ = 2 links two asset returns. (a) Derive the lower tail dependence coefficient. (b) State the upper tail dependence coefficient. (c) Interpret the result.

Show the solution

Clayton copula: C(u,v) = (u^(−θ) + v^(−θ) − 1)^(−1/θ).
On the diagonal: C(u,u) = (2u^(−θ) − 1)^(−1/θ).
Factor out u^(−θ): 2u^(−θ) − 1 = u^(−θ)(2 − u^θ). So C(u,u) = u × (2 − u^θ)^(−1/θ).
Then C(u,u) ÷ u = (2 − u^θ)^(−1/θ). As u→0, u^θ→0 because θ > 0.
So λL = 2^(−1/θ). With θ = 2, λL = 2^(−1/2) = 1 ÷ √2 = 0.7071.
The Clayton copula has no upper tail dependence, so λU = 0.
Interpretation: if one return falls into its extreme low tail, the chance the other does too approaches about 0.71. Large joint gains do not cluster.

Answer: (a) λL = 2^(−1/θ) = 2^(−1/2) ≈ 0.7071. (b) λU = 0. (c) Joint extreme falls are strongly linked; joint extreme gains are asymptotically independent.

Example 2

Two classes of loss are modelled with a Gumbel copula with θ = 1.5. (a) Find λU and λL. (b) Find the θ that would give λU = 0.5. (c) Say whether a Gaussian copula could reproduce this behaviour.

Show the solution

Gumbel copula: λU = 2 − 2^(1/θ), λL = 0.
(a) With θ = 1.5, 1/θ = 2/3. 2^(2/3) = 1.5874. So λU = 2 − 1.5874 = 0.4126. λL = 0.
(b) Set 2 − 2^(1/θ) = 0.5. Then 2^(1/θ) = 1.5.
Take logs: (1/θ) ln 2 = ln 1.5. So 1/θ = 0.405465 ÷ 0.693147 = 0.58496.
So θ = 1 ÷ 0.58496 = 1.7095. This is at least 1, so it is valid.
(c) The Gaussian copula has λU = 0 for |ρ| < 1, so it cannot reproduce positive upper tail dependence. It would understate the chance of joint large losses.

Answer: (a) λU ≈ 0.413, λL = 0. (b) θ ≈ 1.71. (c) No. A Gaussian copula gives λU = 0, so it understates joint large losses; Gumbel or t fits better.

Exam tips

Learn the four-way table by heart: Clayton lower, Gumbel upper, t both, Gaussian and Frank neither. Many multiple-choice questions test only this.
In written answers, define λU and λL with conditional probabilities and a limit before giving any numbers. Examiners reward the definition.
When asked to compare copulas for joint large losses, argue from tail dependence, not from correlation. Name the model risk of a Gaussian copula.
If you derive a coefficient, show the factorisation step before the limit. Do not jump to the formula.
In the computer-based paper, if you simulate from a copula, check tail behaviour by plotting the points. Clusters in a corner show tail dependence in that corner.

Practice questions from Introduction to copulas

Tail Dependence: frequently asked questions

Which copula has lower tail dependence?

The Clayton copula has lower tail dependence, with λL = 2^(−1/θ) for θ > 0. The Student t copula also has lower tail dependence, and it has equal upper tail dependence. The Gaussian, Frank and Gumbel copulas have none in the lower tail.

How do I calculate tail dependence for the Clayton copula?

Write C(u,u) = (2u^(−θ) − 1)^(−1/θ) and divide by u. This gives (2 − u^θ)^(−1/θ), which tends to 2^(−1/θ) as u→0. The upper tail coefficient is 0.

What is the difference between upper and lower tail dependence for Gumbel and Clayton?

Gumbel has upper tail dependence λU = 2 − 2^(1/θ) and no lower tail dependence. Clayton has the opposite: lower tail dependence and none in the upper tail. Pick Gumbel for joint large losses and Clayton for joint low values.

Does a high correlation mean there is tail dependence?

No. The Gaussian copula can have a correlation close to 1 and still have zero tail dependence. Tail dependence is about the limit in the extreme corner, not about the overall strength of dependence.

Why does tail dependence matter for insurers?

It shows how likely large losses are to happen together across lines or assets. A model with no tail dependence can understate capital needs and reinsurance needs in a stress event.