Why Build a Copula From a Generator?
A copula separates marginal loss behavior from dependence. Once marginal distributions have been selected, a copula specifies how the associated probability ranks move together. An Archimedean construction is useful because it expresses a multivariate dependence model through one function, called a generator, and an addition operation on transformed marginal probabilities.
The construction is compact, but validity requires more than substituting numbers into a formula. The generator must have the right boundary behavior, its inverse must be defined over the required range, and it must satisfy a dimension-dependent shape condition. In particular, a function that gives the correct margins is not automatically a copula: it must also be grounded and d-increasing.
A generator that is strictly decreasing from 1 toward zero is often called strict; its zero limit occurs only at infinity. A non-strict generator may reach zero at a finite endpoint and is then zero thereafter. The generalized inverse convention matters at \(u=0\), and also when evaluating a sum that passes the finite endpoint. Under the definition above, \(\psi\) is already zero at and beyond that endpoint.
Conditions for Validity
For a fixed dimension, the standard validity condition is d-monotonicity. This condition is a shape restriction ensuring that the construction is d-increasing, in addition to having uniform margins. For \(d=2\), d-monotonicity means that \(\psi\) is nonnegative, nonincreasing, and convex. For \(d\geq3\), one standard differentiable formulation is that \(\psi\) has derivatives through order \(d-2\), with \((-1)^k\psi^{(k)}(t)\geq0\) for \(k=0,\ldots,d-2\), and that \((-1)^{d-2}\psi^{(d-2)}\) is nonnegative, nonincreasing, and convex. The conditions can be interpreted with the usual one-sided or generalized convexity conventions at boundary points.
Thus, a generator valid in two dimensions need not be valid in three. A stronger condition is available when the same construction must work in every finite dimension.
This distinction prevents an important overstatement: complete monotonicity is not required merely to construct a copula in one specified dimension. Conversely, checking only that \(\psi(0)=1\) and that the margins work does not establish d-increasingness. The shape condition is a substantive part of the validity check.
Why the Margins Work
The margin argument follows directly from the inverse construction. Since \(\psi(0)=1\), the generalized inverse at 1 is zero. Setting all but one input equal to 1 therefore removes their terms from the sum, leaving the original input.
For \(u>0\), continuity and the generalized-inverse definition give \(\psi(\psi^{-1}(u))=u\). At \(u=0\), the boundary convention gives \(\psi(\infty)=0\), or \(\psi(t_*)=0\) for a finite endpoint. The same reasoning applies to each coordinate margin. This proves the uniform margins, but validity still also depends on the d-monotonicity condition in the theorem.
Confirm continuity, nonincrease, \(\psi(0)=1\), and that the generator reaches zero or tends to zero at infinity.
For a fixed \(d\), verify d-monotonicity. For validity in every finite dimension, verify complete monotonicity.
Use the finite zero endpoint or infinity as appropriate when an input is zero; do not assume the ordinary inverse is finite there.
Apply the generator to the sum of the inverse-transformed inputs, then check limiting cases and the copula margins.
Worked Constructions
In applications, the marginal cumulative distribution functions transform losses to probability-scale values. The copula then gives their joint distribution on that scale. The examples below focus on constructing and checking that joint distribution; no claim is made that hypothetical losses or values represent observed insurance data.
Worked Example: Independence From an Exponential Generator
Suppose two hypothetical loss margins have cumulative probabilities \(u=0.70\) and \(v=0.40\) at specified loss thresholds. Use \(\psi(t)=e^{-t}\) to compute the probability that both losses are below their thresholds and the probability that both exceed them. State why the construction is valid.
Method and assumptions. This generator is continuous, strictly decreasing, satisfies \(\psi(0)=1\), and tends to zero at infinity. Its derivatives obey \((-1)^k\psi^{(k)}(t)=e^{-t}\geq0\) for every \(k\), so it is completely monotone and valid in every finite dimension.
Step 1: Find the inverse. Solving \(u=e^{-t}\) gives \(\psi^{-1}(u)=-\ln u\). Therefore, \(\psi^{-1}(0.70)=-\ln(0.70)\) and \(\psi^{-1}(0.40)=-\ln(0.40)\).
Step 2: Construct the joint probability. The sum of the inverse values is \(-\ln(0.70)-\ln(0.40)=-\ln(0.28)\). Thus \(C(0.70,0.40)=e^{-[-\ln(0.28)]}=0.28\).
Step 3: Compute the joint exceedance probability. By inclusion-exclusion, the probability that both probability-scale variables exceed their thresholds is \(1-0.70-0.40+C(0.70,0.40)=1-0.70-0.40+0.28=0.18\).
Check. The joint lower probability \(0.28\) is no greater than either marginal probability, and the joint exceedance probability is nonnegative. The calculation identifies the independence copula \(C(u,v)=uv\), as expected. A common mistake is to use \(1-C(u,v)\) for the probability both exceed: that is the probability at least one exceeds, not both.
Worked Example: A Three-Variable Gumbel Construction
For a hypothetical three-risk model, use the generator \(\psi(t)=\exp(-\sqrt{t})\). Compute the joint probability at marginal probability values \(u_1=e^{-1}\), \(u_2=e^{-2}\), and \(u_3=e^{-2}\). Also verify that the same generator is valid in dimension three.
Method and assumptions. This is the Gumbel generator with parameter \(\theta=2\), of the form \(\exp(-t^{1/\theta})\). For \(\theta\geq1\), this standard generator is completely monotone, so it gives a copula in every finite dimension. The example uses \(\theta=2\).
Step 1: Invert each input. From \(u=\exp(-\sqrt{t})\), take logarithms to obtain \(-\ln u=\sqrt{t}\), hence \(\psi^{-1}(u)=(-\ln u)^2\). Therefore the three inverse values are \(1^2=1\), \(2^2=4\), and \(2^2=4\).
Step 2: Add the transformed values and apply the generator. The sum is \(1+4+4=9\). Thus \(C(e^{-1},e^{-2},e^{-2})=\exp(-\sqrt{9})=e^{-3}\).
Step 3: Report and check. To four decimal places, \(e^{-3}=0.0498\). This joint probability is below each marginal probability: \(0.0498<e^{-2}\approx0.1353<e^{-1}\approx0.3679\). The calculation uses three inverse terms; omitting one would construct a different, two-variable probability rather than the requested trivariate value.
Worked Example: A Finite-Endpoint Generator and Its Boundary
Consider the hypothetical bivariate generator \(\psi(t)=\max(1-t,0)\). At probability-scale inputs \(u=0.80\) and \(v=0.55\), calculate the copula value. Then determine the copula value at \(u=0\), \(v=0.55\), using the finite-endpoint convention.
Method and assumptions. The generator reaches zero at \(t_*=1\). It is continuous, nonincreasing, and convex, so it is 2-monotone and valid as a bivariate generator. This shape condition does not establish validity in dimensions greater than two.
Step 1: Find the inverse for positive inputs. For \(0<u\leq1\), the equation \(u=1-t\) gives \(\psi^{-1}(u)=1-u\). Thus \(\psi^{-1}(0.80)=0.20\) and \(\psi^{-1}(0.55)=0.45\).
Step 2: Apply the construction. The inverse values sum to \(0.20+0.45=0.65\), which is below the endpoint. Therefore \(C(0.80,0.55)=\psi(0.65)=1-0.65=0.35\).
Step 3: Handle a zero input. By convention, \(\psi^{-1}(0)=t_*=1\). The sum for \(C(0,0.55)\) is \(1+0.45=1.45\); the generator is zero there, so \(C(0,0.55)=0\).
Check. The result \(0.35\) is no greater than either margin, and a zero input produces a zero copula value. A common mistake is to treat \(\psi^{-1}(0)\) as an ordinary finite inverse without stating its endpoint convention. The explicit convention makes the boundary calculation unambiguous.
Common Validity Traps
A tempting but invalid shortcut is to accept any continuous, decreasing function that starts at one. For example, \(\psi(t)=0.5+0.5e^{-t}\) is continuous and decreasing and has \(\psi(0)=1\), but it tends to \(0.5\), not zero. For an input \(u=0.25\), the set \(\{t\geq0:\psi(t)\leq u\}\) is empty. Its generalized inverse is therefore not defined as a finite value by the stated construction, and the copula formula cannot be applied to all inputs.
Another trap is to check only the margins. The inverse calculation explains why the margins are uniform, but it does not prove that all rectangle probabilities are nonnegative. That property is captured by d-increasingness, ensured here by the applicable d-monotonicity condition. Finally, never infer validity in every dimension solely from a two-dimensional convexity check.
| Question | Appropriate check | What it establishes |
|---|---|---|
| Does the inverse cover inputs down to zero? | Check that \(\psi\) reaches zero or tends to zero; specify the endpoint convention. | The construction is defined at the boundary. |
| Is the construction valid in dimension \(d\)? | Check d-monotonicity. | Validity for that fixed dimension. |
| Is the construction valid in all finite dimensions? | Check complete monotonicity. | A sufficient condition for validity in every finite dimension. |
| Are the margins uniform? | Set all but one input equal to 1 and use \(\psi^{-1}(1)=0\). | The required one-dimensional margins. |
Check Your Understanding
For each problem, use the generator definition and the appropriate validity or boundary check. Probabilities in the answers are rounded to four decimal places where needed.
- Let \(\psi(t)=e^{-2t}\). Find \(\psi^{-1}(0.25)\), then calculate the bivariate copula value at \((0.25,0.60)\). (Answer: \(\psi^{-1}(0.25)=0.6931\); \(C(0.25,0.60)=0.1500\).)
- For \(\psi(t)=\exp(-\sqrt{t})\), calculate the trivariate copula value at \((e^{-1},e^{-1},e^{-2})\). (Answer: \(e^{-\sqrt{6}}=0.0863\).)
- For \(\psi(t)=\max(1-t,0)\), calculate \(C(0.35,0.40)\) and \(C(0.80,0.55)\). State the dimension in which the convexity check establishes validity. (Answer: \(0\) and \(0.35\), respectively; the check establishes validity in dimension 2.)
- A proposed generator is \(\psi(t)=0.3+0.7e^{-t}\). Can its generalized inverse as defined here cover every input from 0 to 1? Explain using its limit. (Answer: No; it tends to \(0.3\), so inputs below \(0.3\), including zero, have no finite inverse under the stated definition.)
- A generator is verified to be d-monotone for \(d=3\), but no complete-monotonicity check is supplied. What validity conclusion is justified, and what stronger conclusion is not established? (Answer: It is valid as a three-dimensional generator; validity in every finite dimension is not established.)