Copula Densities and the Likelihood Factorization

From a Copula to a Joint Density

A copula specifies how marginal distributions are joined into a joint distribution. For continuous losses, the corresponding density separates into two kinds of contribution: the marginal densities describe the individual loss scales and shapes, while the copula density adjusts for dependence. This factorization is useful both for calculating joint likelihoods and for seeing which parts of a model are affected by a change in the dependence specification. The results and methods here are aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Definition: Let \(X\) and \(Y\) have continuous marginal distribution functions \(F_X\) and \(F_Y\), with densities \(f_X\) and \(f_Y\). Suppose their joint distribution has copula \(C\), so that \(F_{X,Y}(x,y)=C(F_X(x),F_Y(y))\). If the mixed partial derivative \(c(u,v)=\partial^2 C(u,v)/(\partial u\,\partial v)\) exists, it is the copula density. At points where the marginal densities exist, the joint density is \(f_{X,Y}(x,y)=c(F_X(x),F_Y(y))f_X(x)f_Y(y)\).

The arguments supplied to \(c\) are probabilities: first calculate \(u=F_X(x)\) and \(v=F_Y(y)\), then evaluate \(c(u,v)\). The copula density is generally not itself the joint density on the loss scale. The factors \(f_X(x)\) and \(f_Y(y)\) are also required.

Why the Factorization Holds

Assume that the joint distribution is absolutely continuous and that the marginal distributions and copula are sufficiently differentiable at the point being considered. Starting with the copula representation, differentiate first with respect to \(x\), then with respect to \(y\). The chain rule gives the result directly.

\begin{aligned} F_{X,Y}(x,y) &= C(F_X(x),F_Y(y)),\\ \frac{\partial F_{X,Y}}{\partial x} &= C_u(F_X(x),F_Y(y))f_X(x),\\ f_{X,Y}(x,y) &= \frac{\partial}{\partial y} \left[C_u(F_X(x),F_Y(y))f_X(x)\right]\\ &= C_{uv}(F_X(x),F_Y(y))f_X(x)f_Y(y)\\ &= c(F_X(x),F_Y(y))f_X(x)f_Y(y). \end{aligned}

In the second differentiation, \(f_X(x)\) is constant with respect to \(y\), and the chain rule applied to \(F_Y(y)\) supplies the factor \(f_Y(y)\). This explains why the copula density multiplies, rather than replaces, the marginal densities. In \(d\) dimensions, the same chain-rule argument gives \(f_{\boldsymbol X}(\boldsymbol x)=c(F_1(x_1),\ldots,F_d(x_d))\prod_{j=1}^d f_j(x_j)\), when the relevant derivatives and densities exist.

A useful example of a copula with an explicit density is the Farlie–Gumbel–Morgenstern (FGM) copula:

C(u,v)=uv\left[1+\theta(1-u)(1-v)\right], \qquad c(u,v)=1+\theta(1-2u)(1-2v), \qquad |\theta|\leq 1.

The parameter restriction ensures this is a valid copula. Its density is obtained by taking the mixed partial derivative of \(C\). The FGM family will serve here as a concrete calculation tool; the factorization itself applies to any copula with a suitable density.

Worked Example: Joint Density From Exponential Margins

Suppose \(X\) and \(Y\) are hypothetical continuous losses with exponential marginal distributions of rates 1 and 2, respectively. Their dependence is specified by the FGM copula with \(\theta=0.8\). Find \(f_{X,Y}(x,y)\) at \(x=\ln(4/3)\), \(y=\ln 2\). State the density in loss units. The method is to transform the losses through their marginal CDFs, evaluate the copula density, and multiply by both marginal densities; this avoids treating \(c\) as the joint density.

For an exponential distribution with rate \(\lambda\), \(F(t)=1-e^{-\lambda t}\) and \(f(t)=\lambda e^{-\lambda t}\), for \(t\geq 0\). For \(X\), whose rate is 1, \(e^{-x}=e^{-\ln(4/3)}=3/4\). Thus \(F_X(x)=1/4\) and \(f_X(x)=3/4\).

For \(Y\), whose rate is 2, \(e^{-2y}=e^{-2\ln 2}=1/4\). Thus \(F_Y(y)=3/4\) and \(f_Y(y)=2(1/4)=1/2\).

The copula density at the transformed observations is \(c(1/4,3/4)=1+0.8(1-1/2)(1-3/2)=1+0.8(1/2)(-1/2)=0.8\). Therefore, \(f_{X,Y}(x,y)=0.8(3/4)(1/2)=0.3\).

Check. Both marginal density factors and the dependence multiplier have been included, and the evaluated CDFs lie in \([0,1]\). The joint density at this point is \(0.3\) in the corresponding inverse loss units. The common mistake avoided is reporting \(0.8\), which is only the copula-density factor.

Likelihood Factorization for a Sample

Suppose \((x_i,y_i)\), \(i=1,\ldots,n\), are independent observations from the same continuous bivariate model. Independence here is across sampled pairs; \(X\) and \(Y\) within each pair may be dependent. For marginal parameters \(\alpha,\beta\) and copula parameter \(\theta\), the likelihood is the product of the joint densities:

L(\alpha,\beta,\theta) = \prod_{i=1}^{n} c_\theta\!\left(F_X(x_i;\alpha),F_Y(y_i;\beta)\right) f_X(x_i;\alpha)f_Y(y_i;\beta).

Taking logarithms turns the product into a sum:

\ell(\alpha,\beta,\theta) = \sum_{i=1}^{n}\log c_\theta\!\left(F_X(x_i;\alpha),F_Y(y_i;\beta)\right) +\sum_{i=1}^{n}\log f_X(x_i;\alpha) +\sum_{i=1}^{n}\log f_Y(y_i;\beta).

This is an algebraic factorization of the full likelihood, not a claim that marginal and dependence estimation are automatically independent tasks. If the marginal parameters are fixed, the marginal-density terms do not change when comparing copula parameters, although the copula term still depends on the transformed observations. If marginal parameters change, the CDF values supplied to the copula density change too. Consequently, even the dependence contribution can change when the margins change.

Worked Example: Separating Dependence Contributions in a Likelihood

A hypothetical sample consists of three independent pairs with uniform \((0,1)\) margins and observed values \((0.25,0.25)\), \((0.75,0.75)\), and \((0.25,0.75)\). Use an FGM copula with \(\theta=0.8\) to calculate the likelihood and log-likelihood. The margins are specified and fixed, so their densities are 1 on the support. The method is to calculate each copula-density contribution and multiply them; the trap avoided is omitting an observation or using the copula CDF instead of its density.

For uniform \((0,1)\) margins, \(F_X(x)=x\), \(F_Y(y)=y\), and each marginal density is 1 at these observations. For the first pair, the deviations are \(1-2(0.25)=0.5\) and \(0.5\), so \(c(0.25,0.25)=1+0.8(0.5)(0.5)=1.2\).

For the second pair, the deviations are \(-0.5\) and \(-0.5\), giving \(c(0.75,0.75)=1+0.8(-0.5)(-0.5)=1.2\). For the third pair, they are \(0.5\) and \(-0.5\), giving \(c(0.25,0.75)=1+0.8(0.5)(-0.5)=0.8\).

Each joint density equals its copula-density contribution because both marginal densities are 1. Therefore \(L=1.2(1.2)(0.8)=1.152\). The log-likelihood is \(\ell=\log(1.2)+\log(1.2)+\log(0.8)=\log(1.152)\approx 0.1416\), rounded to four decimal places.

Check. With the independence copula, \(\theta=0\), every copula-density contribution would be 1, giving likelihood 1 for these fixed uniform margins. The likelihood of 1.152 is the product of all three dependence contributions, with no marginal-density adjustment needed because the margins are uniform.

When Marginal Parameters Change

The factorization also clarifies an important distinction: the copula-density contribution is separated algebraically from the marginal densities, but it is evaluated at marginal CDF values. Thus a different marginal model or parameter can affect both the marginal-density term and the arguments of the copula density. The following comparison holds the dependence parameter fixed while changing the margins.

Worked Example: A Marginal Change Also Changes the Copula Evaluation

Consider one hypothetical observation \(x=y=\ln(4/3)\). Both margins are exponential with the same rate \(\lambda\), and dependence follows an FGM copula with \(\theta=0.8\). Calculate the joint density for \(\lambda=1\) and for \(\lambda=2\). Round final densities to four decimal places. The method is to recompute both the marginal densities and transformed CDF values under each candidate rate; the mistake avoided is holding the copula-density value fixed when the marginal parameter changes.

For \(\lambda=1\), each marginal CDF at the observation is \(1-e^{-x}=1/4\), and each marginal density is \(e^{-x}=3/4\). The copula density is \(c(1/4,1/4)=1+0.8(1/2)(1/2)=1.2\). Hence the joint density is \(1.2(3/4)(3/4)=1.2(9/16)=27/40=0.6750\).

For \(\lambda=2\), each marginal CDF is \(1-e^{-2x}=1-(3/4)^2=7/16\), and each marginal density is \(2e^{-2x}=2(9/16)=9/8\). The deviations for the copula density are \(1-2(7/16)=1/8\) for both margins. Thus \(c(7/16,7/16)=1+0.8(1/8)(1/8)=1+1/80=81/80\). The joint density is \((81/80)(9/8)(9/8)=6561/5120\approx 1.2814\).

Check. The two results use the same observed losses and the same copula parameter, but different marginal rates. Both the marginal density factors and the CDF inputs to \(c\) changed. A density value greater than 1 is permissible; unlike a probability, a density is not bounded above by 1.

Scope and Common Traps

The formula \(f_{X,Y}=c(F_X,F_Y)f_Xf_Y\) requires continuous margins and a joint distribution with a density of the stated form. It is not a general formula for discrete or mixed distributions. For discrete variables, probabilities are obtained from joint CDF differences rather than by multiplying density terms; when there are atoms, a density-only factorization may not describe the model. The differentiability assumptions also matter at points where densities or derivatives do not exist.

Using the copula CDF instead of its density. The joint density uses the mixed partial derivative \(c\), not \(C\) itself.
Evaluating at raw losses. The copula-density inputs are \(F_X(x)\) and \(F_Y(y)\), not \(x\) and \(y\).
Omitting marginal densities. The copula density adjusts dependence; it does not account for the marginal scales.
Assuming likelihood contributions are unaffected by margin changes. The marginal CDF parameters determine the arguments at which the copula density is evaluated.
Interpreting a density as a probability. A density can exceed 1; probabilities require integration over a region.

Key takeaway: For continuous margins, evaluate the copula density at the marginal CDF values and multiply by every marginal density. For independent sampled pairs, the joint likelihood is the product of these factored joint densities, and its logarithm separates into copula and marginal terms.

Check Your Understanding

For each problem, use the continuous density factorization and keep the transformed CDF values distinct from the observed losses.

Under an FGM copula with \(\theta=0.5\), find \(c(u,v)\) at \(u=0.2\) and \(v=0.6\). (Answer: \(0.9400\))
Let \(X\) and \(Y\) have exponential margins with rates 2 and 1. At \(x=y=\ln 2\), use an FGM copula with \(\theta=0.5\) to find the joint density. (Answer: \(0.2500\))
Two independent sampled pairs have uniform \((0,1)\) margins and observed values \((0.1,0.9)\) and \((0.9,0.1)\). Under an FGM copula with \(\theta=-0.5\), calculate their likelihood. (Answer: \(1.7424\))
At a particular observation, \(F_X(x)=0.3\), \(F_Y(y)=0.8\), \(f_X(x)=0.4\), and \(f_Y(y)=0.25\). Under an FGM copula with \(\theta=0.5\), calculate the joint density. (Answer: \(0.0880\))
For two independent observations, the copula-density and marginal-density values are respectively \((0.9,0.4,0.5)\) and \((1.1,0.2,0.25)\), where each triple lists \(c,f_X,f_Y\). Find the likelihood. (Answer: \(0.0099\))

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