Likelihood-Ratio Tests in Parametric Copula Models

Likelihood-Based Testing of Copula Independence

A copula separates the dependence structure of a multivariate distribution from its marginal distributions. In a parametric copula model, a dependence parameter offers a direct target for testing independence: the null hypothesis sets that parameter to the value corresponding to the independence copula. Likelihood methods are especially useful when the chosen parametric family describes the dependence patterns of interest, but the reference distribution depends on whether the null parameter is an interior point or a boundary point of the parameter space.

We first treat observations $ (U_i,V_i) $ that are independent and identically distributed with a specified copula density $c_\theta$, so their marginal distributions are uniform and known. This assumption isolates the copula likelihood. If margins are estimated from data, the calibration may require additional treatment.

Definition: For a one-parameter copula density $c_\theta$, the log-likelihood from $n$ observations is $$ \ell_n(\theta)=\sum_{i=1}^n \log c_\theta(U_i,V_i). $$ If $\theta_0$ is the parameter value giving independence, the likelihood-ratio statistic for $H_0:\theta=\theta_0$ against an alternative allowing other values is $$ T_{\mathrm{LR}}=2\left\{\sup_{\theta\in\Theta}\ell_n(\theta)-\ell_n(\theta_0)\right\}. $$

For the independence copula, $c_{\theta_0}(u,v)=1$, so $\ell_n(\theta_0)=0$. The supremum is taken over the model's full parameter space, unless the alternative is explicitly one-sided. The likelihood-ratio statistic is nonnegative because the null value is among the values considered under the alternative.

A Gaussian-Copula Example

For a bivariate Gaussian copula with correlation parameter $\rho\in(-1,1)$, let $x_i=\Phi^{-1}(U_i)$ and $y_i=\Phi^{-1}(V_i)$, where $\Phi$ is the standard normal distribution function. The copula density is the bivariate normal density divided by the two standard normal marginal densities. Its log-likelihood is $$ \ell_n(\rho)=-\frac{n}{2}\log(1-\rho^2) -\frac{1}{2(1-\rho^2)} \sum_{i=1}^n\left\{\rho^2(x_i^2+y_i^2)-2\rho x_i y_i\right\}. $$ Independence corresponds to $\rho=0$, an interior point of the parameter space.

Worked Example: Gaussian-Copula Likelihood and Test Statistics

Suppose the transformed observations are $(1,1),(1,1),(-1,-1),(1,-1)$. The cross-products sum to $2$, and the sums of the squared $x_i$'s and $y_i$'s are both $4$. Thus $$ \ell_4(\rho)=-2\log(1-\rho^2)-\frac{4\rho^2-2\rho}{1-\rho^2}. $$

Differentiating gives $$ \ell_4'(\rho)=\frac{2(1-2\rho)(1+\rho^2)}{(1-\rho^2)^2}. $$ The derivative is positive for $\rho<1/2$ and negative for $\rho>1/2$, so the maximum is at $\hat\rho=1/2$. The null log-likelihood is zero, and therefore $$ T_{\mathrm{LR}}=2\ell_4(1/2)=-4\log(3/4)\approx 1.151. $$

At $\rho=0$, the score is $\sum_i x_i y_i=2$. The per-observation Fisher information is $1$, so the full-sample information is $I_4(0)=4$. The score statistic is $2^2/4=1$. A Fisher-information Wald statistic is $4(1/2)^2=1$. These statistics are close here, but they need not agree exactly in a finite sample.

Regular Interior-Null Calibration

The usual chi-square calibration requires regularity. For the result below, assume the observations are i.i.d.; $\theta_0$ is an interior point; the model is identifiable near $\theta_0$; the log density is sufficiently differentiable with integrable derivatives to justify the score and information identities; the score has finite, positive variance; and the maximum-likelihood estimator is consistent and asymptotically normal. These conditions yield a local quadratic likelihood expansion and a central limit theorem for the score.

Theorem: Under these regularity conditions, define the one-observation Fisher information by $$ I_1(\theta_0)=E_{\theta_0}\left[\left\{\frac{\partial}{\partial\theta}\log c_{\theta_0}(U,V)\right\}^2\right]>0, $$ and the full-sample information by $I_n(\theta_0)=nI_1(\theta_0)$. Under $H_0$, the likelihood-ratio statistic, the score statistic $U_n(\theta_0)^2/I_n(\theta_0)$, and the Fisher-information Wald statistic $nI_1(\theta_0)(\hat\theta-\theta_0)^2$ all converge in distribution to $\chi^2_1$, where $U_n(\theta)=\ell_n'(\theta)$.

The information subscripts distinguish the per-observation quantity $I_1$ from the full-sample quantity $I_n=nI_1$. In particular, the score statistic is $U_n^2/[nI_1]$, not $U_n^2/[nI_n]$.

Proof. Under the null, the score central limit theorem gives $$ \frac{U_n(\theta_0)}{\sqrt n}\xrightarrow{d}N\!\left(0,I_1(\theta_0)\right). $$ Write $\Delta_n=U_n(\theta_0)/\sqrt n$. The regular local likelihood expansion, uniformly for bounded $h$, is $$ \ell_n\left(\theta_0+\frac{h}{\sqrt n}\right)-\ell_n(\theta_0) =h\Delta_n-\frac{1}{2}h^2 I_1(\theta_0)+o_p(1). $$ The quadratic on the right is maximized at $h=\Delta_n/I_1(\theta_0)$. Maximizing it and multiplying the resulting log-likelihood difference by $2$ gives $$ T_{\mathrm{LR}}=\frac{\Delta_n^2}{I_1(\theta_0)}+o_p(1). $$ The score statistic is exactly $\Delta_n^2/I_1(\theta_0)$. The likelihood expansion also gives $\sqrt n(\hat\theta-\theta_0)=\Delta_n/I_1(\theta_0)+o_p(1)$, so the Wald statistic has the same leading term. Since $\Delta_n/\sqrt{I_1(\theta_0)}\xrightarrow{d}N(0,1)$, its square converges to $\chi^2_1$. $\square$

For the Gaussian copula, differentiating the log density at $\rho=0$ gives the one-observation score $XY$, where $X$ and $Y$ are independent standard normal variables under the null. Its expectation is zero and its variance is $E[X^2]E[Y^2]=1$, so $I_1(0)=1$. This verifies the information used in the preceding example.

Worked Example With a Farlie–Gumbel–Morgenstern Copula

The Farlie–Gumbel–Morgenstern (FGM) copula has density $c_\theta(u,v)=1+\theta(1-2u)(1-2v)$, with $-1\leq\theta\leq1$. Independence occurs at the interior point $\theta=0$. This simple density makes the score and likelihood calculations explicit.

Worked Example: FGM Maximum Likelihood and Three Statistics

Suppose three observed pairs give $a_i=(1-2U_i)(1-2V_i)$ equal to $1/4,-1/4,-1/16$. For example, the pairs can be $(1/4,1/4)$, $(1/4,3/4)$, and $(1/4,9/16)$. The log-likelihood is $$ \ell_3(\theta)=\log(1+\theta/4)+\log(1-\theta/4)+\log(1-\theta/16). $$ The score at independence is $U_3(0)=\sum_i a_i=-1/16$. The per-observation information at zero is $E_0[a^2]=1/9$, since under independence $U,V$ are independent uniforms and $E[(1-2U)^2]=1/3$. Thus $I_3(0)=3/9=1/3$, and the score statistic is $$ \frac{U_3(0)^2}{I_3(0)}=\frac{1/256}{1/3}=\frac{3}{256}\approx0.0117. $$

To find the maximum-likelihood estimate, set the derivative to zero: $$ \frac{1/4}{1+\theta/4}-\frac{1/4}{1-\theta/4}-\frac{1/16}{1-\theta/16}=0. $$ After multiplying by the positive denominators for $-1<\theta<1$, this reduces to $3\theta^2-32\theta-16=0$. The root in the parameter space is $\hat\theta=(32-\sqrt{1216})/6\approx-0.4785$. The derivative changes from positive to negative at this root, so it is the likelihood maximum. Substitution gives $T_{\mathrm{LR}}\approx0.0301$. The observed information there is $$ J_3(\hat\theta)=\sum_{i=1}^3\frac{a_i^2}{(1+\hat\theta a_i)^2}\approx0.1342. $$ The observed-information Wald statistic is $\hat\theta^2J_3(\hat\theta)\approx0.0307$. The three finite-sample values differ; their common asymptotic reference is $\chi^2_1$ under the regular interior-null conditions.

When Independence Is a Boundary Value

Not every copula parameterization places independence in the interior. For example, the positive-dependence Clayton family has parameter space $\theta\geq0$, with independence obtained by the continuous extension at $\theta=0$. The null is then a boundary point. A symmetric, two-sided normal approximation for the unconstrained estimate is not appropriate: the fitted parameter cannot move below zero.

Theorem: Suppose a scalar parameter has parameter space $\theta\geq0$, the null is $\theta=0$, and the one-sided model satisfies a regular local quadratic likelihood expansion with positive information $I_1(0)$. Under the null, $$ T_{\mathrm{LR}}\xrightarrow{d}\{\max(0,Z)\}^2,\qquad Z\sim N(0,1). $$ This distribution is a mixture assigning probability $1/2$ to zero and probability $1/2$ to a $\chi^2_1$ variable.

Derivation. In the local expansion, feasible local parameters have $h\geq0$. The quadratic $h\Delta_n-\tfrac12h^2I_1(0)$ is maximized at $h=\max(0,\Delta_n/I_1(0))$. Twice its maximum is $\{\max(0,\Delta_n/\sqrt{I_1(0)})\}^2$. The standardized score converges to $Z\sim N(0,1)$. By symmetry, $P(Z\leq0)=1/2$, and on that event the statistic is zero; on $Z>0$, its distribution is the squared positive half of a standard normal, giving the stated mixture. $\square$

Worked Example: Calibrating a One-Sided Boundary Test

For the boundary limit above, a 5% upper-tail test needs a critical value $q$ satisfying $$ \frac12 P(\chi^2_1>q)=0.05. $$ Thus $P(\chi^2_1>q)=0.10$, and $q$ is the 90th percentile of $\chi^2_1$, approximately $2.706$. Using the usual 95th percentile, $3.841$, would give a limiting rejection probability of only $\tfrac12(0.05)=0.025$, not 0.05. This calculation applies only under the stated one-sided regularity conditions; other boundary structures can have different limits.

Choosing and Interpreting the Statistic

Statistic	What it uses	Regular interior-null limit
Likelihood ratio	Difference between the unrestricted and null maximized log-likelihoods	$\chi^2_1$
Score	Null score and null information, without fitting the alternative	$\chi^2_1$
Wald	Distance from the fitted parameter to the null, scaled by estimated uncertainty	$\chi^2_1$

Under regular interior conditions, these tests are asymptotically equivalent, not identical in finite samples. The likelihood-ratio test compares fitted likelihoods; the score test requires fitting only under the null; and the Wald test depends on the estimate and its uncertainty. In a copula family with nuisance parameters, the corresponding score and information must account for nuisance estimation, typically through an efficient score and its information. The one-parameter formulas above do not automatically cover that case.

Common mistake: It is tempting to use a $\chi^2_1$ reference distribution whenever a test has one parameter, and to apply a symmetric Wald interval even when the allowed parameter values are one-sided. The degrees of freedom alone do not establish the null limit. Check whether the null is an interior point, whether the likelihood is regular there, and whether the alternative is constrained. A boundary can produce a mixture limit, as in the one-sided scalar case, while more complicated constraints can produce other limits.

Key takeaway: In a regular one-parameter copula model with independence at an interior point, likelihood-ratio, score, and Wald tests share a $\chi^2_1$ null limit. If independence lies on the parameter-space boundary, derive the calibration from the constrained local likelihood rather than applying the interior result automatically.

Check Your Understanding

Use these questions to check the assumptions and calculations behind the tests.

Why is the independence parameter $\rho=0$ an interior null for the Gaussian copula, while $\theta=0$ is a boundary null in the positive-dependence Clayton family?
Distinguish the per-observation information $I_1(\theta_0)$ from the full-sample information $I_n(\theta_0)$. Write the score statistic using both notations.
In the FGM example, why is the score at independence the sum of the $a_i$'s?
For the one-sided boundary limit, why does the likelihood-ratio statistic equal zero with probability one half?
What additional issue must be considered if the marginal distributions used to construct the copula observations are estimated rather than known?

Statistic	What it uses	Regular interior-null limit
Likelihood ratio	Difference between the unrestricted and null maximized log-likelihoods	\(\chi^2_1\)
Score	Null score and null information, without fitting the alternative	\(\chi^2_1\)
Wald	Distance from the fitted parameter to the null, scaled by estimated uncertainty	\(\chi^2_1\)