Why Compare Three Tests?
A fitted rating coefficient describes an estimated effect; a test asks whether the data provide evidence against a specified no-effect hypothesis. In a GLM, the same hypothesis can be assessed using a Wald test, a score test, or a likelihood-ratio test. These methods differ in which fitted quantities they need and can give different results in finite samples. Under standard regularity conditions, they are asymptotically equivalent.
For example, under a log link, a coefficient \(\beta\) for a rating level gives a multiplicative relativity of \(\exp(\beta)\), relative to the applicable reference under the chosen coding. Testing \(H_0:\beta=0\) is therefore testing whether that coefficient's multiplicative effect is 1. The tests described here concern the coefficient or coefficients specified in the null; they do not establish that an entire rating model is adequate.
The model is assumed to have a correctly specified likelihood, independent observations or cells conditional on the predictors, and identifiable parameters. The usual chi-square approximations also rely on adequate information and regularity conditions, such as parameters not being on problematic boundaries. In a GLM for frequency or loss cost, an offset such as log exposure may be included, but comparisons must use the same observations, response definition, offset, and likelihood specification.
The Three Tests and Their Requirements
The Wald test starts from the unrestricted fit. It compares the estimated restrictions with their null values, scaled by the estimated covariance matrix of the coefficients. If \(\widehat{\boldsymbol{\beta}}\) is the unrestricted estimate and \(\widehat V\) its estimated covariance matrix, the statistic is:
For one coefficient, \(H_0:\beta_j=0\), the Wald statistic is often reported as \(z=\hat\beta_j/\operatorname{SE}(\hat\beta_j)\). Squaring it gives a one-degree-of-freedom chi-square statistic, \(W=z^2\). The Wald test requires an unrestricted estimate and its standard error, or the relevant covariance matrix for a joint test.
The score test evaluates the direction and strength of the likelihood's slope at the restricted fit. It requires fitting the model under the null, but not fitting the unrestricted model. To state the joint-test formula precisely, write the parameters as \((\boldsymbol{\gamma},\boldsymbol{\psi})\), where \(\boldsymbol{\psi}\) contains the \(q\) tested coefficients. Fit the restricted model with \(\boldsymbol{\psi}=\boldsymbol{0}\), estimating nuisance coefficients \(\boldsymbol{\gamma}\). Let \(U_{\psi}\) be the score for \(\boldsymbol{\psi}\) and let \(I\) be the information matrix, partitioned into blocks for \(\boldsymbol{\gamma}\) and \(\boldsymbol{\psi}\). The nuisance-adjusted score and information are:
At the restricted maximum-likelihood estimate, \(U_{\gamma}=0\). The tested score is then \(U_{\psi}\), but its information must still account for estimation of the nuisance parameters. The score statistic is \(U_{\mathrm{eff}}^\mathsf{T}I_{\mathrm{eff}}^{-1}U_{\mathrm{eff}}\), which is approximately chi-square with \(q\) degrees of freedom under the null. In a one-parameter test with no nuisance parameters, this reduces to the squared score divided by its information.
The likelihood-ratio (LR) test compares the maximized likelihood under the unrestricted model with that under the restricted model. Denote their maximized log-likelihoods by \(\ell_{\mathrm{full}}\) and \(\ell_{\mathrm{reduced}}\), respectively. For nested models fitted to the same data under the same likelihood:
where \(q\) is the number of independent restrictions, equivalently the number of additional free parameters in the full model. For a GLM, a difference in residual deviances can give the same LR statistic when the deviances are calculated from the same likelihood and the models are nested. A quasi-likelihood fit alone does not define a full likelihood-ratio test; do not label a deviance comparison an LR test unless the likelihood basis justifies it.
| Test | Fit required | What it measures | Reference distribution |
|---|---|---|---|
| Wald | Unrestricted model and coefficient covariance | Distance of the unrestricted estimate from the null, scaled by uncertainty | Approximately \(\chi^2_q\) |
| Score | Restricted model and score/information at that fit | Slope of the likelihood away from the null, adjusted for nuisance parameters | Approximately \(\chi^2_q\) |
| Likelihood-ratio | Restricted and unrestricted models | Loss in maximized log-likelihood from imposing the null | Approximately \(\chi^2_q\) |
Why the Tests Are Asymptotically Similar
The relationship can be seen by considering a smooth profile log-likelihood \(\ell_p(\boldsymbol{\psi})\), with nuisance parameters optimized for each tested value. For clarity, suppose first that there is one tested coefficient and the null is \(\psi=0\). Near the null, a quadratic approximation gives:
The maximum of this quadratic occurs where its derivative is zero, so \(\hat\psi\approx U_{\mathrm{eff}}/I_{\mathrm{eff}}\). Substituting that value into the quadratic gives the approximate LR statistic:
The expression on the right is the score statistic. The estimated variance of \(\hat\psi\) is approximately \(1/I_{\mathrm{eff}}\), so the Wald statistic is approximately \(\hat\psi^2/(1/I_{\mathrm{eff}})=I_{\mathrm{eff}}\hat\psi^2\) as well. Thus the three statistics agree to the order captured by the local quadratic approximation. In finite samples, the likelihood may not be well approximated by the same quadratic over the relevant range, so the numerical results need not match.
Worked Applications
Worked Example: Wald Test for One Log-Link Rating Effect
A hypothetical insurer fits a frequency GLM with a log link. For a rating indicator, the unrestricted estimate is \(\hat\beta=0.180\), with standard error \(0.075\). Test at the 5% level whether the coefficient is zero. Report the test statistic, approximate p-value, and interpretation as a relativity.
Method. This is a one-coefficient Wald test. The null is \(H_0:\beta=0\), against \(H_1:\beta\ne 0\), so \(q=1\). The standardized estimate is:
Under \(H_0\), \(W\) is approximately \(\chi^2_1\). The two-sided normal-tail p-value is approximately \(0.0164\), equivalently the upper-tail probability for \(\chi^2_1\) at 5.760. Since \(0.0164<0.05\), reject \(H_0\) at the 5% level.
Under the log link, the estimated multiplicative factor is \(\exp(0.180)=1.1972\), or about 1.197. The fitted factor is approximately 19.7% above 1, and the test provides evidence against a coefficient of zero. The common mistake is to compare the coefficient \(0.180\) directly with a chi-square critical value; the Wald chi-square statistic is \(5.760\), not \(0.180\). The result is an approximate test, not proof that the true effect is exactly the estimated factor.
Worked Example: Likelihood-Ratio Test of Two Rating Coefficients
A hypothetical property model compares two nested Poisson GLMs fitted to the same claim counts, exposure offsets, and rating cells. The reduced model constrains two territory coefficients to zero; the full model estimates both. Their maximized log-likelihoods are \(-817.0\) and \(-812.6\), respectively. Test the joint null at the 5% level.
Method. The LR test uses both fits. The full model has two additional free coefficients, so \(q=2\). The statistic is:
For two degrees of freedom, the upper-tail probability is \(\exp(-G^2/2)\). Therefore:
Because \(0.0123<0.05\), reject the joint null. The data provide evidence that at least one of the two territory coefficients differs from zero, conditional on the other terms in the model. The joint test does not identify which coefficient is responsible; that requires separate estimates or additional tests. The common mistake is to use \(2|\ell_{\mathrm{full}}|\), which would give \(1{,}625.2\). The LR statistic is twice the difference in maximized log-likelihoods, here \(8.8\).
Worked Example: Score Test at the Restricted Fit
A hypothetical claim-severity GLM tests two rating coefficients jointly. At the restricted fit, the nuisance-adjusted score vector for the two coefficients is \(U_{\mathrm{eff}}=(3,-1)^\mathsf{T}\), and the corresponding efficient information matrix is:
Calculate the score statistic and decide whether the joint null is rejected at 5%.
Method. The restricted fit has already accounted for the nuisance coefficients. The determinant of the information matrix is \(4(3)-1(1)=11\), so:
Multiplying by the score vector gives:
Thus:
Under the joint null, compare with \(\chi^2_2\). The p-value is \(\exp(-3.364/2)\approx0.186\), which exceeds 0.05, so do not reject the joint null. The common mistake is to treat the two coefficients as independent and divide each squared score by only its diagonal information. The off-diagonal entries represent information shared between the coefficients, so the full matrix calculation is required.
Choosing and Reporting a Test
The appropriate test depends partly on what the analysis has fitted and what question is being asked. A Wald test is convenient when coefficient estimates and their covariance matrix are available. A score test can assess a proposed effect using only the restricted fit, which can be useful when an unrestricted fit is costly or difficult. An LR test directly compares the fit of nested likelihood models, but it requires both fits and a valid common likelihood.
Specify the coefficient or linear combination being tested, and determine the number \(q\) of independent restrictions.
For an LR test, confirm the models are nested and use the same response, observations, offsets, and likelihood specification.
Use the unrestricted estimate and covariance for Wald, restricted-fit score and information for score, or both maximized likelihoods for LR.
Use the appropriate chi-square degrees of freedom, report the p-value or decision, and describe the conclusion in the context of the rating effect.
A test's p-value is evidence against the stated null under the assumed model; it is not the probability that the null is true. Nor does statistical significance alone determine whether a rating change is suitable. The estimated effect, uncertainty, model assumptions, and practical or regulatory constraints remain relevant to a pricing decision.
Common Exam Traps
- Using the wrong degrees of freedom. A joint test of three independent restrictions uses \(q=3\), even if the individual coefficients have different standard errors.
- Confusing a one-sided coefficient test with a two-sided test. The usual test of \(H_0:\beta=0\) against a nonzero alternative is two-sided. A reported \(z\)-statistic must be interpreted with the specified alternative.
- Subtracting likelihoods in the wrong order. The unrestricted maximized log-likelihood should be at least as large as the restricted one; \(2(\ell_{\mathrm{full}}-\ell_{\mathrm{reduced}})\) should therefore be nonnegative, apart from numerical issues.
- Calling every deviance comparison an LR test. The interpretation requires a likelihood-based model comparison with compatible fits; a quasi-likelihood analysis does not automatically supply one.
- Assuming the three finite-sample answers must match. Their agreement is an asymptotic result. Different approximations can produce different statistics and p-values with limited information or estimates far from the null.
Check Your Understanding
For each original problem, identify the test and check the calculation before interpreting the result.
- A hypothetical log-link frequency model gives a coefficient estimate of \(-0.120\) with standard error \(0.050\) for a garage-type indicator. Use a two-sided Wald test of zero at 5%. State the statistic and decision. (Answer: \(z=-2.400\), \(W=5.760\), \(p\approx0.0164\); reject at 5%.)
- In a nested GLM comparison, the reduced model log-likelihood is \(-420.3\), and the full model log-likelihood is \(-417.8\). The full model adds three independent coefficients. Calculate the LR statistic and p-value, and state the 5% decision. (Answer: \(G^2=2[-417.8-(-420.3)]=5.0\), \(df=3\), \(p\approx0.1718\); do not reject at 5%.)
- At a restricted fit, a one-coefficient test has score \(U_{\mathrm{eff}}=2.4\) and efficient information \(I_{\mathrm{eff}}=1.6\). Calculate the score statistic and its approximate p-value. (Answer: \(S=2.4^2/1.6=3.6\), \(df=1\), \(p\approx0.0578\).)
- A log-link GLM estimates a coefficient of \(0.25\) for a charging-zone level. State its multiplicative factor and explain what the test \(H_0:\beta=0\) tests. (Answer: \(\exp(0.25)\approx1.284\); the null tests whether the multiplicative factor equals 1, against a nonzero coefficient.)
- A proposed LR comparison uses the same loss observations and response, but the reduced model omits an exposure offset that is present in the full model. Can the ordinary nested-model LR formula be interpreted as a test of the omitted rating coefficients? Explain. (Answer: No. The models do not hold the offset specification fixed, so the likelihood difference does not isolate the stated coefficient restrictions.)