Choosing Which Predictor Slopes to Relax

Choosing a Relaxation Is a Model-Selection Decision

A partial proportional-odds model allows some predictors to have a common effect across response cutpoints while allowing other predictors’ effects to vary. The practical question is not simply whether the parallel-slopes assumption fails somewhere. It is which predictor or predictor block should receive additional cutpoint-specific slopes. A useful decision combines targeted tests, comparisons of nested models, and knowledge of how the predictors relate to the scientific question.

A small p-value is evidence against a particular restriction, not an automatic instruction to relax that predictor. Testing many predictors can generate a small p-value by chance; correlated predictors can make isolated tests misleading; and a model that fits better may be unnecessarily complicated or scientifically incoherent. The selection process should make these issues explicit.

Definition: Let the ordered response have categories $1,\ldots,J$, with $J\geq 3$. Write the cumulative-logit model as $$ \log\left\{\frac{P(Y\leq j\mid x)}{P(Y>j\mid x)}\right\} =\alpha_j-\sum_{r=1}^{p}x_r\beta_{rj},\qquad j=1,\ldots,J-1. $$ A predictor $x_r$ has a proportional-odds (parallel-slopes) effect when $\beta_{r1}=\cdots=\beta_{r,J-1}$. Relaxing that predictor permits its slope to vary by cutpoint, subject to any constraints required for admissible category probabilities. A partial proportional-odds model relaxes this equality for selected predictors and retains it for the others.

For a categorical predictor represented by $q$ contrasts, the restriction applies to its $q$-dimensional coefficient vector: each contrast has the same coefficient at every cutpoint under proportional odds. A predictor-specific test therefore usually treats all the contrasts for that predictor as one block. Splitting a categorical predictor’s contrasts into separate tests can answer a different question and may produce an awkward model that depends on the chosen coding.

Targeted Tests and Their Degrees of Freedom

Suppose a model already contains a chosen set of terms, and consider relaxing predictor $r$. Let $q_r$ be the number of coefficients used to represent it. Its common-slope version has $q_r$ slope parameters. Its unrestricted cutpoint-specific version has $q_r(J-1)$. Thus the relaxation introduces $$ d_r=q_r(J-2) $$ additional parameters: for each contrast, the equality of $J-1$ slopes imposes $J-2$ independent restrictions.

A predictor-specific score test evaluates those restrictions at the fit under the common-slope null. In general, if $\delta$ is the vector of added slope differences and $\eta$ denotes nuisance parameters, the efficient score and information for $\delta$, evaluated under $\delta=0$, are $$ U_{\delta\cdot\eta}=U_\delta-I_{\delta\eta}I_{\eta\eta}^{-1}U_\eta,\qquad I_{\delta\cdot\eta}=I_{\delta\delta}-I_{\delta\eta}I_{\eta\eta}^{-1}I_{\eta\delta}. $$ The score statistic is $Q=U_{\delta\cdot\eta}^{T}I_{\delta\cdot\eta}^{-1}U_{\delta\cdot\eta}$. The nuisance adjustment matters: it accounts for estimating thresholds and the other model parameters, rather than treating them as known.

Alternatively, fit the common-slope model and the model with predictor $r$ relaxed, leaving the other terms unchanged. The likelihood-ratio statistic is twice the difference in maximized log-likelihoods. Under regularity conditions, it has an asymptotic chi-square reference distribution with $d_r$ degrees of freedom under the null. Score and likelihood-ratio tests can differ in finite samples, but they test the same restriction in their intended setting.

Theorem: Suppose the observations are independent and the null and alternative are nested, identifiable models; the true parameter is an interior point of the null parameter space; the information matrix is nonsingular; and the usual differentiability and regularity conditions for likelihood inference hold. If relaxing a predictor adds $d_r$ free parameters, then under the null the likelihood-ratio statistic converges in distribution to $\chi^2_{d_r}$. For a predictor represented by $q_r$ contrasts, $d_r=q_r(J-2)$.

Count the added parameters.
The unrestricted predictor contributes $q_r(J-1)$ slopes instead of $q_r$, so the difference is $q_r(J-2)$.

Form the nested comparison.
The null sets the added slope differences to zero. The alternative estimates them freely, while retaining the same observations, thresholds, and other terms.

Apply likelihood theory.
Under the stated regularity conditions, Wilks’ theorem gives a chi-square limit with degrees of freedom equal to the number of independent restrictions, here $d_r$. Therefore the observed likelihood-ratio statistic is assessed against $\chi^2_{d_r}$.

Worked Examples of Predictor-Specific Evidence

Worked Example: Testing a Continuous Predictor With Four Categories

Suppose the outcome has $J=4$ ordered categories and a fitted proportional-odds model includes a continuous exposure. Testing whether its slope can vary across cutpoints compares one common slope with three cutpoint-specific slopes. The added degrees of freedom are $$ d=(1)(4-2)=2. $$ Assume the predictor-specific score statistic is $Q=7.82$. Its null reference distribution is $\chi^2_2$. For two degrees of freedom, the upper-tail probability is $\exp(-Q/2)$, so $$ p=\exp(-7.82/2)=\exp(-3.91)\approx 0.0200. $$ This is evidence against a common slope for this exposure, at conventional unadjusted significance levels. It does not establish that the exposure should be the only relaxed predictor: other predictors may also have been tested, and substantive considerations still matter.

Worked Example: Relaxing a Three-Level Predictor as a Block

Let a treatment predictor have three levels, represented by $q=2$ indicator contrasts, and let the response have $J=5$ ordered categories. Relaxing treatment as a whole adds $$ d=q(J-2)=2(5-2)=6 $$ parameters. Suppose the likelihood-ratio statistic comparing the relaxed and common-slope models is $12.59$. Against $\chi^2_6$, this is approximately the 95th percentile, giving $p\approx0.050$.

The test is an omnibus test: it asks whether the treatment coefficient vector is constant across cutpoints. It does not identify which contrast differs or at which cutpoint. Nor should the result be interpreted as two independent decisions about the treatment indicators; the block test respects the coding of the categorical predictor.

Multiple Testing and Dependence

If separate score tests are calculated for $m$ candidate predictors, the collection of p-values is a family of tests. With many tests, the chance of at least one false rejection can exceed the nominal level. A simple family-wise error control is Bonferroni: compare each p-value to $\alpha/m$, or multiply each p-value by $m$ and cap at one. Holm’s step-down procedure is usually less conservative: sort the p-values, compare the smallest to $\alpha/m$, the next to $\alpha/(m-1)$, and continue until a comparison fails.

Result: Bonferroni control keeps the probability of one or more false rejections at most $\alpha$, provided each individual null p-value is valid. If $A_i$ is the event of falsely rejecting the $i$th null, then $$ P\left(\bigcup_{i=1}^{m}A_i\right)\leq\sum_{i=1}^{m}P(A_i)\leq m\frac{\alpha}{m}=\alpha. $$ The first inequality is the union bound, so this argument does not require the tests to be independent.

Worked Example: Applying Holm’s Procedure

Suppose five predictors are screened, with p-values $0.004,\ 0.018,\ 0.031,\ 0.20,$ and $0.62$. At family-wise level $\alpha=0.05$, sort them as shown and apply Holm’s thresholds:

Rank	p-value	Holm threshold	Decision
1	0.004	0.05/5 = 0.010	Reject
2	0.018	0.05/4 = 0.0125	Stop; do not reject this or later nulls
3	0.031	0.05/3 ≈ 0.0167	Not reached
4	0.20	0.05/2 = 0.025	Not reached
5	0.62	0.05	Not reached

Only the first predictor is selected by this family-wise testing rule. The unadjusted p-value $0.018$ would have passed a 0.05 cutoff, but not the second Holm threshold. This is a screening decision under a specified error-control rule, not proof that every other predictor truly has a common slope. In exploratory work, false-discovery-rate procedures may instead be appropriate when the goal is to limit the expected proportion of false findings among rejections.

Predictor-specific tests are generally dependent. Predictors may be correlated, and their score statistics may share information from the same thresholds and observations. Bonferroni and Holm remain valid under dependence when the individual p-values are valid, but can be conservative. A joint test of several relaxations answers whether the block as a whole improves on the restrictions; it does not identify a unique predictor within that block. The question being asked should determine whether to use individual tests, a joint test, or both.

Likelihood Comparisons, Hierarchy, and Parsimony

A likelihood-ratio comparison is meaningful as a nested-model test only when one model is obtained from the other by imposing restrictions and both models use the same response observations and likelihood specification. If several predictors are relaxed together, the degrees of freedom are the sum of their added parameter counts, provided the restrictions are independent. Comparing nonnested candidates by subtracting their log-likelihoods does not produce the usual likelihood-ratio test.

A likelihood improvement should also be weighed against model complexity. For models fitted by maximum likelihood to the same data, AIC is $ -2\ell+2k$, where $k$ is the number of estimated parameters. If a relaxation adds $d$ parameters and increases the maximized log-likelihood by $\Delta\ell$, then $$ \mathrm{AIC}_{\mathrm{relaxed}}-\mathrm{AIC}_{\mathrm{restricted}}=-2\Delta\ell+2d. $$ Thus AIC favors the relaxed model exactly when $\Delta\ell>d$, or equivalently when the likelihood-ratio statistic $2\Delta\ell$ exceeds $2d$. This is a selection criterion, not a significance test. The threshold is not the same as a conventional chi-square critical value.

Worked Example: AIC and a Parsimonious Relaxation

Return to a continuous predictor with $J=4$, for which relaxation adds $d=2$ parameters. Suppose the likelihood-ratio statistic is $6.4$, so $\Delta\ell=6.4/2=3.2$. The AIC difference is $$ \mathrm{AIC}_{\mathrm{relaxed}}-\mathrm{AIC}_{\mathrm{restricted}} =-6.4+2(2)=-2.4. $$ The relaxed model has AIC lower by $2.4$, so AIC favors it among these two candidates. The chi-square test gives $p=\exp(-6.4/2)=\exp(-3.2)\approx0.0408$, also providing unadjusted evidence against the restriction. These are distinct summaries: AIC penalizes complexity according to its criterion, while the test evaluates a null hypothesis under a reference distribution.

Statistical evidence is only one part of choosing a model. If a predictor is a pre-specified confounder or is required by the scientific design, it may need to remain in the model regardless of its slope test. If an interaction is included, model hierarchy commonly requires retaining the corresponding lower-order terms, even if those terms are not individually significant. For a multi-level categorical predictor, retaining or relaxing its contrasts as a block usually preserves a coding-invariant interpretation. These are modeling commitments to document, not conclusions that a p-value can replace.

Substantive knowledge can also identify plausible candidates before examining test results. For example, a predictor whose effect is expected to differ between lower and upper portions of an ordinal scale may warrant a pre-specified relaxation. Such reasoning should be stated in advance when possible. Selecting a model after inspecting many tests and then reporting only the chosen p-value understates the selection process; adjusted inference, validation, or transparent reporting may be needed.

Common Pitfalls and a Practical Decision Sequence

A frequent mistake is to relax every predictor with an unadjusted p-value below 0.05, then describe the final model as if it had been specified before looking at the data. This ignores multiplicity and can create an unnecessarily complicated model. The correction is to define the candidate family, use an appropriate multiplicity strategy, and distinguish exploratory screening from confirmatory testing.

Another mistake is to treat a nonsignificant test as proof that the predictor’s effect is parallel. A test may have little power, especially with sparse categories or limited data. Conversely, a statistically significant test need not justify a complicated model if the estimated departure is small and not substantively useful. Evaluate estimates and uncertainty, sample support, admissibility of the cumulative probabilities, and the purpose of the analysis. A chosen set of relaxed slopes must still yield a coherent ordinal model; a favorable test statistic alone does not guarantee an admissible fit.

A defensible selection record identifies the candidate predictors and blocks, states which tests were used and whether p-values were adjusted, reports the likelihood comparisons and parameter costs, and explains any departures from the statistical screening. Hierarchy and pre-specified scientific priorities should be recorded alongside parsimony, rather than invoked after the fact to justify a convenient result.

Define candidate blocks.
Decide which predictors may be relaxed, how categorical predictors are grouped, and which terms must be retained for design or hierarchy.

Test stated restrictions.
Use predictor-specific score tests or nested likelihood comparisons, with degrees of freedom based on the added slope parameters.

Account for the search.
Apply a family-wise or false-discovery-rate procedure when appropriate, and recognize that correlated tests do not create independent evidence.

Choose and explain the model.
Combine adjusted evidence with substantive priorities, hierarchy, admissibility, and a clear assessment of whether the improved fit justifies added complexity.

Key takeaway: Predictor-specific tests identify evidence against particular parallel-slope restrictions, but they do not select a scientifically defensible model by themselves. Define the candidate blocks, account for multiple testing, compare fit with complexity, and make hierarchy and substantive priorities explicit.

Check Your Understanding

Use the definitions and decision principles above to answer these questions.

For $J=6$ response categories, how many additional slope parameters are introduced by relaxing a predictor represented by three contrasts?
Why is a joint test of all contrasts for a categorical predictor often preferable to separate tests of its coded indicators?
State one advantage of Holm’s procedure and explain why it does not require independent tests.
A relaxed model has a larger likelihood but a higher AIC. What does that imply about the likelihood improvement relative to the parameter cost?
Give one reason a predictor might be retained or relaxed even when its individual test does not support that decision.