Why the Covariate Distribution Is Part of the Estimand
A GLMM can provide a marginal prediction for each covariate profile by integrating over the random-effects distribution. To report one marginal mean or contrast across profiles, however, another choice remains: which covariate profiles, and in what proportions, should be represented? A study-sample average, an average for an external population, and a prediction for one reference profile can all be calculated from the same fitted model. They generally answer different questions.
For example, a standardized treatment risk for a study sample describes a model-based average over that sample’s covariate distribution. If the target is a different population with a different age or disease-severity distribution, the study-sample average need not describe that population. A prediction at one specified age and severity is different again: it describes that profile rather than averaging across a population.
In practice, \(Q\) may be an empirical distribution of observed covariates, a distribution specified for an external target population, or a point mass at a chosen reference profile. The random-effects integration inside \(m(a,x)\) and the covariate averaging over \(Q\) are distinct operations. For all three choices to be comparable, use the same fitted model, outcome scale, treatment values, and random-effects distribution.
These estimates are model-based descriptions of standardized outcomes. A causal interpretation of a treatment contrast requires additional assumptions, including appropriate control of confounding, consistency, positivity, and a correctly specified model for the relevant quantities. Standardization alone does not establish those assumptions.
Three Choices for the Covariate Distribution
| Choice of \(Q\) | What is averaged | Population or profile described | Main consideration |
|---|---|---|---|
| Study-sample distribution | Predictions across observed study covariate profiles | The study sample, as represented by its empirical covariate distribution | May not represent a broader target population |
| External target distribution | Predictions across covariate profiles and proportions specified for the target | The stated external target population | Requires credible target-distribution information and adequate model support |
| Reference profile | One prediction at a fixed covariate value; \(Q\) is a point mass | The specified profile only | Is not a population average unless the target itself is that point profile |
For observed profiles \(x_1,\ldots,x_n\), empirical standardization gives $$ \theta_{\mathrm{sample}}(a)=\frac{1}{n}\sum_{i=1}^{n}m(a,x_i). $$ If the target has discrete profiles \(x_k\) with proportions \(q_k\), where \(q_k\geq 0\) and \(\sum_k q_k=1\), then $$ \theta_Q(a)=\sum_k q_km(a,x_k). $$ For a reference profile \(x^\ast\), the distribution places all its mass at \(x^\ast\), so \(\theta_Q(a)=m(a,x^\ast)\). These formulas use the same covariate-specific predictions but different averaging rules.
Worked Example: Standardizing to the Study Sample
Worked Example: Standardizing to the Study Sample
Hypothetical data: Suppose a fitted binary-outcome GLMM supplies the following random-effects-marginal risks for treatment \(A=1\) and control \(A=0\). The study sample contains 30% younger and 70% older participants. Assume the predictions are valid for both treatment values at both age profiles.
| Age profile | Risk if \(A=0\) | Risk if \(A=1\) |
|---|---|---|
| Younger | 0.20 | 0.10 |
| Older | 0.40 | 0.25 |
The empirical covariate distribution assigns weights 0.30 and 0.70. The standardized control risk is $$ \theta_{\mathrm{sample}}(0)=(0.30)(0.20)+(0.70)(0.40)=0.06+0.28=0.34. $$ The standardized treatment risk is $$ \theta_{\mathrm{sample}}(1)=(0.30)(0.10)+(0.70)(0.25)=0.03+0.175=0.205. $$ Thus the standardized risk difference, treatment minus control, is $$ 0.205-0.34=-0.135. $$ Under this model and this study-sample distribution, the estimated treatment risk is 13.5 percentage points lower than the control risk. The estimate describes an average over the sample’s 30% younger and 70% older covariate distribution; it does not automatically describe a population with different age proportions.
Worked Example: Standardizing to an External Target
Worked Example: Standardizing to an External Target
Hypothetical scenario: Use the same model predictions, but suppose a health-planning question concerns an external population that is 60% younger and 40% older. Assume this target distribution is measured appropriately, and that the fitted model can predict outcomes for both profiles under each treatment level.
For the external target, the control risk is $$ \theta_{\mathrm{target}}(0)=(0.60)(0.20)+(0.40)(0.40)=0.12+0.16=0.28. $$ The treatment risk is $$ \theta_{\mathrm{target}}(1)=(0.60)(0.10)+(0.40)(0.25)=0.06+0.10=0.16. $$ The target-standardized risk difference is $$ 0.16-0.28=-0.12. $$ The model therefore gives a 12.0-percentage-point lower treatment risk for this external target distribution. The difference from the study-standardized result, \(-0.135\), is not an arithmetic inconsistency: the age weights changed, so the estimand changed. Neither estimate is universally preferable; the appropriate one depends on which population the question concerns.
This calculation depends on having target covariate proportions that are relevant to the intended population and predictions that are credible at those profiles. If the target includes covariate values absent or extremely rare in the study data, the model may be extrapolating rather than interpolating. A numerical weighted average does not remove that limitation.
Worked Example: A Reference Covariate Profile
Worked Example: A Reference Covariate Profile
Hypothetical model: Let \(Y\) be a count of events per person-period, \(A\) an indicator of treatment, and age be measured in years. Suppose a random-intercept Poisson GLMM has conditional mean $$ E(Y\mid A,\mathrm{age},b)=\exp\{-0.3+0.2A+0.04(\mathrm{age}-50)+b\}, $$ with \(b\sim N(0,0.3^2)\). Consider the reference profile age \(=60\). Assume this model and random-effects distribution are suitable for the prediction.
At age 60, the age contribution is \(0.04(60-50)=0.40\). For \(A=0\), the linear predictor excluding \(b\) is \(-0.3+0+0.40=0.10\); for \(A=1\), it is \(-0.3+0.20+0.40=0.30\). Integrating over the normal random intercept gives $$ m(a,60)=\exp\{-0.3+0.2a+0.04(60-50)+0.3^2/2\}. $$ Since \(0.3^2/2=0.045\), the reference-profile marginal means are $$ m(0,60)=\exp(0.145)\mathrel{\approx}1.156,\qquad m(1,60)=\exp(0.345)\mathrel{\approx}1.412. $$ The difference is \(1.412-1.156=0.256\) events per person-period, approximately. These are random-effects-marginal predictions for people at age 60, not averages over an age distribution. If the target population includes a range of ages, its standardized mean requires averaging predictions across that population’s age distribution.
Why the Averaging Formula Works—and What It Does Not Fix
The standardization formula follows from averaging the profile-specific marginal mean under the chosen distribution. For a discrete target, partition the population into profiles \(x_k\) with proportions \(q_k\). By the law of total expectation, $$ E_Q(Y^a)=\sum_k P_Q(X=x_k)E(Y^a\mid X=x_k) =\sum_k q_km(a,x_k), $$ where the notation \(Y^a\) is appropriate when the prediction is interpreted as an outcome under setting \(A=a\). Without the additional causal assumptions needed for that interpretation, the same weighted sum remains a model-based standardized prediction, not necessarily a causal mean.
The derivation also makes the key point visible: changing \(q_k\) changes the averaging population even when every \(m(a,x_k)\) stays fixed. Standardization does not correct a misspecified outcome model, create information for unsupported profiles, or make an external covariate distribution representative merely by weighting predictions with it.
Diagnostics, Alternatives, and Limitations
Before calculating a standardized result, state the population it is intended to describe and define the covariates and their distribution precisely. Then check that the fitted model can support predictions at the profiles receiving weight. Useful checks include comparing study and target covariate distributions, identifying target profiles outside the study’s observed range, and examining whether a small number of profiles carry most of the target weight. Where support is poor, report the limitation and consider whether a narrower target or additional data are needed.
The three choices are alternatives for different questions, not competing recipes for one universally correct answer. Study-sample standardization is natural when the target is the study’s observed covariate mix. External standardization is useful when the stated target population differs and its covariate distribution can be specified. A reference profile is useful for a focused prediction at a meaningful covariate value, but can be unrepresentative if presented as a summary of a heterogeneous population.
A conditional GLMM coefficient is another kind of summary: it describes a model parameter on the conditional scale, whereas standardized means and contrasts are response-scale quantities averaged over random effects and, where applicable, covariates. They need not coincide or have the same interpretation. The standard GLM material on link scales and response-scale predictions is useful background, but it does not decide which population distribution should define \(Q\).
Finally, the target distribution itself may be estimated rather than known. Uncertainty in that distribution can matter, in addition to uncertainty in fitted model parameters. A reported estimate should identify the target, covariate distribution, outcome scale, and contrast direction; uncertainty procedures should reflect the design and sources of uncertainty relevant to the analysis.
Common Mistake
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- A study has two covariate profiles with risks under control of 0.10 and 0.30, and under treatment of 0.08 and 0.24. The profiles make up 25% and 75% of the study sample. Calculate the sample-standardized risks and risk difference, treatment minus control.
- For the same predictions, an external target is 75% profile 1 and 25% profile 2. Calculate its standardized risks and risk difference. Why can it differ from the study-standardized result?
- What population does a prediction standardized to a point mass at age 60 describe? Is it an average over all ages?
- Name two checks that help assess whether external standardization is credible.
Solutions. (1) Control: \((0.25)(0.10)+(0.75)(0.30)=0.025+0.225=0.25\). Treatment: \((0.25)(0.08)+(0.75)(0.24)=0.02+0.18=0.20\). The difference is \(0.20-0.25=-0.05\), or 5 percentage points lower under treatment. (2) Control: \((0.75)(0.10)+(0.25)(0.30)=0.15\). Treatment: \((0.75)(0.08)+(0.25)(0.24)=0.12\). The difference is \(0.12-0.15=-0.03\). It differs because the external target assigns different proportions to the profiles. (3) It describes the specified age-60 profile, with random-effects integration as defined by the model; it is not an average over all ages. (4) Compare the target covariate distribution with the study data, and check for target profiles outside or poorly represented in the study’s covariate support. Also examine whether a few profiles dominate the target weights.