Why Use Beta Regression for a Biomarker?
Some longitudinal biomarkers are proportions or normalized measurements that can take any value strictly between zero and one. Examples include a normalized laboratory measure or the fraction of a specified component in a sample. Their range is bounded, and their variability often changes with their mean. A Gaussian model can assign positive probability to impossible values below zero or above one, and a constant residual variance may be a poor description of the data. Beta regression addresses these features by modeling a response on the unit interval with a beta distribution.
In a longitudinal model, the response is recorded repeatedly for subject \(i\) at occasion \(j\), and its conditional distribution may depend on time, covariates, and subject-specific random effects. The beta distribution handles the response’s bounded continuous scale; it does not, by itself, account for within-subject dependence. That dependence must be represented in the longitudinal specification.
Mean and Precision Parameterization
Write \(\alpha_{ij}=\mu_{ij}\phi_{ij}\) and \(\beta_{ij}=(1-\mu_{ij})\phi_{ij}\). The conditional beta density is
The parameter \(\mu_{ij}\) is the conditional mean, while \(\phi_{ij}\) is the precision. In this parameterization, the conditional variance is
Thus, at a fixed mean, increasing precision reduces variance. Precision does not mean that observations are more likely to be near one particular endpoint; it controls concentration around the mean. The variance also depends on the mean: even at the same precision, a mean near 0.5 has a larger variance than a mean near either endpoint.
A standard regression specification uses a logit link for the mean and a log link for precision. Let \(x_{ij}\) be a vector of predictors for the mean and \(w_{ij}\) a vector for precision. In a repeated-measures setting, a subject-specific mean model can include random effects \(b_i\):
Here \(\beta\) and \(\gamma\) are regression coefficient vectors, and \(z_{ij}\) selects the random effects entering the mean. The logit ensures that the fitted mean remains strictly between zero and one; the log link ensures positive precision. A constant-precision model is the special case in which \(\phi_{ij}=\phi\) for all observations. A precision model with covariates is useful when conditional dispersion changes systematically with time or other predictors.
For example, if \(x_{ij}\) includes time, its coefficient describes a change in the log odds of the conditional mean per time unit, holding other modeled terms fixed. It is not a change of that size in the biomarker itself. Likewise, a coefficient in the precision equation describes a change in log precision, not a direct change in the mean.
Worked Examples
Worked Example: Converting Mean and Precision to Beta Shapes
Suppose a hypothetical biomarker measurement has conditional mean \(\mu=0.70\) and precision \(\phi=20\). Find the beta shape parameters and conditional standard deviation.
The shapes are \(\alpha=\mu\phi\) and \(\beta=(1-\mu)\phi\), so
The variance and standard deviation are
The model therefore describes a response with mean 0.70 and standard deviation 0.100 on the zero-to-one scale. The density is supported only between zero and one. These calculations are conditional on the specified mean and precision; a longitudinal model may have different values for different subjects or occasions.
Worked Example: Predicting the Conditional Mean From Time
For a hypothetical subject with random intercept \(b_i=0\), suppose the fitted mean equation is \(\operatorname{logit}(\mu_{ij})=-1+0.8t_{ij}\), where time is measured in weeks. Find the fitted mean at week 2.
First calculate the linear predictor: \(-1+(0.8)(2)=0.6\). Applying the inverse logit gives
The fitted conditional mean at week 2 is approximately 0.646, on the zero-to-one biomarker scale. The time coefficient of 0.8 is a change in log odds per week, not an increase of 0.8 biomarker units per week. The calculation assumes the stated model and conditions on \(b_i=0\); a subject with a nonzero random intercept would generally have a different fitted mean.
Worked Example: A Covariate Changes Precision, Not the Mean
Suppose a hypothetical model has a constant mean of 0.60 for two groups and precision equation \(\log(\phi)=\log(20)+0.5x\), where \(x=0\) for the reference group and \(x=1\) for the second group. Find each group’s precision and conditional variance.
For \(x=0\), \(\log(\phi)=\log(20)\), so \(\phi=20\). The variance is
For \(x=1\), \(\log(\phi)=\log(20)+0.5\), giving \(\phi=20\exp(0.5)\approx 32.97\). Its variance is
The second group has greater precision and lower conditional variance, while both groups have the same modeled mean of 0.60. The variance ratio is \(0.007064/0.01143\approx 0.618\), so the second group’s modeled variance is about 38.2% lower. This is a model-based comparison, not evidence that group membership causes a change.
Assumptions and a Short Justification
Beta regression requires a continuous response strictly inside the unit interval. Conditional on the modeled predictors and random effects, the beta distribution must be an adequate description of the response distribution. In a repeated-measures model, observations are typically treated as conditionally independent given the subject-specific effects and modeled predictors; dependence left over after conditioning can invalidate standard uncertainty calculations. The chosen mean and precision equations must also be adequate: a suitable mean model cannot compensate for a systematic misspecification of dispersion, or vice versa.
The mean and variance expressions follow directly from the beta distribution’s shape parameters. For a beta random variable with shapes \(\alpha\) and \(\beta\), its mean is \(\alpha/(\alpha+\beta)\) and its variance is \(\alpha\beta/[(\alpha+\beta)^2(\alpha+\beta+1)]\). Here \(\alpha+\beta=\phi\), so the mean is \(\mu\phi/\phi=\mu\). Substituting the shapes into the variance gives
This derivation explains why precision controls concentration and why the variance changes with the mean even when precision is constant.
Diagnostics, Alternatives, and Limitations
A fitted beta model should be checked for more than convergence. Compare observed and fitted biomarker distributions across time and important covariate groups, and examine residual patterns for systematic changes in mean or spread. Residuals with trends over time, group-specific spread not captured by the precision equation, or pronounced departures from the fitted distribution suggest that the specification may be inadequate. In repeated measurements, remaining subject-level patterns can indicate that the random-effects structure or other representation of dependence is insufficient.
Exact zeros and ones are not values in the support of the ordinary beta distribution. A recorded endpoint may reflect a genuine boundary value, a detection limit, rounding, or a data-processing convention; these situations are not interchangeable. Check how the measurements were generated and recorded before choosing a model. If zeros or ones are genuine and occur with positive probability, a model designed for boundary masses, such as a zero-inflated or hurdle biomarker model, may be more appropriate. Do not simply insert an endpoint into the beta density, or silently alter the observations without justifying the change.
A Gaussian longitudinal model is an alternative when the response is effectively unbounded or when its conditional distribution is adequately approximated by a Gaussian distribution. It can be convenient, but fitted values or predictive distributions may extend outside zero and one, and a constant residual variance may fail. Beta regression respects the interior support and can represent mean-dependent variance, but it cannot accommodate genuine endpoint mass without an extension. A transformation followed by a Gaussian model is another possibility, but it changes the scale of interpretation and does not automatically solve endpoint problems.
Beta regression also has limits. The standard model is for continuous interior responses, and its adequacy depends on the chosen mean and precision predictors and on the dependence structure used for repeated data. A good fit for the marginal shape at one occasion does not establish that within-subject dependence is correctly represented. In a joint longitudinal and time-to-event analysis, the beta model specifies the longitudinal response distribution; it does not alone specify the event-time model or the relationship between the longitudinal process and event risk.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- A beta-regression model has \(\mu=0.40\) and \(\phi=9\). What are its shape parameters and conditional variance?
- If the mean stays fixed and precision doubles, what happens to the variance? Does the variance necessarily halve?
- A fitted mean equation gives a linear predictor of \(-0.4\). Calculate the fitted conditional mean using the logit link.
- Why is an observed value of exactly 1.00 not directly compatible with the ordinary beta density?
Solutions. (1) The shapes are \(\alpha=(0.40)(9)=3.6\) and \(\beta=(0.60)(9)=5.4\). The variance is \((0.40)(0.60)/(1+9)=0.24/10=0.0240\). (2) The variance changes from \(\mu(1-\mu)/(1+\phi)\) to \(\mu(1-\mu)/(1+2\phi)\). It decreases, but does not generally halve because the denominator changes from \(1+\phi\) to \(1+2\phi\), not from \(\phi\) to \(2\phi\). (3) The inverse logit is \(\exp(-0.4)/[1+\exp(-0.4)]=0.6703/1.6703\approx 0.4013\). (4) The ordinary beta distribution has support only for values strictly between zero and one, so an exact 1.00 has zero probability under that continuous model. The data-generating or recording mechanism should be investigated and a model that accommodates the endpoint considered if appropriate.