Why Model Subject-Specific Baselines?
Repeated biomarker measurements from the same person often tend to be higher or lower than the measurements from another person, even after accounting for treatment and time. A model with only fixed effects gives a mean trajectory for the population, but does not by itself represent this persistent subject-to-subject baseline variation. A random intercept adds a subject-specific deviation from that fixed-effects mean.
The key interpretation is additive: a subject’s random intercept shifts the modeled trajectory up or down by the same amount at every visit. This tutorial focuses on that shift and on how its distribution describes variation in latent subject baselines. It does not quantify the correlation between repeated measurements, which is a separate interpretation.
Model and Assumptions
Let \(Y_{ij}\) be the biomarker measurement for subject \(i\) at visit \(j\), \(A_i\) indicate treatment group (0 for control and 1 for treatment), and \(t_{ij}\) be time in months, coded so that baseline is \(t=0\). For illustration, consider the following linear mixed model:
The terms involving the \(\beta\) coefficients describe the fixed-effects mean trajectory. The subject-specific random intercept is \(b_i\), and \(\varepsilon_{ij}\) is the visit-specific residual error. The fixed-effects mean at baseline is \(\beta_0+\beta_1 A_i\): it is \(\beta_0\) for control and \(\beta_0+\beta_1\) for treatment. At baseline, the random intercept adds \(b_i\) to that mean.
For the calculations below, assume that the residuals are independent across visits and subjects, with \(\varepsilon_{ij}\sim N(0,\sigma^2)\), and that they are independent of all the random intercepts. The random intercepts are also assumed independent of the modeled covariates, conditional on the model’s specification. The model conditions on the fixed covariates and treats their effects as represented by the fixed-effects terms.
These assumptions have distinct roles. Mean zero for \(b_i\) makes the fixed-effects mean the average across the modeled random-intercept distribution. Normality gives a specific distribution for deviations and supports probability calculations under the model. Independence across subjects describes how the subject-level deviations are sampled; independence from the residuals separates persistent subject differences from visit-specific errors. These are modeling assumptions, not facts guaranteed by having repeated measurements.
What the Distribution Says About Baseline
Define the latent baseline level for subject \(i\) as the model’s expected biomarker value at \(t=0\), before adding visit-specific residual error. In the example model it is \(B_i=\beta_0+\beta_1 A_i+b_i\). Conditional on treatment group, adding a constant to a normal random variable shifts its mean without changing its variance. Therefore:
The distribution is centered on the fixed-effects baseline mean for that treatment group, and its spread is determined by \(\tau_0\). For a subject whose intercept is \(b_i=0\), the latent baseline equals the fixed-effects mean. A positive \(b_i\) places the subject above that mean; a negative \(b_i\) places the subject below it. The magnitude is in the biomarker’s measurement units. By contrast, \(\tau_0^2\) has squared biomarker units.
This is a statement about latent subject baselines, not a claim that a single observed baseline measurement must equal \(B_i\). The observed measurement also includes \(\varepsilon_{ij}\), so it can differ from the latent level because of visit-specific error.
Worked Example: Probability of a Baseline Deviation Within 10 Units
Suppose a hypothetical biomarker is measured in units, and a fitted model specifies \(\tau_0=5\) units. What proportion of subjects does the model expect to have a random-intercept deviation between \(-10\) and \(10\) units, inclusive?
Under the model, \(b_i\sim N(0,5^2)\). Standardizing by the standard deviation gives:
Here \(Z\sim N(0,1)\), and \(\Phi\) is its cumulative distribution function. Thus the model assigns approximately 95.45% probability to a subject’s latent baseline being within 10 biomarker units of the fixed-effects baseline mean for that subject’s treatment group. This is a model-based probability, not a statement that a particular realized sample must contain exactly that percentage.
Worked Example: Adding a Subject’s Intercept to the Fixed-Effects Mean
Suppose, hypothetically, that the estimated control baseline mean is 40 biomarker units and the treatment baseline mean is 46 units. A subject has a random intercept of \(b_i=-3\) units. Calculate the model-implied latent baseline for a subject in each group, using this same deviation to illustrate the additive interpretation.
For a control subject, the fixed-effects baseline mean is 40 units. Adding the deviation gives \(40+(-3)=37\) units. For a treatment subject, it is 46 units, so the corresponding latent baseline is \(46+(-3)=43\) units.
The random intercept does not replace the treatment effect: both group means remain separated by 6 units. It shifts each subject’s baseline relative to the appropriate group-specific fixed-effects mean. In an actual model, each subject has their own latent \(b_i\); the shared value here is used only to make the arithmetic comparison clear.
A Short Derivation of the Baseline Distribution
The distribution of the latent baseline follows directly from the random-intercept assumption. Write \(m_i=\beta_0+\beta_1 A_i\), which is fixed once the treatment group and model coefficients are specified. Since \(B_i=m_i+b_i\), the mean and variance are:
A constant shift of a normal variable is normal, so \(b_i\sim N(0,\tau_0^2)\) implies \(m_i+b_i\sim N(m_i,\tau_0^2)\). Substituting \(m_i=\beta_0+\beta_1 A_i\) yields the stated distribution for \(B_i\). This derivation also shows why the random-intercept variance concerns differences in latent baselines, not the spread of observed measurements around a single mean.
Worked Example: Converting a Standard Deviation Into a Baseline Range
Suppose a hypothetical model estimates a treatment-group fixed-effects baseline mean of 120 biomarker units and a random-intercept standard deviation of 6 units. Calculate the model-implied probability that a subject’s latent baseline is within 9 units of the group mean, and find the 90th percentile of the random-intercept deviation.
For the probability, \(b_i\sim N(0,6^2)\). Standardizing 9 units gives \(9/6=1.5\):
Thus the model assigns about 86.64% probability to a latent baseline between \(120-9=111\) and \(120+9=129\) units. For the 90th percentile, the standard-normal 90th percentile is approximately 1.2816, so the corresponding deviation is \(1.2816(6)=7.690\) units. The model-implied 90th percentile of latent baseline is therefore \(120+7.690=127.690\) units. This percentile describes the model’s baseline distribution, not a confidence interval for the fixed-effects mean.
How to Assess the Assumptions and Recognize Limitations
The random-intercept model is useful when there is a plausible persistent subject-level offset, but its simplicity can be a limitation. It assumes that the subject effect is constant over the observed follow-up: it does not allow one subject’s biomarker to change faster than another’s through a subject-specific time effect. If individual changes differ systematically, a constant shift alone may not represent the trajectory variation well.
The normal distribution is an assumption about the latent \(b_i\), not a requirement that every observed biomarker measurement itself have a normal distribution. Still, severe departures in the random-effects distribution can matter for probability statements and model-based inference. In practice, diagnostics may include checking residual patterns against fitted values and time, and examining whether estimated subject effects show conspicuous skewness or extreme values. Estimated subject effects are themselves uncertain, especially for subjects with few measurements, so their empirical distribution should not be treated as a direct, error-free display of the latent distribution.
A random intercept also does not automatically account for every form of dependence or irregularity in the measurements. For example, residual patterns that vary systematically over time may indicate that the residual assumptions are inadequate. A relationship between subject baseline propensity and a covariate not represented appropriately in the fixed-effects model can also undermine the assumed independence between \(b_i\) and the covariates. Diagnostics can flag problems, but they cannot prove the assumptions true.
Two broad alternatives clarify what the random intercept contributes. A fixed-effects-only trajectory omits \(b_i\), and so does not explicitly represent persistent subject-level baseline deviations. Subject-specific fixed intercepts instead assign a separate intercept to each observed subject without modeling those intercepts as draws from a population distribution. That approach can describe the enrolled subjects, but it does not directly provide a distribution for baseline variation or a model-based baseline for a new subject. The random-intercept approach makes that distributional assumption and uses it to describe between-subject baseline variation.
Check Your Understanding
Answer each question, then compare with the solutions below.
- If \(b_i\sim N(0,4^2)\), what are the units of \(b_i\), \(\tau_0\), and \(\tau_0^2\)?
- A fixed-effects baseline mean is 55 units and a subject’s random intercept is \(2.5\) units. What is the subject’s latent baseline under the model?
- If \(b_i\sim N(0,3^2)\), calculate \(P(-6\leq b_i\leq 6)\).
- Does a random intercept allow each subject to have a different rate of biomarker change over time? Explain briefly.
Solutions. (1) \(b_i\) and \(\tau_0\) are in the biomarker’s units; \(\tau_0^2\) is in squared biomarker units. (2) The latent baseline is \(55+2.5=57.5\) units. (3) Standardizing gives \(P(-2\leq Z\leq 2)=\Phi(2)-\Phi(-2)\approx0.9545\). (4) No. A random intercept shifts the trajectory by a constant amount; it does not give subjects different time slopes.