From a Random-Intercept Model to Population Information
Repeated measurements on one subject share features that measurements from different subjects do not. A random-intercept model represents this dependence by giving each subject a shared, unobserved deviation from the population mean. Deriving the resulting covariance is essential: if the shared deviation is ignored, repeated observations can appear to contain more independent information than they do.
This tutorial works through the model’s marginal mean and covariance and then derives a local population Fisher information matrix for a fixed intercept and two variance components. The calculation is evaluated at specified nominal parameter values, as is customary in local design. Because the model here is linear and Gaussian, its marginal moments and Gaussian distribution are exact under the stated assumptions; the matrix is therefore also the marginal Gaussian Fisher information for this particular model.
Here \(Y_{ij}\) is one measured outcome. The model has no occasion-specific fixed effect: conditional on a subject’s random intercept, its expected outcome is the same at every occasion. The design is described by the number of occasions \(m\), and the model parameters are \(\theta=(\beta,\omega^2,\sigma^2)\). Use \(\omega^2\) and \(\sigma^2\), rather than their square roots, as the variance-component parameters in the information matrix.
Marginal Mean and Covariance
Write the \(m\) measurements from one subject as the vector \(Y_i=(Y_{i1},\ldots,Y_{im})^\mathsf{T}\), and let \(\mathbf{1}\) be the \(m\)-vector of ones. Then \(Y_i=\beta\mathbf{1}+b_i\mathbf{1}+\epsilon_i\), where \(\epsilon_i\) is the vector of residuals. Taking expectations over both sources of variation gives the marginal mean. For the covariance, independence of \(b_i\) and \(\epsilon_i\) makes their covariance contribution zero, while the shared random intercept contributes to every pair of occasions:
Thus \(Y_i\sim N(\beta\mathbf{1},V)\), where \(V=\omega^2\mathbf{1}\mathbf{1}^\mathsf{T}+\sigma^2 I_m\). Each measurement has marginal variance \(\omega^2+\sigma^2\), and any two distinct measurements on the same subject have covariance \(\omega^2\). Their marginal correlation is \(\omega^2/(\omega^2+\sigma^2)\). This is the compound-symmetry covariance structure: all measurements have the same variance and all distinct within-subject pairs have the same covariance.
Worked Example: Constructing a Three-Measurement Covariance
Consider a hypothetical study with three measurements per subject. Suppose the nominal population mean is \(\beta=8.5\) mg/L, the random-intercept variance is \(\omega^2=0.25\) (mg/L)\(^2\), and the residual variance is \(\sigma^2=0.36\) (mg/L)\(^2\). Find the marginal mean, covariance, and within-subject correlation.
The mean vector is \(\beta\mathbf{1}\), so each of its three entries is 8.5 mg/L. The marginal variance of one measurement is \(\omega^2+\sigma^2=0.25+0.36=0.61\) (mg/L)\(^2\). The covariance of two distinct measurements is \(\omega^2=0.25\) (mg/L)\(^2\). Therefore,
The correlation between any two distinct measurements is covariance divided by the product of their standard deviations: \(0.25/\sqrt{0.61\times0.61}=0.25/0.61=0.4098\). The positive correlation arises from the shared subject-level intercept, not from a change in the residual variance.
Deriving the Fisher Information
For one subject, write the Gaussian log-likelihood, omitting a constant that does not depend on \(\theta\), as \(-\tfrac12\log|V|-\tfrac12(Y_i-\mu)^\mathsf{T}V^{-1}(Y_i-\mu)\), where \(\mu=\beta\mathbf{1}\). For a multivariate normal model with differentiable mean \(\mu\) and covariance \(V\), the expected Fisher information for parameters \(\theta_a,\theta_b\) is
This expression follows by differentiating the Gaussian log-likelihood and taking the expected product of its scores. The linear score terms in \(Y_i-\mu\) contribute the mean term. The centered quadratic terms contribute the trace term, using the covariance of quadratic forms in a multivariate normal vector. The expected cross-product of a linear term and a centered quadratic term is zero because centered Gaussian variables have zero third moments. These facts also explain why the fixed-mean and covariance parameters have zero expected cross-information in this model.
In this case, \(\partial\mu/\partial\beta=\mathbf{1}\), while the mean does not depend on either variance component. Also, \(\partial V/\partial\omega^2=\mathbf{1}\mathbf{1}^\mathsf{T}\) and \(\partial V/\partial\sigma^2=I_m\). Define \(\lambda=\sigma^2+m\omega^2\). The covariance matrix has eigenvalue \(\lambda\) in the direction \(\mathbf{1}\) and eigenvalue \(\sigma^2\) in each of the \(m-1\) orthogonal contrast directions. Consequently, \(\mathbf{1}^\mathsf{T}V^{-1}\mathbf{1}=m/\lambda\); the trace calculation for the variance components uses one contribution in the mean direction and \(m-1\) in the contrast directions. Combining these results gives the one-subject matrix:
The row and column order is \((\beta,\omega^2,\sigma^2)\). The fixed-intercept information is \(m/\lambda\). For \(N\) independent subjects with the same design and nominal parameters, the population matrix is \(N I(\theta)\), because independent subjects’ log-likelihoods and their expected information add. With differing numbers of occasions, calculate each subject-design contribution using its own \(m\) and add the contributions.
Worked Example: Calculating the One-Subject Information Matrix
Use a hypothetical design with \(m=3\) measurements and nominal values \(\omega^2=0.25\) and \(\sigma^2=0.36\) (mg/L)\(^2\). Calculate the information for one subject in the parameter order \((\beta,\omega^2,\sigma^2)\).
First, \(\lambda=0.36+3(0.25)=1.11\) (mg/L)\(^2\). The fixed-intercept information is \(3/1.11=2.703\) (mg/L)\(^{-2}\). The variance-component entries are
For the residual-variance diagonal, \(\tfrac12[(3-1)/(0.36)^2+1/(1.11)^2] =\tfrac12[15.432+0.8116]=8.122\). The fixed-intercept cross-entries are zero, as shown by the general formula.
The first diagonal entry has units (mg/L)\(^{-2}\), because it is information for a parameter measured in mg/L. The variance-component block has units (mg/L)\(^{-4}\). For 40 independent subjects with this same schedule and nominal values, multiply every entry by 40; for example, the total fixed-intercept information is \(40(2.703)=108.1\) (mg/L)\(^{-2}\). The zero cross-entries do not mean the estimates are always independent in a finite sample; they state that the expected Fisher information has zero cross-terms under this model.
Interpreting the Matrix and Checking the Model
The variance-component block captures how the measurement schedule distinguishes shared between-subject variation from occasion-level residual variation. With only one measurement per subject, \(m=1\), the two variance components enter the marginal variance only through their sum. Indeed, the variance-component block then has identical rows and columns and is rank deficient. There is no information in this design for separating \(\omega^2\) from \(\sigma^2\). At least two repeated measurements are needed to distinguish the covariance of distinct measurements from their additional residual variance.
The formula also makes the role of the model assumptions visible. The model implies equal marginal variances across occasions and equal within-subject covariances. It further implies that measurements from different subjects are independent. Before relying on its information matrix, check whether the scientific setting and data support these features. Strong occasion trends in the mean, changing residual variance, or a pattern of within-subject covariance that is not roughly constant can indicate that the random-intercept model is inadequate. A negative estimated random-intercept variance is not a valid value under the model; it may signal weak information, model mismatch, or estimation difficulty rather than a meaningful negative variance.
A useful diagnostic is to compare observed within-subject patterns with the model’s implied structure: the same-subject covariance should be approximately level across distinct occasion pairs, while between-subject covariance is zero under the stated independence assumption. Also inspect whether the fitted variance components are near a boundary and whether the information matrix is poorly conditioned. A singular or nearly singular matrix indicates that some parameter combinations are not well distinguished by the design and nominal values.
The main alternative simplifications each lose something. Treating every measurement as independent discards the shared intercept and generally overstates information from repeated observations. Reducing each subject to a single average avoids pretending that occasions are independent, but discards the contrast information that helps distinguish \(\omega^2\) from \(\sigma^2\). The random-intercept model retains both the subject-level average and the within-subject contrasts. It is still limited: it cannot represent changing subject effects over time, unequal within-subject covariances, or dependence between subjects. The model and its nominal parameter values should therefore be treated as design assumptions to assess, not as guaranteed descriptions of every repeated-measure setting.
Worked Example: Information in Subject Averages
Consider a hypothetical design with \(N=20\) subjects and \(m=3\) measurements per subject, using \(\omega^2=0.25\) and \(\sigma^2=0.36\) (mg/L)\(^2\). Find the variance of the overall average across all \(20\times3\) measurements under the model, and compare it with the result from incorrectly treating all 60 measurements as independent with marginal variance \(0.61\) (mg/L)\(^2\).
For one subject, the average of three measurements is \(\bar Y_i=\beta+b_i+\bar\epsilon_i\). Its variance is \(\omega^2+\sigma^2/3=0.25+0.36/3=0.37\) (mg/L)\(^2\). Since subjects are independent, averaging these 20 subject averages gives variance \(0.37/20=0.01850\) (mg/L)\(^2\), with standard error \(\sqrt{0.01850}=0.1360\) mg/L. The corresponding fixed-intercept information is the reciprocal, \(20/0.37=54.05\) (mg/L)\(^{-2}\).
If all 60 observations were incorrectly treated as independent, the calculated variance would be \(0.61/60=0.01017\) (mg/L)\(^2\), and the calculated standard error would be \(0.1008\) mg/L. This understates uncertainty because it omits the shared subject-level variation. Repeated measurements reduce the residual contribution to each subject average, but they do not average away the random intercept within that subject.
Common Pitfall and Practical Use
For design use, evaluate the matrix at nominal values of \(\omega^2\) and \(\sigma^2\), confirm that the planned number of repeated measurements matches the assumed schedule, and scale or sum information across subjects as appropriate. This produces a local information calculation: changing the nominal variance components can change the matrix and the apparent precision. The calculation is most useful when its covariance assumptions are credible and the design supplies enough repeated measurements to separate the variance components.
Check Your Understanding
Try each question, then compare your reasoning with the solutions below.
- Under the random-intercept model, what is the covariance between two different measurements from the same subject? What is the covariance between measurements from two independent subjects?
- For \(m=2\), \(\omega^2=0.20\), and \(\sigma^2=0.30\), calculate \(\lambda\) and the fixed-intercept information for one subject.
- Why can a design with one measurement per subject not separate \(\omega^2\) from \(\sigma^2\)?
- For \(m=3\), what is the variance of one subject’s average in terms of \(\omega^2\) and \(\sigma^2\), and why does the random-intercept component remain?
Solutions. 1. The same-subject covariance is \(\omega^2\), because both measurements contain the same \(b_i\); the covariance across independent subjects is zero under the model. 2. \(\lambda=0.30+2(0.20)=0.70\), so the fixed-intercept information is \(m/\lambda=2/0.70=2.857\) in the reciprocal squared units of \(\beta\). 3. With one measurement, the marginal variance is only \(\omega^2+\sigma^2\); there is no distinct-pair covariance to identify \(\omega^2\) separately. 4. The average is \(\beta+b_i+(\epsilon_{i1}+\epsilon_{i2}+\epsilon_{i3})/3\), so its variance is \(\omega^2+\sigma^2/3\). Averaging reduces the independent residual contribution, but the shared \(b_i\) is present in every measurement and remains in the average.