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Population Design and the Population Fisher Matrix · Tutorial 100 of 497

Allocating Subjects Versus Within-Subject Measurements

Learn how subject-level correlation changes the information gained from additional subjects versus additional measurements, and how the preferred allocation depends on the parameter and study costs.

Advanced 10 min read

What You'll Learn

  • Derive the information for a population intercept with a random-intercept covariance structure.
  • Compare balanced allocations with the same number of observations.
  • Evaluate subject-versus-measurement choices under a specified cost budget.
  • Explain why repeated measurements can be especially informative about within-subject slopes.
  • Recognize assumptions and diagnostics that can make simple allocation comparisons misleading.

Why the Allocation Matters

A study can often spend resources in two different ways: recruit another independent subject, or collect another measurement from someone already enrolled. These choices are not interchangeable. Measurements from the same subject can share random effects, so they are generally correlated; a new subject adds another independent realization of those effects. Which choice is more informative depends on the parameter of interest, the measurement schedule, and the costs.

The comparison below uses a specified approximate mixed-effects model to make these distinctions explicit. The calculation is local: for a nonlinear response model, its mean is understood as linearized around a nominal parameter value, and the resulting marginal distribution is approximated as Gaussian. The goal is to compare allocations under that approximation, not to claim that every nonlinear mixed-effects model has the same allocation rule.

Definition: A subject is an independent sampling unit with its own random effects. A within-subject measurement is an additional observation on an already enrolled subject. Under a marginal model, measurements from one subject may be correlated, whereas different subjects are assumed independent. Population information is the sum of the information contributions from independent subjects.

A Random-Intercept Approximation

Consider the following linear mixed-effects model for subject \(i\) at measurement occasion \(j\):

$$ Y_{ij}=\beta_0+b_i+\varepsilon_{ij}, \qquad b_i\sim N(0,\omega^2), \qquad \varepsilon_{ij}\sim N(0,\sigma^2). $$

Assume that the random intercepts \(b_i\) are independent across subjects, the residuals are independent across subjects and occasions, and random intercepts and residuals are independent. For the calculations that follow, \(\omega^2\) and \(\sigma^2\) are treated as known, positive variances, and the target is the population mean \(\beta_0\). These assumptions make it possible to isolate the effect of allocation on information for the mean.

If subject \(i\) contributes \(m\) measurements, their marginal mean vector is \(\beta_0\mathbf{1}\) and their marginal covariance matrix is \(V_m=\omega^2\mathbf{1}\mathbf{1}^{\mathsf T}+\sigma^2 I_m\), where \(\mathbf{1}\) is an \(m\)-vector of ones and \(I_m\) is the \(m\)-by-\(m\) identity matrix. The covariance has diagonal entries \(\omega^2+\sigma^2\) and off-diagonal entries \(\omega^2\). Thus, repeated measurements are correlated even though the residual errors are independent.

For a Gaussian model with known covariance and a mean that depends on \(\beta_0\), the Fisher information for \(\beta_0\) from one subject is the mean derivative transposed, multiplied by the inverse covariance, and multiplied by the mean derivative. Here that derivative is \(\mathbf{1}\), so the subject contributes \(\mathbf{1}^{\mathsf T}V_m^{-1}\mathbf{1}\). A short calculation gives a useful closed form.

Result: Under this model, one subject with \(m\) measurements contributes \(I_{\text{subject}}(\beta_0;m)=m/(\sigma^2+m\omega^2)\) information about \(\beta_0\). With \(N\) independent subjects having the same schedule, total information is \(I_{\text{total}}=N m/(\sigma^2+m\omega^2)\). These formulas concern information for the mean with the variance components held known.

Why the Formula Has Diminishing Returns

To verify the result, note that \(V_m\mathbf{1}=(\sigma^2+m\omega^2)\mathbf{1}\). Therefore \(V_m^{-1}\mathbf{1}=\mathbf{1}/(\sigma^2+m\omega^2)\), and the information is

$$ \mathbf{1}^{\mathsf T}V_m^{-1}\mathbf{1} =\frac{\mathbf{1}^{\mathsf T}\mathbf{1}}{\sigma^2+m\omega^2} =\frac{m}{\sigma^2+m\omega^2}. $$

When \(\omega^2=0\), observations within a subject are independent and each adds \(1/\sigma^2\) information for the mean. When \(\omega^2>0\), the information from one subject approaches \(1/\omega^2\) as \(m\) grows: increasingly many readings estimate that subject's level more precisely, but cannot eliminate the subject-to-subject deviation \(b_i\). For \(N\) independent subjects, the total information continues to increase with \(N\), because each new subject supplies another independent random intercept.

For a Gaussian mean parameter with known covariance, the variance of the maximum-likelihood estimator of \(\beta_0\) is the reciprocal of total information. Thus, if two candidate designs differ only in their allocation, this variance provides a direct comparison of their local precision under the model. It is not a guarantee about performance if the covariance structure or other assumptions are wrong.

Worked Example: Four Measurements on One Subject or Four Subjects

Suppose, hypothetically, that the residual variance is \(\sigma^2=16\) response-units squared and the random-intercept variance is \(\omega^2=9\) response-units squared. Compare four measurements on one subject with one measurement on each of four independent subjects. Both designs use four measurements in total.

With four measurements on one subject, the information is \(4/(16+4\times9)=4/52=0.07692\) per response-unit squared. The corresponding variance for estimating \(\beta_0\) is \(1/0.07692=13\) response-units squared, and its standard error is \(\sqrt{13}=3.606\) response units.

With one measurement on each of four subjects, each subject contributes \(1/(16+9)=0.04\) per response-unit squared. Total information is \(4\times0.04=0.1600\) per response-unit squared. The variance is \(1/0.1600=6.25\) response-units squared, and the standard error is \(2.500\) response units.

For this target and these assumptions, four independent subjects provide more information than four measurements concentrated on one subject. The difference arises because all four measurements on the single subject share the same random intercept. The calculation does not establish that extra measurements are never useful; it answers a specific question about the population mean.

Adding a Cost Constraint

An allocation comparison is incomplete when subject recruitment and measurement costs differ. Let \(c_s\) be the cost of enrolling a subject and \(c_m\) the cost of each measurement. A balanced design with \(N\) subjects and \(m\) measurements per subject costs \(N(c_s+m c_m)\) and has the information given above. With a fixed budget, a larger \(m\) may be worthwhile if it avoids expensive recruitment, even though each additional measurement has diminishing information for the mean.

Worked Example: Comparing Allocations Under a Budget

Use the hypothetical variances \(\sigma^2=16\) and \(\omega^2=9\), and suppose recruiting a subject costs 8 budget units while each measurement costs 1 unit. With a budget of 40 units, compare balanced designs with \(m=1,2,4,\) or \(8\) measurements per subject. Assume the remaining budget, if any, cannot be used to create a different schedule.

For \(m=1\), each subject costs \(8+1=9\) units, so 4 subjects can be enrolled. Information is \(4\times1/(16+9)=0.1600\) per response-unit squared.

For \(m=2\), each subject costs \(8+2=10\) units, so 4 subjects can be enrolled. Information is \(4\times2/(16+2\times9)=8/34=0.2353\) per response-unit squared.

For \(m=4\), each subject costs 12 units, so 3 subjects can be enrolled. Information is \(3\times4/(16+4\times9)=12/52=0.2308\) per response-unit squared. For \(m=8\), each subject costs 16 units, so 2 subjects can be enrolled. Information is \(2\times8/(16+8\times9)=16/88=0.1818\) per response-unit squared.

Among these four candidate balanced designs, \(m=2\) gives the greatest information under the stated budget. This is not a universal optimum: the result depends on the chosen candidates, costs, target parameter, and assumed variances. A different recruitment cost or measurement cost could change the ranking.

The Target Parameter Can Change the Answer

The random-intercept calculation is specifically about \(\beta_0\). If the target is a within-subject change or slope, repeated measurements at informative times can be especially valuable. Consider the extension \(Y_{ij}=\beta_0+\beta_1t_{ij}+b_i+\varepsilon_{ij}\), with the same random-intercept and residual assumptions. Two measurements at times \(-1\) and \(+1\) have a difference in which \(b_i\) cancels. This is a different information question from estimating the population intercept.

Worked Example: Repeated Measurements for a Population Slope

Suppose hypothetically that \(\sigma^2=16\) and \(\omega^2=9\), and that each measurement costs the same. Compare four measurements arranged as two subjects with observations at \(t=-1\) and \(t=+1\) for each, against four independent subjects with one observation each, two at each time. The target is the slope \(\beta_1\), with an intercept \(\beta_0\) also estimated.

For a subject measured at both times, the difference is \(Y_{i,+1}-Y_{i,-1}=2\beta_1+\varepsilon_{i,+1}-\varepsilon_{i,-1}\). The random intercept cancels, and the difference has variance \(2\sigma^2=32\). Dividing the difference by 2 gives a slope contrast with variance \(32/4=8\), so each paired subject contributes \(1/8=0.1250\) information for \(\beta_1\). Two such subjects contribute \(0.2500\) per squared slope unit.

In the one-measurement-per-subject design, each marginal observation has variance \(\omega^2+\sigma^2=25\). There are two observations at each time, so the time values are balanced and the intercept and slope columns are orthogonal. The information for the slope is \(\sum_i(t_i-\bar{t})^2/25=4/25=0.1600\) per squared slope unit, since \(\bar{t}=0\) and each of the four time values has square 1.

Under these assumptions, the paired design provides more information for the slope: 0.2500 rather than 0.1600, despite using the same number of measurements. The paired contrast removes the random intercept, while the single-observation design is exposed to variation between subjects. The conclusion is target-specific and presumes that the measurements at both times are feasible and the linear time effect is appropriate.

Diagnostics, Alternatives, and Limitations

These comparisons are useful for planning only when the model and the allocation being compared are stated clearly. In practice, inspect whether estimated within-subject correlations are compatible with the assumed covariance, whether residual variability changes over time or with the mean, and whether the proposed schedule samples the parts of the response where the target parameter affects the mean. In a nonlinear model, a local linearization can be poor when the response changes sharply over the scheduled times or nominal parameter values are uncertain. Information calculations should then be treated as local approximations rather than definitive predictions.

A key limitation of the closed-form mean calculation is that it treats \(\omega^2\) and \(\sigma^2\) as known. If variance components are also targets, measurements within a subject can help distinguish residual variation from between-subject variation; the formula above does not quantify that benefit. Similarly, unequal schedules, missing observations, covariates, and random slopes require their own subject-level information calculations. The broad principle remains that independent subjects and repeated measurements contribute through different covariance structures, not as interchangeable independent records.

A main alternative to a balanced allocation is an unequal or adaptive schedule, for example assigning more measurements to subjects at times expected to be informative for a slope. Such a design may improve information for a particular target but can complicate comparisons and may be sensitive to model misspecification. The balanced calculations here are transparent benchmarks, not a substitute for evaluating the actual candidate schedules under the intended model.

  • Pitfall: counting every measurement as an independent subject. This ignores the shared random intercept and overstates information for the population mean. Use the subject's marginal covariance, then sum contributions across independent subjects.
  • Pitfall: concluding that more subjects always dominate repeated measurements. That conclusion does not follow for slope or variance-component targets; paired measurements can remove shared subject effects from contrasts.
  • Pitfall: comparing allocations without specifying cost. Equal numbers of measurements need not have equal recruitment and measurement costs. State the budget and feasible schedules before ranking designs.
Key takeaway: For a population mean under a random-intercept model, additional subjects reduce uncertainty by adding independent subject-level information, while repeated measurements have diminishing returns when subjects share a random effect. For within-subject changes, repeated measurements can be more informative. Allocation decisions therefore require a stated target, covariance model, schedule, and cost.

Check Your Understanding

Use the model and assumptions stated above. Each question is followed by its solution.

  1. With \(\sigma^2=4\) and \(\omega^2=6\), calculate the information for \(\beta_0\) from one subject with three measurements.
  2. Under the same variances, compare the information for \(\beta_0\) from two subjects with two measurements each and four subjects with one measurement each.
  3. Why does the random-intercept variance cancel from the paired difference at times \(-1\) and \(+1\)?
  4. State one reason the random-intercept formula may be insufficient when planning a population study.

Solutions. (1) \(I=3/(4+3\times6)=3/22=0.1364\) per response-unit squared. (2) Two subjects with two measurements each give \(2\times2/(4+12)=4/16=0.2500\); four subjects with one measurement each give \(4/(4+6)=0.4000\). Thus the four-subject allocation has greater information for the mean. (3) Both measurements contain the same \(b_i\), so their difference includes \(b_i-b_i=0\). (4) For example, the formula treats variance components as known; if estimating them is important, the design must also be assessed for information about those variance parameters.

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