Identifying Occasion Variability With Replicated Periods

Why Repeated Periods Help

A subject’s pharmacokinetic parameter may differ from the population average for a stable reason, such as persistent differences in physiology. The same subject may also differ from one treatment period to another, even when the treatment and protocol are intended to be the same. A third source of variation is measurement or estimation noise. Repeated periods make it possible to distinguish the stable subject contribution from variation that changes by period, but they do not automatically distinguish occasion variation from residual noise.

The key is to compare observations from the same subject across periods. Such comparisons remove the subject’s stable effect. Replicate measurements within a period provide a separate way to assess residual variation. The contrasts and assumptions below are stated for a scalar quantity on a scale where an additive model is reasonable, such as a log-transformed individual parameter estimate. They do not assert that raw concentration observations themselves follow this simple model.

Definition: Let $x_{ijr}$ denote replicate $r$ for subject $i$ in treatment period (occasion) $j$. Use the model $$ x_{ijr}=\mu+\eta_i+\kappa_{ij}+\epsilon_{ijr}. $$ Here $\mu$ is the population mean, $\eta_i$ is the stable between-subject effect, $\kappa_{ij}$ is the between-occasion effect for subject $i$ in period $j$, and $\epsilon_{ijr}$ is residual error. Assume these effects have mean zero, with variances $\omega^2$, $\pi^2$, and $\sigma^2$, respectively. For the calculations in this tutorial, assume the three sources are mutually independent; occasion effects are independent across periods, and residual errors are independent across replicates and periods.

These assumptions make the variance decomposition precise. Two observations from the same subject in different periods share $\eta_i$, but not $\kappa_{ij}$ or $\epsilon_{ijr}$. Two replicates in the same subject-period share both $\eta_i$ and $\kappa_{ij}$, but have different residual errors. The intended interpretation also requires comparable treatment and measurement conditions across periods, and no unmodelled period trend or carryover effect that is being mistaken for random occasion variation.

Contrasts That Remove Stable Subject Differences

Suppose there is one observation in each of two periods for the same subject. Subtract the second from the first:

D_i=x_{i1}-x_{i2} =(\kappa_{i1}-\kappa_{i2})+(\epsilon_{i1}-\epsilon_{i2}).

The population mean and the subject effect cancel. This is why a between-period contrast within a subject is more informative about occasion variation than a comparison between unrelated subjects. However, the contrast does not cancel residual error: both period-specific residual terms remain. Under the model,

\operatorname{Var}(D_i)=2\pi^2+2\sigma^2 =2(\pi^2+\sigma^2).

Thus, the variability of paired-period differences alone identifies the sum of the occasion and residual variances, not $\pi^2$ by itself. This is an important limit: repeating periods removes between-subject variability, but some additional information is needed to separate occasion variability from residual noise.

Key result: If each period mean averages $R$ independent replicate observations, its residual variance is $\sigma^2/R$. The difference between two such means from the same subject has variance $2\pi^2+2\sigma^2/R$. If $\sigma^2$ is known or estimated independently, the occasion variance can be recovered by subtracting the residual contribution and dividing by two.

The result follows directly by expanding the difference of period means. Each mean contains the same $\mu+\eta_i$, so those terms cancel. Its occasion effect has variance $\pi^2$, while the average of $R$ independent residual errors has variance $\sigma^2/R$. The two periods are independent with respect to their occasion and residual effects, so the variance of their difference is the sum of the two period variances. A contrast can therefore isolate the occasion component from the subject component, but separating it from residual variation requires replication or other information about $\sigma^2$.

Worked Example: Interpreting a Paired-Period Contrast

Suppose, hypothetically, that the log-scale variance components are $\pi^2=0.0900$ and $\sigma^2=0.0400$, in squared log-scale units. Each period has one observation. Find the variance and standard deviation of the paired-period difference, and determine whether that difference isolates occasion variability.

By the paired-contrast formula, $$ \operatorname{Var}(D_i)=2(\pi^2+\sigma^2) =2(0.0900+0.0400)=0.2600. $$

Therefore, the standard deviation of the difference is $\sqrt{0.2600}=0.5099$ log-scale units. The occasion contribution to the contrast variance is $2\pi^2=0.1800$, while the residual contribution is $2\sigma^2=0.0800$. The contrast removes the stable subject effect, but it does not isolate occasion variation: its variance includes both components.

If each period instead averages $R=4$ independent replicates, the contrast variance becomes $$ 2\pi^2+2\sigma^2/4 =2(0.0900)+2(0.0400/4)=0.2000. $$ Averaging reduces residual noise, but does not remove the occasion component.

Using Replicates to Estimate Residual Variation

When two replicates are collected in the same subject-period, their difference is $$ x_{ij1}-x_{ij2}=\epsilon_{ij1}-\epsilon_{ij2}. $$ The shared mean, subject effect, and occasion effect cancel. Under independent residual errors, the difference has mean zero and variance $2\sigma^2$. Consequently, the squared within-period replicate differences can estimate residual variance. With $M$ such pairs, a method-of-moments estimate is the mean squared difference divided by two:

\widehat{\sigma^2} =\frac{1}{2M}\sum_{m=1}^{M}(x_{m1}-x_{m2})^2.

This formula assumes the pairs are formed within the same subject and period and that the residual errors in the pair are independent with a common variance. If replicate errors are correlated or their variances differ substantially, this simple estimate need not be appropriate.

Worked Example: Estimating Residual Variance From Replicate Pairs

In a hypothetical study, four within-period replicate differences on a log scale are $0.20$, $-0.10$, $0.30$, and $-0.20$. Assume the pairs share subject and occasion effects, while their residual errors are independent and have a common variance. Estimate $\sigma^2$.

The sum of squared differences is $$ 0.20^2+(-0.10)^2+0.30^2+(-0.20)^2 =0.0400+0.0100+0.0900+0.0400=0.1800. $$

With $M=4$, the estimate is $$ \widehat{\sigma^2}=\frac{0.1800}{2(4)}=0.02250. $$

This estimates the residual variance in squared log-scale units. The factor of two is essential: each replicate difference contains two independent residual errors, so its variance is $2\sigma^2$, not $\sigma^2$.

Estimating Occasion Variation From Within-Subject Variation

With one observation in each of $J$ periods for subject $i$, define the subject’s period average $\bar{x}_{i\cdot}=J^{-1}\sum_{j=1}^{J}x_{ij}$. Centering each period observation around this average removes both $\mu$ and $\eta_i$. The remaining deviations contain occasion and residual effects. Under the stated assumptions, the expected within-subject sample variance is $\pi^2+\sigma^2$. Therefore, if residual variance is estimated from within-period replication, subtracting that estimate gives a method-of-moments estimate of occasion variance.

S_i^2=\frac{1}{J-1}\sum_{j=1}^{J}(x_{ij}-\bar{x}_{i\cdot})^2, \qquad E(S_i^2)=\pi^2+\sigma^2.

Pooling the within-subject sums of squares across $N$ subjects with the same number $J$ of periods gives

\widehat{\pi^2} = \frac{\sum_{i=1}^{N}\sum_{j=1}^{J}(x_{ij}-\bar{x}_{i\cdot})^2} {N(J-1)} -\widehat{\sigma^2}.

This is a component estimate, not a guarantee that every finite data set will yield a positive value. Its interpretation relies on the replicated periods being exchangeable under the model and on a valid estimate of residual variance. With only one observation per period and no independent residual information, the within-subject variation still combines occasion and residual sources.

Worked Example: Subtracting Residual Noise From Period Variation

Consider hypothetical log-scale observations from two subjects, each measured once in three periods. Subject A has $10.0, 10.4, 9.6$; subject B has $12.0, 12.2, 11.8$. Suppose within-period replicate data provide $\widehat{\sigma^2}=0.0400$. Estimate the occasion variance by pooling the within-subject sums of squares.

The period mean for A is $10.0$. Its sum of squared deviations is $$ (10.0-10.0)^2+(10.4-10.0)^2+(9.6-10.0)^2 =0+0.1600+0.1600=0.3200. $$

The period mean for B is $12.0$, and its sum of squared deviations is $$ (12.0-12.0)^2+(12.2-12.0)^2+(11.8-12.0)^2 =0+0.0400+0.0400=0.0800. $$

There are $N(J-1)=2(2)=4$ within-subject degrees of freedom. The pooled within-subject variance is $(0.3200+0.0800)/4=0.1000$. Subtracting the residual estimate gives $$ \widehat{\pi^2}=0.1000-0.0400=0.0600 $$ in squared log-scale units. The observations’ different subject means do not enter this calculation because each subject is centered separately. With only two subjects, this is an illustrative, imprecise estimate rather than a claim of high precision.

Assumptions, Diagnostics, and Alternatives

The contrasts are useful only when their structure matches the data. Check that the same subject contributes observations to the periods being contrasted, that the periods are defined consistently, and that no systematic change in dose, protocol, or time is being absorbed into the occasion component. Inspect subject-period summaries and paired differences for unusual trends or values. A persistent increase or decrease across periods is not well represented by independent, mean-zero occasion effects.

A negative estimate from subtracting residual variance is possible because estimates fluctuate. For example, if estimated within-subject variance is $0.030$ and estimated residual variance is $0.040$, the method-of-moments result is $-0.010$, which cannot be a variance. This signals that the data do not provide clear evidence for a positive occasion component under this calculation; it does not establish that the true component is negative. Simply replacing the result by zero may be useful for a constrained descriptive report, but hides the sampling uncertainty and should not be mistaken for a precise estimate.

There are two main contrast-based routes. Replicate measurements within a period estimate residual variability directly, after which between-period contrasts or centered period variation can estimate occasion variability. Alternatively, if residual variance is known from credible external information, paired-period contrasts can be corrected for its contribution. Neither route works from paired-period contrasts alone when both components are unknown. A hierarchical mixed-effects model can estimate the components jointly and accommodate unbalanced observations, but its conclusions still depend on the model structure and the information in the repeated-period data. Contrasts are valuable because they reveal exactly which effects cancel; they are not a substitute for checking the assumptions.

Common mistake: “The same-subject period difference cancels between-subject variability, so its variance is the between-occasion variance.” The correction is that the difference also contains residual error from both periods: $\operatorname{Var}(D_i)=2(\pi^2+\sigma^2)$. Estimate or otherwise know the residual contribution before attributing the remaining variability to occasions.

Check Your Understanding

Answer each question, then compare with the brief solutions.

For one observation in each of two periods, which random effect cancels from the same-subject difference, and which two sources remain?
If $\pi^2=0.0500$ and $\sigma^2=0.0200$, what is the variance of a paired-period difference with one observation per period?
Why does a difference between two replicates collected in the same subject-period contain no occasion effect?
A within-subject variance estimate is $0.0700$, and an independent residual variance estimate is $0.0250$. What is the method-of-moments occasion variance estimate, and what does it represent?
What does a negative estimate obtained by subtracting residual variance from within-subject variance imply?

Solutions. (1) The stable subject effect $\eta_i$ cancels; the two occasion effects and two residual errors remain. (2) $2(0.0500+0.0200)=0.1400$, in the corresponding squared units. (3) The replicates share the same $\mu$, $\eta_i$, and $\kappa_{ij}$, so subtraction removes all three; only their residual errors differ. (4) $0.0700-0.0250=0.0450$, an estimate of occasion variance after removing the estimated residual contribution. (5) It indicates that finite-sample variation or estimation uncertainty has produced a within-subject estimate below the residual estimate; it is not evidence for a physically negative variance, and the component is not clearly resolved by this calculation.