Estimating Period Effects in Crossover Analysis

Why Estimate a Period Contrast?

In a two-period crossover, each participant receives both treatments, one in each period. Outcomes can differ between periods for reasons not attributable to treatment—for example, a systematic change in measurements over time. A period contrast quantifies this difference while accounting for which treatment was given in each period and for the sequence to which each participant was assigned. It is a model-based adjustment, not automatically evidence that time itself caused the observed difference.

For pharmacokinetic outcomes analyzed on the log scale, the contrast is measured in log units. Its exponentiation gives a ratio of adjusted geometric means for period 2 relative to period 1, conditional on the crossover model. The estimate is distinct from the treatment contrast, which compares the formulations.

Model and Definition

Consider a complete two-period, two-treatment crossover with sequences TR and RT. T denotes the test treatment and R the reference treatment. Let \(Y_{ij k}\) be the log-transformed outcome for participant \(j\) in sequence \(i\) during period \(k\). A standard additive model is

Y_{ijk}=\mu+S_i+U_{j(i)}+\pi_k+\tau_{L(i,k)}+\varepsilon_{ijk}.

Here \(\mu\) is an overall mean, \(S_i\) is the fixed sequence effect, \(U_{j(i)}\) is a participant effect nested within sequence, \(\pi_k\) is the period effect, and \(\tau_{L(i,k)}\) is the treatment effect for the treatment given in that sequence and period. The participant effect accounts for stable differences among participants. The residual \(\varepsilon_{ijk}\) represents remaining within-participant variation.

The model used for the simple contrast assumes an additive treatment and period structure: there is no treatment-by-period interaction, and no carryover effect that would make the current-period outcome depend on the previous treatment beyond the terms shown. For the usual pooled standard error and t-based interval, assume the within-participant period differences are independent between participants, approximately normal within each sequence, and have a common variance across sequences. Complete observations in both periods are required for the paired-difference calculation.

Definition: The period contrast is \(\delta=\pi_2-\pi_1\), the period 2 effect minus the period 1 effect. Define each participant’s paired difference as \(D=Y_2-Y_1\). Under the additive crossover model, the mean paired difference in sequence TR is \(\delta-\tau\), and in sequence RT it is \(\delta+\tau\), where \(\tau=\tau_T-\tau_R\). Therefore, the adjusted estimate of the period contrast is

\widehat{\delta}=\frac{\bar D_{TR}+\bar D_{RT}}{2}.

The two sequence-specific mean differences contain the period contrast with opposite signs of the treatment contrast. Averaging them cancels the treatment contribution. Differencing within a participant also removes the participant’s stable effect and the sequence effect, because both remain constant across that participant’s two periods.

Deriving the Estimate and Its Uncertainty

The sequence equations are \(\bar D_{TR}\approx\delta-\tau\) and \(\bar D_{RT}\approx\delta+\tau\). Adding them gives \(\bar D_{TR}+\bar D_{RT}\approx2\delta\), which yields the stated estimator. Subtracting them instead isolates the treatment contrast; that is not the target here.

Let \(n_{TR}\) and \(n_{RT}\) be the numbers of complete participants in the two sequences. Let \(s_D^2\) be the pooled within-sequence sample variance of the paired differences:

s_D^2= \frac{(n_{TR}-1)s_{D,TR}^2+(n_{RT}-1)s_{D,RT}^2} {n_{TR}+n_{RT}-2}.

Under the common-variance assumption, the sequence means are independent and have variances \(\sigma_D^2/n_{TR}\) and \(\sigma_D^2/n_{RT}\). Since the period estimate is half their sum, its estimated standard error and degrees of freedom are

SE(\widehat{\delta})=\frac{s_D}{2}\sqrt{\frac{1}{n_{TR}}+\frac{1}{n_{RT}}}, \qquad df=n_{TR}+n_{RT}-2.

A t-based confidence interval is \(\widehat{\delta}\pm t_{1-\alpha/2,df}SE(\widehat{\delta})\). For testing \(H_0:\delta=0\), the statistic is \(t=\widehat{\delta}/SE(\widehat{\delta})\), with the stated degrees of freedom under the model assumptions. The confidence interval and test concern a period contrast; they do not establish that a period effect is clinically important or causal.

Worked Example: A Positive Period Contrast in Balanced Sequences

Suppose a hypothetical crossover has four complete participants per sequence. Their paired log-outcome differences \(D=Y_2-Y_1\) are \(0.00, 0.04, 0.08, 0.12\) in sequence TR and \(0.04, 0.08, 0.12, 0.16\) in sequence RT. These are illustrative log-scale data, not results from a real study. The complete paired observations meet the design requirement. The example also assumes the additive model and common within-sequence variance.

The sequence means are \(\bar D_{TR}=(0.00+0.04+0.08+0.12)/4=0.0600\) and \(\bar D_{RT}=(0.04+0.08+0.12+0.16)/4=0.1000\) log units. Thus,

\widehat{\delta}=\frac{0.0600+0.1000}{2}=0.0800\text{ log units}.

For each sequence, the squared deviations from its mean sum to \(0.0080\). Thus each sample variance is \(0.0080/(4-1)=0.002667\) log-units squared, and the pooled variance is also \(0.002667\). The pooled standard deviation is \(s_D=\sqrt{0.002667}=0.05164\) log units. Therefore,

SE(\widehat{\delta})=\frac{0.05164}{2}\sqrt{\frac14+\frac14} =0.01826\text{ log units},\qquad df=4+4-2=6.

The statistic for \(H_0:\delta=0\) is \(0.0800/0.01826=4.382\). A 95% confidence interval, using \(t_{0.975,6}=2.447\), is \(0.0800\pm2.447(0.01826)\), or approximately \(0.0353\) to \(0.1247\) log units. The estimate is positive: under the model, the adjusted period 2 geometric mean is higher. Exponentiating gives \(\exp(0.0800)=1.083\), or about 8.3% higher; the exponentiated interval is approximately 1.036 to 1.133. This describes a period contrast, not a test-versus-reference comparison.

Worked Example: Unequal Sequence Sizes

In a second hypothetical complete crossover, there are five participants in TR and seven in RT. Suppose the paired-difference means are \(0.0500\) and \(0.0100\) log units, respectively, and the pooled within-sequence standard deviation is \(0.0300\) log units. Assume the same additive model, common variance, and independent participant differences. Unequal sequence sizes do not prevent estimation, provided both sequences are represented.

The adjusted period contrast and its standard error are

\widehat{\delta}=\frac{0.0500+0.0100}{2}=0.0300\text{ log units},

SE=\frac{0.0300}{2}\sqrt{\frac15+\frac17} =0.0150\sqrt{\frac{12}{35}} =0.008783\text{ log units}.

The degrees of freedom are \(5+7-2=10\), and the t statistic for a zero period contrast is \(0.0300/0.008783=3.415\). Using \(t_{0.975,10}=2.228\), a 95% confidence interval is \(0.0300\pm2.228(0.008783)\), or approximately \(0.01043\) to \(0.04957\) log units. Exponentiating the estimate gives \(\exp(0.0300)=1.030\), corresponding to an estimated period 2 geometric mean about 3.05% higher under the model. The estimate remains a period contrast; it does not imply that every participant’s outcome increased by that amount.

What the Estimate Means—and When It Fails

In a log-scale analysis, \(\exp(\delta)\) is the model-based ratio of adjusted geometric means in period 2 versus period 1, with treatment and sequence accounted for as specified. A period effect can reflect systematic temporal differences, study conditions, or other period-related influences. The model alone does not identify which mechanism produced it.

The contrast relies on information from both sequences. If only one sequence is observed, its mean paired difference combines period and treatment effects. For TR, for example, the observed mean difference estimates \(\delta-\tau\), not \(\delta\) alone. This is not repaired merely by increasing the number of participants in that one sequence.

Worked Example: Why Both Sequences Are Needed

Suppose a hypothetical study has complete participants only in sequence TR, and their mean paired difference is \(0.0400\) log units. Under the model this supplies one equation, \(\delta-\tau=0.0400\). It cannot determine both unknown contrasts. For instance, \(\delta=0.0400,\tau=0\) satisfies the equation, as does \(\delta=0.1000,\tau=0.0600\). Both fit the same observed mean difference but imply different period effects. Thus the period contrast adjusted for treatment is not identifiable from this information alone.

A direct comparison of raw period averages is generally not equivalent to the adjusted estimate: treatment allocation differs by period within each sequence, and participant composition may also differ between sequences. A complete-case paired-difference calculation is transparent for this balanced two-period structure. A fitted crossover model can provide the corresponding adjusted estimate and can accommodate more complex covariance or incomplete observations when its assumptions and estimation method are appropriate. It is not safe to treat the two periods as independent samples, because measurements from the same participant are related.

The simple pooled t calculation can be unreliable when the paired-difference distributions are strongly non-normal in small samples, the variances differ substantially between sequences, or the additive model is unsuitable. Inspect sequence-specific differences and the fitted model’s residual behavior; later diagnostics should investigate residual assumptions and treatment-by-period interaction directly. A single period estimate cannot diagnose or rule out such departures.

Common Mistakes

Using a single sequence mean as the period estimate. In TR, \(\bar D_{TR}\) estimates \(\delta-\tau\), not \(\delta\). Combine the two sequence means as \((\bar D_{TR}+\bar D_{RT})/2\).
Forgetting the factor of one-half in the standard error. The estimate is half the sum of two independent sequence means, so its variance is one quarter of their summed variances.
Calling the period contrast a treatment effect. The contrast compares periods after adjustment; it is not the test-versus-reference comparison.
Interpreting an exponentiated contrast causally. It is a model-based period ratio. The estimate alone does not establish why outcomes differ between periods.
Assuming imbalance makes the calculation impossible. Unequal sequence sizes are allowed in the formula, but both sequences must be represented and the variance assumptions must be reasonable.

Key takeaway: With complete two-period data in both sequences, within-participant differences remove stable subject and sequence effects. Averaging the TR and RT mean differences cancels the treatment contrast, leaving the estimated period contrast \(\widehat{\delta}=(\bar D_{TR}+\bar D_{RT})/2\).

Check Your Understanding

Work through each question, then compare your reasoning with the solutions below.

If \(\bar D_{TR}=0.0200\) and \(\bar D_{RT}=0.0600\) log units, find the period contrast estimate and explain what it represents.
Why does differencing the two periods for each participant remove the sequence effect and stable participant effect?
For \(n_{TR}=5\), \(n_{RT}=7\), and \(s_D=0.0300\), calculate the standard error and degrees of freedom.
What can be concluded about the adjusted period contrast if only TR participants are observed and their mean paired difference is known?

Solutions. (1) \(\widehat{\delta}=(0.0200+0.0600)/2=0.0400\) log units. It estimates the adjusted difference between period 2 and period 1 effects, not the treatment contrast. (2) Both effects are constant across a given participant’s two periods, so subtracting \(Y_1\) from \(Y_2\) cancels them. (3) \(SE=(0.0300/2)\sqrt{1/5+1/7}=0.008783\) log units, and \(df=5+7-2=10\). (4) The TR mean difference combines the period and treatment contrasts as \(\delta-\tau\); without information from RT or other identifying assumptions, \(\delta\) cannot be separated from \(\tau\).