Transporting Geographic External Controls Across Calendar Time

Why Transportability Matters

An external control group can add information when a current randomized trial has few control participants. But an external group from another region or calendar period may not represent the current trial’s control outcomes. Differences in age, disease severity, clinical practice, outcome measurement, or background care can make a direct comparison misleading. The question is not simply how many external observations are available; it is how much evidence they provide about the control outcomes that would be expected in the current trial.

This tutorial develops a normal-outcome model for making that comparison explicit. It separates adjustment for measured differences from uncertainty about remaining transportability. The calculations are illustrative, not a recommendation to borrow a fixed amount in every application.

Definition: The target population is the population whose treatment effect the trial is intended to estimate. An external control is a control group observed outside the current randomized trial. A transported external control mean is an estimate of the outcome mean that the external controls would have had if standardized to the target population and relevant calendar period. Transportability is the assumption that, after the stated adjustments, the external data are informative about that target mean.

Adjustment Before Borrowing

Adjustment starts with the estimand and the populations it refers to. For example, if the target is the mean treatment effect among patients eligible for the current trial, an external cohort with a different age or disease-severity distribution should not automatically be compared using its crude mean. A direct standardization can reweight outcome means within measured covariate strata to match the target population’s stratum proportions. This requires that the strata are defined comparably, that relevant strata are represented in the external data, and that the outcome is measured consistently.

Calendar time can matter even after standardization. Changes in supportive care, diagnostic criteria, or outcome assessment may shift control outcomes. A time adjustment should be based on substantive knowledge or suitable data, and its uncertainty should be represented. An adjustment cannot repair unmeasured differences merely by being included in a model.

Worked Example: Standardizing Across Regions and Time

Suppose a hypothetical external cohort reports a continuous outcome score at 12 weeks. The target trial population is 60% lower-severity and 40% higher-severity patients. In the external region, the corresponding mean scores are 5 and 9 points, but the external cohort is 80% lower-severity and 20% higher-severity.

The external cohort’s crude mean is $0.80(5)+0.20(9)=5.8$ points. That mean reflects the external cohort’s case mix, not the target trial’s. Standardizing to the target proportions gives

\bar{Y}_{E,\mathrm{std}}=0.60(5)+0.40(9)=6.6\text{ points}.

Suppose, based on a prespecified analysis of comparable historical data, the expected outcome under current-period care is 1.2 points higher than in the external calendar period. The transported mean is then $6.6+1.2=7.8$ points. This arithmetic assumes that the severity-specific means are applicable to the target population and that the estimated calendar shift applies to it. The 1.2-point shift is not known without error; that uncertainty must be accounted for in the borrowing model rather than treated as certainty.

A Model for Borrowing With Transport Uncertainty

Consider independent estimates of the treatment-arm mean and two control means: the current concurrent control mean and the transported external control mean. Let $\bar{Y}_T$, $\bar{Y}_C$, and $\bar{Y}_E$ denote these estimates after any stated adjustment. Let $v_T$, $v_C$, and $v_E$ be their sampling variances. Define the target treatment effect as $\theta=\mu_T-\mu_C$, in the outcome’s units; positive values favor treatment when higher outcomes are beneficial.

The model assumes each estimate is approximately normal with its stated mean and sampling variance, and that the arm estimates are independent. It also assumes that the external estimate has been adjusted to the target population and period, while allowing a residual transportability discrepancy. Represent that discrepancy by a mean-zero normal term with variance $\tau^2$. A larger $\tau^2$ means less confidence that the adjusted external mean represents the current target control mean.

Model and borrowing rule: For a borrowing level $0<\lambda\leq 1$, use external-control variance $v_E/\lambda+\tau^2$. The value $\lambda=1$ uses the full prespecified external likelihood contribution, while smaller values reduce its precision. At $\lambda=0$, exclude the external controls. The combined control estimate and its variance are $$ \bar{Y}_{C,\lambda}= \frac{\bar{Y}_C/v_C+\bar{Y}_E/(v_E/\lambda+\tau^2)} {1/v_C+1/(v_E/\lambda+\tau^2)},\qquad v_{C,\lambda}= \frac{1}{1/v_C+1/(v_E/\lambda+\tau^2)}. $$ For $\lambda=0$, define $\bar{Y}_{C,0}=\bar{Y}_C$ and $v_{C,0}=v_C$. The treatment-effect estimate is $D_\lambda=\bar{Y}_T-\bar{Y}_{C,\lambda}$, with sampling variance $V_\lambda=v_T+v_{C,\lambda}$.

The added $\tau^2$ is important: even as $\lambda$ increases, transport uncertainty prevents the external estimate from receiving unlimited precision. This formulation is a simple, transparent sensitivity model, not a universal model for every outcome or data source. In particular, the assumed independence and the chosen transport variance require justification.

Posterior Calculation and Justification

Use a normal prior $\theta\sim N(m_0,s_0^2)$, independent of the data. Given the control estimates and the model above, the contrast satisfies $D_\lambda\mid\theta\sim N(\theta,V_\lambda)$. Combining this likelihood with the prior gives a normal posterior.

\theta\mid D_\lambda\sim N(m_\lambda,s_\lambda^2),\qquad s_\lambda^2=\left(\frac{1}{s_0^2}+\frac{1}{V_\lambda}\right)^{-1},\qquad m_\lambda=s_\lambda^2\left(\frac{m_0}{s_0^2}+\frac{D_\lambda}{V_\lambda}\right).

To justify the result, the part of the posterior density depending on $\theta$ is proportional to

\exp\left\{-\frac{1}{2}\left[\frac{(\theta-m_0)^2}{s_0^2} +\frac{(D_\lambda-\theta)^2}{V_\lambda}\right]\right\}.

Expanding the two squares, the coefficient of $\theta^2$ is $1/s_0^2+1/V_\lambda$, and the coefficient of the linear term is $-2(m_0/s_0^2+D_\lambda/V_\lambda)$. Completing the square therefore gives posterior variance $s_\lambda^2=(1/s_0^2+1/V_\lambda)^{-1}$ and posterior mean $m_\lambda=s_\lambda^2(m_0/s_0^2+D_\lambda/V_\lambda)$, as stated. This result depends on the normal likelihood and prior; it does not make the transport assumptions true.

Worked Example: No Borrowing Versus Partial Borrowing

In a hypothetical trial, the treatment-arm mean is 12.0 points and the current concurrent control mean is 8.0 points. Each estimate has standard error 0.95 points, so $v_T=v_C=0.95^2=0.9025$ points squared. A transported external mean is 7.8 points with standard error 0.60 points, so $v_E=0.36$ points squared. Set $\tau^2=2.25$ points squared and use the prior $\theta\sim N(0,100)$ points squared.

With no borrowing, $D_0=12.0-8.0=4.0$ points and $V_0=0.9025+0.9025=1.805$ points squared. Thus the posterior mean is $100(4.0)/(100+1.805)=3.929$ points, and the posterior standard deviation is $\sqrt{100(1.805)/(100+1.805)}=1.332$ points. The normal 95% credible interval is approximately $3.929\pm1.96(1.332)=[1.318,6.540]$ points.

Now use $\lambda=1$. The external effective variance is $0.36+2.25=2.61$ points squared. The current and external control precisions are $1/0.9025=1.1080$ and $1/2.61=0.3831$ per point squared. Therefore, $\bar{Y}_{C,1}=(8.0(1.1080)+7.8(0.3831))/(1.1080+0.3831)=7.949$ points, and $v_{C,1}=1/(1.1080+0.3831)=0.6706$ points squared. It follows that $D_1=12.0-7.949=4.051$ points and $V_1=0.9025+0.6706=1.5731$ points squared.

The posterior mean is $100(4.051)/(100+1.5732)=3.989$ points, and the posterior standard deviation is $\sqrt{100(1.5732)/(100+1.5732)}=1.245$ points. The 95% credible interval is approximately $3.989\pm1.96(1.245)=[1.549,6.429]$ points. Under this model, full external borrowing slightly narrows the interval and changes the posterior mean only modestly. The interval is a posterior probability statement conditional on the model and prior, not a guarantee that the true effect lies within its endpoints.

Worked Example: Comparing Borrowing Levels

Keep the data and assumptions from the preceding example, but set $\lambda=0.25$. The external effective variance is $0.36/0.25+2.25=3.69$ points squared, giving external precision $1/3.69=0.2710$ per point squared. The combined control estimate is $(8.0(1.1080)+7.8(0.2710))/(1.1080+0.2710)=7.961$ points, with variance $1/(1/0.9025+1/3.69)=0.7251$ points squared.

Thus $D_{0.25}=12.0-7.961=4.039$ points and $V_{0.25}=0.9025+0.7251=1.6276$ points squared. The posterior mean is $100(4.039)/(100+1.6275)=3.974$ points, the posterior standard deviation is $\sqrt{100(1.6275)/(100+1.6275)}=1.266$ points, and the 95% credible interval is approximately $[1.493,6.455]$ points.

The resulting summaries are close across no, partial, and full borrowing in this example. That stability is reassuring only with respect to this specified range of $\lambda$, prior, adjustment, and transport variance. It does not establish that unmeasured geographic or calendar differences are absent.

Sensitivity, Diagnostics, and Alternatives

Worked Example: A Conclusion Sensitive to Transportability

Consider a different hypothetical trial, with treatment mean 10.0 points, current control mean 8.0 points, and standard errors 0.95 points for both. Suppose a transported external control mean is 5.5 points, its standard error is 0.60 points, and the assumed transport variance is only $0.25$ points squared. Use the same $N(0,100)$ prior. These values deliberately create a large disagreement between external and current controls.

With no borrowing, $D_0=2.0$ points and $V_0=1.805$ points squared. The posterior mean is $100(2.0)/(101.805)=1.965$ points, with standard deviation $\sqrt{100(1.805)/101.805}=1.332$ points and 95% credible interval approximately $[-0.645,4.574]$ points.

With $\lambda=1$, the effective external variance is $0.36+0.25=0.61$ points squared. Its precision is $1/0.61=1.6393$, compared with current-control precision $1/0.9025=1.1080$. The combined control estimate is $(8.0(1.1080)+5.5(1.6393))/(1.1080+1.6393)=6.508$ points, with variance $1/2.7473=0.3640$ points squared. Then $D_1=10.0-6.508=3.492$ points and $V_1=0.9025+0.3640=1.2665$ points squared.

The posterior mean is $100(3.492)/(101.2665)=3.448$ points, and the posterior standard deviation is $\sqrt{100(1.2665)/101.2665}=1.118$ points. Its 95% credible interval is approximately $[1.256,5.640]$ points. The posterior mean shifts by about 1.48 points between no borrowing and full borrowing. This is a conclusion sensitive to the external-control assumption; the discrepancy deserves substantive investigation, not a choice of borrowing level based only on which result is preferred.

Useful diagnostics begin with the data-generating context. Compare eligibility criteria, baseline severity, outcome definitions, follow-up windows, missingness, and background care across sources. Examine calendar trends and region-specific patterns rather than relying on an overall external mean. Check whether standardization requires extrapolation into covariate strata with little or no external representation. Large differences between concurrent and transported controls are a warning that adjustment, measurement, or residual transportability may be inadequate.

A practical sensitivity analysis varies the calendar adjustment, $\tau^2$, and $\lambda$, and reports how posterior means and intervals change. The no-borrowing analysis is the direct concurrent-control benchmark. An unadjusted external analysis can help show the consequence of ignoring measured differences, but it is not a defensible primary analysis when those differences are material. Regression standardization or stratified models can handle measured covariates more flexibly than a single adjusted mean, while hierarchical models can represent variation across regions or periods. Those approaches still require assumptions and sufficient data; added model complexity does not remove the need to assess transportability.

Common Pitfalls and Limitations

Treating sample size as transportability. A large external cohort may estimate its own mean precisely while being a poor guide to the current target control mean. Sampling precision and transport uncertainty are distinct.
Calling a calendar adjustment certain. A point correction does not remove uncertainty in the estimated time shift. Its uncertainty should contribute to the model’s uncertainty, for example through $\tau^2$ or a richer model.
Assuming standardization handles every difference. It addresses only measured, consistently defined variables supported by the available data. It cannot guarantee control of unmeasured confounding or changes in outcome measurement.
Interpreting a stable sensitivity analysis as proof. Similar results across selected borrowing levels only show robustness to those specified choices, not to every plausible source of bias.

The normal summary model is most useful when arm means and their sampling variances are adequate summaries, and when the treatment and control estimates can reasonably be treated as independent. Clustered data, repeated measures, binary outcomes, time-to-event outcomes, or uncertainty in the standardization and calendar adjustment may require a model that represents those features directly. If the external and current controls differ substantially, a simple variance inflation may not adequately describe why they differ.

Key takeaway: Transported controls contribute information only through explicit assumptions linking their adjusted outcomes to the current trial’s target population and calendar period. Report those assumptions, include residual transport uncertainty, and show whether the posterior treatment effect changes materially across defensible borrowing levels.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

Why can a precisely estimated external mean still be unreliable for a current trial’s control mean?
In the model, what happens to the external effective variance as $\lambda$ decreases toward zero? What does $\tau^2$ represent?
A full-borrowing analysis gives a posterior mean treatment effect of 3.4 points, while the no-borrowing mean is 2.0 points. What should be examined before interpreting the difference?
Why is a standardized external mean not automatically a valid transported mean?

Solutions. 1. Its sampling variance may be small, but geographic, case-mix, calendar, or measurement differences can make it unrepresentative of the target control outcome. 2. The term $v_E/\lambda$ grows without bound as $\lambda$ approaches zero, so external precision vanishes; $\tau^2$ represents residual uncertainty about transport after the stated adjustment. 3. Examine the populations, outcome definitions, calendar shifts, covariate support, and the external-current control discrepancy, then assess sensitivity to plausible transport variances and borrowing levels. 4. Standardization addresses measured differences only under appropriate measurement, overlap, and model assumptions; residual and unmeasured differences can remain.