Why Trace an Individual Replicate?
A simulation summary can reveal that an operating characteristic is unexpectedly high or low, but it cannot by itself show where the problem arose. A single replicate provides a concrete record to inspect: the randomization probabilities used at each step, the assignments made, the outcomes generated, and the analysis result returned. Tracing that record helps distinguish a coding or data-handling error from a valid but unusual realization of the specified design.
The goal is not to decide whether a design is good or bad from one trial. Random variation can make individual replicates look extreme, especially when arm counts are small. The goal is to establish whether the data and output for that replicate are internally consistent with the simulation rules and the analysis procedure.
What to Record and Reconcile
Consider a two-arm trial with a fixed planned sample size \(N\). For participant \(i\), let \(A_i=T\) denote assignment to treatment and \(A_i=C\) denote assignment to control. Let \(\pi_i\) be the conditional probability of treatment assignment immediately before participant \(i\) is randomized. In a Bernoulli allocation step, the assignment is generated using a random draw \(U_i\sim\operatorname{Uniform}(0,1)\): treatment is assigned when \(U_i<\pi_i\), and control otherwise. In a response-adaptive trial, \(\pi_i\) can depend on information observed before that assignment; it must not depend on future outcomes.
For each replicate, check the chain in order. Confirm that each \(\pi_i\) is a valid probability, that the allocation draw produces the recorded assignment, and that the adaptive rule used only information available at that time. Then reconcile the number assigned to each arm with \(N\), verify that each outcome belongs to the correct participant and arm, and recompute the reported analysis from those records. If an inconsistency appears, the earliest point where it occurs is usually the most informative place to investigate.
The trace should also preserve the scenario identifier, replicate identifier, and random seed or random-number stream information. These do not establish that the code is correct, but they allow the same run to be reproduced and compared with the stored record. A useful audit does not rely on a single final summary or on manually reconstructed data that may itself contain transcription errors.
Worked Example: Checking Assignments Against Recorded Draws
Worked Example: Checking Assignments Against Recorded Draws
Suppose a hypothetical six-participant replicate records the following treatment probabilities and uniform random draws. Assume the probabilities shown were generated by the specified adaptive rule from the history available before each assignment.
| Participant \(i\) | \(\pi_i\) | \(U_i\) | Assignment in trace | Assignment implied by rule |
|---|---|---|---|---|
| 1 | 0.50 | 0.31 | T | T |
| 2 | 0.50 | 0.72 | C | C |
| 3 | 0.60 | 0.44 | T | T |
| 4 | 0.70 | 0.65 | C | T |
| 5 | 0.80 | 0.12 | T | T |
| 6 | 0.90 | 0.95 | C | C |
Apply the assignment rule to each row. For participant 1, \(0.31<0.50\), so treatment is implied. For participant 2, \(0.72\not<0.50\), so control is implied. The same check gives treatment for participant 3, treatment for participant 4, treatment for participant 5, and control for participant 6.
The trace disagrees at participant 4: \(0.65<0.70\), so the recorded draw implies treatment, not control. The implied arm counts are four treatment and two control; the trace’s assignments give three treatment and three control. Both totals equal six, so a simple sample-size check would not find this discrepancy. The participant-level check does.
A mismatch can result from an assignment being overwritten, an incorrect comparison operator, a draw being linked to the wrong participant, or a saved probability that differs from the one actually used. Inspect the code and stored values at this first mismatch. Do not infer that the randomization algorithm is wrong merely because one particular assignment sequence is uncommon: it is the conflict between the stored draw, stored probability, and recorded assignment that is diagnostic here.
Recompute the Analysis From the Replicate Data
Once assignments and outcomes are reconciled, independently recompute the analysis quantity that is supposed to be reported. For a difference in sample means, the treatment-minus-control estimate is \(\bar{x}_T-\bar{x}_C\). With independent arms, the usual plug-in standard error for a Welch-style comparison is
where \(n_T\) and \(n_C\) are the arm sizes and \(s_T^2\) and \(s_C^2\) are the sample variances calculated within the arms using denominator \(n_T-1\) or \(n_C-1\), respectively. This formula describes a particular analysis calculation; it is not automatically an appropriate inferential procedure for every outcome or sample size. When auditing a simulation, first reproduce the analysis the code claims to implement, then assess separately whether its assumptions suit the scenario.
Worked Example: Recalculating a Difference in Means
Consider a hypothetical replicate with four outcomes in each arm. Control outcomes are \(1,1,0,0\); treatment outcomes are \(1,1,1,0\). The outcome is binary, recorded as 1 for success and 0 for failure. We check a reported treatment-minus-control difference using the sample variances and standard error above.
The sample means are
Thus the estimated difference is \(0.75-0.50=0.25\), or 25 percentage points in success proportions. For control, the squared deviations from 0.50 sum to \(4(0.50^2)=1.00\), so \(s_C^2=1.00/(4-1)=0.3333\). For treatment, the squared deviations from 0.75 sum to \(3(0.25^2)+(-0.75)^2=0.75\), so \(s_T^2=0.75/(4-1)=0.25\).
Substitution gives
If the analysis uses the standard normal reference to test a zero difference, its statistic is \(z=0.25/0.3819\approx0.655\), with a two-sided standard normal p-value of approximately 0.513. This calculation checks the arithmetic under that stated reference; with only four observations per arm, it should not be presented as a generally reliable inferential test for a binary outcome. In particular, a single replicate’s p-value is not evidence about the simulation’s overall Type I error rate.
A common coding error is to divide the treatment sum of squared deviations by 4 instead of by \(4-1\). That gives \(0.1875\), not the sample variance \(0.25\), and then understates the standard error. The audit should reproduce the exact variance convention used by the claimed analysis, rather than treating a familiar-looking output as proof that the computation is correct.
Worked Example: Tracing a Reversed Contrast
Worked Example: Tracing a Reversed Contrast
In another hypothetical replicate, five participants are assigned to treatment and three to control. Treatment outcomes are \(1,1,1,1,0\); control outcomes are \(1,0,0\). The analysis specification says that the reported effect is treatment minus control in success proportions.
From the records, \(\hat{p}_T=4/5=0.80\) and \(\hat{p}_C=1/3\approx0.3333\). Therefore, the specified contrast is
Suppose the stored analysis output is \(-0.4667\). The magnitude agrees with the raw data, but the sign is reversed. This points to a contrast-orientation or arm-indexing mismatch: the output may be calculating control minus treatment, or the labels may have been switched between data generation and analysis. Verify the arm mapping and the declared contrast before interpreting the negative sign as a treatment disadvantage.
A useful check is to calculate the arm-specific summaries directly from the participant records and compare them with both the analysis inputs and the final contrast. If the summaries already disagree, investigate data selection or arm labels. If the summaries match but only the contrast differs, inspect the subtraction order and output naming. The arithmetic locates the discrepancy; it does not, by itself, identify which line of code caused it.
Unusual Results Are Not Necessarily Failures
A replicate can be extreme and still be valid. Adaptive assignment can produce unequal arm sizes, and random outcomes can cluster by chance. An unusually large estimated effect, an empty arm in a design that permits such an event, or a wide interval may therefore warrant inspection without proving a bug. Conversely, an output that looks plausible can still be wrong if its arm labels, denominator, or analysis inputs are inconsistent.
Use a trace to ask concrete questions: Are all recorded probabilities between 0 and 1? Do assignment counts sum to the planned sample size? Does each outcome appear once and belong to its assigned participant? Did the probability for a participant use only prior information? Can the analysis estimate and standard error be reconstructed? Does a reported interval contain its point estimate when its stated construction requires that? An answer that conflicts with the recorded data is a reproducibility problem to resolve before interpreting aggregate performance.
For an extreme but internally consistent replicate, compare it with the scenario’s data-generating mechanism and allocation rule. The replicate may be rare, but rarity alone does not establish a defect. Re-run it with the same random stream to check reproducibility, then inspect nearby replicates or targeted tests to determine whether the behavior is systematic. A single trace is a diagnostic tool, not a replacement for summaries across replicates.
Replicate tracing complements, rather than replaces, broader validation. General code validation tests whether the implementation follows the intended design; aggregate summaries show the pattern across many replicates. Tracing is most useful when those checks flag a surprising result and the question becomes, “What happened in this particular run?” It is also limited by what was logged: if probabilities, assignments, or analysis inputs were not saved, the exact source of a discrepancy may be impossible to reconstruct after the run.
Check Your Understanding
Answer each question, then compare with the solutions below.
- A participant has treatment probability \(0.40\), random draw \(0.52\), and recorded assignment T. What assignment does the stated Bernoulli rule imply, and what should be checked?
- In a replicate with planned \(N=20\), the arm counts are 11 and 8. What is the immediate reconciliation issue?
- For control outcomes \(0,0,1,1\), what are the sample mean and sample variance?
- An internally consistent replicate has an unusually large estimated treatment effect. Does that alone establish a coding error? Give one appropriate next check.
Solutions. 1. Since \(0.52\not<0.40\), the rule implies control. Check whether the recorded assignment, draw, and probability are linked correctly and whether the actual assignment used that rule. 2. The counts total 19, not 20, so one participant is missing, unassigned, or omitted from the count; reconcile the participant records before analyzing the replicate. 3. The mean is \((0+0+1+1)/4=0.50\). Squared deviations sum to \(1.00\), so the sample variance is \(1.00/(4-1)=0.3333\). 4. No. First verify that probabilities, assignments, outcomes, and analysis inputs are internally consistent; then reproduce the run or inspect other replicates to assess whether the extreme value is a rare valid realization or part of a systematic pattern.