What an Allocation-Path Plot Shows
An allocation-path plot follows a trial’s randomization as participants enter. It may show the probability of assigning each new participant to each arm, the assignments that actually occurred, or both. These quantities answer different questions: the probability path describes the randomization rule at each decision, while the allocation path records the realized history of the trial.
That distinction matters in an adaptive trial. An arm’s assignment probability can change after accumulating information, but a probability change does not itself assign anyone to that arm. Conversely, a run of assignments can make an arm’s cumulative count rise quickly even when its probability has not changed. A plot should therefore be read by checking its axes, legend, and definition of each plotted quantity before interpreting its shape.
For a set of active arms, the assignment probabilities at a decision sum to one: \(\sum_j\pi_{jk}=1\). The indicators also sum to one for each enrolled participant: \(\sum_j I_{jk}=1\). But an individual \(\pi_{jk}\) is not the same as the cumulative proportion \(N_j(m)/m\). The former applies to the next assignment; the latter summarizes all assignments so far.
Many plots show probabilities as a step function: the line holds at one value until the randomization rule is updated, then jumps to a new value. Cumulative counts also rise in steps, but each participant contributes exactly one assignment, so only the selected arm’s count rises at that assignment. Cumulative proportions usually move more gradually because each new assignment changes the denominator as well as, sometimes, the numerator.
Reading Shifts and Persistence
An early shift is a visible change in assignment probability or in the subsequent rate of accumulation of assignments. To interpret it, locate when the change occurs and determine which quantity changed. A probability jump indicates a change in the rule for assigning future participants. A change in the slope of a cumulative-count path indicates a change in the observed pace of assignments to that arm, which may reflect a probability change, random variation, or both.
A shift is persistent when the probability remains at its new level, or when the cumulative allocation pattern continues to reflect the changed rule, over a meaningful later portion of the plotted trial. Persistence is not established by a single steep segment or a brief run of assignments. Nor does a persistent probability shift mean that cumulative proportions immediately reach the new probability: earlier assignments remain in the cumulative totals.
Before drawing conclusions, establish whether the horizontal axis is participant index or calendar time. With participant index, each step corresponds to an assignment. With calendar time, the spacing between assignments can vary with recruitment. A flat count path during a period with no enrollment does not by itself indicate that an arm was suspended or removed.
Worked Example: A Two-Arm Probability Shift
Consider a hypothetical two-arm trial with arms A and B. For the first four assignments, the probability of A is \(0.50\); for the next four, it is \(0.75\), with B receiving the remaining probability. Suppose the observed assignments are B, A, B, A, A, A, B, A. Find the cumulative counts and interpret the paths.
Probability path. For assignments 1–4, \((\pi_{A,k},\pi_{B,k})=(0.50,0.50)\). For assignments 5–8, \((\pi_{A,k},\pi_{B,k})=(0.75,0.25)\). The step at assignment 5 is an early increase in A’s chance of assignment.
Realized counts. In assignments 1–4, A was assigned twice and B twice, so \(N_A(4)=2\) and \(N_B(4)=2\). In assignments 5–8, A was assigned three times and B once. Thus \(N_A(8)=2+3=5\) and \(N_B(8)=2+1=3\). The final cumulative proportions are \(5/8=0.625\) for A and \(3/8=0.375\) for B.
The probability for A after the update is \(0.75\), but its final cumulative proportion is \(0.625\), because the first four participants were assigned under the earlier rule and the later assignments did not all go to A. The upward probability step is persistent through the final assignment in this example; eight assignments alone do not establish why the rule changed or whether the treatment is beneficial.
Why the Probability Path Does Not Determine the Count Path
Assignments are random outcomes generated under the stated probabilities. Therefore a cumulative count path can depart substantially from what a viewer might expect from looking only at the probability line, especially early in a trial. The precise expectation statement requires care when the probabilities adapt to previous assignments.
The result follows by writing the increase as a sum of assignment indicators and conditioning each indicator on the information available just before its assignment. Since \(\mathcal{H}_m\) is contained in \(\mathcal{H}_{k-1}\) for \(k>m\), the law of iterated expectation gives \(E(I_{jk}\mid\mathcal{H}_m)=E\{E(I_{jk}\mid\mathcal{H}_{k-1})\mid\mathcal{H}_m\}=E(\pi_{jk}\mid\mathcal{H}_m)\). Summing these equalities gives the stated result.
If the probabilities for those assignments are fixed independently of the assignments being counted, the expectation simplifies to their sum. In an adaptive design, however, future probabilities may themselves depend on earlier assignments or outcomes. One must not condition on the entire realized sequence of adaptive probabilities and then automatically treat that sequence as fixed: it may contain information about the assignments whose counts are being considered.
Worked Example: Expected Assignments Under a Fixed Rule
Suppose a hypothetical trial has a fixed allocation rule for the next six participants: \(\pi_A=0.60\) and \(\pi_B=0.40\) at every assignment. Find the expected increase in each arm’s count and compare it with one possible realized result.
Because the probabilities are fixed, the expected increase for A is \(6(0.60)=3.60\) participants, and for B it is \(6(0.40)=2.40\). The expected counts need not be whole numbers; they are averages over repetitions of the six-assignment process.
Suppose the actual six assignments are A, A, B, A, B, B. The realized increases are three for A and three for B. They differ from the expectations by \(3-3.60=-0.60\) and \(3-2.40=0.60\), respectively. This is compatible with the rule: each assignment was randomized, so a particular sequence need not reproduce the probabilities exactly.
A plotted probability of \(0.60\) for A should therefore be read as the conditional chance at each decision under this example’s fixed rule, not as a promise that exactly 60% of the next six participants will be assigned to A.
Recognizing Arm Removal
A permanently removed arm can leave two clear signatures. Its assignment probability becomes zero after the removal point, and its cumulative count remains constant thereafter. The other active arms’ probabilities must then sum to one, assuming randomization continues among those arms. A plot may mark the removal time explicitly; use that marker and the design definition rather than inferring removal solely from a stretch with no assignments to that arm.
A temporary suspension is different. During a suspension, the arm may have probability zero for a period but can become eligible for assignment again. A permanent removal is represented by probability zero for every subsequent decision. Since both can produce a flat cumulative-count segment, the probability path and the status information are important for distinguishing them.
Cumulative proportions need special care at removal. If total enrollment continues, the removed arm’s count stays fixed while the overall denominator grows, so its cumulative proportion falls. That decline is arithmetic, not evidence that the arm’s treatment effect worsened after removal. The remaining arms’ cumulative counts continue to rise as participants are assigned to them.
Worked Example: A Removed Arm in a Three-Arm Trial
In a hypothetical platform trial, arms A, B, and C receive the first ten participants. At participant 11, C is permanently removed. Before removal, the probabilities are \((0.40,0.35,0.25)\); afterward, the probabilities for A and B are \((0.60,0.40)\), and C’s probability is zero. At removal, the cumulative counts are \(N_A(10)=4\), \(N_B(10)=3\), and \(N_C(10)=3\). Among the next five participants, suppose the observed assignments are A, B, A, A, B.
After those five assignments, \(N_A(15)=4+3=7\), \(N_B(15)=3+2=5\), and \(N_C(15)=3\). The total is \(7+5+3=15\), as required. The cumulative proportions are \(7/15=0.4667\) for A, \(5/15=0.3333\) for B, and \(3/15=0.2000\) for C.
The probability path for C drops from \(0.25\) to zero and stays there, while its cumulative count stays at three. Its cumulative proportion decreases from \(3/10=0.3000\) at participant 10 to \(3/15=0.2000\) at participant 15, even though no new C assignments occur. This combination of a permanent zero probability and a flat count path is consistent with removal.
Comparing the Main Plot Types
Different plot types make different features easiest to see. A probability plot emphasizes the randomization rule; a cumulative-count plot emphasizes the realized number assigned; and a cumulative-proportion plot emphasizes the realized share of enrollment to date. A plot that overlays these quantities should label them distinctly and use scales that do not invite direct numerical comparison between unlike quantities.
| Plotted quantity | What it answers | Important caution |
|---|---|---|
| Assignment probability \(\pi_{jk}\) | What is the chance of assigning the next participant to arm \(j\) at this decision? | It is not the proportion already assigned to that arm. |
| Cumulative count \(N_j(m)\) | How many participants have actually been assigned to arm \(j\) so far? | A flat segment may reflect no enrollment, not a change in arm status. |
| Cumulative proportion \(N_j(m)/m\) | What share of all participants so far were assigned to arm \(j\)? | It incorporates earlier assignments and can lag behind a new probability. |
A single simulated path is useful for seeing how the rule and actual assignments interact, but it is only one possible realization. Multiple simulated paths can show the range of histories produced by the design; their spread is not itself a probability path. In particular, do not interpret a cloud of cumulative-count curves as treatment-effect uncertainty unless the plot explicitly represents that quantity.
Checks, Pitfalls, and Limitations
Before interpreting a plot, check that the probabilities for active arms add to one at each decision, that cumulative counts add to the number assigned, and that cumulative proportions use the stated denominator. Check whether the horizontal axis is participant index or calendar time, and whether a vertical line marks an interim decision, suspension, or permanent removal. If the displayed paths do not satisfy basic accounting identities, investigate the plotting convention, missing assignments, or data processing before drawing a design conclusion.
- Confusing probability with realized share. A probability of \(0.70\) for the next assignment does not mean that 70% of participants already enrolled were assigned to that arm.
- Calling every flat count path a removal. No enrollment during a calendar-time interval also produces a flat count. A zero probability sustained after a documented removal is stronger evidence of permanent ineligibility.
- Expecting exact proportionality in a short run. Random assignments can produce counts that differ from probability-weighted expectations. A short path cannot establish that the randomization rule was followed incorrectly.
- Reading a cumulative proportion as a current probability. A cumulative proportion includes all prior assignments, so it changes more slowly than a probability updated at an interim.
Allocation-path plots describe assignment behavior, not by themselves the clinical effectiveness of treatments. They also cannot establish whether a probability shift was justified without information about the adaptation rule and the information available at the time. The plot is a diagnostic summary of the assignment history; interpretation remains tied to its definitions, timing, and design context.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- At participant 20, an arm’s probability for the next assignment rises from \(0.30\) to \(0.65\). Does this mean that 65% of the first 20 participants received that arm? Explain.
- An arm has cumulative count 8 at participant 12 and count 8 at participant 18. What can be concluded from the count path alone?
- After permanent removal of arm C at participant 10, total enrollment reaches 20 and C’s count remains 4. What is C’s cumulative proportion at participant 20?
- Why is it generally unsafe to calculate an adaptive arm’s expected future count by conditioning on the entire realized sequence of its future assignment probabilities?
Solutions. (1) No. The value \(0.65\) is the conditional assignment probability for the next participant under the current rule. The first 20 participants’ cumulative proportion must be calculated from their actual count, \(N_j(20)/20\). (2) The arm received no assignments among participants 13 through 18. The count path alone cannot establish whether the arm was removed, temporarily suspended, or simply not selected by chance; inspect the probability and status paths. (3) Its cumulative proportion is \(4/20=0.2000\), or 20%. The count remains fixed while the total enrollment denominator increases. (4) In an adaptive design, future probabilities can depend on earlier assignments. Conditioning on their full realized sequence can therefore reveal information about the assignments being counted, so the probabilities cannot automatically be treated as fixed. The safe conditional calculation conditions on information available before each assignment, or uses the relevant expectations of future probabilities given current information.