Why the Response-Rate Estimator Changes the Design
A doubly adaptive biased coin design (DBCD) uses a target allocation share and current allocation counts to calculate the probability of assigning the next participant to each arm. When the target depends on estimated response variability, the response-rate estimator is part of the allocation rule: changing it can change the target, the next assignment probability, and eventually the sequence of assignments. The effect is not just a difference in how a response rate is reported.
To make that connection precise, consider two arms and a binary outcome, coded 1 for response and 0 otherwise. Suppose the target map allocates in proportion to the estimated standard deviation of that outcome. This is the variance-based target used throughout the examples below. Other target maps would give different calculations; the estimator comparison here holds the target map and coin rule fixed.
Here \(\gamma>0\) controls the correction for the difference between current allocation shares and the target. The probability for arm 2 is \(1-\pi_1\). A target share is not itself an assignment probability: \(\rho_1\) enters the coin rule together with \(n_1,n_2,\) and \(\gamma\).
Assumptions and a Useful Sensitivity Result
The plug-in standard deviation above is appropriate when outcomes are binary and the estimated response rate is intended to represent the response probability for that arm. The examples assume that outcomes are observed before the next assignment, that the stated estimator uses exactly the responses specified for it, and that the response data are recorded consistently. They do not require the two response probabilities to be equal. They do require both arms to have defined estimates and, for the displayed DBCD probability, positive allocation counts and a target share strictly between 0 and 1. If an estimate is on a boundary, that condition may fail.
The allocation formula also makes the immediate influence of an estimator choice transparent. Holding the counts and \(\gamma\) fixed, define \(A=(n_2/n_1)^\gamma\). Then \(\pi_1=A\rho_1/[A\rho_1+(1-\rho_1)]\). Differentiating with respect to the target share gives
Thus, at a fixed decision point, a larger estimated target share for arm 1 gives a larger assignment probability for arm 1. The sensitivity depends on both the counts and the target. Across later decision points, however, counts may differ too, so this one-step monotonicity does not imply that every later probability or cumulative allocation difference will keep the same direction.
The formula follows by dividing the numerator and denominator in the DBCD expression by \(n_1^{-\gamma}\): the first term acquires the factor \((n_2/n_1)^\gamma\), while the second becomes \(1-\rho_1\). Differentiating the resulting ratio gives the positive derivative above. This is a local comparison: it holds when counts, \(\gamma\), and all other parts of the design are held fixed.
One History, Three Estimators
A cumulative estimator uses all observed responses in an arm. A rolling-window estimator uses only a specified number of that arm’s most recent responses. A smoothed estimator combines the observed data with a stated prior or other regularization rule. These are distinct procedures; a rolling window is not just a more recent display of the cumulative estimate, and smoothing is not a change to the DBCD coin rule.
Worked Example: Comparing Three Estimators at the Same Decision
Hypothetical data. At one decision, \(n_1=5\) and \(n_2=4\). In arm 1, the ordered outcomes are 1, 0, 1, 1, 1; in arm 2 they are 0, 1, 0, 0. Let \(\gamma=1\). Compare the cumulative estimate, a four-response rolling window in each arm, and a Beta(1,1) smoothed estimate. The same observed outcomes and current counts are used in all three calculations.
Cumulative estimator. The response rates are \(\hat p_1=4/5=0.8000\) and \(\hat p_2=1/4=0.2500\). The plug-in standard deviations are \(\sqrt{0.8000(0.2000)}=0.4000\) and \(\sqrt{0.2500(0.7500)}=0.4330\). Therefore \(\rho_1=0.4000/(0.4000+0.4330)\approx0.4802\). Substitution into the DBCD rule gives
So the next participant has about a 42.50% probability of assignment to arm 1 and a 57.50% probability of assignment to arm 2.
Four-response rolling window. The four most recent arm 1 outcomes are 0, 1, 1, 1, giving \(\hat p_1=3/4=0.7500\). The four arm 2 outcomes give \(\hat p_2=1/4=0.2500\). Both plug-in standard deviations are \(\sqrt{0.75(0.25)}=\sqrt{0.1875}\approx0.4330\), so \(\rho_1=0.5000\). The allocation probability is
The rolling-window rule therefore gives a 44.44% probability of assignment to arm 1 at this decision.
Beta(1,1) smoothing. With a Beta(1,1) prior for each arm’s response probability, the posterior mean is \((x_k+1)/(n_k+2)\), where \(x_k\) is the number of responses. Thus \(\hat p_1=5/7\approx0.7143\) and \(\hat p_2=2/6=1/3\). The corresponding plug-in standard deviations are \(\sqrt{(5/7)(2/7)}=\sqrt{10}/7\approx0.4518\) and \(\sqrt{(1/3)(2/3)}=\sqrt{2/9}\approx0.4714\). Hence \(\rho_1\approx0.4894\), and
The three next-assignment probabilities differ even though the underlying observed outcomes, current counts, target map, and \(\gamma\) were held fixed. The difference comes from the estimator definition.
How a One-Step Difference Creates Different Paths
An allocation path is the sequence of assignments and resulting counts as the trial proceeds. To see how estimator choice can change that path, use the three probabilities from the previous example and the same hypothetical uniform random draw \(U=0.4300\). Under the usual inverse-probability rule, assign to arm 1 if \(U<\pi_1\), and otherwise to arm 2. This is a device for comparing the rules, not a claim that different designs would generally observe the same outcomes after their assignments diverge.
Worked Example: The Same Random Draw, Different Assignments
For the cumulative estimator, \(\pi_1\approx0.4250\), so \(0.4300\) is not less than \(\pi_1\): the next assignment is to arm 2. For the rolling-window estimator, \(\pi_1\approx0.4444\), so the same draw assigns to arm 1. For the Beta-smoothed estimator, \(\pi_1\approx0.4340\), so it also assigns to arm 1. One common random draw therefore separates the cumulative-rule path from the other two at this decision.
Now follow each of the first two branches one more step, using explicitly hypothetical outcomes. In the cumulative branch, suppose the arm 2 assignment yields a response. The updated cumulative rates are \(\hat p_1=4/5=0.8000\) and \(\hat p_2=2/5=0.4000\). Their plug-in standard deviations are \(0.4000\) and \(\sqrt{0.24}\approx0.4899\), so \(\rho_1=0.4000/(0.4000+0.4899)\approx0.4495\). The counts are now 5 and 5; with \(\gamma=1\), equal counts make \(\pi_1=\rho_1\approx0.4495\).
For the rolling-window branch, suppose the arm 1 assignment yields a nonresponse. The last four arm 1 outcomes are now 1, 1, 1, 0, so the window rate remains \(3/4=0.7500\); the arm 2 window rate remains \(1/4=0.2500\). Thus the target is still \(\rho_1=0.5000\). The updated counts are 6 and 4, giving
The next probabilities, approximately 0.4495 in the cumulative branch and 0.4000 in the rolling-window branch, reflect both the estimator and the different count histories. These calculations illustrate a conditional path, not an unconditional prediction: later assignments and responses are random.
Boundary Estimates, Diagnostics, and Design Choices
A plug-in variance can be zero when an estimated response rate is 0 or 1. If this makes the target share 0 or 1, the DBCD formula assigns zero probability to one arm, even when the design is intended to continue sampling both arms. This is a boundary failure of that estimator-target combination, not an arithmetic error. Smoothing can avoid such a boundary for finite data, but it changes the estimated target and must be treated as part of the design.
Worked Example: A Boundary Estimate and Its Smoothed Alternative
Suppose, hypothetically, that both arms have \(n_1=n_2=3\). Arm 1 has three responses and arm 2 has one response. Under the cumulative estimator, \(\hat p_1=1\) and \(\hat p_2=1/3\). The plug-in standard deviations are \(0\) and \(\sqrt{(1/3)(2/3)}=\sqrt{2/9}\approx0.4714\), giving \(\rho_1=0\). With positive equal counts, the DBCD formula then gives \(\pi_1=0\): arm 1 will not receive the next assignment.
Under a Beta(1,1) smoothed estimator, the rates become \(\hat p_1=4/5=0.8000\) and \(\hat p_2=2/5=0.4000\). The standard deviations are \(0.4000\) and \(\sqrt{0.24}\approx0.4899\), so \(\rho_1\approx0.4495\). Since the counts are equal, \(\pi_1\approx0.4495\). Smoothing has prevented the zero assignment probability, but it has also replaced the cumulative target with a different target. It is a deliberate design choice, not a harmless arithmetic adjustment.
Useful diagnostics follow directly from the calculation. At each decision, check that the estimator uses the intended responses, that its window or prior is as specified, that the estimated rates lie in the intended range, and that the target shares sum to one. Then verify the allocation probability using the current counts and the chosen \(\gamma\). Plotting estimates and probabilities over time can reveal abrupt changes when a window drops an observation, as well as boundary estimates or unexpected reversals. A probability that differs from the target is not by itself an error: the DBCD correction is designed to account for current allocation shares.
Cumulative estimates use more data and are usually less sensitive to one new response, but can adapt slowly if response behavior changes. Rolling windows can respond faster to recent outcomes, but discard older information and can make estimates volatile, especially with short windows. Smoothing can stabilize small-sample and boundary estimates, but the prior or regularization strength affects the target. None is universally preferable. The appropriate choice depends on the intended response model and design objective; the exact estimator, initialization, window length or prior, and handling of boundary cases should be explicit.
Common Mistake
A frequent error is to calculate an estimated target share and use it directly as the next assignment probability—for example, to say that \(\rho_1=0.4802\) means a 48.02% chance of assignment to arm 1. That ignores the current allocation counts and the DBCD correction. In the first worked example, \(\rho_1\approx0.4802\), but the assignment probability is about 0.4250 because the counts are 5 and 4 and \(\gamma=1\). Another error is to attribute every later difference solely to the estimator: once two rules assign different participants, their count histories—and possibly the outcomes available to their estimators—can differ too. Separate the one-step comparison at fixed data from the full path comparison.
Check Your Understanding
Answer each question, then compare with the brief solutions below.
- With \(\rho_1=0.60\), \(n_1=6\), \(n_2=4\), and \(\gamma=1\), what is the next assignment probability for arm 1?
- Why can a rolling-window estimate differ from a cumulative estimate even when both are calculated from the same arm’s observed responses?
- For two equal allocation counts, what is the DBCD assignment probability for arm 1 when \(\rho_1=0.35\), regardless of the positive value of \(\gamma\)?
- What design consequence can follow from a cumulative binary-response estimate equal to 1 in one arm and an interior estimate in the other?
Solutions. (1) \(\pi_1=4(0.60)/[4(0.60)+6(0.40)]=2.4/4.8=0.5000\). (2) The rolling window excludes responses older than its specified window, whereas the cumulative estimator includes all observed responses; the data used are different. (3) Equal counts make the count correction cancel, so \(\pi_1=\rho_1=0.3500\). (4) The arm’s plug-in standard deviation can be zero, producing a zero target share and hence zero assignment probability for that arm under this target map and coin rule.