Tutorials › Response-Adaptive Randomization › Design Specification for a Reproducible DBCD

Doubly Adaptive Biased Coin Designs · Tutorial 190 of 501

Design Specification for a Reproducible DBCD

Learn to write and audit a complete DBCD specification, from the response-based target map and initialization to the corrected allocation probabilities and reproducible random assignment.

Advanced 10 min read

What You'll Learn

  • State a complete DBCD rule, including its response estimator, target map, initialization, and assignment mechanism.
  • Distinguish the target allocation shares from the probabilities used to randomize the next participant.
  • Calculate target shares and corrected assignment probabilities from hypothetical response and allocation data.
  • Explain what the correction exponent does and how explicit constraints change a design.
  • Identify implementation details that must be recorded for reproducibility.

From a DBCD Idea to an Executable Specification

A description such as “allocate adaptively toward the arms with better estimated outcomes” does not determine a unique trial. It leaves open which outcomes count as responses, how response rates are estimated, how estimates become target shares, when adaptation begins, and how strongly current allocation imbalances are corrected. Different choices can produce different probabilities at the very next assignment.

A reproducible doubly adaptive biased coin design (DBCD) therefore needs a fully specified sequence of calculations. At each step, observed responses update an estimator, the estimator determines a target allocation vector, and a correction rule combines that target with current allocation shares. A random draw from the resulting probabilities assigns the next participant.

Definition: A complete DBCD specification states (1) the response outcome and response estimator, (2) the target map from response estimates to target shares, (3) initialization and the point at which adaptation begins, (4) the allocation-ratio correction and its exponent \(\gamma\), (5) all allocation or probability constraints, and (6) the procedure that converts assignment probabilities into an arm assignment.

Notation and the Sequential Rule

Suppose there are \(K\) treatment arms. Just before participant \(t+1\) is assigned, \(N_k(t)\) is the number assigned to arm \(k\) among the first \(t\) participants, and \(S_k(t)\) is the number of observed responses on that arm. For binary responses, each observed outcome is \(Y_i\in\{0,1\}\), where 1 denotes a response under a prespecified response definition. The allocation share is \(\pi_k(t)=N_k(t)/t\), when \(t>0\). It describes assignments already made; it is not the target share for the next assignment.

Let \(\hat p_k(t)\) be the prespecified estimate of the response probability on arm \(k\), calculated from responses observed by time \(t\). A target map \(R\) turns the vector of estimates and any prespecified design inputs into target shares \(\rho_k(t)\), with \(\rho_k(t)\geq 0\) and \(\sum_{k=1}^K\rho_k(t)=1\). The DBCD assignment probability is

$$ q_k(t+1) = \frac{\rho_k(t)\left\{\rho_k(t)/\pi_k(t)\right\}^{\gamma}} {\displaystyle\sum_{j=1}^K \rho_j(t)\left\{\rho_j(t)/\pi_j(t)\right\}^{\gamma}}, \qquad k=1,\ldots,K. $$

This expression assumes the target shares and current allocation shares are positive wherever the ratio is used, and that the denominator is finite and positive. A design must ensure that these conditions hold before using the formula. The exponent \(\gamma\geq0\) controls the strength of the correction: when an arm’s current share is below its target, its target-to-current ratio is greater than 1. The correction then increases its weight relative to the uncorrected target weight. At \(\gamma=0\), the probabilities equal the target shares, provided the formula is defined.

Specify the Estimator, Target, and Constraints

The target map should follow from the trial’s stated objective, not from a vague preference for an arm that appears better. For example, if the objective is to estimate arm-specific binary response means with equal per-participant costs and minimize a prespecified weighted sum of their sampling variances, a Neyman-type target can be appropriate. The exact objective and target map must still be stated: other objectives can imply different targets.

For a concrete specification in the examples below, use a Beta(1,1) smoothing rule for each binary response rate:

$$ \hat p_k(t)=\frac{S_k(t)+1}{N_k(t)+2}. $$

This estimator is defined even if an arm has recorded no responses or no failures. Use the illustrative equal-cost target map \(\tilde\rho_k(t)\propto\sqrt{\hat p_k(t)\{1-\hat p_k(t)\}}\), which assigns shares in proportion to estimated response standard deviations. To impose a prespecified target-share floor \(\varepsilon\), define \(\rho_k(t)=\varepsilon+(1-K\varepsilon)\tilde\rho_k(t)\), where \(0\leq\varepsilon<1/K\). This is an explicit part of the target map: it preserves a positive share while keeping the shares’ sum equal to 1. A different floor, estimator, or scientific objective would define a different design.

For initialization, suppose \(m_0\geq1\) participants are assigned to each arm using a stated randomization procedure, and adaptation begins only after these \(Km_0\) assignments and their required responses have been recorded. During initialization the design must say how assignments are made, how incomplete responses are handled, and whether enrollment pauses until those responses are available. Once adaptation begins, \(\pi_k(t)\) is positive because each arm has at least \(m_0\) assignments.

Worked Calculations

Worked Example: Estimating the Target Shares

Consider a hypothetical three-arm trial after initialization with \(N=(4,4,4)\) assignments and \(S=(3,2,1)\) responses. Use the Beta(1,1) estimator, the equal-cost target map above, and \(\varepsilon=0.05\).

The estimated response rates are \(\hat p_1=(3+1)/(4+2)=4/6=0.6667\), \(\hat p_2=(2+1)/(4+2)=3/6=0.5000\), and \(\hat p_3=(1+1)/(4+2)=2/6=0.3333\). The three estimated variances are \(2/9\), \(1/4\), and \(2/9\). Their square roots are approximately \(0.4714\), \(0.5000\), and \(0.4714\), whose sum is approximately \(1.4428\). Thus the unconstrained target shares are approximately \(\tilde\rho=(0.3266,0.3468,0.3266)\).

Apply the stated floor transformation, not an informal clipping step: \(\rho_k=0.05+0.85\tilde\rho_k\). This gives \(\rho\approx(0.3276,0.3448,0.3276)\), with shares summing to 1 up to rounding. The target is slightly more balanced than the unconstrained shares because the floor is part of the design.

Worked Example: Correcting for Unequal Allocation Counts

Now consider a different hypothetical point in a three-arm trial. Let \(N=(10,5,5)\), \(S=(6,2,2)\), use the same estimator and target map, and set \(\gamma=1\), \(\varepsilon=0.05\). The allocation shares are \(\pi=(0.5000,0.2500,0.2500)\). The estimates are \(7/12=0.5833\), \(3/7=0.4286\), and \(3/7=0.4286\).

The estimated standard deviations are approximately \(0.4930,0.4949,0.4949\). Normalizing gives unconstrained shares approximately \((0.3325,0.3338,0.3338)\), and applying the floor transformation gives \(\rho\approx(0.3326,0.3337,0.3337)\).

With \(\gamma=1\), the unnormalized weights are \(\rho_k^2/\pi_k\): \(0.3326^2/0.5000\approx0.2213\), \(0.3337^2/0.2500\approx0.4454\), and \(0.3337^2/0.2500\approx0.4454\). Their sum is approximately \(1.1121\). Normalizing gives \(q\approx(0.1990,0.4005,0.4005)\). The target shares are close to equal, but the first arm’s allocation share is twice the others, so its next-assignment probability is reduced below its target share.

For a reproducible draw, specify that a recorded uniform random number \(U\) on \([0,1)\) is compared with cumulative probabilities in arm order. For example, if \(U=0.55\), the cumulative probabilities are approximately \(0.1990\), \(0.5995\), and \(1\); the assignment is to arm 2. This is one hypothetical draw, not a deterministic rule for choosing the arm with the largest probability.

Worked Example: Updating After the Assignment

Continue the preceding hypothetical sequence: the participant was assigned to arm 2 and had response \(Y=1\). The updated totals are \(N=(10,6,5)\) and \(S=(6,3,2)\). The updated estimates are \(7/12=0.5833\), \(4/8=0.5000\), and \(3/7=0.4286\).

The corresponding standard deviations are approximately \(0.4930,0.5000,0.4949\). After normalizing and applying the same \(\varepsilon=0.05\) transformation, the updated target is approximately \(\rho=(0.3316,0.3356,0.3327)\). The new allocation shares are \(\pi=(10/21,6/21,5/21)\approx(0.4762,0.2857,0.2381)\).

For \(\gamma=1\), the next unnormalized weights are approximately \(0.3316^2/0.4762=0.2310\), \(0.3356^2/0.2857=0.3943\), and \(0.3327^2/0.2381=0.4649\). They sum to approximately \(1.0902\), so the next assignment probabilities are approximately \((0.2119,0.3617,0.4264)\). The low current share on arm 3 raises its probability above its target share; the new response also changes the estimates and hence the target itself.

Why the Correction Has This Form

The correction is easiest to audit by comparing two arms. Dividing their probabilities cancels the common normalizing denominator:

$$ \frac{q_i(t+1)}{q_j(t+1)} = \frac{\rho_i(t)}{\rho_j(t)} \left\{ \frac{\rho_i(t)/\pi_i(t)} {\rho_j(t)/\pi_j(t)} \right\}^{\gamma}. $$

If the target shares are equal and arm \(i\) is underrepresented relative to arm \(j\), then \(\pi_i<\pi_j\), so the ratio in braces exceeds 1. For \(\gamma>0\), this factor raises the relative probability of assigning to arm \(i\). The correction is not a guarantee that observed counts will match targets at every time: random assignments can create temporary departures. Nor does the formula alone determine the target; it only combines a target vector with current allocation shares.

Constraints, Comparison, and Diagnostics

A reproducible specification must say whether constraints apply to the target shares, the assignment probabilities, or both. The floor used in the examples constrains \(\rho\), not automatically \(q\). If the design also requires \(q_k\geq\delta\), it must state an exact probability transformation and its conditions—for example, \(q_k^{*}=\delta+(1-K\delta)q_k\) with \(0\leq\delta<1/K\). This is a modified randomization rule and should be reported as such. Maximum sample sizes, temporary arm unavailability, or other restrictions also require explicit rules for recalculating probabilities; there is no universal adjustment that leaves the original design unchanged.

Compared with fixed randomization, a DBCD uses observed responses and allocation history to update future probabilities. Compared with allocation directly at the target (\(\gamma=0\)), positive \(\gamma\) adds correction for current imbalance. Increasing \(\gamma\) strengthens that correction, but it does not alter the response estimator or target map. These are distinct design choices and should not be conflated.

A practical audit should check, at each assignment, that the response data include only outcomes available under the stated rule; all \(N_k(t)\) and \(S_k(t)\) agree with the assignment and outcome log; target shares are nonnegative and sum to 1; the allocation shares are positive; all constraints are applied in the specified order; and the final \(q_k\) values are nonnegative and sum to 1. Record unrounded intermediate values for calculations, the random-number generator and seed or stream, arm order for cumulative probabilities, and the draw used. Such details make the realized sequence auditable without treating a particular random sequence as part of the probability formula.

A common mistake is to say, “arm 1’s target share is 0.3326, so the next participant must have a 0.3326 probability of going to arm 1.” That ignores the allocation-ratio correction. In the second example the target share is about \(0.3326\), but the corrected probability is about \(0.1990\). The target describes the desired allocation proportions; \(q\) is the probability used for the next random assignment. Another mistake is to clip an estimated target after calculation without specifying how the remaining shares are adjusted. A reproducible design defines a transformation, such as the floor mixture used here, and applies it consistently.

Key takeaway: A DBCD is reproducible only when its estimator, target map, initialization, correction exponent, constraints, and random draw procedure are all explicit. Reporting target shares alone is not enough to reconstruct assignment probabilities.

Check Your Understanding

Answer each question from the stated rules. Solutions follow the questions.

  1. Why are target shares and current allocation shares different quantities in the DBCD rule?
  2. With \(\gamma=0\), what are the assignment probabilities when the formula is defined?
  3. Under the Beta(1,1) estimator, an arm has \(N_k=3\) assignments and \(S_k=0\) responses. Calculate \(\hat p_k\).
  4. Why must a probability floor be specified as part of the design rather than silently applied after calculating \(q\)?

Solutions. (1) \(\rho_k\) is the target produced by the response-based target map; \(\pi_k=N_k/t\) summarizes assignments already made. The DBCD uses both to calculate \(q_k\), the probability of the next assignment. (2) Setting \(\gamma=0\) makes every correction factor equal to 1, so \(q_k=\rho_k/\sum_j\rho_j=\rho_k\). (3) \(\hat p_k=(0+1)/(3+2)=1/5=0.2000\). Smoothing keeps this estimate away from zero. (4) A floor changes the probabilities and therefore changes the randomization rule. The exact transformation and its constraints must be known to calculate the same \(q\) values and reproduce the design.

Clinical Biostats