Worked Example: Binary Responses and a Moving Target

Why a Moving Target Requires Sequential Calculation

In a doubly adaptive biased coin design (DBCD), the probability of assigning the next patient to an arm depends on two evolving quantities: the target allocation shares and the trial’s current allocation shares. With binary outcomes, each observed response or nonresponse also changes the estimated response rates. When the target is defined using those estimates, a single outcome can therefore alter both the target and the next assignment probability.

The calculations below use one prespecified, illustrative target map: allocate target share to each arm in proportion to its estimated response rate. This is a transparent response-favoring policy for demonstrating the updates; it is not claimed to be universally optimal. A different scientific objective could require a different target map.

Definition: For two arms $A$ and $B$, let $N_A,N_B$ be the numbers assigned so far, and let $S_A,S_B$ be their numbers of binary responses. The empirical response estimates are $\hat p_A=S_A/N_A$ and $\hat p_B=S_B/N_B$. In this example, the target shares are $\rho_A=\hat p_A/(\hat p_A+\hat p_B)$ and $\rho_B=1-\rho_A$. The current allocation shares are $q_A=N_A/(N_A+N_B)$ and $q_B=1-q_A$. For tuning parameter $\gamma\geq 0$, the DBCD assignment probabilities are $$ \pi_A=\frac{\rho_A(\rho_A/q_A)^\gamma}{\rho_A(\rho_A/q_A)^\gamma+\rho_B(\rho_B/q_B)^\gamma}, \qquad \pi_B=1-\pi_A. $$

At each step, these probabilities are calculated using only information available before the next assignment. The assignment is then randomized, its binary outcome is observed, and the response estimates, target shares, and current allocation shares are updated. In the worked example, all starting counts and subsequent outcomes are hypothetical.

Assumptions for the Worked Sequence

Suppose each patient has a binary outcome coded as response or nonresponse. For the illustration, assume responses are independent across patients conditional on treatment, each arm has a stable response probability during the sequence, and the outcome is observed before the next patient is assigned. These assumptions make the arm-specific empirical response rates interpretable as estimates of fixed arm response probabilities. They do not establish that one arm causes better outcomes; causal interpretation in a trial also depends on its design and conduct.

Begin with an initial allocation of four patients to each arm. The hypothetical starting data are $S_A=2$ responses among $N_A=4$, and $S_B=3$ among $N_B=4$. Thus both arms have positive response estimates, the target map is defined, and the current shares are positive. Set $\gamma=1$, and suppose there is no additional assignment-probability floor active in this example. Because both estimated rates are positive, both target shares and both assignment probabilities are positive at each displayed step.

Worked Example: Five Patients Move the Target

Starting from the hypothetical data above, follow five successive assignments. The listed outcome is observed after the assignment shown. The response estimates and target are then updated before calculating the next assignment probability.

Before patient 1. The estimates are $\hat p_A=2/4=0.5000$ and $\hat p_B=3/4=0.7500$, giving $\rho_A=0.5000/(0.5000+0.7500)=0.4000$ and $\rho_B=0.6000$. The current shares are $q_A=q_B=0.5000$. Hence $$ \pi_A=\frac{0.4000(0.4000/0.5000)}{0.4000(0.4000/0.5000)+0.6000(0.6000/0.5000)} =\frac{0.3200}{1.0400}=0.3077. $$ The probability of assigning patient 1 to $B$ is $0.6923$. Suppose the patient is assigned to $B$ and does not respond.

Before patient 2. Now $N_A=4,S_A=2,N_B=5,S_B=3$, so $\hat p_A=0.5000$ and $\hat p_B=0.6000$. The updated target is $\rho_A=5/11=0.4545$, while the current share is $q_A=4/9=0.4444$. Substitution gives $$ \pi_A=\frac{(5/11)((5/11)/(4/9))} {(5/11)((5/11)/(4/9))+(6/11)((6/11)/(5/9))} =\frac{1125}{2421}=0.4647. $$ Suppose patient 2 is assigned to $A$ and responds.

Before patient 3. Arm $A$ now has $3$ responses among $5$ patients, and arm $B$ has $3$ among $5$. Both estimates are $0.6000$, so both targets are $0.5000$. The current shares are also $0.5000$, giving $\pi_A=0.5000$. Suppose patient 3 is assigned to $A$ and does not respond.

Before patient 4. The estimates are now $\hat p_A=3/6=0.5000$ and $\hat p_B=3/5=0.6000$, so the target returns to $\rho_A=5/11=0.4545$. But the current share is $q_A=6/11=0.5455$. The target for $A$ is below its current share, and $$ \pi_A= \frac{(5/11)((5/11)/(6/11))} {(5/11)((5/11)/(6/11))+(6/11)((6/11)/(5/11))} =\frac{125}{341}=0.3666. $$ Suppose patient 4 is assigned to $B$ and responds.

Before patient 5. The estimates are $\hat p_A=0.5000$ and $\hat p_B=4/6=0.6667$, so $\rho_A=3/7=0.4286$. The current share is $q_A=6/12=0.5000$. Therefore $$ \pi_A=\frac{(3/7)((3/7)/(1/2))} {(3/7)((3/7)/(1/2))+(4/7)((4/7)/(1/2))} =\frac{18}{50}=0.3600. $$ Suppose patient 5 is assigned to $B$ and does not respond.

After patient 5, the rates are $\hat p_A=3/6=0.5000$ and $\hat p_B=4/7=0.5714$. The new target is $\rho_A=7/15=0.4667$, and the current share is $q_A=6/13=0.4615$. The next probability is $$ \pi_A= \frac{(7/15)((7/15)/(6/13))} {(7/15)((7/15)/(6/13))+(8/15)((8/15)/(7/13))} =\frac{4459}{9451}=0.4718. $$ Thus the target and assignment probability have both changed over the sequence. The five realized assignments do not have to match any of those probabilities exactly.

Why the Rule Uses Both Shares

The ratio $\rho_k/q_k$ compares an arm’s target share with its current share. The DBCD weight for arm $k$ is $\rho_k(\rho_k/q_k)^\gamma$; normalizing the two weights gives probabilities that sum to one. For two arms, the odds form makes the two influences especially visible:

\frac{\pi_A}{\pi_B} = \left(\frac{\rho_A}{\rho_B}\right)^{\gamma+1} \left(\frac{q_B}{q_A}\right)^\gamma.

To obtain this identity, divide the unnormalized weight for $A$ by that for $B$: $\{\rho_A(\rho_A/q_A)^\gamma\}/\{\rho_B(\rho_B/q_B)^\gamma\}$. Collecting the powers of the target and current shares gives the displayed expression. Because $\pi_A+\pi_B=1$, these odds determine the probabilities: if the odds are $r$, then $\pi_A=r/(1+r)$.

When $\gamma=0$, the allocation probability is the target share, $\pi_A=\rho_A$. For positive $\gamma$, the current shares also matter: an arm below its target share receives a correction in its favor, while an arm above its target share receives a correction against it. This describes the direction of adjustment, not a promise that the next patient will move the realized share toward the target.

Worked Example: Updating an Estimate Changes the Target

Suppose, hypothetically, arm $A$ has $2$ responses among $4$ patients and arm $B$ has $3$ among $4$. The target for $A$ is $0.5000/(0.5000+0.7500)=0.4000$. If the next patient is assigned to $A$ and responds, its estimate becomes $3/5=0.6000$, while $B$'s remains $0.7500$. The new target for $A$ is $$ \rho_A=\frac{0.6000}{0.6000+0.7500}=\frac{4}{9}=0.4444. $$

If instead that patient does not respond, the estimate for $A$ becomes $2/5=0.4000$, and its target becomes $$ \rho_A=\frac{0.4000}{0.4000+0.7500}=\frac{8}{23}=0.3478. $$ The two possible outcomes lead to different target shares. In either case, the current allocation share also changes because another patient was assigned to $A$; the next DBCD probability must use both updated quantities.

Reading the Updates and Checking the Calculation

A useful check is to keep the three kinds of quantities separate:

Response estimate: the observed response fraction within an arm, such as $S_A/N_A$.
Target share: the fraction prescribed by the chosen target map, such as $\rho_A$.
Assignment probability: the chance of assigning the next patient to that arm, such as $\pi_A$.

They need not be equal. In particular, $\pi_A$ is not the fraction of patients who will ultimately receive $A$. It is a conditional probability for one assignment, calculated at a particular state of the trial. After that assignment and outcome, the state may change.

Common calculation checks include verifying that $N_A$ and $N_B$ each increase by one after assignment, that only the assigned arm’s response count increases after a response, and that the updated rates use the updated denominators. Also check that the two target shares and the two assignment probabilities each sum to one. A target or probability outside $[0,1]$, or a pair that does not sum to one, signals an arithmetic or implementation error.

A frequent mistake is to calculate the target once from the starting estimates and keep using it throughout. That is incorrect when the target map uses current response estimates: every observed outcome can change the target. Another mistake is to set $\pi_A=\rho_A$ automatically. That equality holds for $\gamma=0$, but for positive $\gamma$ the current allocation shares enter the rule as well.

Comparison and Limitations

Under fixed equal randomization, each patient is assigned to either arm with probability $0.5$, regardless of the response history. Under direct target-probability randomization, the next assignment probability is the current target share itself. A DBCD with positive $\gamma$ instead combines the target with the current allocation shares, correcting in the direction of target tracking. These approaches therefore answer different allocation-policy choices; they do not make the response estimates more accurate by definition.

The example’s empirical response-rate target is sensitive to the observations that have accumulated, especially when counts are small. Its clinical interpretation also depends on the prespecified target map: favoring arms in proportion to estimated response rates is not automatically the right objective for every trial. The calculation further assumes outcomes are available before the next update. Delayed or missing outcomes require a design-specific rule for what information is used. If an estimate is zero, the illustrative proportional target can reach a boundary; this example avoids that case by starting with positive response counts. In practice, the target map, initialization, and any required allocation constraints must be specified so that the rule remains well-defined.

Key takeaway: At each patient, update the binary-response estimates from observed data, recompute the prespecified target, calculate current allocation shares, and then apply the DBCD rule. A changing target is part of the sequential design, not a calculation error.

Check Your Understanding

Answer each question using the stated two-arm DBCD rule. Solutions follow the questions.

With $\gamma=1$, target shares $\rho_A=0.4000,\rho_B=0.6000$, and current shares $q_A=q_B=0.5000$, calculate $\pi_A$.
With $\gamma=1$, $\rho_A=0.5500,\rho_B=0.4500$, and $q_A=0.4000,q_B=0.6000$, calculate $\pi_A$.
Arm $A$ has $2$ responses among $4$ patients and arm $B$ has $3$ among $4$. Under the response-rate target map, what is $\rho_A$ if the next patient assigned to $A$ responds?
Why is a next-patient assignment probability not the same quantity as the target share or the eventual realized allocation share?

Solutions. (1) The weights are $0.4000(0.4000/0.5000)=0.3200$ and $0.6000(0.6000/0.5000)=0.7200$, so $\pi_A=0.3200/(0.3200+0.7200)=0.3077$. (2) The weights are $0.5500(0.5500/0.4000)=0.75625$ and $0.4500(0.4500/0.6000)=0.3375$, giving $\pi_A=0.75625/1.09375=0.6914$. (3) The updated estimate for $A$ is $3/5=0.6000$, while $B$'s estimate is $3/4=0.7500$; hence $\rho_A=0.6000/(0.6000+0.7500)=0.4444$. (4) The target is prescribed by the target map, the assignment probability is the conditional chance for the next patient at the current state, and the realized share is the fraction assigned after a sequence of random outcomes. They are distinct quantities and need not coincide.

Worked Example: Binary Responses and a Moving Target

What You'll Learn