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Doubly Adaptive Biased Coin Designs · Tutorial 146 of 501

Proving the Restoring Direction of the Biased Coin

You will prove how the two-arm DBCD probability responds to allocation imbalance when the estimated target is held fixed, and identify the assumptions and limits of that conclusion.

Advanced 10 min read

What You'll Learn

  • Write the two-arm DBCD probabilities using a fixed interior target and the current allocation proportions.
  • Prove by comparing odds that an underrepresented arm receives probability above its target when the correction exponent is positive.
  • Show why an overrepresented arm receives probability below its target, and what happens at exact balance.
  • Check numerical examples using normalized DBCD weights.
  • Distinguish the direction of the next assignment probability from a guarantee about future realized allocations.
  • Identify boundary states and changing targets that invalidate the simple fixed-target comparison.

What “Restoring Direction” Means

A doubly adaptive biased coin (DBCD) uses both a target allocation and the allocation already realized to determine the probability of the next assignment. This creates a feedback effect: if an arm currently has a smaller share than its target, the rule should raise that arm’s next-assignment probability above the target. The claim is about the conditional probability of the next assignment, not a guarantee that the next patient will be assigned to that arm or that future allocation proportions will move monotonically.

Here we prove that direction algebraically for two arms while holding the estimated target fixed. Freezing the target matters: if the target changes at the same time as the allocation proportions, a comparison with the current target alone does not establish whether the assignment rule is correcting imbalance or following a moving target.

Definition: Let \(\hat\rho_1\) and \(\hat\rho_2=1-\hat\rho_1\) be the fixed target allocation proportions for two arms, with \(0<\hat\rho_1<1\). After \(n\) assignments, let \(\hat\pi_{n,1}=N_1(n)/n\) and \(\hat\pi_{n,2}=N_2(n)/n=1-\hat\pi_{n,1}\) be the realized allocation proportions, assumed positive. For \(\gamma>0\), the two-arm DBCD assigns the next patient to arm \(k\) with probability proportional to \(\hat\rho_k(\hat\rho_k/\hat\pi_{n,k})^\gamma\).

The positivity conditions ensure that both ratios in the rule are defined and that both weights are positive. In this tutorial, the target estimate is treated as fixed during the comparison; write \(r=\hat\rho_1\) and \(q=\hat\pi_{n,1}\). Thus \(1-r=\hat\rho_2\) and \(1-q=\hat\pi_{n,2}\). Let \(p_1\) denote the conditional probability of assigning the next patient to arm 1, given the current trial history and these fixed values.

$$ p_1 = \frac{r(r/q)^\gamma} {r(r/q)^\gamma+(1-r)\{(1-r)/(1-q)\}^\gamma}. $$

The denominator normalizes the two positive weights so that the assignment probabilities sum to one. The central question is whether \(p_1\) lies above or below \(r\), the fixed target share for arm 1. Comparing probabilities directly involves a fraction; comparing their odds gives a simpler algebraic route.

Proof by Comparing Odds

Because both weights are positive, \(0<p_1<1\), so the odds \(p_1/(1-p_1)\) are well defined. The odds of assignment to arm 1 under the DBCD rule are the ratio of its weight to arm 2’s weight:

$$ \begin{aligned} \frac{p_1}{1-p_1} &= \frac{r(r/q)^\gamma} {(1-r)\{(1-r)/(1-q)\}^\gamma}\\ &= \frac{r}{1-r} \left\{\frac{r(1-q)}{q(1-r)}\right\}^{\gamma}. \end{aligned} $$

The target odds for arm 1 are \(r/(1-r)\). For any two probabilities strictly between zero and one, comparing the probabilities is equivalent to comparing their odds: the function \(x/(1-x)\) is strictly increasing on that interval. Consequently, \(p_1>r\) precisely when the DBCD odds exceed the target odds. Dividing both sides of that odds comparison by the positive target odds gives the following condition:

$$ \left\{\frac{r(1-q)}{q(1-r)}\right\}^{\gamma}>1. $$

Since \(\gamma>0\), raising a positive number to the power \(\gamma\) preserves whether it is greater than, equal to, or less than one. The expression inside braces is greater than one exactly when \(r(1-q)>q(1-r)\). Expanding both sides and cancelling the common term \(-rq\) reduces this to \(r>q\). The same steps give equality when \(r=q\) and the reverse inequality when \(r<q\).

Theorem: Under the fixed-target two-arm DBCD rule, suppose \(0<r<1\), \(0<q<1\), and \(\gamma>0\). If \(q<r\), so arm 1 is underrepresented relative to its target, then \(p_1>r\). If \(q=r\), then \(p_1=r\). If \(q>r\), then \(p_1<r\). Equivalently, the next-assignment probability is above target for the underrepresented arm and below target for the overrepresented arm.

This conclusion is a statement about the direction of the probability adjustment. It does not say how large the adjustment must be, nor does it guarantee that the realized next assignment will reduce the imbalance. The next patient is still assigned randomly according to the calculated probabilities.

Worked Examples

Worked Example: An Underrepresented Arm

Suppose the fixed estimated target for arm 1 is \(r=0.60\), its current allocation share is \(q=0.45\), and \(\gamma=1\). These are hypothetical values. Both arms have positive allocation shares, and the target is interior, so the rule is defined. Because \(0.45<0.60\), arm 1 is underrepresented.

The arm 1 and arm 2 weights are, respectively,

$$ \begin{aligned} w_1&=0.60(0.60/0.45)^1=0.80,\\ w_2&=0.40(0.40/0.55)^1\approx0.2909. \end{aligned} $$

After normalization, the probability of assigning the next patient to arm 1 is

$$ p_1=\frac{0.80}{0.80+0.2909}\approx0.7333. $$

Thus the next-assignment probability for arm 1 is about 0.733, above its fixed target of 0.60, as the theorem predicts. This is a probability for the next assignment, not a prediction that the next patient will certainly receive arm 1.

Worked Example: An Overrepresented Arm

Suppose arm 1 has fixed target \(r=0.40\), current allocation share \(q=0.55\), and \(\gamma=1\). Treat these as hypothetical values. Both allocation shares are positive, the target is strictly between zero and one, and the fixed-target conditions hold. Since \(0.55>0.40\), arm 1 is overrepresented.

The DBCD weights are

$$ \begin{aligned} w_1&=0.40(0.40/0.55)^1\approx0.2909,\\ w_2&=0.60(0.60/0.45)^1=0.80. \end{aligned} $$

The normalized next-assignment probability for arm 1 is

$$ p_1=\frac{0.2909}{0.2909+0.80}\approx0.2667. $$

The probability, about 0.267, is below arm 1’s target of 0.40. Equivalently, arm 2 is underrepresented: its current share is \(0.45\), below its target of \(0.60\), and its next-assignment probability is about \(1-0.2667=0.7333\), above that target.

Worked Example: Exact Balance

Let the fixed target for arm 1 be \(r=0.30\), and suppose its current share is also \(q=0.30\). Use the hypothetical value \(\gamma=2\). The positive shares are \(0.30\) and \(0.70\), so the allocation rule is defined.

At exact balance, each target-to-share ratio is one. The two weights and the normalized probability are therefore

$$ \begin{aligned} w_1&=0.30(0.30/0.30)^2=0.30,\\ w_2&=0.70(0.70/0.70)^2=0.70,\\ p_1&=\frac{0.30}{0.30+0.70}=0.30. \end{aligned} $$

The next-assignment probability equals the target at balance. This check is useful because a purported restoring rule should not introduce a directional adjustment when the current allocation already matches the fixed target.

Why the Result Holds and What It Does Not Establish

The proof depends on a simple relationship between target odds and current allocation odds. In the arm 1 expression, the multiplier on the target odds is

$$ \left\{\frac{r(1-q)}{q(1-r)}\right\}^{\gamma}. $$

When \(q<r\), the numerator-to-denominator comparison inside braces is greater than one, so the multiplier is greater than one. The DBCD therefore raises arm 1’s odds above its target odds, and hence raises its probability above its target probability. When \(q>r\), the multiplier is less than one, reversing the direction. The result follows from normalization and the ratio correction; it does not require a claim about outcomes or about the quality of the target estimate.

The fixed-target condition is important in applications. An adaptive design may update its target estimate as response data accrue. The theorem applies at a given allocation state when that target is held fixed for the comparison. If the target is updated between assignments, the next probability may change both because the realized allocation has changed and because the target itself has moved. The fixed-target proof alone does not separate those effects or guarantee correction relative to a later target.

Likewise, “restoring” describes the direction of the conditional assignment probability relative to target. Random variation can still move the realized allocation share farther from target on a particular assignment. The probability is not a deterministic correction to the observed count.

Diagnostics, Alternatives, and Failure Modes

A practical algebra check is to calculate the current share \(q\), compare it with the fixed target \(r\), and then verify that the computed \(p_1-r\) has the corresponding sign. The odds form provides a second check without requiring a full normalization:

$$ \frac{p_1}{1-p_1} = \frac{r}{1-r} \left\{\frac{r(1-q)}{q(1-r)}\right\}^{\gamma}. $$

For \(r,q\in(0,1)\) and \(\gamma>0\), the factor in braces must be above one if and only if \(q<r\). If a calculation instead gives a probability below target in that case, check that the target and allocation-share values have not been swapped, that the two weights were normalized correctly, and that the same target was used for both arms.

The comparison with simpler approaches helps clarify what is specific to the DBCD feedback. Under simple randomization to a fixed target, the next-assignment probability for arm 1 is always \(r\), regardless of \(q\); it does not adjust for current allocation imbalance. The DBCD rule instead uses the current shares in its correction weights. At exact balance it agrees with the target probability, while away from balance it adjusts in the restoring direction under the theorem’s conditions. This comparison concerns the direction, not a claim that one design is uniformly preferable.

Several situations fall outside the proof’s assumptions:

  • A boundary allocation share. If \(q=0\) or \(q=1\), a target-to-share ratio is undefined. The theorem does not prescribe how to initialize or modify the design at such a state.
  • A boundary target. If \(r=0\) or \(r=1\), target odds are not finite and the interior-target argument does not apply.
  • No positive correction exponent. The direction proof uses \(\gamma>0\). Without that condition, the odds multiplier need not be greater than one when the arm is underrepresented.
  • A moving target. If the target changes during the comparison, the result is not a proof about correction relative to one fixed target.
  • Confusing probability with outcome. A probability above target raises the chance of the next assignment to that arm; it does not determine the realized assignment.

These are limits of this particular statement, not evidence that the DBCD formula is algebraically inconsistent. They identify which conditions must be checked before using the restoring-direction conclusion.

Key takeaway: For an interior, fixed two-arm target, positive allocation shares, and \(\gamma>0\), the DBCD probability for an arm is above its target exactly when that arm’s current allocation share is below target; it equals target at balance and is below target when the arm is overrepresented.

Check Your Understanding

Answer each question using the fixed-target two-arm setting. Full solutions follow the questions.

  1. If arm 1 has target share \(0.35\) and current share \(0.25\), is its next-assignment probability above, equal to, or below target, assuming \(\gamma>0\)? State the conditions needed for the theorem.
  2. Suppose the target and current share for arm 1 are both \(0.70\). What is its next-assignment probability under the DBCD rule?
  3. For a fixed target \(r=0.60\), current share \(q=0.75\), and \(\gamma=1\), calculate the DBCD probability for arm 1 and compare it with the target.
  4. Explain why a next-assignment probability above target does not guarantee that the realized allocation share will be closer to target after one assignment.

Solutions. (1) It is above target: \(0.25<0.35\). The theorem requires an interior fixed target, positive current shares for both arms, and \(\gamma>0\). (2) It equals \(0.70\), because both target-to-share ratios equal one and the normalized weights are \(0.70\) and \(0.30\). (3) The weights are \(w_1=0.60(0.60/0.75)=0.48\) and \(w_2=0.40(0.40/0.25)=0.64\). Thus \(p_1=0.48/(0.48+0.64)=0.4286\), below the target \(0.60\), as expected because arm 1 is overrepresented. (4) The rule randomizes: even when arm 1’s probability is above target, the next patient can be assigned to arm 2. The theorem concerns the conditional assignment probability, not the realized outcome of that random assignment.

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