Reading the Allocation Rule as Target Tracking

Reading a Target-Tracking Rule

In a doubly adaptive biased coin design (DBCD), two quantities describe different things. The target share says what fraction of assignments the design is aiming toward at the current stage. The current share records the fraction already assigned to each arm. The next patient's assignment probability is neither quantity by itself: it is computed by combining the target with a correction based on the gap between target and current allocation.

This distinction matters in practice. If an arm's target share is 60%, that does not mean the next patient must have a 60% chance of receiving it. If that arm is already underrepresented relative to its target, the DBCD rule can give it a next-assignment probability above 60%. Conversely, if it is overrepresented, its next-assignment probability can be below 60%.

Definition: Immediately before the \((n+1)\)st assignment, let \(N_1\) and \(N_2\) be the numbers assigned to arms 1 and 2, with \(n=N_1+N_2\). Their current shares are \(x_1=N_1/n\) and \(x_2=N_2/n\). Let \(\rho_1\) and \(\rho_2\) be the target shares used for this assignment, where \(\rho_1+\rho_2=1\). The DBCD next-assignment probabilities are proportional to \(\rho_k(\rho_k/x_k)^\gamma\) for arm \(k\), where \(\gamma\geq 0\) is the correction exponent.

For the formula in this tutorial, the target shares and current shares are both strictly between 0 and 1, and \(n>0\), so the ratios are defined. The target is treated as given at the moment of assignment; in an adaptive design it may have been obtained from information available up to that point. This calculation describes the randomization probability conditional on the available history, not a guarantee about the treatment received by any particular patient.

Separate the Target From the Correction

For arm 1, the rule first gives a weight \(\rho_1\) and multiplies it by a correction factor \((\rho_1/x_1)^\gamma\). A target share larger than the current share makes this factor greater than 1 when \(\gamma>0\); a target share smaller than the current share makes it less than 1. But the final probability also depends on arm 2's weight, because the two weights must be normalized to sum to 1.

\pi_{n+1,1} = \frac{\rho_1(\rho_1/x_1)^\gamma} {\rho_1(\rho_1/x_1)^\gamma+\rho_2(\rho_2/x_2)^\gamma}, \qquad \pi_{n+1,2}=1-\pi_{n+1,1}.

Here \(\pi_{n+1,1}\) denotes the probability of assigning the next patient to arm 1, conditional on the information available just before assignment. The weight \(\rho_1\) is the target component; \((\rho_1/x_1)^\gamma\) is the current-share correction. Neither component alone is the final assignment probability.

The odds form makes the joint contribution especially clear. Dividing the unnormalized arm 1 weight by the unnormalized arm 2 weight gives:

\frac{\pi_{n+1,1}}{\pi_{n+1,2}} = \frac{\rho_1}{\rho_2} \left(\frac{\rho_1/x_1}{\rho_2/x_2}\right)^\gamma = \frac{\rho_1}{\rho_2} \left(\frac{\rho_1/\rho_2}{x_1/x_2}\right)^\gamma.

Thus, the next-assignment odds are the target odds multiplied by a correction comparing the target odds with the current allocation odds. Equivalently, taking logarithms gives:

\log\left(\frac{\pi_{n+1,1}}{\pi_{n+1,2}}\right) = (1+\gamma)\log\left(\frac{\rho_1}{\rho_2}\right) - \gamma\log\left(\frac{x_1}{x_2}\right).

This identity is a way to read the rule, not a claim that the next assignment will make the observed shares equal to the targets. The random assignment changes the counts by only one, and its effect on later shares depends on subsequent assignments as well.

Worked Calculations

Worked Example: An Underrepresented Arm With a Higher Target

Suppose the target shares are \(\rho_1=0.60\) and \(\rho_2=0.40\). After 50 assignments, 20 patients have received arm 1 and 30 have received arm 2, so \(x_1=20/50=0.40\) and \(x_2=30/50=0.60\). For this hypothetical calculation, take \(\gamma=1\). Find the probability that the next patient is assigned to arm 1.

The arm 1 weight is \(0.60(0.60/0.40)^1=0.60(1.5)=0.90\). The arm 2 weight is \(0.40(0.40/0.60)^1=0.40(0.6667)=0.2667\). Normalizing the weights gives:

\pi_{51,1} = \frac{0.90}{0.90+0.2667} = 0.7714.

The next patient therefore has about a 77.1% probability of assignment to arm 1 and a 22.9% probability of assignment to arm 2. The target share for arm 1 is 60%, but its current share is only 40%, so the correction raises its next-assignment probability above its target share. This remains a probability for one assignment, not a statement that 77.1% of all patients will ultimately receive arm 1.

Worked Example: An Overrepresented Arm With a Lower Target

Suppose a different hypothetical trial has targets \(\rho_1=0.40\), \(\rho_2=0.60\), and current shares \(x_1=0.70\), \(x_2=0.30\). Let \(\gamma=1.5\). Calculate the probability of assigning the next patient to arm 1.

The correction ratios are \(0.40/0.70=0.5714\) for arm 1 and \(0.60/0.30=2\) for arm 2. The corresponding weights are:

w_1=0.40(0.5714)^{1.5}\approx 0.1728, \qquad w_2=0.60(2)^{1.5}\approx 1.697.

After normalization,

\pi_{n+1,1} = \frac{0.1728}{0.1728+1.697} \approx 0.0924.

The next-patient probability for arm 1 is about 9.24%, well below its 40% target share. Its current share, 70%, is already above target, while the other arm is below target; the correction reduces arm 1's relative weight. The result does not mean the target has changed to 9.24%.

Worked Example: How the Target Changes the Joint Calculation

Consider a hypothetical allocation with current shares \(x_1=x_2=0.50\), \(\gamma=2\), and targets \(\rho_1=0.70\), \(\rho_2=0.30\). Find the probability of assigning the next patient to arm 1, then compare it with the case where both target shares are 0.50.

With the 70:30 target, the weights are \(0.70(0.70/0.50)^2=0.70(1.4)^2=1.372\) and \(0.30(0.30/0.50)^2=0.30(0.6)^2=0.108\). Therefore:

\pi_{n+1,1} = \frac{1.372}{1.372+0.108} = 0.9270.

If the target is instead 50:50 while the current shares remain 50:50, each weight is \(0.50(1)^2=0.50\), giving \(\pi_{n+1,1}=0.50/(0.50+0.50)=0.5000\). The comparison shows that the target and current shares jointly enter the calculation: holding the current shares fixed while changing the target changes both weights and the normalized probability.

What Can Be Read From the Odds?

Because the two assignment probabilities sum to 1, arm 1 is more likely than arm 2 precisely when its odds exceed 1. The odds identity supplies a direct check:

\pi_{n+1,1}>\pi_{n+1,2} \quad\Longleftrightarrow\quad \frac{\rho_1}{\rho_2} \left(\frac{\rho_1/\rho_2}{x_1/x_2}\right)^\gamma>1.

This condition depends on both the target odds and the current odds. It is not enough to inspect whether arm 1's target is above one half, nor is it enough to inspect its current share alone. The two weights must be compared, or the odds condition evaluated. For example, with the 60:40 target and 40:60 current allocation in the first example, the odds are \(0.90/0.2667=3.375\), which exceed 1; hence \(\pi_{51,1}>0.5\), as the calculated probability confirms.

A useful special case is \(x_1=\rho_1\) and \(x_2=\rho_2\). Both correction ratios then equal 1, so the weights reduce to the targets themselves and \(\pi_{n+1,1}=\rho_1\). When current shares differ from target shares, the correction modifies the next probability; the amount and direction of that modification can be checked directly from the formula.

Assumptions, Diagnostics, and Alternatives

The calculation requires a current share for each arm and positive target shares. If \(n=0\), current shares cannot be computed. If an arm has no assignments yet, its current share is zero and the ratio in the rule is undefined. A design must therefore specify how assignments are initialized before using this form of the update. A zero target share also makes the usual positive-target calculation inappropriate and requires a separately specified rule.

Before interpreting a reported probability, check that the inputs refer to the same assignment time, that \(x_1+x_2=1\) and \(\rho_1+\rho_2=1\), and that both unnormalized weights are positive under the stated assumptions. Then check that the normalized probabilities sum to 1 and lie between 0 and 1. Computing the odds as a second route is a useful arithmetic diagnostic. Disagreement between the odds calculation and the normalized-weight calculation usually signals a reversed ratio, a mismatched arm index, or an incorrect normalization.

A common mistake is to reason, “The target for arm 1 is 60%, so the next patient must have a 60% chance of assignment to arm 1.” That substitutes the target share for the next-assignment probability. The correction is to calculate both arms' weights using their current shares and then normalize. Another mistake is to treat a probability above 50% as proof that the arm has the higher target. The current-share correction can affect which arm has the larger next probability, so the combined weights—not the target alone—settle that comparison.

One alternative is to assign directly with probabilities \(\rho_1\) and \(\rho_2\), without a current-share correction. That makes each assignment probability equal to the target at that moment, but the formula does not use the observed current allocation to adjust the next assignment. The DBCD rule instead combines the target with current-share information. This distinction is about the form of the randomization rule; neither calculation by itself guarantees an exact final sample split.

Key takeaway: Read each arm's DBCD weight as its target share multiplied by a correction comparing target share with current share. Normalize the two weights to obtain the next-assignment probabilities, and use the odds form to check which arm is more likely.

Check Your Understanding

For each question, identify the target and current shares before interpreting the next-assignment probability.

With target shares 0.50 and 0.50, current shares 0.60 and 0.40, and \(\gamma=1\), what is the next-assignment probability for arm 1?
If current shares equal target shares, what are the next-assignment probabilities under the stated rule?
Why can a 60% target share not be reported automatically as a 60% next-patient assignment probability?
What problem occurs if the current share of one arm is zero, and what must the design do?

Solutions. For question 1, the weights are \(0.50(0.50/0.60)=0.4167\) and \(0.50(0.50/0.40)=0.6250\). Thus \(\pi_{n+1,1}=0.4167/(0.4167+0.6250)=0.4000\). For question 2, both correction ratios equal 1, so the probabilities equal the target shares. For question 3, the next probability also depends on both arms' current shares and on the correction exponent; it must be calculated from and normalized across both weights. For question 4, division by a zero current share is undefined, so the design must specify an initialization or other rule before applying this formula.