Deriving the Next Expected Allocation Proportion

From an Assignment Probability to an Expected Share

A DBCD uses the current treatment counts and an estimated target allocation to choose the probability of the next assignment. That probability does not itself equal the expected treatment share after the assignment: the new share also includes participants already assigned. The key calculation is to combine the two quantities while conditioning on the information available before the next assignment.

Let \(N_{n,k}\) be the number assigned to treatment \(k\) after \(n\) assignments, and let \(\pi_{n,k}=N_{n,k}/n\) be its current allocation proportion when \(n>0\). Write \(\hat\rho_{n,k}\) for the estimated target proportion for treatment \(k\), and let \(\gamma\geq 0\) be the DBCD tuning parameter. The estimated targets sum to one. In the standard interior version of the rule, the conditional probability of assigning the next participant to treatment \(k\) is

q_{n+1,k} = \frac{\hat\rho_{n,k} \left(\hat\rho_{n,k}/\pi_{n,k}\right)^\gamma} {\displaystyle\sum_{j=1}^{K}\hat\rho_{n,j} \left(\hat\rho_{n,j}/\pi_{n,j}\right)^\gamma}.

Here \(K\) is the number of treatments. This formula assumes positive current shares and positive estimated targets, so the ratios and resulting weights are defined. The \(q_{n+1,k}\) are probabilities: they are nonnegative and sum to one. In the one-step calculation, they are evaluated using the current state; they are not recalculated using data that do not yet exist.

Definition: Let \(\mathcal F_n\) denote all information available after \(n\) assignments, including the allocation counts and the response data used to form the current estimated target. The conditional expected allocation proportion for treatment \(k\) after the next assignment is \(E(\pi_{n+1,k}\mid\mathcal F_n)\). It averages over the next randomized assignment while holding the current information fixed.

Deriving the One-Step Conditional Expectation

Define \(I_{n+1,k}\) to equal 1 if the next participant is assigned to treatment \(k\), and 0 otherwise. Conditional on \(\mathcal F_n\), this indicator has expectation \(q_{n+1,k}\), because it is a Bernoulli indicator with success probability \(q_{n+1,k}\). The updated count is \(N_{n+1,k}=N_{n,k}+I_{n+1,k}\), and the updated share divides that count by \(n+1\). Therefore,

\begin{aligned} E(\pi_{n+1,k}\mid\mathcal F_n) &= E\left(\frac{N_{n,k}+I_{n+1,k}}{n+1}\,\middle|\,\mathcal F_n\right)\\ &= \frac{N_{n,k}+E(I_{n+1,k}\mid\mathcal F_n)}{n+1}\\ &= \frac{N_{n,k}+q_{n+1,k}}{n+1}\\ &= \frac{n\pi_{n,k}+q_{n+1,k}}{n+1}. \end{aligned}

Subtracting the current share gives the conditional expected change:

E(\pi_{n+1,k}-\pi_{n,k}\mid\mathcal F_n) = \frac{q_{n+1,k}-\pi_{n,k}}{n+1}.

This is the one-step drift. It is positive when the next-assignment probability exceeds the current share, negative when it is smaller, and zero when they agree. The factor \(1/(n+1)\) matters: even a substantial difference between \(q_{n+1,k}\) and \(\pi_{n,k}\) is diluted by the participants already assigned. This identity is exact; it does not require a large-sample approximation.

The conditional expectation is also a convex combination of the current share and the next-assignment probability: \(E(\pi_{n+1,k}\mid\mathcal F_n)=\frac{n}{n+1}\pi_{n,k}+\frac{1}{n+1}q_{n+1,k}\). Both coefficients are nonnegative and sum to one. Thus the conditional expectation lies between those two values, though an actual randomized next share can fall on either side of its conditional expectation.

Worked Example: A Two-Arm Update with Unequal Shares

Suppose, hypothetically, that \(n=10\), with 4 participants assigned to treatment 1 and 6 to treatment 2. The current shares are \((0.4,0.6)\), the estimated target is \((0.6,0.4)\), and \(\gamma=1\). Calculate the next assignment probability for treatment 1 and its conditional expected share after one assignment.

The two unnormalized DBCD weights are

w_1=0.6\left(\frac{0.6}{0.4}\right)=0.9, \qquad w_2=0.4\left(\frac{0.4}{0.6}\right)=0.2667.

Their sum is \(1.1667\), so the next assignment probability for treatment 1 is \(q_{11}=0.9/1.1667\approx 0.7714\). The next share is random, but its conditional expectation is

E(\pi_{11}\mid\mathcal F_{10}) =\frac{4+0.7714}{11} \approx 0.4338.

The expected share rises from \(0.4000\) to about \(0.4338\), a conditional expected increase of \(0.0338\). Equivalently, the drift formula gives \((0.7714-0.4000)/11\approx 0.0338\). It would be incorrect to report \(0.7714\) as the expected next share: it is the probability of the next assignment, not the share among all 11 participants.

How the Target and Current Share Enter

The assignment probability \(q_{n+1,k}\) is determined by the DBCD weights, while the expected share also depends on \(N_{n,k}\) and \(n\). These roles should be kept distinct. In particular, an estimated target is not generally the assignment probability: the design adjusts assignment probabilities in response to the difference between target and current shares.

When \(\gamma=0\), the weights reduce to \(\hat\rho_{n,k}\), so \(q_{n+1,k}=\hat\rho_{n,k}\). For positive \(\gamma\), the rule reweights those targets according to the target-to-current-share ratios. The conditional expectation formula remains unchanged; only the value of \(q_{n+1,k}\) changes with \(\gamma\) and the state.

Worked Example: A Stronger Adjustment with Gamma Equal to Two

Suppose, hypothetically, that there have been \(n=8\) assignments, with 5 to treatment 1 and 3 to treatment 2. The current shares are \((0.625,0.375)\). Let the estimated target be \((0.7,0.3)\) and set \(\gamma=2\). Find the conditional expected share for treatment 1 after the next assignment.

The weights are

\begin{aligned} w_1&=0.7\left(\frac{0.7}{0.625}\right)^2 =0.7(1.2544)=0.87808,\\ w_2&=0.3\left(\frac{0.3}{0.375}\right)^2 =0.3(0.64)=0.192. \end{aligned}

The total weight is \(1.07008\), so \(q_{91}=0.87808/1.07008\approx 0.8206\). Thus

E(\pi_{91}\mid\mathcal F_8) =\frac{5+0.8206}{9} \approx 0.6467.

The conditional expected share is about \(0.6467\), above the current \(0.6250\), since \(q_{91}>0.6250\). It is below the target \(0.7000\). That is a consequence of the one-step averaging formula, not an arithmetic discrepancy: the next participant is only one of nine in the updated total.

Several Treatments and a Conservation Check

The same argument applies separately to each treatment in a \(K\)-arm design. Since exactly one participant is assigned, the indicators satisfy \(\sum_{k=1}^K I_{n+1,k}=1\). Hence the next shares sum to one, and their conditional expectations must also sum to one:

\sum_{k=1}^{K} E(\pi_{n+1,k}\mid\mathcal F_n) = \frac{\sum_{k=1}^{K}N_{n,k}+\sum_{k=1}^{K}q_{n+1,k}}{n+1} = \frac{n+1}{n+1}=1.

This is a useful arithmetic check. It does not imply that every component moves toward its target in that single step: the conditional drift for each treatment depends on \(q_{n+1,k}-\pi_{n,k}\), and all the assignment probabilities must sum to one.

Worked Example: Three Treatments with Unequal Current Shares

Suppose, hypothetically, that \(n=12\), with counts \((6,4,2)\), current shares \((0.5,1/3,1/6)\), estimated target \((0.5,0.3,0.2)\), and \(\gamma=1\). Calculate the next probabilities and the conditional expected allocation proportions.

The unnormalized weights are target squared divided by current share:

w_1=\frac{0.5^2}{0.5}=0.5,\qquad w_2=\frac{0.3^2}{1/3}=0.27,\qquad w_3=\frac{0.2^2}{1/6}=0.24.

The total is \(1.01\), giving \(q_{13}\approx(0.4950,0.2673,0.2376)\). Applying the count-plus-probability formula componentwise,

\begin{aligned} E(\pi_{13,1}\mid\mathcal F_{12})&=(6+0.4950)/13\approx0.4996,\\ E(\pi_{13,2}\mid\mathcal F_{12})&=(4+0.2673)/13\approx0.3283,\\ E(\pi_{13,3}\mid\mathcal F_{12})&=(2+0.2376)/13\approx0.1721. \end{aligned}

The rounded expected shares sum to \(1.0000\). Treatment 2's expected share is slightly below its current share because \(q_{13,2}<1/3\); treatment 3's is above its current share because \(q_{13,3}>1/6\). The calculation concerns the expected proportions after one assignment, not a guarantee about the realized allocation.

Assumptions, Diagnostics, and Common Errors

The derivation requires that exactly one participant is assigned at the next step, that the assignment follows the stated conditional probabilities, and that the probabilities are calculated from information available at time \(n\). The current counts must sum to \(n\), the estimated targets must be valid normalized proportions, and the weights must be defined. For the displayed standard rule, positive current shares and positive targets ensure finite, positive weights. If a design uses floors, smoothing, or another boundary convention, its actual assignment probabilities should be substituted for \(q_{n+1,k}\); the count-plus-probability identity still applies.

A practical diagnostic is to check the inputs and the output before interpreting the drift. Confirm that current shares and estimated targets each sum to one, all weights are finite and nonnegative, and normalized probabilities sum to one. Then verify that the expected next shares sum to one and that the computed change agrees with \((q_{n+1,k}-\pi_{n,k})/(n+1)\). If these checks fail, likely causes include a denominator or count error, inconsistent rounding, or using the target as though it were the assignment probability.

A common mistake is to say, “The next assignment probability is 0.77, so the expected allocation share after the next step is 0.77.” This confuses the chance that one incoming participant receives a treatment with that treatment's share among all participants. The correction is to add the assignment probability to the existing expected count and divide by \(n+1\). Another error is to treat a plug-in target as fixed across time without qualification. It is fixed only when conditioning on \(\mathcal F_n\); later response data may change the estimated target and hence later assignment probabilities.

For comparison, under a fixed target-proportional assignment rule the next assignment probability would equal the target share, whereas DBCD generally modifies it using the current allocation share. In either case, the conditional expected updated share is obtained from the same count-plus-probability calculation. This one-step result does not establish long-run convergence, nor does it say that the realized share follows its conditional expectation at each step.

Check Your Understanding

Use the one-step conditional expectation identity to answer each question. Solutions are included for checking your calculations.

At \(n=9\), treatment 1 has count 3 and next assignment probability \(q_{10,1}=0.6\). What is its conditional expected share after the next assignment?
If the current share is \(0.45\) and the next assignment probability is \(0.30\), is the one-step conditional drift positive or negative? Give its magnitude when \(n=19\).
Why must the conditional expected shares across all treatments sum to one?
In a two-arm design with \(\gamma=0\) and estimated target \((0.65,0.35)\), what is the next assignment probability vector under the stated rule?

Solutions. (1) \((3+0.6)/10=0.36\). (2) The drift is \((0.30-0.45)/(19+1)=-0.0075\), so it is negative, with magnitude \(0.0075\). (3) The current counts sum to \(n\), and the next-assignment probabilities sum to one; therefore the expected next counts sum to \(n+1\), and dividing by \(n+1\) gives one. (4) With \(\gamma=0\), each weight equals its estimated target, so the probability vector is \((0.65,0.35)\).