Unequal Allocation in Promising-Zone Re-Estimation

Why Allocation Changes the Re-Estimation Calculation

A total sample size does not, by itself, determine the precision of a treatment comparison. With unequal numbers of participants in the two arms, the variance of the estimated treatment difference depends on both arm sizes. Consequently, a promising-zone increase must be translated into information using the intended allocation, not by simply scaling the total sample size as though the arms were equally sized.

Consider a continuous endpoint measured in the treatment and control arms. The calculations below use a normal model with known common variance, independent observations, and a positive treatment effect defined as the treatment mean minus the control mean. These assumptions make the information and conditional-power calculations explicit. In practice, the variance and analysis model must be appropriate for the endpoint and design; if a variance estimate is used, its uncertainty also matters.

Definition: Let $n_T$ and $n_C$ be the treatment and control sample sizes, $N=n_T+n_C$, and $r=n_T/n_C$ the treatment-to-control allocation ratio. Under independent normal outcomes with common known variance $\sigma^2$, the estimated mean difference has variance $\sigma^2(1/n_T+1/n_C)$. Its information is $$ I=\frac{1}{\sigma^2(1/n_T+1/n_C)} =\frac{n_Tn_C}{\sigma^2(n_T+n_C)} =\frac{N r}{\sigma^2(1+r)^2}. $$

Information has units equal to the reciprocal of the squared endpoint units. For example, if the endpoint is measured in millimetres of mercury (mmHg), $I$ is measured in $1/(\text{mmHg})^2$. The last expression shows the effect of allocation: for fixed $N$ and $\sigma^2$, information depends on $r$. Equal allocation maximizes information per participant when the two arms have the same variance and participant costs.

From Target Power to Unequal-Allocation Sample Size

For a planned treatment difference $\delta>0$, the usual large-sample normal design calculation for a two-sided test at level $\alpha$ uses the critical value $z_{1-\alpha/2}$. If the target power is $1-\beta$, the required information is

I_{\mathrm{req}}=\frac{(z_{1-\alpha/2}+z_{1-\beta})^2}{\delta^2}.

This follows because, under the planned effect, the standardized treatment-difference statistic has mean $\delta\sqrt{I}$ and variance 1. The dominant upper-tail contribution to power reaches $1-\beta$ when $\delta\sqrt{I}-z_{1-\alpha/2}=z_{1-\beta}$. The opposite-tail contribution is generally negligible for a useful positive effect and conventional power targets; the formula is the standard normal approximation.

Substituting the unequal-allocation information formula gives the continuous total sample-size requirement:

N_{\mathrm{req}} =I_{\mathrm{req}}\sigma^2\frac{(1+r)^2}{r}.

This is a continuous requirement, not yet a feasible pair of arm sizes. If $r$ is rational, an exact ratio can be achieved by choosing integer counts in the corresponding proportion and taking a sufficiently large multiple. If $r$ is irrational, no pair of positive integer counts can have that exact ratio. In that case, choose a feasible integer allocation that approximates the desired ratio, then recalculate its information and verify that it meets the target.

Worked Example: Required Sample Size With a 2:1 Allocation

Suppose a hypothetical two-arm trial measures systolic blood pressure in mmHg. Assume independent normal outcomes with known common standard deviation $\sigma=10$ mmHg, plan for a treatment-minus-control difference of $\delta=5$ mmHg, and use a two-sided $\alpha=0.05$ test with 90% power. The intended allocation is 2 treatment participants for each control participant.

The normal quantiles are $z_{0.975}=1.960$ and $z_{0.90}=1.2816$. Thus

I_{\mathrm{req}} =\frac{(1.960+1.2816)^2}{5^2} =\frac{10.508}{25} =0.4203\;1/(\text{mmHg})^2.

Here $r=2$, so the continuous total requirement is

N_{\mathrm{req}} =0.4203(10^2)\frac{(1+2)^2}{2} =0.4203(100)(4.5) =189.1.

An exact 2:1 allocation requires $n_T=2k$, $n_C=k$, and therefore $N=3k$. The smallest such total at least 189.1 is 192, giving $n_T=128$ and $n_C=64$. Recalculate information from these integer counts:

I=\frac{128(64)}{100(128+64)} =\frac{8192}{19200} =0.4267\;1/(\text{mmHg})^2.

This exceeds the required 0.4203. The normal-approximation power is approximately $\Phi(5\sqrt{0.4267}-1.960)=\Phi(1.306)=0.904$, with a negligible opposite-tail contribution. Thus the rounded allocation meets the stated 90% target under the model assumptions.

Conditional Power Under Unequal Allocation

At an interim analysis, let $I_1$ denote the information already observed and $I_2$ the planned future information. Assume independent increments: the future score contribution is independent of the interim data and, under a specified planning effect $\delta$, has mean $\delta I_2$ and variance $I_2$. Let $Z_1$ be the observed interim standardized statistic. For a fixed final critical value $c=z_{1-\alpha/2}$, conditional power for the positive treatment direction is

CP(\delta) =\Phi\!\left( \frac{Z_1\sqrt{I_1}+\delta I_2-c\sqrt{I_1+I_2}} {\sqrt{I_2}} \right).

The expression inside $\Phi$ is the conditional mean of the final standardized statistic, measured relative to the final critical value and in units of the future standard deviation. This calculation conditions on the observed interim statistic and specifies the effect used to model future data. It is not a probability that the planning effect is true. Here the critical value is treated as fixed; designs with interim efficacy boundaries require a boundary-aware calculation.

Under a 2:1 allocation, if the final arm counts are $n_T,n_C$ and the interim counts are $n_{T1},n_{C1}$, then calculate $I_1$ from the interim counts and $I_2$ from the additional participants in each arm. In general, $I_2$ is not the information for the final total by itself: it is the information contributed by the future observations.

Worked Example: Checking Whether an Interim Result Is Promising

Continue with a hypothetical blood-pressure endpoint with $\sigma=10$ mmHg, planning effect $\delta=5$ mmHg, two-sided $\alpha=0.05$, and 2:1 allocation. At interim, 40 treatment and 20 control participants have been observed, and the estimated treatment-minus-control difference is 4.00 mmHg. The original plan is to finish with 96 treatment and 48 control participants.

The model assumptions are independent normal observations with common known variance $10^2$, and the fixed-normal-critical-value calculation is used. Interim information and standard error are

I_1=\frac{40(20)}{100(60)}=0.1333, \qquad SE_1=10\sqrt{\frac1{40}+\frac1{20}}=\sqrt{7.5}=2.739\text{ mmHg}.

Thus $Z_1=4.00/2.739=1.461$. The future counts under the original plan are 56 treatment and 28 control participants, so

I_2=\frac{56(28)}{100(84)}=0.1867, \qquad I_1+I_2=0.3200.

Using $c=1.960$, the conditional-power argument is

\frac{1.461\sqrt{0.1333}+5(0.1867)-1.960\sqrt{0.3200}} {\sqrt{0.1867}} \approx \frac{0.5334+0.9335-1.1087}{0.4320} =0.829.

Therefore $CP\approx\Phi(0.829)=0.796$. If the prespecified promising zone is 0.80 to 0.95, this result is below the zone; the rule does not call for a promising-zone increase. The comparison is with the prespecified interval, not with an informal impression that the interim difference is favorable.

Translating a Promising-Zone Target Into Future Information

If the rule calls for increasing the sample size to reach a target conditional power $q$, set the argument of $\Phi$ equal to $z_q$ and solve for future information $I_2$:

\frac{Z_1\sqrt{I_1}+\delta I_2-c\sqrt{I_1+I_2}} {\sqrt{I_2}}=z_q.

The equation generally requires numerical solution. For each candidate $I_2>0$, evaluate the left-hand side and compare it with $z_q$. Once a solution is selected, convert the resulting final information $I_1+I_2$ to a continuous total sample size using the planned allocation ratio, then choose feasible integer arm counts and recalculate conditional power. Do not round the information first and assume the rounded counts preserve the target exactly.

Worked Example: Increasing to a 90% Conditional-Power Target

Use the interim data and assumptions in the preceding example, but suppose the prespecified promising zone includes the observed conditional power and calls for an increase to a target $q=0.90$. The observed interim score contribution is $Z_1\sqrt{I_1}=0.5333$. Numerical solution of

\frac{0.5333+5I_2-1.960\sqrt{0.1333+I_2}}{\sqrt{I_2}} =1.2816

gives $I_2\approx0.2820\;1/(\text{mmHg})^2$. Hence the continuous final information is approximately $0.1333+0.2820=0.4153$, corresponding under $r=2$ to

N_{\mathrm{cont}} =0.4153(100)(4.5) \approx186.9.

The smallest total divisible by three and at least this large is 189, with 126 treatment and 63 control participants. Since 40 and 20 participants have already been observed, the future allocation is 86 treatment and 43 control participants. Its future information is

I_2=\frac{86(43)}{100(129)} =\frac{3698}{12900} =0.2867, \qquad I_1+I_2=0.4200.

Recalculating with these feasible counts gives a conditional-power argument of

\frac{0.5333+5(0.2867)-1.960\sqrt{0.4200}}{\sqrt{0.2867}} \approx\frac{0.6964}{0.5354} =1.301,

so the conditional power is approximately $\Phi(1.301)=0.903$. The integer allocation therefore reaches or slightly exceeds the 90% target.

As a check on the calculation, a more conservative final total of 201 would give 134 treatment and 67 control participants. With 40 and 20 already observed, the future counts would be 94 and 47, giving $I_2=4418/(14100)=0.3133$ and final information $0.4467$. Substitution in the same formula gives an argument of about 1.412 and conditional power about 0.921. That is a feasible, more conservative choice, but it is not the smallest allocation needed for the stated 90% target.

Integer Allocation, Diagnostics, and Alternatives

An exact allocation ratio is a design constraint on integer counts, not a guarantee that every continuous sample-size result can be implemented without rounding. For rational $r$, reduce the ratio to integer terms and use the smallest adequate multiple. For irrational $r$, exact equality is impossible; choose a feasible approximation and recalculate information from the actual counts. The approximation should be evaluated against the design target, rather than judged only by how close the displayed ratio looks.

Worked Example: Approximating an Irrational Allocation Ratio

Suppose a hypothetical study seeks a treatment-to-control ratio $r=\sqrt{2}$ and has a total of 120 participants, with common known standard deviation $\sigma=8$ units. No integer arm counts can achieve exactly $\sqrt{2}$, because a ratio of two integers is rational. A nearby allocation is 70 treatment and 50 control participants, with ratio 1.400.

The actual information is

I=\frac{70(50)}{8^2(120)} =\frac{3500}{7680} =0.4557\;(\text{endpoint units})^{-2}.

For comparison, the continuous allocation implied by $r=\sqrt{2}$ at $N=120$ has information $120\sqrt{2}/[64(1+\sqrt{2})^2]\approx0.4549$. The feasible 70:50 allocation is a close approximation and has slightly greater information. The design decision should use the actual integer-count information, especially if the target lies near a threshold.

A useful diagnostic is to recompute all quantities in the final allocation: $n_T$, $n_C$, $N$, $I_1$, $I_2$, final information, and conditional power. Check that the future arm counts are nonnegative, that the total does not fall below the current enrollment, and that the prescribed allocation is respected as closely as feasible. Large discrepancies between a continuous target and the resulting integer calculation can signal an incorrect conversion from information to sample size or an unintended change in the allocation ratio.

The principal alternative is equal allocation. Under equal variances and equal per-participant costs, it yields greater information for a fixed total sample size than unequal allocation. Unequal allocation may nevertheless be selected for operational or scientific reasons. Whatever the reason, its information penalty or gain must be accounted for in the design calculation. A fixed sample-size design avoids an interim size change, while a promising-zone design uses interim data to adapt the planned future information according to a prespecified rule. Neither approach removes the need to state the planning effect and variance assumptions clearly.

Using total sample size as information. This ignores the arm sizes and variance. Use $I=n_Tn_C/[\sigma^2(n_T+n_C)]$.
Applying the final allocation to interim data. Calculate interim information from the participants observed at interim and future information from the additional participants.
Rounding without checking. A continuous target does not ensure that the chosen integer counts achieve the same information. Recalculate conditional power after rounding.
Assuming any desired ratio is exactly feasible. Exact integer allocation is possible only for rational ratios; otherwise approximate and evaluate the actual allocation.
Trusting a numerical root without substitution. Insert the candidate future information back into the conditional-power equation. A root should reproduce the stated target to the chosen precision.

Key takeaway: Under a common-variance normal model, unequal allocation changes information through both arm sizes. A promising-zone increase should be calculated in information units, converted to feasible arm counts, and checked again using those actual counts.

Check Your Understanding

Answer each question, then compare with the solutions below.

With $n_T=90$, $n_C=60$, and $\sigma=12$, calculate the information and allocation ratio.
For $r=3$, $\sigma=8$, and required information $I_{\mathrm{req}}=0.50$, calculate the continuous total sample size.
Why must future information, rather than final information alone, be used in the conditional-power numerator and denominator?
Can a treatment-to-control ratio of $\pi$ be achieved exactly with positive integer arm sizes? What should be done in practice?

Solutions. (1) $r=90/60=1.5$, and $I=90(60)/[144(150)]=5400/21600=0.2500$ per squared endpoint unit. (2) $N=I_{\mathrm{req}}\sigma^2(1+r)^2/r=0.50(64)(16/3)=170.7$ participants continuously; integer arm counts must then be chosen and checked. (3) The interim contribution is already observed and fixed when conditioning; only future observations are random, so their variance is $I_2$ on the score scale and the future contribution has mean $\delta I_2$. (4) No. Since $\pi$ is irrational and a ratio of positive integers is rational, exact equality is impossible; use a feasible integer approximation and recalculate information and conditional power from those counts.