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Unequal Allocation in Gray-Test Design

Learn how to adjust a balanced Gray-test sample-size target for an allocation ratio and when practical constraints can justify unequal allocation.

Advanced 12 min read

What You'll Learn

  • Calculate the information loss and sample-size inflation from a specified allocation ratio.
  • Convert a balanced total sample-size target into continuous group-size targets.
  • Distinguish equal-headcount efficiency from cost-adjusted allocation efficiency.
  • Derive an optimal allocation ratio when groups have different per-person score variances or recruitment costs.
  • Check feasibility, rounding, and assumptions before using an allocation-based power calculation.

Why Allocation Changes the Design

A Gray-test design specifies not only the target power and significance level, but also how participants will be divided between the two groups. If the groups are expected to contribute comparable information per participant, a balanced allocation gives the greatest information for a fixed total sample size. An unequal ratio can still be appropriate when recruitment is constrained or when per-person information and recruitment costs differ, but its consequences should be made explicit.

The calculation below adjusts an approximate sample-size target for a two-group Gray-test comparison. It does not replace the assumptions used to obtain that target: the target effect, competing-event incidence patterns, censoring conditions, test weighting, significance level, and power must still represent the planned study. The introductory material on approximate sample-size calculation provides the balanced target; here the focus is how allocation changes that target.

Definition: Let \(n_A\) and \(n_B\) be the planned sample sizes in groups A and B, and let \(N=n_A+n_B\). Define the allocation ratio as \(r=n_B/n_A\). The allocation proportion in group B is \(q=n_B/N=r/(1+r)\). Let \(N_{\mathrm{eq}}\) denote the approximate total sample size for the same design assumptions under equal allocation.

For the familiar allocation adjustment, assume independent participants, a fixed total sample size, and comparable per-person contributions to the Gray score variance in the two groups. Also assume the same target alternative and follow-up conditions as those used to obtain \(N_{\mathrm{eq}}\). Under this approximation, the score’s noncentrality parameter is proportional to the product of the two group sizes divided by their sum. The allocation adjustment is therefore based on \(n_A n_B/N\), rather than on the larger group size alone.

$$ \frac{n_A n_B}{N} = Nq(1-q) = N\frac{r}{(1+r)^2} $$

At equal allocation, \(q=1/2\), so this quantity is \(N/4\). At another allocation, retaining the same approximate noncentrality requires matching the information to the balanced design’s target. This gives the sample-size inflation factor and the corresponding group sizes:

$$ \frac{N}{4}\frac{4r}{(1+r)^2}=\frac{N_{\mathrm{eq}}}{4}, \qquad N=N_{\mathrm{eq}}\frac{(1+r)^2}{4r}, \qquad n_A=\frac{N}{1+r},\quad n_B=\frac{rN}{1+r}. $$
Key result: Under the comparable-information approximation, the relative information at total \(N\), compared with equal allocation at the same \(N\), is \(4r/(1+r)^2\). The total sample-size multiplier needed to recover the balanced target is its reciprocal, \((1+r)^2/(4r)\). Both factors equal 1 when \(r=1\).

Deriving the Allocation Penalty

The information comparison follows directly from the group-size product. Since \(n_A=N/(1+r)\) and \(n_B=rN/(1+r)\),

$$ \frac{4n_A n_B}{N^2} = \frac{4\{N/(1+r)\}\{rN/(1+r)\}}{N^2} = \frac{4r}{(1+r)^2}. $$

This ratio cannot exceed 1, because \((1+r)^2-4r=(r-1)^2\geq 0\). Equality holds only when \(r=1\). Thus, with comparable per-person information and a fixed total, moving away from balance reduces this approximation to information. The result is symmetric in \(r\) and \(1/r\): a 1:4 split and a 4:1 split have the same penalty, though the group labels are reversed.

The argument is an approximation for design, not an exact finite-sample identity for every Gray-test setting. A Gray test uses a weighted score and its variance; differences in censoring, score variability, or event patterns across groups can invalidate the comparable-information premise. In those circumstances, use group-specific variance or information inputs rather than applying the simple multiplier uncritically.

Worked Example: Adjusting a Balanced Target for a 4:1 Ratio

Suppose the approximate balanced design requires \(N_{\mathrm{eq}}=784.9\) participants for its specified Gray-test alternative and power. The planned ratio is four participants in group B for every one in group A, so \(r=4\). Assume the comparable-information approximation is reasonable.

The information fraction at a fixed total is

$$ \frac{4r}{(1+r)^2} = \frac{4(4)}{(1+4)^2} = \frac{16}{25} = 0.64. $$

Thus the design has 64% of the balanced allocation’s approximate information at the same total. The required total is inflated by \(1/0.64=1.5625\):

$$ N = 784.9\frac{25}{16} = 1226.40625. $$

The continuous group-size targets are

$$ n_A=\frac{1226.40625}{5}=245.28125\approx245.3, \qquad n_B=\frac{4(1226.40625)}{5}=981.125\approx981.1. $$

The continuous targets sum to \(1226.40625\), as required. For integer recruitment, rounding each target upward gives 246 in group A and 982 in group B, for 1,228 participants total. These rounded counts are close to, but not exactly, a 4:1 ratio. A design that must preserve an exact 4:1 ratio could instead use 246 and 984, totaling 1,230. Final integer choices should be checked against recruitment limits and the power target.

When Unequal Allocation Can Be Efficient

“Efficient” depends on what is being held fixed. With a fixed total number of participants and equal per-person information, equal allocation is most efficient. But groups may differ in score variability, information contribution, or recruitment cost. Under a suitable approximation, those differences can make an unequal ratio more efficient for a fixed variance target or budget.

To make this distinction precise, suppose the variance of a contrast-like score quantity is approximated by

$$ \operatorname{Var}(\widehat{\Delta}) \approx \frac{v_A}{n_A}+\frac{v_B}{n_B}, $$

where \(v_A\) and \(v_B\) are positive per-participant variance contributions for groups A and B. This is a planning approximation, not a claim that every Gray-test variance has this exact form. If \(v_A=v_B\), minimizing the expression at fixed \(N\) gives equal allocation. If they differ, the variance-minimizing allocation can differ from 1:1.

Now suppose a participant in group A costs \(c_A\) and one in group B costs \(c_B\). For a fixed variance target, minimize total recruitment cost \(c_A n_A+c_B n_B\). A Lagrange multiplier calculation gives the cost-efficient ratio:

$$ \frac{n_B}{n_A} = \sqrt{\frac{v_Bc_A}{v_Ac_B}}. $$

The formula balances marginal reduction in variance against marginal recruitment cost. It is not a universal Gray-test allocation rule: the \(v\)'s must reflect the planned score information under the relevant design assumptions, and the costs must be meaningful and sufficiently stable.

Worked Example: Accounting for Different Variances and Costs

Suppose a design approximation gives \(v_A=1.44\) and \(v_B=0.81\), in common score-variance units per participant. Recruitment costs are \(c_A=4\) and \(c_B=1\), in cost units per participant. Assume these inputs are suitable for the planned comparison and there are no group-size caps.

The cost-efficient ratio is

$$ r=\frac{n_B}{n_A} = \sqrt{\frac{(0.81)(4)}{(1.44)(1)}} = \sqrt{2.25} = 1.5. $$

Thus the model recommends 1.5 participants in group B per participant in group A, or a 2:3 allocation. The more costly group A receives the smaller share, while the lower per-person variance contribution in group B is also accounted for. This result minimizes approximate total cost for a fixed target variance under the stated model; it does not by itself guarantee a particular Gray-test power. That power still depends on the score alternative and the validity of the variance approximation.

Worked Example: A Recruitment Constraint Favors a 1:2 Split

Suppose a balanced design requires \(N_{\mathrm{eq}}=400\) participants, and group B can recruit at most twice as many participants as group A. Under equal per-person information, using the maximum permitted ratio \(r=2\) gives

$$ \text{inflation factor} = \frac{(1+2)^2}{4(2)} = \frac{9}{8} = 1.125, \qquad N=400(1.125)=450. $$

The group sizes are \(n_A=450/3=150\) and \(n_B=2(450)/3=300\). At a total of 450, this split recovers the balanced target’s approximate information. If recruitment instead capped the study at 400 total, the same 1:2 allocation would retain only \(8/9\) of the information available under equal allocation at 400. The recruitment constraint can make the ratio feasible or necessary, but it does not erase its information cost.

Practical Checks and Failure Modes

Before applying an allocation multiplier, check that the balanced target and the proposed ratio refer to the same Gray-test design: the same target alternative, test weighting, follow-up assumptions, censoring conditions, significance level, and power. The allocation calculation changes the division of participants; it does not repair an unrealistic assumed effect or a mismatch between planned and expected follow-up.

  • Check the ratio definition. Here \(r=n_B/n_A\). Reversing the group labels changes \(r\) to \(1/r\), though the simple information penalty remains the same.
  • Check the direction of the adjustment. For any \(r\ne1\), the information fraction must be below 1 and the sample-size multiplier above 1.
  • Check the arithmetic and totals. Group targets must sum to \(N\); rounding may alter both the total and the exact ratio.
  • Check group-specific information. If censoring or score contributions differ materially by group, comparable information per participant may be implausible. A single imbalance factor may then misstate power.
  • Check feasibility. A mathematically optimal ratio may violate recruitment caps, minimum group sizes, or other design constraints.

A common mistake is to multiply the balanced total by \(4r/(1+r)^2\). That is the information fraction at a fixed total, not the sample-size multiplier. For example, at \(r=2\), the information fraction is \(8/9\), so the total must be multiplied by \(9/8\), not by \(8/9\). Another mistake is to claim that a 1:1 split is always cost-efficient. It is optimal for a fixed total under equal per-person information, but cost or variance differences can change the best allocation when the objective is a fixed variance or budget.

The simple calculation is useful for translating a balanced target into an initial unequal-allocation plan and for understanding the penalty of imbalance. Its limitations matter: it does not account automatically for unequal group-specific score variances, changes in the alternative under different group sizes, or complex recruitment restrictions. When those factors are important, compare candidate designs using an appropriate group-specific information calculation rather than treating the multiplier as exact.

Check Your Understanding

Try each question, then compare your reasoning with the solutions.

  1. A balanced design needs 500 participants. With \(r=3\), what are the information fraction, the required continuous total, and the two group sizes?
  2. At a fixed total, why does equal allocation maximize the information fraction \(4r/(1+r)^2\)?
  3. For \(v_A=1\), \(v_B=4\), \(c_A=1\), and \(c_B=1\), what cost-efficient ratio \(n_B/n_A\) follows from the variance-cost approximation?
  4. A study can enroll only 300 participants, with \(r=1/2\). Under equal per-person information, what fraction of the balanced allocation’s information does this provide?

Solutions. (1) The information fraction is \(4(3)/(1+3)^2=12/16=0.75\). The total is \(500/0.75=666.7\). The continuous targets are \(n_A=666.7/4=166.7\) and \(n_B=3(666.7)/4=500.0\), approximately. (2) Since \((1+r)^2-4r=(r-1)^2\geq0\), the fraction is at most 1, with equality at \(r=1\). (3) The ratio is \(\sqrt{(4)(1)/(1)(1)}=2\), so group B receives twice the sample size of group A under this approximation. (4) The fraction is \(4(1/2)/(1+1/2)^2=2/2.25=0.8889\), or about 88.9% of the balanced information at the same total.

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