Allocation Ratios and Information per Participant

Why Allocation Changes Information

In a two-group mean comparison, the total number of participants does not by itself determine the precision of the estimated mean difference. The split between groups matters: a small group contributes a relatively imprecise estimate of its mean, and that imprecision carries into the difference. This tutorial quantifies that effect for a fixed treatment-to-control allocation ratio and shows how to plan a total sample size for a target information level.

The earlier material on Information in Two-Sample Mean Comparisons defines information through the variance of the estimated mean difference. Here we use that relationship to compare allocation ratios and derive the sample-size calculation. The result is an expected, design-stage calculation; it does not depend on observed treatment outcomes.

Model and assumptions: Suppose the treatment observations \(Y_{Ti}\) are independent and identically distributed as \(N(\mu_T,\sigma^2)\), and the control observations \(Y_{Ci}\) are independent and identically distributed as \(N(\mu_C,\sigma^2)\). The groups are independent of one another, the common variance \(\sigma^2\) is treated as fixed for planning, and each participant contributes one outcome measured on the same scale. The estimand is the mean difference \(\delta=\mu_T-\mu_C\), estimated by \(\hat{\delta}=\bar{x}_T-\bar{x}_C\).

Under these assumptions, the estimated difference has variance \(\sigma^2/n_T+\sigma^2/n_C\), where \(n_T\) and \(n_C\) are the treatment and control sample sizes. Define its information as the reciprocal of this variance. Information therefore increases as the variance decreases. Its units are the inverse of the squared outcome units; for example, if outcomes are measured in millimetres, information is in inverse square millimetres.

\mathcal{I}_{\delta} =\frac{1}{\operatorname{Var}(\hat{\delta})} =\frac{1}{\sigma^2/n_T+\sigma^2/n_C} =\frac{n_Tn_C}{\sigma^2(n_T+n_C)}.

This expression is the expected information for estimating the mean difference under the stated model. In a normal model with known variance, it is also the Fisher information for \(\delta\). If the common variance is unknown and estimated from the data, the same expression remains a useful planning calculation when a variance value is specified in advance, but the realized standard error will also depend on the variance estimate.

Information Per Participant at a Fixed Ratio

Let \(N=n_T+n_C\) be the total sample size and let \(r=n_T/n_C\) be the treatment-to-control allocation ratio. Thus, \(r=2\) means two treatment participants for every control participant. Since \(n_T=rN/(1+r)\) and \(n_C=N/(1+r)\), substituting into the information formula gives information as a function of the total sample size and ratio.

\mathcal{I}_{\delta} =\frac{N}{\sigma^2}\frac{r}{(1+r)^2}, \qquad \frac{\mathcal{I}_{\delta}}{N} =\frac{r}{\sigma^2(1+r)^2}.

The second expression is the information per participant. It depends on both the allocation ratio and the planning variance. For a fixed variance, it shows directly how much precision is expected, on average, from each additional participant when the same ratio is maintained. An allocation ratio and its reciprocal have the same information per participant: for example, \(r=2\) and \(r=1/2\) are equally efficient under this model. The roles of treatment and control can be exchanged without changing the variance of the difference.

Why Equal Allocation Maximizes Information

When participants have the same variance in both groups and the total sample size is fixed, equal allocation maximizes information. One way to see this is to compare the ratio-specific information per participant with its value at \(r=1\).

\frac{r}{(1+r)^2}\leq\frac{1}{4} \quad\Longleftrightarrow\quad 4r\leq(1+r)^2 \quad\Longleftrightarrow\quad 0\leq(r-1)^2.

The final inequality holds for every positive allocation ratio, and equality occurs only when \(r=1\). Thus, for a fixed \(N\) and \(\sigma^2\), equal allocation gives information \(N/(4\sigma^2)\), the largest possible under this model. The relative information efficiency of ratio \(r\), compared with equal allocation at the same total \(N\), is

\operatorname{Efficiency}(r) =\frac{\mathcal{I}_{\delta}(r)}{\mathcal{I}_{\delta}(1)} =\frac{4r}{(1+r)^2}.

This efficiency is a proportion, not a power. For instance, an efficiency of \(0.889\) means that the unequal allocation supplies about 88.9% of the information provided by equal allocation with the same total number of participants, under the model assumptions. It does not mean that power is 88.9% of the equal-allocation power.

Worked Example: Comparing Equal and Two-to-One Allocation

Suppose a hypothetical study plans to compare treatment and control means with a common outcome standard deviation of \(10\) units and a total of \(120\) participants. Compare equal allocation with a treatment-to-control ratio of \(2:1\). Assume independent groups, a common variance of \(10^2=100\) square units, and the same outcome measurement in both groups.

With equal allocation, \(n_T=n_C=60\). The variance of the estimated difference is

\operatorname{Var}(\hat{\delta}) =\frac{100}{60}+\frac{100}{60} =3.333\ \text{square units}, \qquad \mathcal{I}_{\delta}=\frac{1}{3.333}=0.3000\ \text{inverse square units}.

With a \(2:1\) ratio, \(n_T=80\) and \(n_C=40\). The variance and information are

\operatorname{Var}(\hat{\delta}) =\frac{100}{80}+\frac{100}{40} =1.25+2.50 =3.750\ \text{square units}, \qquad \mathcal{I}_{\delta}=\frac{1}{3.750}=0.2667\ \text{inverse square units}.

The unequal allocation provides \(0.2667/0.3000=0.8889\), or about 88.9%, of the information under equal allocation. This agrees with \(4r/(1+r)^2=8/9\) for \(r=2\). Both calculations use the stated common standard deviation and independent-group model; if the groups had different variances, this comparison would need to be recalculated under a different variance model.

Planning a Sample Size for Target Information

The information formula can be rearranged to find the total sample size needed to reach a chosen target \(\mathcal{I}_{\text{target}}\) at a specified ratio. Solving \( \mathcal{I}_{\text{target}}=N r/\{\sigma^2(1+r)^2\}\) for \(N\) gives the continuous total required before rounding.

N_{\text{required}} =\mathcal{I}_{\text{target}}\sigma^2\frac{(1+r)^2}{r}.

This calculation requires a planning value for \(\sigma^2\), the target information, and the intended ratio. The ratio determines how the total is divided: \(n_T=rN/(1+r)\) and \(n_C=N/(1+r)\). Since participant counts must be whole numbers, the continuous result is not automatically a feasible design. After rounding, calculate information from the actual group sizes and confirm that it meets or exceeds the target. If it does not, increase the total and check again.

Worked Example: Finding Group Sizes for a Target Information Level

A hypothetical mean-comparison study has a planning standard deviation of \(8\) units, a target information of \(0.250\) inverse square units, and a planned treatment-to-control ratio of \(1.5:1\). Assume independent observations and a common variance of \(8^2=64\) square units. Calculate a feasible integer allocation.

First calculate the continuous total:

N_{\text{required}} =0.250(64)\frac{(1+1.5)^2}{1.5} =16\frac{6.25}{1.5} =66.67.

The corresponding continuous group sizes are \(n_T=1.5(66.67)/2.5=40.00\) and \(n_C=66.67/2.5=26.67\). Because these are not both integers, try nearby integer allocations. With \(40\) treatment and \(27\) control participants, the total is \(67\) and the ratio is approximately \(1.481:1\). The achieved information is

\mathcal{I}_{\delta} =\frac{40(27)}{64(40+27)} =\frac{1080}{4288} =0.2519\ \text{inverse square units}.

This allocation exceeds the target. To check whether a smaller total could meet it near the planned ratio, consider \(40\) treatment and \(26\) control participants, for a total of \(66\):

\mathcal{I}_{\delta} =\frac{40(26)}{64(66)} =\frac{1040}{4224} =0.2462\ \text{inverse square units}.

That allocation falls short. Therefore \(40\) treatment and \(27\) control participants meet the target among these nearby allocations. The final check uses the actual integer counts, rather than assuming that the rounded continuous sample size automatically attains the target.

Applying the Ratio Formula and Checking a Design

A quick comparison can be useful when a total sample size is fixed but several ratios are under consideration. The following calculation uses the same assumptions as the model: independent groups, a common variance, and a common outcome scale. It illustrates that increasing the larger group does not fully compensate for leaving the smaller group unchanged.

Worked Example: Information at a Three-to-One Ratio

Suppose a hypothetical study plans \(100\) participants, a treatment-to-control ratio of \(3:1\), and a common standard deviation of \(12\) units. The implied group sizes are \(n_T=75\) and \(n_C=25\), and the common variance is \(144\) square units. The information is

\mathcal{I}_{\delta} =\frac{75(25)}{144(100)} =\frac{1875}{14400} =0.1302\ \text{inverse square units}.

For comparison, equal allocation with the same total would give \(n_T=n_C=50\), hence

\mathcal{I}_{\delta} =\frac{50(50)}{144(100)} =\frac{2500}{14400} =0.1736\ \text{inverse square units}.

The three-to-one design provides \(0.1302/0.1736=0.7500\), or 75.0%, of the equal-allocation information. The ratio formula gives the same result: \(4(3)/(1+3)^2=12/16=0.7500\). The calculation does not say whether a three-to-one allocation is appropriate overall; it only quantifies its precision cost for this mean difference under the specified model.

Choosing and Diagnosing an Allocation

When treatment and control participants have the same outcome variance and the same per-participant contribution to the analysis, equal allocation is the information-efficient choice for a fixed total sample size. Unequal allocation can still be reasonable when there are practical constraints, such as limited availability of control participants, a need to expose more participants to a treatment for safety or implementation reasons, or different recruitment costs. In that case, the information calculation makes the precision cost visible rather than deciding the design by itself.

Several checks help reveal when the calculation is being misapplied:

Check the ratio direction. State whether \(r\) means treatment divided by control. If the groups are reversed, \(r\) becomes \(1/r\), but information should remain unchanged.
Check the variance inputs. The formula assumes a common variance. A large difference in variability between groups, or a planning variance that is implausible for the outcome scale, undermines the calculation.
Check the units. If the outcome is measured in units, \(\sigma^2\) is in square units and information is in inverse square units. Mixing standard deviations and variances produces an incorrect result.
Recalculate after rounding. The target is met only if the information computed from the final integer group sizes is at least the target.
Check independence and the analysis model. Dependence among observations or an analysis whose variance differs from \(\sigma^2/n_T+\sigma^2/n_C\) changes the information. The simple formula is not a correction for clustering, missing outcomes, or other departures.

A common mistake is to conclude that doubling the treatment group while holding the control group fixed doubles the information, because the total sample size has increased substantially. The correct calculation uses both group sizes: information is \(n_Tn_C/\{\sigma^2(n_T+n_C)\}\). As \(n_T\) grows with \(n_C\) fixed, information approaches \(n_C/\sigma^2\); it cannot grow without bound because uncertainty in the control mean remains. Another mistake is to treat the efficiency ratio as a power ratio. Efficiency compares information at fixed total sample size; power also depends on the effect, the test, and the significance level.

Scope and Limitations

These formulas apply to a two-group comparison of means under the stated common-variance model. They are useful for relating allocation, sample size, and precision, including when a design is expressed through an information target. They do not by themselves determine a sample size for a desired power, choose an allocation based on costs or unequal variances, or account for dependence or incomplete outcomes. If the design assumptions change, the variance of the estimated contrast must be derived for that design before computing information.

For a group sequential design, the planned information level is a precision scale for the mean contrast; allocation determines how quickly a given total enrollment contributes to that scale. This calculation alone does not specify interim analyses or stopping rules. Its practical value is more focused: it lets a designer compare feasible ratios and verify the information delivered by the planned group sizes.

Check Your Understanding

Try each question, then compare your work with the solutions below.

With a common standard deviation of \(5\) units and a treatment-to-control ratio of \(1:1\), what is the information per participant?
For a fixed total sample size and common variance, explain why ratios \(r=2\) and \(r=1/2\) have the same information.
A design has \(n_T=30\), \(n_C=20\), and a common standard deviation of \(4\) units. Calculate the information for the mean difference.
With \(\sigma=6\), target information \(0.200\) inverse square units, and equal allocation, calculate the continuous total sample size required. Give a feasible equal-allocation total that reaches the target.

Solutions. (1) At \(r=1\), information per participant is \(1/(4\sigma^2)=1/(4\cdot25)=0.0100\) inverse square units. (2) Replacing \(r\) with \(1/r\) leaves \(r/(1+r)^2\) unchanged, since \((1/r)/(1+1/r)^2=r/(1+r)^2\); exchanging group labels therefore does not change the variance of the difference. (3) The common variance is \(16\) square units, so \(\mathcal{I}_{\delta}=30(20)/\{16(50)\}=600/800=0.7500\) inverse square units. (4) For equal allocation, \(N_{\text{required}}=0.200(36)(4)=28.8\). The smallest feasible even total above this is \(30\), giving \(15\) per group and information \(15(15)/\{36(30)\}=225/1080=0.2083\) inverse square units, which exceeds the target.