Precision Versus Correlation When Enlarging the Control

Why Control Size Changes Both Precision and Dependence

In a multi-arm trial, several treatment arms may each be compared with the same control group. Increasing the control sample size can improve the precision of every treatment-control comparison. It also changes how strongly those comparisons move together, because each contrast contains the same estimated control mean. These are related design consequences, but they are not the same quantity: a smaller covariance does not by itself tell us the correlation, and neither quantity alone describes the standard error of an individual contrast.

This tutorial quantifies those changes under the standard common-variance model. The calculations concern sampling variances and dependence, not the choice of an interim selection rule or a final testing procedure.

Model and assumptions: There are \(K\) treatment groups and one shared control group. For group \(g\), observations are independent with mean \(\mu_g\) and common variance \(\sigma^2\), and observations from different groups are independent. Group \(g\) has fixed sample size \(n_g\), with sample mean \(\bar{x}_g\). For treatment \(i\), define the treatment-minus-control contrast \(D_i=\bar{x}_i-\bar{x}_0\). The variance and covariance formulas below follow from these independence and common-variance assumptions. Normal outcomes are not needed for these variance and covariance formulas; they are needed for exact normal sampling distributions.

Variance, Covariance, and Correlation as the Control Grows

Because the treatment and control sample means are independent, the variance of their difference is the sum of their variances. For two different treatment arms, the shared term is the control mean: the treatment means are independent of each other and of the control mean, while the same random control mean is subtracted in both contrasts.

\operatorname{Var}(D_i)=\sigma^2\left(\frac{1}{n_i}+\frac{1}{n_0}\right), \qquad \operatorname{Cov}(D_i,D_j)=\frac{\sigma^2}{n_0}\quad (i\ne j).

The standard error of \(D_i\), when \(\sigma\) is known, is the square root of its variance. The correlation of two same-direction treatment-minus-control contrasts is their covariance divided by the product of their standard deviations. Substitution gives

\operatorname{Corr}(D_i,D_j) = \frac{\sigma^2/n_0} {\sigma^2\sqrt{(1/n_i+1/n_0)(1/n_j+1/n_0)}} = \frac{\sqrt{n_i n_j}}{\sqrt{(n_0+n_i)(n_0+n_j)}}.

Thus, with treatment sample sizes fixed, increasing \(n_0\) reduces each contrast variance and the shared-control covariance. It also reduces the correlation between any two same-direction contrasts. The reduction in correlation is not identical to the reduction in covariance: correlation standardizes covariance by both contrast standard deviations.

Key result: Under the stated model, enlarging the shared control improves the marginal precision of every treatment-control contrast and weakens the positive dependence between same-direction contrasts. These changes do not make the contrasts independent at any finite control sample size.

Why the Precision Gains Diminish

Holding \(n_i\) fixed, the part of \(\operatorname{Var}(D_i)\) that changes with the control size is \(\sigma^2/n_0\). Its derivative with respect to \(n_0\) is \(-\sigma^2/n_0^2\). The variance therefore falls as the control grows, but the absolute rate of decrease gets smaller as \(n_0\) becomes larger. Even an infinitely large control cannot eliminate the treatment-sample contribution \(\sigma^2/n_i\).

The correlation also decreases monotonically. Taking the derivative of its logarithm, with \(n_i\) and \(n_j\) fixed, gives

\frac{d}{dn_0}\log\{\operatorname{Corr}(D_i,D_j)\} = -\frac{1}{2}\left(\frac{1}{n_0+n_i}+\frac{1}{n_0+n_j}\right)<0.

This proves the decrease for positive sample sizes. As \(n_0\) becomes very large, the correlation approaches zero, while the variance of each contrast approaches \(\sigma^2/n_i\). The control contribution becomes small relative to the treatment contributions; it does not vanish from the design at any finite \(n_0\).

Worked Example: Comparing Two Shared-Control Sizes

Hypothetical data: Suppose two treatment arms have sample sizes \(n_1=48\) and \(n_2=72\), and the common outcome variance is \(\sigma^2=64\) squared outcome units. Compare a shared control with \(n_0=24\) and one with \(n_0=30\). The groups are assumed independent and to have the stated common variance.

With \(n_0=24\), the variances and standard errors are

\operatorname{Var}(D_1)=64\left(\frac{1}{48}+\frac{1}{24}\right)=4.000, \qquad SE(D_1)=\sqrt{4.000}=2.000\text{ outcome units};

\operatorname{Var}(D_2)=64\left(\frac{1}{72}+\frac{1}{24}\right)=3.556, \qquad SE(D_2)=\sqrt{3.556}=1.886\text{ outcome units}.

The covariance is \(64/24=2.667\) squared outcome units. The correlation, a dimensionless quantity, is

\operatorname{Corr}(D_1,D_2) = \frac{\sqrt{48\cdot72}}{\sqrt{(24+48)(24+72)}} = 0.7071.

With \(n_0=30\), the corresponding calculations are

\operatorname{Var}(D_1)=64\left(\frac{1}{48}+\frac{1}{30}\right)=3.467, \qquad SE(D_1)=\sqrt{3.466667}=1.862\text{ outcome units};

\operatorname{Var}(D_2)=64\left(\frac{1}{72}+\frac{1}{30}\right) =64(0.0472222\ldots)=3.022222\ldots, \qquad SE(D_2)=\sqrt{3.022222\ldots}=1.738\text{ outcome units}.

The covariance is now \(64/30=2.133\) squared outcome units, and the correlation is

\operatorname{Corr}(D_1,D_2) = \frac{\sqrt{48\cdot72}}{\sqrt{(30+48)(30+72)}} = 0.6591.

The larger control reduces both standard errors, the covariance, and the correlation. For example, the second contrast’s standard error falls from 1.886 to 1.738 outcome units. The contrasts remain positively correlated; they have not become independent.

Comparing Control Sizes for a Precision Requirement

A design comparison can start with a desired standard error for an individual contrast. The variance formula can be rearranged to find the minimum control size that meets the requirement, provided the treatment size and variance are specified. This calculation addresses marginal precision only: it does not by itself determine a suitable overall design or account for all costs and inferential goals.

Worked Example: Finding a Control Size for a Target Standard Error

Hypothetical design: One treatment arm will have \(n_1=40\), the common outcome variance is \(\sigma^2=100\) squared units, and the target is a standard error no greater than 2.000 outcome units for its treatment-minus-control contrast. Assume independent groups and a common variance as in the model.

The target requires the contrast variance to be at most \(2^2=4\). Therefore,

100\left(\frac{1}{40}+\frac{1}{n_0}\right)\le 4 \quad\Longrightarrow\quad \frac{1}{n_0}\le 0.015 \quad\Longrightarrow\quad n_0\ge 66.666\ldots.

Since a sample size must be an integer, the smallest qualifying control size is \(n_0=67\). At this size,

\operatorname{Var}(D_1)=100\left(\frac{1}{40}+\frac{1}{67}\right)=3.99254, \qquad SE(D_1)=\sqrt{3.99254}=1.998\text{ outcome units}.

For illustration, suppose a second treatment arm has \(n_2=80\). Its standard error at the same control size is \(\sqrt{100(1/80+1/67)}=1.656\) outcome units. The covariance between the two contrasts is \(100/67=1.493\) squared units, and their correlation is

\operatorname{Corr}(D_1,D_2) = \frac{\sqrt{40\cdot80}}{\sqrt{(67+40)(67+80)}} = 0.4510.

The target is met for the first contrast, and the second contrast is positively correlated with it. The standard error requirement alone does not guarantee a particular correlation or establish that this design is optimal.

What Changes With Unequal Treatment Sizes?

A larger control does not affect every contrast by exactly the same amount. For fixed \(\sigma^2\), the absolute reduction in a contrast’s variance when control size changes from \(n_0=a\) to \(n_0=b>a\) is

\sigma^2\left(\frac{1}{a}-\frac{1}{b}\right).

This absolute reduction is the same for each contrast under the common-variance model, but the relative reduction depends on the treatment sample size, because the starting variance includes \(1/n_i\). A contrast with a small treatment group has a larger treatment contribution and may therefore receive a smaller proportional improvement from enlarging the control.

Worked Example: Same Variance Reduction, Different Proportional Gains

Hypothetical data: Let \(\sigma^2=36\) squared outcome units, with two treatment sizes \(n_1=30\) and \(n_2=90\). Compare a control size of 30 with a control size of 60. Assume independent groups and a common variance.

For the smaller treatment arm, the variance changes from

36\left(\frac{1}{30}+\frac{1}{30}\right)=2.400 \quad\text{to}\quad 36\left(\frac{1}{30}+\frac{1}{60}\right)=1.800.

Its standard error falls from \(\sqrt{2.400}=1.549\) to \(\sqrt{1.800}=1.342\) outcome units. This is a 25.00% reduction in variance. For the larger treatment arm, the variance changes from

36\left(\frac{1}{90}+\frac{1}{30}\right)=1.600 \quad\text{to}\quad 36\left(\frac{1}{90}+\frac{1}{60}\right)=1.000.

Its standard error falls from \(\sqrt{1.600}=1.265\) to \(1.000\) outcome unit, a 37.50% reduction in variance. In both cases, the variance decreases by \(36(1/30-1/60)=0.600\) squared units. The relative change differs because the initial variances differ.

The correlation between the two contrasts falls from \(\sqrt{30\cdot90}/\sqrt{(30+30)(30+90)}=0.6124\) to \(\sqrt{30\cdot90}/\sqrt{(60+30)(60+90)}=0.4472\). The larger control improves both marginal precision and the dependence profile.

Interpretation, Alternatives, and Limitations

A shared control can be more sample-efficient than assigning a separate control group to every treatment: the same control observations contribute to multiple comparisons. The cost is dependence among those comparisons. With separate, independent control groups under the same independent-group model, two contrasts using different controls have no shared random control mean, so their covariance is zero. That comparison presumes the groups and controls are genuinely independent; it is not a universal guarantee of independence in every trial.

Under unequal group variances, the common-\(\sigma^2\) expressions do not apply as written. If group \(g\) has variance \(\sigma_g^2\), independent observations still give \(\operatorname{Var}(D_i)=\sigma_i^2/n_i+\sigma_0^2/n_0\) and, for contrasts sharing the same control, \(\operatorname{Cov}(D_i,D_j)=\sigma_0^2/n_0\). The resulting correlation must be calculated using those variances; the simpler sample-size-only expression above no longer applies.

In practice, the assumed variances may be estimated, the realized sample sizes may differ from the planned sizes, and missing outcomes or clustering may alter the variance structure. A calculation based on a common variance is therefore a design-model comparison, not a diagnostic proof that the assumptions hold. The introductory material on shared-control dependence establishes why the shared mean induces dependence; the key design point here is that changing \(n_0\) changes both its size and its relative importance.

Common pitfall: “The covariance gets smaller when the control grows, so the contrasts become independent.” The first part is true under the model, but the conclusion does not follow. Independence would require zero covariance for jointly normal contrasts, and the covariance \(\sigma^2/n_0\) remains positive for every finite \(n_0\). Correlation decreases too, but reaches zero only in the limiting case as the control size grows without bound.

Before relying on a comparison, check that all contrasts use the same orientation, that the control is in fact shared, and that the independence and variance assumptions match the design. Also distinguish units: a contrast standard error is in outcome units, whereas a covariance is in squared outcome units and a correlation has no units.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

With \(\sigma^2=25\), \(n_i=50\), and \(n_0=50\), calculate the variance and standard error of \(D_i\).
Two treatment arms have sizes \(n_i=40\) and \(n_j=90\). What is the covariance of their contrasts when \(\sigma^2=36\) and \(n_0=30\)?
Under the common-variance model, does increasing the control sample size change the absolute variance reduction by the same amount for each contrast, if treatment sizes are fixed? Explain.
Why does the correlation of two shared-control contrasts tend to zero as \(n_0\) grows, even though both contrasts still use the same control group?

Solutions. (1) \(\operatorname{Var}(D_i)=25(1/50+1/50)=1.000\) squared outcome units, so \(SE(D_i)=1.000\) outcome unit. (2) The covariance is \(\sigma^2/n_0=36/30=1.200\) squared outcome units; treatment sample sizes do not enter this covariance under the model. (3) Yes. Increasing the control from \(a\) to \(b\) reduces each variance by \(\sigma^2(1/a-1/b)\), regardless of the treatment size. The proportional reduction can differ because each contrast has a different starting variance. (4) As the control grows, its contribution to each contrast’s variance and their shared covariance shrinks. The treatment-specific contributions remain, so the shared component becomes negligible relative to the total contrast standard deviations, making the correlation approach zero.