Shared-Control Efficiency for Binary Outcomes

Why Binary Outcomes Change the Calculation

For a continuous outcome, the variance of a treatment–control difference depends on the outcome variances in the two groups. With a binary outcome, the corresponding quantities are determined by event probabilities: a Bernoulli outcome with event probability $p$ has variance $p(1-p)$. This makes the comparison straightforward to calculate, but the required control size depends on the event probabilities assumed for planning.

The central efficiency point is that a common control group contributes its information to each treatment–control comparison. It does not make the control contribution to any one comparison disappear. Rather, it avoids recruiting a separate control group for every treatment arm. To quantify that saving correctly, first calculate the variance of one difference in proportions, then find how many control participants each comparison needs.

Definition: Let $p_C$ be the control event probability and $p_j$ the event probability in treatment arm $j$. Let $\hat p_C$ and $\hat p_j$ be the observed event proportions from $n_C$ and $n_j$ independent participants, respectively. The estimated treatment-minus-control difference is $\hat p_j-\hat p_C$. Under independent Bernoulli outcomes and fixed group sizes, its variance is $$ \operatorname{Var}(\hat p_j-\hat p_C) = \frac{p_j(1-p_j)}{n_j} + \frac{p_C(1-p_C)}{n_C}. $$

Deriving the Variance of a Difference

For a group of $n$ independent participants with binary outcomes $X_i$, where $X_i=1$ denotes an event and $X_i=0$ denotes no event, the sample proportion is $\hat p=n^{-1}\sum_{i=1}^n X_i$. Each $X_i$ has variance $p(1-p)$. Independence therefore gives the variance of the sample proportion:

\operatorname{Var}(\hat p) = \operatorname{Var}\left(\frac{1}{n}\sum_{i=1}^n X_i\right) = \frac{1}{n^2}\sum_{i=1}^n \operatorname{Var}(X_i) = \frac{p(1-p)}{n}.

If treatment and control participants are independent, their sample proportions are independent. The variance of their difference is the sum of their variances, which yields the stated formula. This is an exact variance under the Bernoulli model; using it as a basis for a normal-approximation confidence interval or test additionally requires the approximation to be adequate.

In a shared-control design, each comparison uses the same $\hat p_C$. Thus the individual comparison variance has the same form as in a two-arm trial with those group sizes, but different treatment–control differences are not independent. For two distinct treatment arms $j$ and $k$, independence of the treatment groups and their independence from control imply:

\operatorname{Cov}(\hat p_j-\hat p_C,\hat p_k-\hat p_C) = \operatorname{Var}(\hat p_C) = \frac{p_C(1-p_C)}{n_C}.

The positive covariance arises because both differences subtract the same random control proportion. It matters when considering the comparisons jointly; it does not alter the variance formula for either individual difference.

Worked Example: Estimating the Standard Error From Observed Data

In a hypothetical two-arm comparison, 72 of 180 treatment participants and 60 of 200 control participants experience the event. Calculate the observed difference in proportions and its estimated standard error.

The observed proportions are $\hat p_T=72/180=0.400$ and $\hat p_C=60/200=0.300$. The estimated treatment-minus-control difference is $0.400-0.300=0.100$, or 10.0 percentage points.

For an estimated standard error, substitute the observed proportions into the variance formula:

\widehat{\operatorname{Var}}(\hat p_T-\hat p_C) = \frac{0.400(1-0.400)}{180} + \frac{0.300(1-0.300)}{200} = 0.001333+0.001050 = 0.002383.

Therefore, $SE=\sqrt{0.002383}=0.04882$, or about 4.88 percentage points. The variance calculation uses estimated event probabilities because the true probabilities are unknown. The observed counts provide more than five events and non-events in each group, so a basic normal approximation is not immediately undermined by very sparse counts. That check is useful but does not guarantee good approximation in every setting.

Control Size for a Target Variance

For planning, suppose the event probabilities and treatment-group size are specified. Write the variance for comparison $j$ as a treatment contribution plus a control contribution. If a maximum variance $V^*$ is desired, the control size must make the sum no greater than $V^*$. The boundary cases need care, especially when the control event probability is zero or one.

a_j=\frac{p_j(1-p_j)}{n_j}, \qquad b=p_C(1-p_C), \qquad \operatorname{Var}(\hat p_j-\hat p_C)=a_j+\frac{b}{n_C}.

Planning rule: If $b>0$, a finite control-group size can achieve variance at most $V^*$ only if $V^*>a_j$. When that condition holds, choose $$ n_C\geq \frac{b}{V^*-a_j}, $$ rounding up to the next whole participant. If $b=0$, the control contribution is zero for every positive $n_C$, so the target is met exactly when $a_j\leq V^*$; if $a_j>V^*$, no control size can meet it.

When $b>0$, the control term $b/n_C$ is strictly positive for every finite $n_C$. Consequently, if $V^*=a_j$, no finite control size achieves the target, even though increasingly large control groups make the variance approach $a_j$. If the control event probability is at a boundary, $p_C=0$ or $p_C=1$, then $b=0$, and this limiting argument does not apply: the control contribution is identically zero.

Worked Example: Planning a Shared Control for Three Arms

Consider a hypothetical design with three treatment arms, each planned to enroll $n_j=400$ participants. Assume $p_C=0.200$, $p_j=0.350$ for every treatment arm, and a target variance $V^*=0.00100$ for each treatment–control difference. Find the minimum common control size and compare total recruitment with three separate two-arm trials using the same group sizes and target.

The treatment contribution for each comparison is $a_j=0.350(0.650)/400=0.2275/400=0.00056875$. The control variance term has numerator $b=0.200(0.800)=0.1600$. Since $V^*-a_j=0.00100-0.00056875=0.00043125>0$, the planning rule applies:

n_C\geq \frac{0.1600}{0.00043125}=371.0145\ldots

Thus, at least 372 control participants are required. Checking the rounded value gives $0.00056875+0.1600/372=0.0009989$, which is below the target. The shared-control design enrolls $3(400)+372=1{,}572$ participants.

If each treatment instead had a separate two-arm trial with its own 372-person control group, the total would be $3(400+372)=2{,}316$. Under these stated assumptions and equal per-comparison targets, sharing the control avoids $2(372)=744$ control participants. The saving is from not duplicating control groups; the 372-person control contribution is still present in each shared-control comparison.

Worked Example: A Boundary Control Probability

Suppose a hypothetical treatment arm has $p_T=0.300$ and $n_T=210$, while the planned control event probability is $p_C=0$. The target variance is $V^*=0.00100$. Does a finite control group meet the target?

Here $b=p_C(1-p_C)=0$, while the treatment contribution is

a_T=\frac{0.300(0.700)}{210} =\frac{0.210}{210} =0.00100.

For every positive finite $n_C$, the total variance is $a_T+b/n_C=0.00100+0=0.00100$. Therefore, any positive control size meets the target exactly. A zero denominator in the usual rearranged formula must not be treated as proof that no finite control size works: the control contribution is zero here.

If instead $n_T=200$, then the treatment contribution is $0.210/200=0.00105$, greater than the target. Because the control contribution remains zero, no choice of $n_C$ can reduce the total variance to $0.00100$. This illustrates both possibilities when $b=0$: the target is met for every positive control size if $a_T\leq V^*$, and it is unattainable if $a_T>V^*$.

Several Treatment Arms and Separate Trials

For a shared control, the same $n_C$ must meet the target for every comparison. If $b>0$, each arm must have $V^*>a_j$, and the common control size must be at least the largest arm-specific requirement. If $b=0$, check the treatment contribution in every arm; the shared control size cannot change those contributions.

Worked Example: The Most Demanding Arm Sets the Shared Control Size

A hypothetical study has two treatment arms, each with $n_j=300$, and a common control event probability $p_C=0.250$. Suppose $p_1=0.400$, $p_2=0.500$, and each comparison has target variance $V^*=0.00100$. Calculate the shared-control requirement and compare it with separate controls.

For arm 1, $a_1=0.400(0.600)/300=0.000800$. Since $b=0.250(0.750)=0.1875$, its control requirement is

n_C\geq \frac{0.1875}{0.00100-0.000800} =937.5,

so arm 1 requires at least 938 control participants. For arm 2, $a_2=0.500(0.500)/300=0.0008333$, giving

n_C\geq \frac{0.1875}{0.00100-0.0008333\ldots} =1{,}125.

The common control must therefore contain at least 1,125 participants. At that size, the arm 2 variance equals the target and the arm 1 variance is $0.000800+0.1875/1{,}125=0.0009667$, also within target. Separate two-arm trials would require 938 and 1,125 control participants, respectively, for a combined 2,063 controls. One shared control requires 1,125, avoiding 938 control participants while meeting both per-comparison targets.

The comparison with separate trials is clearest when the treatment-group sizes, planning probabilities, and precision targets are aligned. For one comparison, sharing does not lower the control size required by the variance formula. Across several comparisons, it can avoid repeating that control recruitment. With differing arms, calculate each separate control requirement and use the largest requirement for a common control; the separate design instead sums its control requirements.

The shared estimates are correlated because they use the same control proportion. Treating them as independent comparisons would therefore be incorrect for joint calculations. That dependence is distinct from the per-comparison variance and should not be confused with a reduction in the control term. Separate trials have separate control estimates, while a shared-control protocol has one estimate reused across its contrasts.

Assumptions, Diagnostics, and Limitations

The calculations assume fixed group sizes, a binary outcome observed for every participant, independent outcomes within and across randomized groups, and a common control event probability for all comparisons. They also presume that the planning probabilities are suitable for the population and outcome definition of interest. If observations are clustered, repeated, dependent, or affected by substantial missingness, the simple Bernoulli variance may not describe the data.

Check the scale and inputs. Probabilities must lie from 0 to 1, group sizes must be positive, and the target variance must be in squared-proportion units. The standard error is the square root of variance and is in proportion units.
Check sparse expected counts. The variance formula remains the Bernoulli variance, but a normal approximation can be poor when expected events or non-events are very few. The adequacy of an inferential approximation is a separate question from the variance calculation.
Check each arm separately. A treatment arm with a large $a_j$ can determine the shared-control requirement or make the target impossible, even when other arms meet it comfortably.
Check boundary probabilities explicitly. At $p_C=0$ or $p_C=1$, the control variance contribution is zero. Do not divide by $V^*-a_j$ without first checking this case and the feasibility condition.
Revisit the planning probabilities. The variance and resulting sample size can change substantially when event rates differ from their planning values. The calculation is conditional on those values, not a guarantee of achieved precision under every possible event rate.

A common mistake is to say that a shared control makes every treatment comparison more precise because the control is reused. Reuse avoids duplicated recruitment across comparisons, but for one comparison the variance is still $p_j(1-p_j)/n_j+p_C(1-p_C)/n_C$. Another mistake is to conclude that a zero or negative denominator always makes the target impossible. That conclusion is valid only when $p_C(1-p_C)>0$; when the control variance is zero, feasibility depends directly on whether the treatment contribution is at most the target.

Key takeaway: For binary outcomes, each comparison’s variance is the sum of the treatment and control proportion variances. A shared control can prevent duplicate control recruitment across comparisons, but its required size is set by the most demanding arm, and boundary event probabilities require an explicit feasibility check.

Check Your Understanding

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

A treatment group has $n_T=250$ and planning event probability $p_T=0.400$; a control group has $n_C=300$ and $p_C=0.200$. Calculate the variance and standard error of the difference in proportions.
For a treatment contribution $a=0.000600$, control variance numerator $b=0.1600$, and target $V^*=0.00100$, what is the minimum integer control size?
If $p_C=1$ and the treatment contribution is $0.00080$, can a target variance of $0.00080$ be met with a positive finite control size? Explain.
Two treatment arms have separate control requirements of 500 and 800 participants under the same per-comparison variance target. How many controls are required with one shared group, and how many with separate trials?

Solutions. (1) The treatment contribution is $0.400(0.600)/250=0.000960$; the control contribution is $0.200(0.800)/300=0.0005333$. The variance is $0.0014933$, and the standard error is $\sqrt{0.0014933}=0.03864$, or 3.864 percentage points. (2) Since $V^*-a=0.000400>0$, $n_C\geq0.1600/0.000400=400$; the minimum is 400. (3) Yes. Here $b=p_C(1-p_C)=0$, so the total variance equals the treatment contribution, $0.00080$, for every positive finite control size. (4) The shared group must meet the larger requirement, so it needs 800 controls; separate trials need $500+800=1{,}300$ controls in total.