Deriving Treatment-Effect Contamination For Binary Outcomes

Why a Blinded Pooled Rate Can Be Misleading

With a binary outcome, the variance within an arm is determined by its event probability: a Bernoulli outcome with event probability \(p\) has variance \(p(1-p)\). But a blinded analysis combines participants from both randomized groups. If their event probabilities differ, the variance across the pooled participants reflects not only variation within each group, but also the difference between the groups. That additional component is treatment-effect contamination.

This matters when a pooled rate is used to estimate a nuisance variance for planning. The pooled variance may be larger than the weighted average of the arm-specific variances, even though no treatment labels have been examined. The difference is not evidence about the treatment effect: it is the mathematical consequence of mixing groups with different event probabilities.

The derivation below assumes one fully observed binary outcome per participant, independent participants, and a fixed treatment allocation fraction. It concerns variance decomposition, not a method for estimating the treatment effect from blinded data.

Definition: Let \(Y\) be a binary outcome, with \(Y=1\) denoting an event. Let \(p_1=P(Y=1\mid A=1)\) and \(p_0=P(Y=1\mid A=0)\) be the event probabilities in the treatment and control groups. Write \(\Delta=p_1-p_0\) for the risk difference, and let \(\pi\) be the fraction allocated to treatment, so the control fraction is \(1-\pi\). The pooled event probability is \(q=\pi p_1+(1-\pi)p_0\). The within-group pooled variance is \(\pi p_1(1-p_1)+(1-\pi)p_0(1-p_0)\); the variance of the pooled Bernoulli mixture is \(q(1-q)\).

Deriving the Contamination Component

Imagine selecting one participant at random from the pooled trial population. Let \(A=1\) indicate treatment assignment and \(A=0\) control assignment. Then \(P(A=1)=\pi\). Conditional on the assignment, \(Y\) is Bernoulli with probability \(p_1\) or \(p_0\). Applying the law of total variance separates variability within assignment groups from variability between their means.

Average the within-group variances.
Conditional on treatment, the variance is \(p_1(1-p_1)\); conditional on control, it is \(p_0(1-p_0)\). Weight these by their allocation fractions.

Calculate the variance of the group means.
The conditional means are \(p_1\) and \(p_0\), whose weighted mean is \(q\). Their between-group variance is \(\pi(1-\pi)(p_1-p_0)^2\).

Add the components.
The total mixture variance is the within-group component plus the between-group component.

\begin{aligned} q(1-q) &= \operatorname{Var}(Y)\\ &= E\{\operatorname{Var}(Y\mid A)\} +\operatorname{Var}\{E(Y\mid A)\}\\ &= \pi p_1(1-p_1)+(1-\pi)p_0(1-p_0) +\pi(1-\pi)(p_1-p_0)^2. \end{aligned}

The final term is nonnegative, and it is zero if the allocation is entirely to one group or if the two event probabilities are equal. For an actual two-arm trial with both groups represented, any nonzero risk difference makes the pooled mixture variance exceed the weighted within-group variance. This is the precise sense in which the pooled variance is contaminated by the treatment effect.

Result: If a risk difference \(\Delta\) is prespecified, the weighted within-group variance implied by a pooled rate \(q\) is \(q(1-q)-\pi(1-\pi)\Delta^2\). The amount subtracted is the between-group component. It is not generally correct to subtract \(\Delta^2/4\) unless allocation is equal, \(\pi=1/2\).

Worked Examples

Worked Example: Quantifying Contamination With Equal Allocation

Suppose a hypothetical trial allocates half its participants to each group, and the event probabilities are \(p_1=0.30\) and \(p_0=0.20\). Find the pooled event probability, pooled mixture variance, weighted within-group variance, and contamination component.

Here \(\pi=0.50\) and \(\Delta=0.30-0.20=0.10\). The pooled event probability is \(q=0.50(0.30)+0.50(0.20)=0.25\), so the pooled mixture variance is \(q(1-q)=0.25(0.75)=0.1875\).

The weighted within-group variance is \(0.50[0.30(0.70)]+0.50[0.20(0.80)]=0.50(0.21)+0.50(0.16)=0.1850\). The between-group component is \(0.50(0.50)(0.10)^2=0.0025\).

Thus \(0.1850+0.0025=0.1875\), as required. In this example, mixing the groups raises the variance by 0.0025 relative to the weighted within-group variance. The calculation describes the mixture; it does not estimate either probability from blinded trial data.

For re-estimation, the arm-specific event probabilities are often not available under masking. If the pooled rate and a prespecified risk difference are available, the allocation-weighted mixture equation can be solved for the two probabilities. The risk-difference convention must be stated: here positive \(\Delta\) means a higher event probability under treatment.

p_1=q+(1-\pi)\Delta,\qquad p_0=q-\pi\Delta.

Substitution verifies both identities: \(p_1-p_0=\Delta\), and \(\pi p_1+(1-\pi)p_0=q\). These equations recover rates consistent with the assumed risk difference and observed pooled rate, provided both recovered rates lie between zero and one.

Worked Example: Unequal Allocation and Reconstructed Rates

Suppose a hypothetical blinded trial has treatment allocation fraction \(\pi=0.40\) and pooled event rate \(q=0.28\). The planning assumption is a risk difference \(\Delta=0.12\). Reconstruct the two rates, check the variance decomposition, and calculate the variance of the estimated risk difference for a total sample size \(n\).

The implied treatment rate is \(p_1=0.28+(1-0.40)(0.12)=0.352\), and the implied control rate is \(p_0=0.28-0.40(0.12)=0.232\). Both are valid probabilities, and their difference is \(0.352-0.232=0.120\). Their weighted average is \(0.40(0.352)+0.60(0.232)=0.1408+0.1392=0.2800\).

The pooled mixture variance is \(0.28(0.72)=0.2016\). The weighted within-group variance is \(0.40[0.352(0.648)]+0.60[0.232(0.768)]\) \(=0.40(0.228096)+0.60(0.178176)=0.198144\). The between-group component is \(0.40(0.60)(0.12)^2=0.003456\), and \(0.198144+0.003456=0.201600\).

For independent groups with \(n_1=\pi n\) and \(n_0=(1-\pi)n\), the variance of the estimated risk difference is \(p_1(1-p_1)/n_1+p_0(1-p_0)/n_0\). Expressed per total sample size, this is \([0.228096/0.40+0.178176/0.60]/n=(0.57024+0.29696)/n=0.8672/n\). This is the variance for the difference in sample proportions under the stated model. The pooled mixture variance alone is not that contrast variance.

From the Pooled Rate to a Contrast Variance

The between-group correction identifies the weighted average of the arm-specific Bernoulli variances. A risk-difference estimator, however, weights each arm variance by the inverse of that arm's sample size. Therefore, when allocation is unequal, one must not simply insert the corrected weighted average into a formula that requires the two arm-specific variances.

Under the assumptions of independent Bernoulli outcomes and fixed group sizes, the sample proportions have variances \(p_1(1-p_1)/n_1\) and \(p_0(1-p_0)/n_0\). Independence across arms gives the variance of their difference by addition. Substituting the reconstructed probabilities gives the expression below.

\operatorname{Var}(\hat p_1-\hat p_0) = \frac{[q+(1-\pi)\Delta][1-q-(1-\pi)\Delta]}{\pi n} + \frac{[q-\pi\Delta][1-q+\pi\Delta]}{(1-\pi)n}.

This calculation uses the assumed \(\Delta\), not a treatment-effect estimate extracted from masked outcomes. If the planning calculation uses a different variance convention or an alternative model, its variance formula must be derived under that model rather than substituted for this one.

Worked Example: Checking an Assumed Effect Against the Pooled Rate

Suppose a hypothetical trial has \(\pi=0.50\), pooled event rate \(q=0.08\), and a prespecified risk difference \(\Delta=0.20\). Determine whether these inputs imply valid arm-specific rates.

The reconstruction gives \(p_1=0.08+(1-0.50)(0.20)=0.18\) and \(p_0=0.08-0.50(0.20)=-0.02\). The control probability is negative, so the inputs cannot describe two Bernoulli groups with that pooled rate, allocation, and risk difference. A calculation that mechanically subtracts the contamination term would give \(0.08(0.92)-0.50(0.50)(0.20)^2=0.0736-0.0100=0.0636\), which is positive but does not make the assumed arm rates valid.

The appropriate conclusion is an incompatibility requiring review of the pooled-rate estimate, the assumed risk difference, or the allocation fraction. The positive corrected variance is not evidence that the inputs are coherent, and replacing a negative reconstructed rate with zero would silently change the assumptions.

Assumptions, Diagnostics, and Alternatives

The decomposition is exact for the stated two-component Bernoulli mixture. Applying it to an interim pooled estimate requires care: the outcome must be defined consistently across groups; the allocation fraction used should reflect the pooled participants contributing to that estimate; and observations should satisfy the independence assumption. Clustered or repeated outcomes do not generally have the simple Bernoulli variance used here. Differential missingness or incomplete follow-up can also make the observed pooled rate a poor estimate of the target event probability.

Useful checks include confirming that \(0<\pi<1\), that the sign convention for \(\Delta\) is explicit, and that both reconstructed rates satisfy \(0\le p_1,p_0\le1\). Equivalently, the assumed risk difference must be compatible with the observed pooled rate and allocation. A negative corrected within-group variance is an immediate warning, but a nonnegative value is not sufficient: as the previous example shows, the implied arm rates can still be invalid.

The correction also depends on the prespecified risk difference. If that assumption is wrong, the corrected variance may not represent the actual within-group variance. The pooled event rate itself has sampling variability, especially with few observed events, so a plug-in correction can be unstable even when the assumed rates are feasible. These issues should be assessed rather than hidden by arbitrary truncation or rounding.

The main alternatives answer different questions. Using \(q(1-q)\) without correction treats the pooled participants as one Bernoulli population and retains the between-group component. Using the correction removes the component implied by the prespecified \(\Delta\), while maintaining masking. Using separate arm-specific event rates can estimate the arm variances directly, but requires access to treatment labels and is therefore not a blinded pooled-rate approach. None of these approaches can make a wrong allocation assumption, incompatible inputs, or an unsuitable independence model harmless.

Common Mistake

A frequent error is to say, “The pooled rate is \(q\), so the variance in each arm is \(q(1-q)\).” That treats both arms as if they shared the pooled probability and ignores the difference between their event rates. When the true rates differ, \(q(1-q)\) includes the positive term \(\pi(1-\pi)\Delta^2\). The correction removes that term from the weighted pooled variance; it does not automatically give each arm variance, nor does it by itself give the variance of a risk-difference estimator. For the latter, reconstruct or otherwise obtain both arm rates and apply the arm-specific variance formula.

Key takeaway: A blinded binary-outcome mixture has variance equal to the weighted within-arm Bernoulli variance plus \(\pi(1-\pi)\Delta^2\). Subtracting this component is justified only for a prespecified, feasible risk difference and the stated allocation; a risk-difference variance still requires the arm-specific variances and sample sizes.

Check Your Understanding

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

With \(\pi=0.50\) and \(\Delta=0.08\), what is the between-group contamination term?
For \(q=0.30\), \(\pi=0.25\), and \(\Delta=0.10\), reconstruct \(p_1\) and \(p_0\). Are they valid probabilities?
Why is the corrected weighted within-group variance not, by itself, the variance of the estimated risk difference?
For \(\pi=0.50\), \(p_1=p_0\), and any valid common event probability, what is the contamination term, and why?

Solutions. (1) \(\pi(1-\pi)\Delta^2=0.50(0.50)(0.08)^2=0.0016\). (2) \(p_1=0.30+0.75(0.10)=0.375\), and \(p_0=0.30-0.25(0.10)=0.275\); both lie between zero and one. (3) The contrast variance weights each arm variance by its inverse sample size, so the two arm-specific variances and allocation sizes are needed. (4) The term is \(0.50(0.50)(0)^2=0\), because equal group means contribute no between-group variation.