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Blinded Re-Estimation · Tutorial 374 of 699

Blinded Pooled-Rate Estimation For Binary Outcomes

Learn to calculate a blinded pooled event rate, assess its assumptions, and use allocation and an explicitly assumed risk difference to obtain planning rates and variances.

Intermediate 11 min read

What You'll Learn

  • Calculate the pooled event-rate estimate and its standard error from masked Bernoulli outcomes.
  • Distinguish a common event probability from the allocation-weighted rate observed when treatment risks differ.
  • Explain why the masked pooled rate alone cannot identify both treatment-arm probabilities.
  • Map a pooled rate and assumed risk difference to planning probabilities using the allocation fractions.
  • Calculate and compare risk-difference planning variances under different allocations.
  • Recognize when a single pooled rate is an inadequate nuisance estimate.

Why a Masked Pooled Rate Needs Careful Interpretation

For a binary outcome, each participant either experiences the event or does not. In a blinded sample-size re-estimation, the treatment labels may be hidden while the outcomes remain available. The total number of events can then inform planning, but it does not automatically estimate the event probability in either treatment arm. What it estimates depends on whether a common event probability is a reasonable model and, if the arm risks differ, on the allocation.

This distinction matters because the pooled rate is often used to estimate a nuisance quantity for planning. Under a common-risk model, the rate estimates that shared probability. Under unequal risks, it estimates an allocation-weighted average instead. If a planning calculation needs two arm-specific probabilities, the masked rate alone is insufficient; an assumed treatment contrast and the allocation are also needed.

Definition: Suppose \(N=n_1+n_0\) participants have binary outcomes, with \(n_1\) assigned to treatment 1 and \(n_0\) to treatment 0. Let \(X\) be the total number of events, without treatment labels. The blinded pooled event-rate estimate is \(\hat{p}_{\mathrm{pool}}=X/N\). If both arms share a common event probability \(p\), this estimates \(p\). If the arm probabilities are \(p_1\) and \(p_0\), its target is the allocation-weighted probability \((n_1p_1+n_0p_0)/N\).

Estimation Under a Common-Probability Model

Assume that outcomes are observed for all included participants, each outcome is Bernoulli, and outcomes are independent with the same event probability \(p\). The allocation counts are fixed and known, but the treatment labels attached to individual outcomes are masked. Under this common-probability model, the total event count satisfies \(X\sim\operatorname{Binomial}(N,p)\). The likelihood for \(p\), apart from a factor that does not depend on \(p\), is \(p^X(1-p)^{N-X}\).

For \(0<X<N\), differentiating the log likelihood gives \(X/p-(N-X)/(1-p)\). Setting this equal to zero yields the maximum likelihood estimate below. The same result applies at \(X=0\) or \(X=N\), where the likelihood is maximized at the boundary.

$$ \hat{p}_{\mathrm{pool}}=\frac{X}{N} $$

Under the model, \(E(\hat{p}_{\mathrm{pool}})=p\) and \(\operatorname{Var}(\hat{p}_{\mathrm{pool}})=p(1-p)/N\). Substituting the estimate for the unknown probability gives the usual estimated standard error, \(\sqrt{\hat{p}_{\mathrm{pool}}(1-\hat{p}_{\mathrm{pool}})/N}\). This standard error describes sampling variation under the common-probability model; it does not include uncertainty about whether that model is appropriate.

Worked Example: Estimating a Common Event Probability

In a hypothetical masked interim sample, 36 of 240 participants experienced the event. Assume complete outcome ascertainment and, for this calculation, a common event probability for both arms. Estimate the event probability and its standard error.

The pooled estimate is \(36/240=0.1500\), or 15.00%. The estimated standard error is

$$ SE(\hat{p}_{\mathrm{pool}}) =\sqrt{\frac{0.1500(1-0.1500)}{240}} =\sqrt{0.00053125} =0.02305. $$

Thus the estimated common event probability is 15.00%, with an estimated standard error of 2.305 percentage points. This is an estimate of a probability, not a comparison of treatment effects. The calculation is justified only to the extent that the common-risk and independent-outcome assumptions are reasonable.

What the Pooled Rate Estimates When Risks Differ

If treatment 1 and treatment 0 have different event probabilities, write them as \(p_1\) and \(p_0\). With fixed arm sizes and independent Bernoulli outcomes within and between arms, the expected total event count is \(n_1p_1+n_0p_0\). Consequently,

$$ E(\hat{p}_{\mathrm{pool}}) =\frac{n_1p_1+n_0p_0}{N} =w_1p_1+w_0p_0, \qquad w_1=\frac{n_1}{N},\quad w_0=\frac{n_0}{N}. $$

So the pooled rate estimates the allocation-weighted average risk, not generally either arm's risk and not generally their simple average. With unequal allocation, the larger arm contributes more to the pooled rate. Masking prevents the analyst from assigning the observed events to arms, so the aggregate rate by itself cannot recover both \(p_1\) and \(p_0\).

For planning a risk-difference comparison, define the assumed contrast as \(\Delta=p_1-p_0\), with positive values indicating a higher event probability in treatment 1. If a pooled rate is used as a working estimate of the weighted average and a planning contrast \(\Delta\) is specified, the corresponding planning probabilities are

$$ p_1=\hat{p}_{\mathrm{pool}}+w_0\Delta, \qquad p_0=\hat{p}_{\mathrm{pool}}-w_1\Delta. $$

These expressions preserve the weighted average: \(w_1p_1+w_0p_0=\hat{p}_{\mathrm{pool}}\). They are a planning parameterization, not estimates of the hidden arm-specific event rates. The assumed contrast must be stated, the resulting probabilities must lie between 0 and 1, and the allocation fractions must match the design being planned.

Worked Example: Mapping a Pooled Rate to Planning Probabilities

Suppose a hypothetical interim sample has \(n_1=120\) and \(n_0=180\), with 42 total events among 300 participants. Treatment labels are masked. For planning, use the observed pooled rate and assume a risk difference of \(\Delta=0.0800\), where \(\Delta=p_1-p_0\).

The pooled rate is \(42/300=0.1400\). The allocation fractions are \(w_1=120/300=0.4000\) and \(w_0=180/300=0.6000\). The planning probabilities are therefore

$$ p_1=0.1400+(0.6000)(0.0800)=0.1880, \qquad p_0=0.1400-(0.4000)(0.0800)=0.1080. $$

As a check, the weighted average is \((0.4000)(0.1880)+(0.6000)(0.1080)=0.1400\). For independent Bernoulli outcomes, the planning variance of the estimated risk difference is the sum of the two arm-specific binomial variances divided by their respective sample sizes:

$$ \operatorname{Var}(\hat{p}_1-\hat{p}_0) =\frac{p_1(1-p_1)}{n_1}+\frac{p_0(1-p_0)}{n_0} =\frac{0.1880(0.8120)}{120}+\frac{0.1080(0.8920)}{180} =0.001807. $$

The corresponding planning standard error is \(\sqrt{0.001807}=0.04251\), or 4.251 percentage points. These are planning calculations based on an assumed contrast; the masked interim data do not establish that the two arm probabilities are 18.80% and 10.80%.

Allocation Changes the Planning Variance

Even when the total sample size and pooled rate are fixed, allocation affects how precisely a risk difference can be estimated. The risk-difference variance depends on both arm sizes and both planning probabilities. Thus, the pooled rate is not enough to calculate that variance unless the allocation is known and the planning contrast (or another way to specify both arm probabilities) is supplied.

Worked Example: Comparing Two Allocations

Suppose a hypothetical trial plans to enroll 300 participants, and the pooled rate used for planning is 0.2500. Compare balanced allocation with a 2:1 allocation to treatment 1, assuming \(\Delta=p_1-p_0=0.1000\). Assume independent Bernoulli outcomes and no missing outcomes.

With balanced allocation, \(w_1=w_0=0.5000\) and \(n_1=n_0=150\). The planning probabilities are \(p_1=0.2500+(0.5000)(0.1000)=0.3000\) and \(p_0=0.2500-(0.5000)(0.1000)=0.2000\). The risk-difference variance is

$$ \frac{0.3000(0.7000)}{150}+\frac{0.2000(0.8000)}{150} =0.002467, \qquad SE=\sqrt{0.002467}=0.04967. $$

With 2:1 allocation, \(n_1=200\), \(n_0=100\), \(w_1=2/3\), and \(w_0=1/3\). The planning probabilities are \(p_1=0.2500+(1/3)(0.1000)=0.2833\) and \(p_0=0.2500-(2/3)(0.1000)=0.1833\). They preserve the same weighted rate: \((2/3)(0.2833)+(1/3)(0.1833)\approx0.2500\). The variance is

$$ \frac{0.2833(0.7167)}{200}+\frac{0.1833(0.8167)}{100} \approx0.002513, \qquad SE\approx\sqrt{0.002513}=0.05012. $$

Under these assumptions, the 2:1 allocation has a slightly larger planning standard error for the risk difference than balanced allocation. This comparison depends on the specified contrast and probabilities; it is not a general claim that one allocation is always preferable.

Assumptions, Diagnostics, and Alternatives

Before using a pooled rate in a re-estimation, check what information the masked data contain and what probability the calculation is intended to represent. A simple pooled estimate is most directly interpretable when each participant contributes one observed binary outcome, the denominator is the number of participants with outcomes, and the outcome probabilities are sufficiently stable across participants and relevant design strata. If outcomes are missing, \(X/N\) is not the event proportion among all enrolled participants unless the denominator and missingness handling have been defined accordingly.

Useful diagnostics include checking the event count against the analysis denominator, confirming the allocation fractions, and comparing the observed pooled rate with the range considered plausible at design. A rate near 0 or 1, very few events, or substantial variation across sites or strata can make a single pooled rate imprecise or unrepresentative. If follow-up differs between participants, the simple proportion may also fail to represent the intended time-specific risk. These problems should prompt review of the estimand and model rather than automatic use of the same pooled formula.

The main alternatives answer different questions. An unblinded analysis can estimate \(p_1\) and \(p_0\) separately from arm-specific event counts, but requires access to treatment labels and is not a blinded pooled-rate calculation. An external or design-stage event-rate assumption avoids using interim outcomes, but may not reflect the current trial population. A common-risk pooled estimate is simple and uses all masked outcomes efficiently under its model; it cannot reveal treatment-specific risks or verify that the common-risk assumption holds.

Common pitfall: Treating the masked pooled rate as the control-arm event probability, or as both arm probabilities, even when the planning contrast is nonzero. The correction is to state the target clearly: under a common-risk model, the pooled rate estimates the shared \(p\); with unequal risks, it estimates \(w_1p_1+w_0p_0\). To construct two planning probabilities from that weighted rate, specify the allocation and an assumed risk difference, then check that the resulting probabilities are feasible.

A pooled-rate estimate is a nuisance-parameter input, not a treatment-effect estimate. Blinding does not by itself guarantee that a sample-size update is appropriate: the update must use a defensible probability model, prespecified planning assumptions, and a re-estimation procedure compatible with the trial design. In particular, do not interpret the assumed contrast used to obtain planning probabilities as information learned from masked outcomes.

Check Your Understanding

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

  1. A hypothetical masked sample has 24 events among 160 participants. Under a common-risk model, calculate the pooled estimate and its estimated standard error.
  2. For a planned 2:1 allocation to treatment 1, a pooled planning rate is 0.3000 and the assumed risk difference is \(\Delta=0.0600\). Find the two planning probabilities and verify their weighted average.
  3. Under a common event probability of 0.2000, what is the estimated standard error of the pooled rate for \(N=250\), using the plug-in formula?
  4. Can a masked pooled rate alone identify both arm-specific probabilities when treatment risks differ? Explain.

Solutions. (1) The estimate is \(24/160=0.1500\). Its estimated standard error is \(\sqrt{0.1500(0.8500)/160}=\sqrt{0.000796875}=0.02823\), or 2.823 percentage points. (2) Here \(w_1=2/3\) and \(w_0=1/3\). Thus \(p_1=0.3000+(1/3)(0.0600)=0.3200\), and \(p_0=0.3000-(2/3)(0.0600)=0.2600\). Their weighted average is \((2/3)(0.3200)+(1/3)(0.2600)=0.3000\). (3) The estimated standard error is \(\sqrt{0.2000(0.8000)/250}=\sqrt{0.0006400}=0.02530\). (4) No. The pooled rate supplies one weighted-average quantity, whereas two arm probabilities are unknown. Allocation and an assumed contrast can provide a planning parameterization, but the resulting arm probabilities are assumptions for planning, not separately identified estimates from the masked total.

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