Using A Pooled Event Rate For Odds-Ratio Planning

Why a Pooled Rate Can Support Odds-Ratio Planning

In a blinded internal-pilot re-estimation, treatment labels are not used to estimate separate event rates. The pooled event rate can still inform planning, but it does not by itself determine the control and treatment probabilities. To obtain those probabilities, combine the pooled rate with the target odds ratio used in the design. The result is a pair of planning probabilities consistent with both quantities.

This approach updates a nuisance quantity—the event probability—while retaining the planned treatment effect on the odds-ratio scale. It does not estimate the treatment odds ratio from the masked data. The re-estimated sample size is therefore conditional on the target odds ratio remaining the effect used for planning.

Definition: Let \(p_0\) be the control event probability and \(p_1\) the treatment event probability. Define the treatment-to-control odds ratio as \(\theta = [p_1/(1-p_1)]/[p_0/(1-p_0)]\). If the planned allocation is \(n_1/n_0=r\), the allocation-weighted pooled event probability is \(q=(p_0+rp_1)/(1+r)\). With masked data, the observed pooled rate estimates \(q\); the target value of \(\theta\) supplies the additional information needed to plan using \(p_0\) and \(p_1\).

The allocation ratio matters because the pooled rate is weighted by the number of participants in each arm. When allocation is equal, \(q\) is the simple average of the two probabilities. When treatment allocation is twice control allocation, for example, the treatment probability contributes twice as much to the pooled rate as the control probability.

Recovering the Arm-Specific Probabilities

Starting from the definition of the odds ratio, solve for the treatment probability as a function of the control probability:

p_1=\frac{\theta p_0}{1-p_0+\theta p_0} =\frac{\theta p_0}{1+(\theta-1)p_0}.

Substitute this expression into the allocation-weighted pooled-rate equation. Write \(a=\theta-1\) and \(C=(1+r)q\). Then the control probability must satisfy:

C=p_0+\frac{r\theta p_0}{1+ap_0} \quad\Longrightarrow\quad a p_0^2+\left(1+r\theta-aC\right)p_0-C=0.

For \(\theta\ne 1\), solve this quadratic and take the root in the interval from 0 to 1; then calculate \(p_1\) from the odds-ratio equation. If \(\theta=1\), the arms have equal planning probabilities, so \(p_0=p_1=q\). There is a unique admissible solution: as \(p_0\) increases from 0 to 1, the corresponding \(p_1\) increases from 0 to 1, and the weighted average \((p_0+rp_1)/(1+r)\) increases continuously from 0 to 1. Thus each pooled rate strictly between 0 and 1 corresponds to one pair of probabilities for a fixed positive \(\theta\) and \(r\).

Worked Example: Recovering Rates With Equal Allocation

Suppose a masked internal-pilot sample has a pooled event rate of \(q=0.20\). Planning assumes equal allocation, \(r=1\), and a treatment-to-control odds ratio of \(\theta=2\). Find the two planning probabilities.

Here \(a=1\) and \(C=(1+1)(0.20)=0.40\). The quadratic becomes:

p_0^2+(1+2-0.40)p_0-0.40=0, \qquad p_0^2+2.60p_0-0.40=0.

The admissible root is:

p_0=\frac{-2.60+\sqrt{(2.60)^2+4(0.40)}}{2} =\frac{-2.60+\sqrt{8.36}}{2} \approx 0.1457.

Then \(p_1=2p_0/(1+p_0)\approx 2(0.1457)/1.1457=0.2543\). As a check, \((0.1457+0.2543)/2=0.2000\), and the treatment-to-control odds ratio is approximately 2. The planning probabilities are about 14.57% in control and 25.43% in treatment.

Updating Sample Size for a Two-Sided Test

A common planning approximation tests the log odds ratio with a two-sided Wald test. Assume independent participants, a binary outcome measured once per participant, fixed allocation ratio \(r\), and the planning probabilities \(p_0\) and \(p_1\). The approximation also requires enough events and non-events for the estimated log odds ratio to be reasonably close to normal. It does not include a continuity correction or an allowance for loss to follow-up.

For \(n_0=n\) controls and \(n_1=rn\) treatment participants, the estimated log odds ratio has approximate variance:

\operatorname{Var}(\widehat{\log\theta}) \approx \frac{1}{n} \left[ \frac{1}{p_0}+\frac{1}{1-p_0} +\frac{1}{r}\left(\frac{1}{p_1}+\frac{1}{1-p_1}\right) \right] =\frac{A}{n}.

Here \(A\) denotes the bracketed quantity. The variance expression follows by adding the approximate variances of the four log cell-count contributions to the estimated log odds ratio: each arm contributes the reciprocal of its expected event count and its expected non-event count. In particular, the treatment contributions are divided by \(r\), because that arm has \(rn\) participants.

Under the planned alternative, the mean of the estimated log odds ratio is \(\delta=\log(\theta)\). A two-sided Wald test at level \(\alpha\) rejects when the absolute Wald statistic exceeds \(z_{1-\alpha/2}\), the \(1-\alpha/2\) standard-normal quantile. For target power \(1-\beta\), the usual normal approximation places the alternative mean at least \(z_{1-\alpha/2}+z_{1-\beta}\) standard errors from zero. Since its noncentrality is \(\delta/\sqrt{A/n}\), solving for \(n\) gives:

n_0\approx \frac{A\left(z_{1-\alpha/2}+z_{1-\beta}\right)^2}{[\log(\theta)]^2}, \qquad n_1=rn_0.

This calculation uses the variance under the planning alternative and is an approximate power calculation. The required size is for the complete trial, not merely the participants still to be enrolled. Round up to meet the target, preserve the planned allocation as closely as possible, and separately account for anticipated attrition if the design requires it.

Worked Example: Equal-Allocation Sample-Size Update

Use the probabilities from the first example to plan a two-sided test with \(\alpha=0.05\), power \(0.80\), and \(\theta=2\). Assume the re-estimation targets equal allocation and that there is no attrition adjustment.

For equal allocation, \(r=1\), so:

A= \frac{1}{0.1457}+\frac{1}{0.8543} +\frac{1}{0.2543}+\frac{1}{0.7457} \approx 13.31.

Using \(z_{0.975}=1.960\), \(z_{0.80}=0.8416\), and \(\log(2)=0.6931\):

n_0\approx n_1 = \frac{13.31(1.960+0.8416)^2}{(0.6931)^2} \approx 217.5.

Round up to 218 participants per arm, or 436 total. For example, if 90 participants per arm have already been enrolled, the target implies 128 additional participants per arm, assuming the final allocation can remain equal. The calculation is a planning approximation, not a claim that the realized trial will have exactly 80% power.

Unequal Allocation and a Second Check

For unequal allocation, use the same allocation ratio in both the pooled-rate equation and the variance expression. Changing \(r\) changes both the recovered arm-specific probabilities and the amount of information contributed by each arm. A frequent error is to use an unweighted average for \(q\), then apply an unequal-allocation variance formula. Those two steps describe different designs.

Worked Example: Re-Planning With Twice as Many Treatment Participants

Suppose \(q=0.20\), the target treatment-to-control odds ratio is 2, and the planned allocation ratio is \(r=2\). Recover the arm probabilities and calculate the approximate sample size for \(\alpha=0.05\) and power \(0.80\).

Now \(C=(1+2)(0.20)=0.60\), so the quadratic is:

p_0^2+(1+2(2)-0.60)p_0-0.60=0, \qquad p_0^2+4.40p_0-0.60=0.

The admissible root is \(p_0=[-4.40+\sqrt{(4.40)^2+4(0.60)}]/2\approx0.1324\). Therefore:

p_1=\frac{2(0.1324)}{1+0.1324}\approx0.2338, \qquad \frac{0.1324+2(0.2338)}{3}\approx0.2000.

The variance coefficient for \(n_0=n\) is:

A= \left(\frac{1}{0.1324}+\frac{1}{0.8676}\right) +\frac{1}{2}\left(\frac{1}{0.2338}+\frac{1}{0.7662}\right) \approx 11.50.

Thus \(n_0\approx11.50(1.960+0.8416)^2/(0.6931)^2\approx187.8\). Round up to 188 controls and allocate 376 treatment participants. This is 564 participants in total. The weighted-rate check uses the 1:2 control-to-treatment allocation, not an equal-arm average.

Assumptions, Diagnostics, and Alternatives

Before applying the update, verify that the pooled rate corresponds to the participants and allocation used in the planning calculation, and that the target odds ratio is expressed as treatment odds divided by control odds. Confirm that the outcome definition and follow-up period have not changed. The procedure assumes a common binary-outcome probability within each arm and uses independent Bernoulli outcomes; clustering or repeated outcomes require a different variance model.

Several checks can reveal a calculation problem:

Reconstruct the pooled rate: verify that \((p_0+rp_1)/(1+r)\) matches the masked rate.
Check the odds-ratio direction: calculate \([p_1/(1-p_1)]/[p_0/(1-p_0)]\). If the result is \(1/\theta\), the arm labels or odds-ratio definition have been reversed.
Inspect the cell counts implied by the plan: very small expected event or non-event counts can make the normal Wald approximation unreliable. A calculated sample size should not be treated as a remedy for that limitation.
Check allocation consistently: use the same \(r\) in the mixture equation, variance, and final arm totals. If observed enrollment departs from the target ratio, the actual pooled rate may not represent the planned mixture without adjustment.

The approach is useful when retaining treatment masking is important and the target odds ratio is part of the prespecified design. It differs from planning with a fixed control event rate: here the control rate is inferred from the pooled rate and assumed odds ratio. It also differs from using unmasked arm-specific rates, which can directly inform both probabilities but reveals treatment-specific interim information. On the risk-difference scale, a pooled rate and a target difference lead to a different probability-recovery calculation; do not substitute a risk difference for the odds ratio in these equations.

The update has important limits. It inherits the assumed odds ratio, so it does not protect against a true treatment effect that differs from the design target. Masked pooling also cannot identify that discrepancy. In addition, Wald sample-size calculations can be inaccurate when expected cell counts are low or probabilities are close to 0 or 1; in those settings, a design-specific method with more reliable small-sample performance may be needed. Prespecify the nuisance-estimation rule, the effect assumption, and how the resulting total sample size will be used.

Common pitfall: Treating the pooled rate as the control rate, or splitting it evenly between arms, ignores the assumed treatment effect. For instance, with equal allocation, pooled rate 0.20, and odds ratio 2, setting both arm rates to 0.20 contradicts the target odds ratio. Instead, solve the weighted mixture and odds-ratio equations together, then verify both quantities.

Check Your Understanding

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

With equal allocation, pooled event probability \(q=0.30\), and target odds ratio \(\theta=1\), what are the arm-specific planning probabilities?
For \(q=0.20\), \(\theta=2\), and \(r=1\), explain why the pooled probability is not the control probability.
For a treatment-to-control odds ratio of 0.5, is the treatment planning probability above or below the control probability? Explain.
In the sample-size formula, why is the treatment-arm variance contribution divided by \(r\)?
If 80 participants per arm have accrued and the updated target is 218 per arm, how many additional participants per arm are required, before any attrition allowance?

Solutions. (1) With \(\theta=1\), \(p_0=p_1=q=0.30\). (2) The target odds ratio of 2 implies \(p_1=2p_0/(1+p_0)\), so the two arm probabilities differ; for equal allocation, their average, not either arm rate, is 0.20. (3) The treatment probability is below the control probability: multiplying the control odds by 0.5 gives smaller odds and hence a smaller probability. (4) With \(rn\) treatment participants, the variance of the treatment contribution is proportional to \(1/(rn)=(1/r)(1/n)\). (5) The target is a total of 218 per arm, so \(218-80=138\) additional participants per arm are required.