Bayesian Assurance for Equivalence Study Size

Why Average Power Over Planning Uncertainty?

A conventional power calculation evaluates the probability of success at specified values for the true test-to-reference ratio and within-subject variability. Those values are planning assumptions, not known facts. If several combinations are plausible, a design can have high power under one combination and much lower power under another. Bayesian assurance summarizes this uncertainty by averaging conditional power over a prior distribution for the unknown quantities.

Assurance is useful when the design question is not simply, “What is the power if the true ratio is exactly this and the variability is exactly that?” Instead, it asks, “Before observing the study data, how often would this design meet its equivalence criterion, averaged over the planning distribution?” The answer depends on the prior distribution as well as the test, model, and sample size; it is not a property of sample size alone.

Definition: Let $\theta$ be the true mean log test-to-reference ratio, $\sigma$ the within-subject standard deviation of the subject-level log contrasts, and $n$ the number of evaluable subjects. If $\pi(\theta,\sigma)$ is their joint prior distribution and $\operatorname{Power}(n;\theta,\sigma)$ is the conditional probability of passing the equivalence test, then the assurance is $$ A(n)=\int\!\!\int \operatorname{Power}(n;\theta,\sigma)\,\pi(\theta,\sigma)\,d\theta\,d\sigma =E_{\pi}\!\left[\operatorname{Power}(n;\theta,\sigma)\right]. $$ For a discrete prior with scenario weights $w_j$, this becomes $A(n)=\sum_j w_j\operatorname{Power}(n;\theta_j,\sigma_j)$, where $w_j\geq 0$ and $\sum_j w_j=1$.

Conditional Power for an Equivalence Test

For a transparent planning calculation, consider a normal approximation for the estimated mean log contrast $\bar{x}$: $$ \bar{x}\mid\theta,\sigma \sim N\!\left(\theta,\frac{\sigma^2}{n}\right), \qquad SE=\frac{\sigma}{\sqrt{n}}. $$ Here $\bar{x}$ and $\theta$ are on the natural-log scale, and $\sigma$ is the standard deviation of subject-level log contrasts, also on the log scale. The model assumes independent, identically distributed subject-level contrasts with a common variance $\sigma^2$, a fixed evaluable sample size, and a known $\sigma$ for the conditional calculation. In practice, $\sigma$ is estimated; the normal calculation is therefore a planning approximation, particularly at small sample sizes.

Let the equivalence limits on the log scale be $-\Delta$ and $+\Delta$. For the common ratio limits 0.80 and 1.25, $\Delta=\log(1.25)$, and $\log(0.80)=-\Delta$. At a one-sided significance level $\alpha$, the two one-sided tests (TOST) require rejecting both boundary null hypotheses: $\theta\leq-\Delta$ and $\theta\geq\Delta$. Write $z_{1-\alpha}$ for the $1-\alpha$ quantile of the standard normal distribution.

Result: Under the stated normal model, TOST passes exactly when $$ -\Delta+z_{1-\alpha}SE < \bar{x} < \Delta-z_{1-\alpha}SE. $$ Thus, when this interval is nonempty, conditional power is $$ \operatorname{Power}(n;\theta,\sigma) = \Phi\!\left(\frac{\Delta-z_{1-\alpha}SE-\theta}{SE}\right) - \Phi\!\left(\frac{-\Delta+z_{1-\alpha}SE-\theta}{SE}\right), $$ where $\Phi$ is the standard normal cumulative distribution function. If $\Delta\leq z_{1-\alpha}SE$, the acceptance interval is empty or has zero probability, so the power is zero.

The formula follows directly from the probability that the normally distributed $\bar{x}$ falls between the two TOST cutoffs. Standardizing each cutoff by subtracting its mean $\theta$ and dividing by $SE$ gives the upper-tail cumulative probability minus the lower-tail cumulative probability. This derivation also shows why the true value matters: even at the same $n$ and $\sigma$, moving $\theta$ toward or beyond an equivalence boundary changes the chance that $\bar{x}$ falls in the acceptance interval.

With $\alpha=0.05$, $z_{1-\alpha}=z_{0.95}\approx1.645$. The corresponding two one-sided tests are often described through a 90% confidence interval, but the calculation here is the probability of passing both tests—not the probability that the parameter is in an interval. For a study requiring multiple endpoints, the assurance must reflect the actual joint success rule; averaging separate endpoint powers does not by itself give the probability that all endpoints pass.

Worked Discrete-Prior Calculation

Worked Example: Averaging Four Planning Scenarios

Suppose a hypothetical crossover study uses ratio limits 0.80 to 1.25, $\alpha=0.05$, and the normal approximation above. Consider four equally weighted prior scenarios: $(\theta,\sigma)=(0,0.16),(0.06,0.16),(0,0.24),(0.06,0.24)$. The values of $\sigma$ are log-scale standard deviations, not coefficients of variation. The prior assigns probability $1/4$ to each scenario. We compare $n=13$ and $n=14$ evaluable subjects.

The margin is $\Delta=\log(1.25)=0.2231$. For each scenario, compute $SE=\sigma/\sqrt{n}$, then substitute in the conditional-power formula. For example, at $n=13,\theta=0,\sigma=0.16$, $SE=0.16/\sqrt{13}=0.04438$. The acceptance interval for $\bar{x}$ is approximately $(-0.1501,0.1501)$, so its conditional power is approximately 0.9993.

Applying the same calculation to all four scenarios gives:

True log ratio $\theta$	Within-subject SD $\sigma$	Power at $n=13$	Power at $n=14$
0.00	0.16	0.9993	0.9996
0.06	0.16	0.9789	0.9850
0.00	0.24	0.9123	0.9333
0.06	0.24	0.7854	0.8128

Since the scenario weights are equal, the assurance at $n=13$ is the arithmetic mean: $$ A(13)=\frac{0.9993+0.9789+0.9123+0.7854}{4} =\frac{3.6759}{4}=0.918975\approx0.9190. $$ At $n=14$: $$ A(14)=\frac{0.9996+0.9850+0.9333+0.8128}{4} =\frac{3.7307}{4}=0.932675\approx0.9327. $$ For an assurance target of 0.93, $n=13$ falls below the target and $n=14$ exceeds it under this prior and model. A full integer sample-size search would evaluate the candidate sizes using the same assumptions; assurance is not guaranteed to increase with $n$ for every possible prior, especially if it assigns weight to true values outside the equivalence limits.

The conclusion is conditional on the specified four-point prior. It does not mean there is a 93.27% probability that the true ratio is equivalent. Rather, before data are collected, the prior-weighted probability that the study passes the test is about 93.27%, assuming the model and planning distribution are appropriate.

How Prior Choices Change the Design

The prior distribution is part of the assurance calculation and should be justified using relevant earlier evidence, expert elicitation, or a transparent planning convention. It can describe uncertainty in $\theta$, in $\sigma$, or in both. A joint prior can also represent dependence: for example, plausible values of the true ratio and variability need not be independent. When using a discrete prior, weights encode how much probability is assigned to each scenario; they are not merely a list of equally important “what-if” cases unless equal weighting is intended.

Worked Example: Changing the Prior Weights

Use the four conditional powers at $n=14$ from the previous calculation, but assign weights 0.40, 0.30, 0.20, and 0.10 in the table’s row order. These weights still sum to 1, but put more prior weight on the two lower-variability scenarios.

The assurance is the weighted sum: $$ A(14)=0.40(0.9996)+0.30(0.9850)+0.20(0.9333)+0.10(0.8128) $$ $$ =0.39984+0.29550+0.18666+0.08128=0.96328\approx0.9633. $$ The resulting assurance is higher than the equal-weight value of 0.9327. The conditional powers have not changed; only the prior weighting has changed. This is why assurance should always be reported with its prior assumptions, not as though it were an assumption-free property of the design.

Worked Example: Prior Weight Outside the Equivalence Limits

Now consider an additional planning scenario with $\theta=0.20$, $\sigma=0.16$, and $n=14$. Its true ratio is $\exp(0.20)\approx1.221$, which is inside the upper ratio limit of 1.25 but close to it. Here $SE=0.16/\sqrt{14}\approx0.04276$, and the acceptance interval for $\bar{x}$ is approximately $(-0.1528,0.1528)$. Its conditional power is $$ \Phi\!\left(\frac{0.1528-0.20}{0.04276}\right) -\Phi\!\left(\frac{-0.1528-0.20}{0.04276}\right) \approx \Phi(-1.104)-\Phi(-8.25)\approx0.135. $$

Suppose, illustratively, that 10% of prior probability is assigned to this scenario and the remaining 90% is distributed equally across the original four. The assurance becomes $$ A(14)=0.90(0.93265)+0.10(0.135) =0.852885\approx0.8529. $$ A true value near a boundary can therefore reduce assurance substantially. If prior probability instead lies outside the equivalence limits, its conditional power tends toward zero as $n$ grows, because increasingly precise estimates concentrate around a value that cannot satisfy equivalence. More subjects do not make an inequivalent true parameter equivalent.

Assumptions, Diagnostics, and Alternatives

A defensible assurance calculation makes its assumptions explicit. In this normal approximation, the subject-level log contrasts are independent and share a common variance; the mean estimator is normal with standard error $\sigma/\sqrt{n}$; the endpoint, equivalence limits, significance level, and success rule are specified in advance; and $n$ represents evaluable subjects. The calculation does not automatically account for dropout, missing or nonevaluable profiles, deviations from the planned analysis, or uncertainty in estimating $\sigma$. Those features need to be incorporated in the planning model rather than silently ignored.

Several checks can reveal a fragile or incorrect calculation:

Check the scale. The ratio limits must be converted to log limits before applying the log-scale model. Do not insert a ratio such as 1.05 as though it were a log contrast.
Check the variability definition. The $\sigma$ in this formula is the SD of the modeled subject-level contrast. It must not be replaced with a different variance component or a coefficient of variation without an appropriate conversion.
Check the acceptance interval. If $\Delta-z_{1-\alpha}SE\leq0$, the interval is empty or has zero width. A reported positive power in that case signals a formula or implementation error.
Check scenario weights. They must be nonnegative and sum to one. Recalculate a few conditional powers independently before trusting a weighted average.
Check prior sensitivity. Report how assurance changes under reasonable alternatives for the location, spread, and weights of the prior. A single precise-looking result can conceal substantial dependence on subjective assumptions.

The principal alternative is a conventional power calculation at one or more fixed values of $\theta$ and $\sigma$. It is easier to explain and is useful for showing sensitivity, but does not average over uncertainty. A robust or conservative design can instead require a target power across a specified range of planning values; this emphasizes protection in less favorable scenarios rather than their prior probability. Assurance answers a different question: it averages according to the chosen prior. These approaches are complementary, not interchangeable.

For small samples, the normal formula may be inaccurate because variability is estimated and the relevant test uses a $t$ distribution and finite-sample degrees of freedom. A practical calculation can evaluate the appropriate finite-sample conditional power at each prior point and then average those powers. More complex models or study rules may require numerical integration or simulation. Regardless of computational method, uncertainty in the prior and model remains; numerical precision cannot remove it.

Common Mistake

Pitfall: “The assurance is 0.93, so there is a 93% probability that the true test-to-reference ratio is equivalent.” This confuses a prior-averaged probability of passing a specified study procedure with a probability statement about the parameter itself. Assurance averages the sampling probability of passing over the planning prior. A posterior probability of equivalence is a different quantity, calculated after observing data using a specified prior and likelihood.

Check Your Understanding

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

For $n=16$, $\sigma=0.20$, and $\alpha=0.05$, calculate $SE$ and the upper endpoint of the TOST acceptance interval when $\Delta=\log(1.25)$. Give the endpoint on the log scale.
In one sentence, interpret an assurance of 0.85 without making a probability claim about the true parameter.
At a fixed sample size, two scenarios have conditional powers 0.90 and 0.60. Their prior weights are 0.70 and 0.30. Calculate assurance.
Why can assurance decrease if a revised prior assigns weight to true ratios outside the equivalence limits?

Solutions. (1) $SE=0.20/\sqrt{16}=0.0500$. The upper endpoint is $\Delta-z_{0.95}SE=0.2231-1.645(0.0500)=0.1409$ on the log scale. (2) Under the stated prior, model, and design, the prior-averaged probability that the study passes its equivalence test is 0.85. (3) $A=0.70(0.90)+0.30(0.60)=0.63+0.18=0.81$. (4) Scenarios outside the limits have low conditional power, which reduces the weighted average; with increasing precision, a true value outside the limits is increasingly unlikely to produce an estimate within the acceptance region.