Power Under Point and Mixture Design Alternatives

Two Questions About Design Performance

A trial’s success rule may be Bayesian: for example, declare success when the posterior probability that the treatment effect exceeds a clinically meaningful threshold is greater than a prespecified value. Before the trial, a separate question is how often that rule would declare success under specified assumptions about the true treatment effect. The answer depends on which effect is assumed.

A point alternative specifies one fixed true effect. Its power describes performance at that particular value. A discrete mixture of design alternatives specifies several possible effects and assigns each a weight; its power averages performance across those scenarios. These summaries answer different design questions. Neither is universally preferable, and a mixture average can hide weak performance in an important scenario.

Definition: For a fixed design \(d\), let \(S\) denote the event that the prespecified Bayesian decision rule declares trial success. The point power at true effect \(\theta\) is \(q(\theta;d)=P(S\mid\theta,d)\), where the probability is over the trial data generated under that fixed effect and the design assumptions. For discrete alternatives \(\theta_1,\ldots,\theta_J\) with nonnegative weights \(w_1,\ldots,w_J\) summing to 1, the mixture power is \(\sum_{j=1}^{J}w_jq(\theta_j;d)\).

The definition holds the decision rule and design fixed. If there are nuisance parameters, such as an outcome variance, the calculation must also specify whether power is conditional on fixed nuisance values or averaged over a stated distribution or set of scenarios. The same convention must be used across the alternatives being compared.

From a Bayesian Decision Rule to Point Power

Consider a continuous treatment effect \(\theta\), where positive values favor treatment, and a clinically meaningful threshold \(\theta_{\mathrm{clin}}\). Suppose the rule declares success when the posterior probability of exceeding that threshold is greater than \(\gamma\):

S=\left\{P(\theta>\theta_{\mathrm{clin}}\mid Y)>\gamma\right\}.

Here \(Y\) represents the trial data, and \(\gamma\) is fixed before examining those data. To calculate point power, first determine which possible data sets satisfy this posterior rule. Then calculate the probability, under the sampling model at the chosen true effect, that the data fall in that success region. Thus Bayesian power is a sampling probability for a specified decision rule; it is not the posterior probability of success for the observed trial.

The following normal model provides a complete calculation. Assume the trial statistic satisfies \(\hat{\theta}\mid\theta\sim N(\theta,s^2)\), with known standard error \(s\), and the analysis prior is \(\theta\sim N(m_0,s_0^2)\). The normal likelihood and prior give a normal posterior:

\theta\mid\hat{\theta}\sim N\left( \frac{s_0^2\hat{\theta}+s^2m_0}{s_0^2+s^2}, \frac{s_0^2s^2}{s_0^2+s^2} \right).

This result assumes the stated sampling model, known \(s\), and the stated normal prior. In particular, the prior is the one used in the analysis rule. Define \(a=s_0^2/(s_0^2+s^2)\) and \(v=s_0^2s^2/(s_0^2+s^2)\). The posterior mean is \(m_0+a(\hat{\theta}-m_0)\), and the posterior standard deviation is \(\sqrt{v}\).

For a one-sided success criterion, the posterior probability exceeds \(\gamma\) precisely when the posterior mean exceeds \(\theta_{\mathrm{clin}}+z_\gamma\sqrt{v}\), where \(z_\gamma\) is the \(\gamma\) quantile of the standard normal distribution. Since \(a>0\), this condition can be rearranged to obtain a threshold \(c\) for the observed estimate:

P(\theta>\theta_{\mathrm{clin}}\mid\hat{\theta})>\gamma \quad\Longleftrightarrow\quad \hat{\theta}> m_0+\frac{\theta_{\mathrm{clin}}+z_\gamma\sqrt{v}-m_0}{a} =:c.

Under a fixed true effect \(\theta\), the sampling model gives \(\hat{\theta}\sim N(\theta,s^2)\). Therefore the probability of exceeding the decision threshold is

q(\theta;d)=P(\hat{\theta}>c\mid\theta,d) =1-\Phi\left(\frac{c-\theta}{s}\right) =\Phi\left(\frac{\theta-c}{s}\right),

where \(\Phi\) is the standard normal cumulative distribution function. This derivation separates two ingredients that are sometimes conflated: the analysis prior and posterior rule determine \(c\), while the sampling distribution under the design alternative determines the probability of crossing it.

Worked Example: Point Power at a Target Effect

Suppose \(\hat{\theta}\mid\theta\sim N(\theta,0.5^2)\), the analysis prior is \(\theta\sim N(0,1)\), and the trial declares success if \(P(\theta>0\mid\hat{\theta})>0.95\). Calculate power at a true treatment effect of 1.0 outcome units. Assume the normal sampling model and known standard error are appropriate, and that the prior and rule are fixed as stated.

Here \(s=0.5\), \(m_0=0\), \(s_0^2=1\), \(\theta_{\mathrm{clin}}=0\), and \(z_{0.95}=1.6449\). The posterior variance and coefficient are

v=\frac{1(0.25)}{1+0.25}=0.2,\qquad a=\frac{1}{1.25}=0.8.

The success rule requires \(\hat{\theta}>c\), with

c=0+\frac{0+1.6449\sqrt{0.2}-0}{0.8} =\frac{1.6449(0.4472)}{0.8} \approx 0.9195.

At the point alternative \(\theta=1.0\), the statistic has distribution \(N(1.0,0.5^2)\), so

q(1.0;d)=P(\hat{\theta}>0.9195\mid\theta=1.0) =\Phi\left(\frac{1.0-0.9195}{0.5}\right) =\Phi(0.1610) \approx 0.5639.

Thus, under these assumptions, about 56.4% of repeated trials with a true effect of 1.0 outcome units would meet this Bayesian success rule. This is not a statement that the posterior probability of benefit in any one trial is 56.4%.

Mixture Power Across Discrete Alternatives

A mixture summary uses a specified set of plausible true effects. Its weights must be nonnegative and add to one. The calculation is a weighted average of the point powers—not a new posterior calculation and not a change to the trial’s success rule. If the true effect is itself modeled as being drawn from this discrete distribution, the average is the probability of success under that marginal data-generating model. Otherwise, the weights can be understood as design-scenario weights for summarizing performance.

Worked Example: Averaging Power Over Three Scenarios

Use the trial model and success rule from the previous calculation. Suppose the design scenarios are true effects of 0.6, 1.0, and 1.4 outcome units, with weights 0.25, 0.50, and 0.25. These weights sum to 1, and the same known standard error, prior, and success rule apply in all three scenarios.

The threshold remains \(c=0.9195\). Applying \(q(\theta;d)=\Phi((\theta-c)/0.5)\) gives

\begin{aligned} q(0.6;d)&=\Phi\left(\frac{0.6-0.9195}{0.5}\right) =\Phi(-0.6390)\approx 0.2614,\\ q(1.0;d)&=\Phi\left(\frac{1.0-0.9195}{0.5}\right) =\Phi(0.1610)\approx 0.5639,\\ q(1.4;d)&=\Phi\left(\frac{1.4-0.9195}{0.5}\right) =\Phi(0.9610)\approx 0.8317. \end{aligned}

The mixture power is

0.25(0.2614)+0.50(0.5639)+0.25(0.8317) =0.0654+0.2819+0.2079 \approx 0.5552.

The design therefore has mixture power of about 55.5% under these specific scenario weights. Each point power answers a conditional question; the mixture value answers the weighted-average question. Because the middle scenario has half the weight, the mixture result is close to, but not identical to, point power at 1.0.

Mixture power is generally not the point power at the weighted-average effect. The reason is that \(q(\theta;d)\) is usually nonlinear in \(\theta\). Averaging its values and evaluating it at an average input are different operations.

Worked Example: Why the Average Effect Does Not Give Mixture Power

Again use \(c=0.9195\) and \(s=0.5\). Consider two alternatives: \(\theta_1=0.4\) with weight 0.70 and \(\theta_2=1.2\) with weight 0.30. The point powers are

q(0.4;d)=\Phi\left(\frac{0.4-0.9195}{0.5}\right) =\Phi(-1.0390)\approx0.1495, \qquad q(1.2;d)=\Phi\left(\frac{1.2-0.9195}{0.5}\right) =\Phi(0.5610)\approx0.7127.

The mixture power is

0.70(0.1495)+0.30(0.7127) =0.1047+0.2138 \approx0.3185.

The weighted-average effect is \(0.70(0.4)+0.30(1.2)=0.64\) outcome units. Point power at that effect is instead

q(0.64;d)=\Phi\left(\frac{0.64-0.9195}{0.5}\right) =\Phi(-0.5590)\approx0.2880.

The values, 0.3185 and 0.2880, differ. A power calculation at the average effect does not preserve the weighted average of the scenario-specific success probabilities.

Choosing and Comparing Summaries

Point power is useful when a single effect is the primary design target—for example, a clinically important effect at which the team wants to evaluate the chance of meeting its success rule. It is easy to communicate and makes explicit which effect supports the reported performance. But it says nothing by itself about performance at smaller, larger, or otherwise plausible effects.

Mixture power is useful when design performance across several plausible scenarios is the question of interest. It can summarize a range of effects in one number, provided the scenarios and weights are justified and reported. A mixture average is not a guarantee of acceptable performance in each scenario: a high value for a favorable scenario can offset low power where the trial team or decision makers have particular concerns.

Summary	Question answered	Key design choice	Main caution
Point power \(q(\theta;d)\)	How often does the rule succeed if the true effect equals this specified value?	The single effect value	May not represent performance elsewhere
Mixture power \(\sum_j w_jq(\theta_j;d)\)	What is the weighted average success probability across these alternatives?	The scenarios and their weights	Can conceal poor performance in an important scenario
Prior-averaged assurance	What is success probability averaged over a specified design prior for unknown effects?	The design prior and its interpretation	It is not automatically the same as an arbitrary discrete scenario average

The last row is included to distinguish mixture power from assurance. Assurance is a prior-averaged success probability: its averaging distribution is a design prior, with the role described in the earlier material on assurance. A finite collection of scenarios can be used as a discrete design prior, in which case the resulting average has that interpretation. But weights chosen simply to represent planning scenarios should not be called prior probabilities unless that is how they are intended and justified.

A useful design report can provide both a point-power curve or selected point values and a mixture summary. The point values show how performance changes across effects; the mixture gives one weighted summary for the stated scenarios. If scenarios are few, report their individual powers alongside the mixture so readers can see whether the average is masking a consequential weak case.

Diagnostics, Assumptions, and Common Pitfalls

Before interpreting either summary, check that the calculation matches the intended trial and decision rule. In the normal illustration, the standard error is known and constant, the sampling distribution is normal, the analysis prior is normal, and the success rule is a one-sided posterior probability threshold. If the actual design has an estimated variance, a time-to-event endpoint, or a different decision rule, the threshold and sampling probabilities must be derived for that design rather than borrowed from the illustration.

For mixture power, also check that every weight is nonnegative, that the weights sum to 1, and that the scenarios use a consistent definition of the treatment effect and nuisance assumptions. A practical diagnostic is to tabulate each scenario’s weight and point power before computing the weighted sum. This makes it easier to detect a misplaced weight, a scenario omitted from the average, or a high average that conceals low power in a clinically relevant case.

Using the analysis prior as scenario weights without justification. The analysis prior determines the posterior decision rule; mixture weights determine which true-effect scenarios contribute to the design summary. They can be the same distribution if that is deliberate, but one does not automatically determine the other.
Calling mixture power the power at the mean effect. This is wrong whenever the power function is nonlinear. Calculate each scenario’s power first, then weight those probabilities.
Reporting only the mixture average. A favorable average does not establish adequate performance at every alternative. Show scenario-specific values when any one scenario matters for the design decision.
Changing assumptions across scenarios without saying so. If, for example, the variance changes with the true effect, each point power must use the corresponding sampling distribution. State whether the mixture averages over effect alone or over effect and nuisance parameters.

The central design choice is not simply which formula to use. It is which performance question the trial team needs answered. A point alternative makes one target explicit; a mixture makes both the range of alternatives and the trade-off encoded by its weights explicit. The assumptions, decision rule, scenario values, and weights should all be reported so the summary can be interpreted and reproduced.

Check Your Understanding

Use the definitions and calculations above to answer each question. Solutions follow each item.

A trial has point powers 0.30 and 0.70 at two effects. What is mixture power if the corresponding weights are 0.25 and 0.75?
Solution: The weights sum to 1, so the mixture power is \(0.25(0.30)+0.75(0.70)=0.075+0.525=0.600\).
In the normal example, why does the posterior success rule reduce to a threshold on \(\hat{\theta}\)?
Solution: The posterior mean is \(m_0+a(\hat{\theta}-m_0)\) with \(a>0\), while its standard deviation is fixed by the known \(s\) and prior variance. The posterior probability exceeds \(\gamma\) exactly when this mean exceeds \(\theta_{\mathrm{clin}}+z_\gamma\sqrt{v}\). Since the coefficient of \(\hat{\theta}\) is positive, rearranging gives the equivalent condition \(\hat{\theta}>c\).
For two equally weighted effects, is mixture power necessarily the point power at their midpoint?
Solution: No. Mixture power is \(\{q(\theta_1;d)+q(\theta_2;d)\}/2\), while power at the midpoint is \(q((\theta_1+\theta_2)/2;d)\). These are equal only in special cases; the normal-model calculation above demonstrates that a nonlinear power function need not preserve averages.
A design team gives three scenarios weights 0.2, 0.3, and 0.4. Can these be used directly as mixture weights?
Solution: Not as stated, because they sum to 0.9 rather than 1. The team must explain the missing 0.1—for example, by adding an omitted scenario—or deliberately renormalize the weights if that matches the intended question. Renormalizing gives weights \(0.2/0.9\), \(0.3/0.9\), and \(0.4/0.9\); that changes their interpretation and should not be done silently.
Why might a design report both point powers and mixture power?
Solution: Point powers show performance conditional on particular true effects, including important weak-performance scenarios. Mixture power provides a single weighted summary across the stated alternatives. Reporting both makes the average interpretable and reveals whether it conceals poor performance in a scenario that matters.