Why Predictive Success Depends on Future Information
During a trial, the treatment-effect estimate is uncertain, and a fixed number of participants remain to provide additional information. Predictive success asks how likely the final estimate is to meet a specified criterion, averaging over both uncertainty about the true treatment effect and the future sampling variability. It is therefore not the same as asking whether the current estimate already exceeds the criterion, or whether the current posterior probability of benefit is large.
This tutorial considers a continuous treatment effect with a normal posterior and a normal likelihood for the remaining data. The future sample sizes are fixed. To make the success event concrete, let larger values of the effect be better, and suppose trial success means the final posterior mean exceeds a prespecified boundary \(c\), measured in the same units as the treatment effect. This is a criterion on the final estimate; other terminal decision rules require defining a different success event.
Model for the Remaining Data
Assume the final analysis uses the same treatment-effect estimand as the current analysis. For a two-arm trial with independent normally distributed outcomes and known arm-specific standard deviations, write \(Z\) for the treatment-minus-control estimate based only on the remaining participants. If \(n_{T,r}\) and \(n_{C,r}\) remain in the treatment and control groups, respectively, then
Here \(Z\) is an unbiased future-data estimate conditional on \(\Delta\), and \(\tau^2\) is its sampling variance. The model assumes the future observations are independent of the observed data conditional on \(\Delta\), and that the remaining sample sizes and analysis plan are fixed. The posterior \(N(m,v)\) already incorporates the prior and the observed data; it is not an additional likelihood to be counted again.
The final posterior combines the current posterior with the likelihood from \(Z\). Because both are normal with known variances, its posterior mean is a precision-weighted average:
Deriving the Predictive Distribution
The final posterior mean is not known at the interim analysis because it depends on future data. To find its distribution given the current data, first average the future-data model over the current posterior for \(\Delta\). Then transform that future estimate into the final posterior mean.
Since \(Z\mid\Delta\sim N(\Delta,\tau^2)\) and \(\Delta\mid D\sim N(m,v)\), the predictive distribution is \(Z\mid D\sim N(m,v+\tau^2)\). The variances add because \(Z=\Delta+\epsilon\), where the future sampling error \(\epsilon\) has variance \(\tau^2\) and is independent of \(\Delta\) given the current information.
The update formula gives \(m_f=m+\frac{v}{v+\tau^2}(Z-m)\). Thus the final posterior mean is a linear transformation of the predictive future estimate.
A linear transformation of a normal variable is normal. Therefore \(m_f\mid D\sim N\left(m,\frac{v^2}{v+\tau^2}\right)\), and \(P(m_f>c\mid D)=\Phi\left(\frac{m-c}{v/\sqrt{v+\tau^2}}\right)\), where \(\Phi\) is the standard normal cumulative distribution function.
The variance in this predictive distribution can also be checked directly. Since \(Z\mid D\) has variance \(v+\tau^2\), the random part of \(m_f=m+\frac{v}{v+\tau^2}(Z-m)\) has variance
This is distinct from the final posterior variance \(v_f=v\tau^2/(v+\tau^2)\). The predictive variance describes how much the final posterior mean may vary across possible future data sets. The final posterior variance describes remaining uncertainty about \(\Delta\) after one particular future data set has been observed. Confusing these two variances gives an incorrect predictive probability.
Worked Calculations
Worked Example: A Two-Arm Trial With Equal Remaining Groups
Suppose the treatment effect is measured in points, with positive values favoring treatment. At an interim analysis, the posterior is \(\Delta\mid D\sim N(4,4)\), so \(m=4\) points and \(v=4\) points squared. Assume known outcome standard deviations of 10 points in both arms, with 50 participants remaining per arm. Success is defined as a final posterior mean greater than 3 points.
First calculate the sampling variance of the estimate from the remaining participants:
The predictive standard deviation of the final posterior mean is
Thus \(m_f\mid D\sim N(4,2)\), where 2 is the predictive variance in points squared. The predictive probability of crossing 3 points is
Under these assumptions, the chance that the final posterior mean exceeds 3 points is about 76.0%. This is not a 76.0% probability that the true treatment effect exceeds 3 points; it is a probability over the as-yet-unobserved future data, conditional on the current posterior and the planned remaining sample.
Worked Example: Less Precise Remaining Data
Suppose \(m=2.5\) points and \(v=4\) points squared. Each arm has 36 participants remaining, and the known outcome standard deviation is 12 points in each arm. The terminal success boundary is \(c=3\) points.
The variance of the remaining-data treatment-effect estimate is
The predictive standard deviation of the final posterior mean is
Therefore,
The estimated chance of exceeding the 3-point boundary is about 33.2%. The current posterior mean is below the boundary, but future evidence can move the final posterior mean in either direction. The result depends on the remaining sample through \(\tau^2\), not solely on the interim posterior.
Worked Example: A Strong Interim Result Can Still Have Low Predictive Success
Suppose the current posterior is \(\Delta\mid D\sim N(1,1)\), with effects measured in points. There are 25 participants remaining in each arm, the known standard deviation is 10 points in both arms, and success requires a final posterior mean above 1.5 points.
The remaining-data variance is \(\tau^2=100/25+100/25=8\) points squared. The predictive standard deviation of the final posterior mean is
The predictive success probability is
Although the current posterior mean is positive, there is only about a 6.68% predictive chance that the final posterior mean will exceed 1.5 points. The remaining sample is relatively informative, so the final posterior mean is expected to stay near its current value rather than move far enough to cross the higher boundary.
Assumptions, Diagnostics, and Alternatives
The calculation is most useful when its model matches the trial’s design and analysis. In particular, the estimand must remain the same, the outcome model must be adequately normal with known variances, and the remaining group sizes must be fixed. For a two-arm comparison, the formula for \(\tau^2\) must use the planned remaining sample in each arm and the correct arm-specific variances. The treatment-effect orientation and the direction of the success boundary must also agree.
Potential warning signs include highly skewed or heavy-tailed outcomes, influential outliers, variance estimates that are materially uncertain, a changing allocation plan, substantial anticipated missing data, or a success definition that depends on more than the final posterior mean. In those settings, the normal calculation may not represent the actual predictive distribution. A richer model may be needed to represent uncertain variances, missingness, or changes to the design. Simulation can also be used when an appropriate model is available but the predictive probability is not available in closed form.
A tempting shortcut is to treat the current posterior mean \(m\) as if it were the known true effect and generate future data only from \(Z\mid \Delta=m\). That omits uncertainty in \(\Delta\mid D\). The shortcut can make the predictive distribution too narrow and the estimated chance of crossing a boundary too extreme. For example, in the first worked example, it would give a predictive standard deviation for \(m_f\) of \(1.000\) point rather than the correct \(1.414\) points. The current posterior probability of exceeding \(c\) is another different quantity: it concerns \(\Delta\) given current data, not whether the final estimate will cross the boundary after more data arrive.
This method also differs from calculating the predictive distribution of the future-only estimate \(Z\). In the first example, \(Z\mid D\sim N(4,8)\), while the final posterior mean has distribution \(m_f\mid D\sim N(4,2)\). The future-only estimate is noisier because it does not combine the existing information with the remaining data. Use the distribution corresponding to the quantity that actually defines success.
Using the Result Responsibly
Predictive success is conditional on the current model, current data, and planned remaining sample. It describes the chance of meeting a defined future criterion if those assumptions and the plan hold. It does not by itself determine whether a trial should continue: that action depends on the consequences of decisions, the clinical value of the effect, and the decision rule. This tutorial’s calculation is specifically for a fixed remaining sample and a final posterior-mean boundary; uncertain sample size or a different terminal rule requires a correspondingly different predictive calculation.
Check Your Understanding
For each question, use the normal-normal model above and interpret probabilities as conditional on the current data and the stated future design.
- The current posterior is \(N(2,1)\), the remaining-data variance is \(\tau^2=3\), and success means \(m_f>2.5\). Find the predictive standard deviation and predictive success probability.
- With \(m=3\), \(v=4\), and \(\tau^2=5\), what is the predictive distribution of the final posterior mean? State its mean and variance.
- For a two-arm trial, each arm has 40 participants remaining and known standard deviation 8 units. Find \(\tau^2\). If \(m=1\), \(v=4\), and \(c=1.5\), calculate predictive success.
- Explain why the current posterior probability \(P(\Delta>c\mid D)\) is not the same as \(P(m_f>c\mid D)\).
Solutions. (1) The predictive standard deviation is \(v/\sqrt{v+\tau^2}=1/\sqrt{4}=0.5000\). Thus \(P(m_f>2.5\mid D)=\Phi((2-2.5)/0.5)=\Phi(-1)\approx 0.1587\). (2) The predictive mean is \(m=3\), and the variance is \(v^2/(v+\tau^2)=16/9=1.778\), so \(m_f\mid D\sim N(3,1.778)\). (3) The remaining-data variance is \(8^2/40+8^2/40=3.2\) units squared. The predictive standard deviation is \(4/\sqrt{4+3.2}=1.491\) units, so success probability is \(\Phi((1-1.5)/1.491)=\Phi(-0.3354)\approx 0.369\). (4) The first probability concerns the unknown treatment effect given current data. The second concerns the future final posterior mean across possible remaining data sets. They involve different random quantities and answer different questions.