Predictive Success With a Fixed Number of Remaining Participants

Why the Number Remaining Matters

At an interim analysis, the observed participants’ outcomes are known, but the remaining participants’ outcomes are not. If the number of participants still to be enrolled and assessed is fixed, the final estimate is a weighted combination of the current data and a random future average. The weight and uncertainty contributed by that future average depend on how many participants remain.

This tutorial derives that predictive distribution for a continuous endpoint under a normal model. It then uses the distribution to calculate the probability that a final estimate will satisfy a Bayesian go rule. The calculation is conditional on the interim data and on the planned number of remaining participants; it is not a probability that the treatment effect itself is above a threshold.

Model and assumptions: Let \(Y_i\) be a continuous endpoint for participant \(i\), and let \(\theta\) be its population mean. Assume \(Y_i\mid\theta\sim N(\theta,\sigma^2)\), independently across participants, with known \(\sigma^2\). Use the prior \(\theta\sim N(\mu_0,\tau_0^2)\). At interim, \(n\) participants have endpoint data with sample mean \(\bar y\); exactly \(m\) more participants will have endpoint data, so the final sample size is \(N=n+m\). The model assumes the endpoint is measured on the same scale for all participants, there are no unmodelled dependencies, and the specified number \(m\) have usable outcomes.

The normal-normal updating formulas give an interim posterior \(\theta\mid D_n\sim N(\mu_n,\tau_n^2)\), where \(D_n\) denotes all interim data. Under the assumptions above, the data affect this posterior through \(\bar y\):

\tau_n^2=\left(\frac{1}{\tau_0^2}+\frac{n}{\sigma^2}\right)^{-1}, \qquad \mu_n=\tau_n^2\left(\frac{\mu_0}{\tau_0^2}+\frac{n\bar y}{\sigma^2}\right).

Let \(\bar Y_{\text{future}}\) be the average of the \(m\) future outcomes, when \(m>0\). Conditional on \(\theta\), this average has mean \(\theta\) and variance \(\sigma^2/m\). The final estimate is not just this future average: it combines it with the observed interim mean,

\bar Y_N=\frac{n\bar y+m\bar Y_{\text{future}}}{N}.

The Predictive Distribution of the Final Estimate

The predictive distribution averages over both sources of uncertainty remaining at interim: uncertainty about \(\theta\), represented by the posterior, and the sampling variability of the future participants’ outcomes. Conditional on \(\theta\) and the interim data, the final estimate has mean \((n\bar y+m\theta)/N\) and variance \(m\sigma^2/N^2\). Averaging over the posterior for \(\theta\) gives:

Predictive distribution: For \(m>0\), under the stated model, \( \bar Y_N\mid D_n\sim N\left(\frac{n\bar y+m\mu_n}{N},\frac{m\sigma^2+m^2\tau_n^2}{N^2}\right) \). The first variance term reflects future outcome variability; the second reflects posterior uncertainty about the mean.

A short derivation makes the two variance components explicit. Write \(\bar Y_N\) as the sum of the fixed interim contribution and the future contribution. Its predictive mean is

E(\bar Y_N\mid D_n) =\frac{n\bar y+mE(\bar Y_{\text{future}}\mid D_n)}{N} =\frac{n\bar y+m\mu_n}{N}.

For the variance, the law of total variance separates variability conditional on \(\theta\) from variability in the conditional mean as \(\theta\) varies over its posterior:

\begin{aligned} \operatorname{Var}(\bar Y_N\mid D_n) &=E\{\operatorname{Var}(\bar Y_N\mid\theta,D_n)\mid D_n\} +\operatorname{Var}\{E(\bar Y_N\mid\theta,D_n)\mid D_n\}\\ &=\frac{m\sigma^2}{N^2}+\frac{m^2\tau_n^2}{N^2}. \end{aligned}

The \(m^2\tau_n^2\) term is important: future participants share the same unknown population mean. Treating their outcomes as if that mean were known would omit this source of predictive uncertainty. When \(m=0\), no outcomes remain to be predicted. The final estimate is already \(\bar y\), so its predictive distribution is a point mass at \(\bar y\), not a normal distribution with positive variance.

Worked Example: Finding the Predictive Distribution

Hypothetical trial. A continuous symptom score is analysed in one group, with lower values indicating less symptom burden. Suppose the prior is \(\theta\sim N(0,20^2)\), the outcome standard deviation is known to be \(10\) points, and \(n=20\) participants have a mean score of \(\bar y=4\) points. Exactly \(m=10\) participants remain.

First calculate the interim posterior variance and mean:

\tau_n^2=\left(\frac{1}{400}+\frac{20}{100}\right)^{-1} =\frac{1}{0.2025}=4.938\ \text{points}^2, \qquad \mu_n=4.938\left(\frac{20(4)}{100}\right)=3.951\ \text{points}.

The final sample size is \(N=20+10=30\). The predictive mean of the final estimate is \( (20(4)+10(3.951))/30=3.984 \) points. Its predictive variance and standard deviation are

\operatorname{Var}(\bar Y_{30}\mid D_{20}) =\frac{10(100)+10^2(4.938)}{30^2} =1.660\ \text{points}^2, \qquad SD(\bar Y_{30}\mid D_{20})=\sqrt{1.660}=1.288\ \text{points}.

Thus, conditional on these interim data and ten remaining participants, \(\bar Y_{30}\) has a normal predictive distribution with mean \(3.984\) points and standard deviation \(1.288\) points. This describes plausible values of the final estimate, not the posterior distribution of \(\theta\).

Predictive Success for a Posterior Go Rule

A common final decision rule is to declare a go decision when the final posterior probability that the mean exceeds a target \(\delta\) is at least \(q\). Here \(\delta\) is expressed in endpoint units and \(q\) is a specified probability threshold. With the normal model and a fixed final sample size \(N\), the final posterior variance is \(v_N=(1/\tau_0^2+N/\sigma^2)^{-1}\), regardless of the observed outcomes. If the final sample mean is \(\bar y_N\), the posterior mean is \(v_N(\mu_0/\tau_0^2+N\bar y_N/\sigma^2)\).

The posterior go rule can therefore be written as a cutoff on the final sample mean. Let \(z_q\) be the \(q\) quantile of the standard normal distribution. Since the posterior probability of exceeding \(\delta\) is \( \Phi((\mu_N-\delta)/\sqrt{v_N}) \), the rule is met exactly when \(\mu_N\geq\delta+z_q\sqrt{v_N}\). Solving this inequality for \(\bar y_N\) gives:

\bar y_N\geq c_N, \qquad c_N=\frac{\sigma^2}{N} \left\{ \frac{\delta+z_q\sqrt{v_N}}{v_N} -\frac{\mu_0}{\tau_0^2} \right\}.

The predictive success probability is the probability, under the interim posterior predictive distribution, that the final estimate reaches this cutoff. If \(m>0\), write \(\mu_{\mathrm{pred}}=(n\bar y+m\mu_n)/N\) and \(s_{\mathrm{pred}}^2=(m\sigma^2+m^2\tau_n^2)/N^2\). Then:

P(\text{final go}\mid D_n) =P(\bar Y_N\geq c_N\mid D_n) =1-\Phi\left(\frac{c_N-\mu_{\mathrm{pred}}}{s_{\mathrm{pred}}}\right).

Worked Example: Calculating Predictive Success

Use the hypothetical data and model from the previous example. Suppose success means a final posterior probability of at least \(q=0.90\) that the mean score exceeds \(\delta=0\) points. With \(N=30\), the final posterior variance is

v_{30}=\left(\frac{1}{400}+\frac{30}{100}\right)^{-1} =\frac{1}{0.3025}=3.306\ \text{points}^2, \qquad \sqrt{v_{30}}=1.818\ \text{points}.

The standard normal 0.90 quantile is \(z_{0.90}=1.282\). Because the prior mean is zero, the final sample-mean cutoff is

c_{30}=\frac{100}{30} \left\{\frac{0+1.282(1.818)}{3.306}-0\right\} =2.350\ \text{points}.

From the predictive distribution calculated above, \(\mu_{\mathrm{pred}}=3.984\) points and \(s_{\mathrm{pred}}=1.288\) points. Therefore,

P(\text{final go}\mid D_{20}) =1-\Phi\left(\frac{2.350-3.984}{1.288}\right) =\Phi(1.268) \approx 0.898.

Given the interim data and ten remaining participants, the probability of meeting this specified final go rule is about \(89.8\%\). This is a predictive probability for the trial’s future result under the model; it is not the posterior probability that the mean exceeds zero now.

How the Fixed Number Remaining Changes the Result

Changing \(m\) changes more than the predictive standard deviation. It changes the weight assigned to future outcomes in the final estimate, the final sample size used in the posterior rule, and therefore the cutoff \(c_N\). Predictive success need not increase just because more participants will be observed. For example, if current evidence already meets a go rule, additional outcomes can create a chance of falling below the final cutoff. Conversely, more information can increase success when current evidence is below the cutoff and the expected future data are favourable. There is no universal direction; calculate the predictive probability for the actual planned \(m\).

Worked Example: Comparing Five, Ten, and Twenty Remaining Participants

Hypothetical comparison. Retain the interim data, prior and outcome standard deviation from the first example, and use the same final go rule as the second example: posterior probability at least \(0.90\) that \(\theta>0\). Compare fixed remaining counts \(m=5,10,20\). For each count, recompute both the predictive distribution and the final cutoff.

For \(m=5\), \(N=25\). The predictive mean is \((80+5(3.951))/25=3.990\) points, and the predictive variance is \( (5(100)+25(4.938))/25^2=0.998 \) points squared, giving a standard deviation of \(0.999\) points. The final posterior variance is \(v_{25}=1/(0.0025+0.25)=3.960\) points squared. The cutoff is \( c_{25}=(100/25)\{1.282\sqrt{3.960}/3.960\}=2.576 \) points. Hence success probability is \( \Phi((3.990-2.576)/0.999)=\Phi(1.416)\approx0.922 \).

For \(m=10\), the previous example gave predictive mean \(3.984\) points, predictive standard deviation \(1.288\) points, and cutoff \(2.350\) points. Thus success probability is approximately \(0.898\).

For \(m=20\), \(N=40\). The predictive mean is \((80+20(3.951))/40=3.975\) points. The predictive variance is \( (20(100)+400(4.938))/40^2=2.485 \) points squared, giving a standard deviation of \(1.576\) points. The final posterior variance is \(v_{40}=1/(0.0025+0.40)=2.484\) points squared, and the cutoff is \( c_{40}=(100/40)\{1.282\sqrt{2.484}/2.484\}=2.033 \) points. The success probability is \( \Phi((3.975-2.033)/1.576)=\Phi(1.232)\approx0.891 \).

In this example, the predictive success estimates are about \(92.2\%\), \(89.8\%\), and \(89.1\%\) for five, ten, and twenty remaining participants. They are not ordered simply by the number remaining. Each enrollment plan produces a different final estimate distribution and a different decision cutoff.

Assumptions, Diagnostics, and Alternatives

The closed-form calculation is appropriate only when the model represents the design and endpoint well enough for the decision being considered. Check that the remaining count really is fixed, that future participants’ endpoint data will be observed as assumed, and that the normal sampling model and common known variance are reasonable. For a treatment-effect estimate, the same derivation applies when the estimate has the specified normal likelihood and the interim and future information combine in the stated way; the variance and allocation structure must match that design.

Model mismatch. Strong skewness, heavy tails, outliers, or substantial heterogeneity may make the normal predictive distribution inaccurate. Inspect the endpoint’s scale and plausible outcome patterns; do not rely on a normal formula merely because the endpoint is numeric.
Uncertain variance. The formulas here condition on known \(\sigma^2\). If variance is estimated or has a posterior distribution, integrating over that uncertainty changes the predictive distribution; the displayed normal calculation is not exact as written.
Incomplete outcomes or changing enrollment. A fixed number enrolled is not necessarily a fixed number with observed endpoints. If missingness, dropout, or accrual changes the number of usable outcomes, the assumed final sample size may be wrong.
Dependence or changing measurement. Correlated outcomes, site effects, or changes in endpoint measurement can invalidate the independent common-mean model and its variance calculation.

One alternative is plug-in conditional power: future outcomes are projected using a single assumed value of \(\theta\), often the interim estimate. That approach can be useful for a specified scenario, but it does not average over posterior uncertainty about \(\theta\). The Bayesian predictive calculation does. Direct simulation can also approximate predictive success in more complex models; it is useful when an analytic distribution is unavailable, but it introduces simulation error and requires diagnostics of its own.

Common pitfall: It is incorrect to calculate predictive uncertainty using only \(m\sigma^2/N^2\) and say that the unknown mean is effectively known because interim data have been collected. Future outcomes depend on the same uncertain \(\theta\), so the full variance is \((m\sigma^2+m^2\tau_n^2)/N^2\). Omitting the posterior-uncertainty term generally makes the predictive distribution too narrow and can overstate predictive success.

Check Your Understanding

Answer each question using the model and definitions in this tutorial. Solutions follow the questions.

If \(m=0\), what is the predictive distribution of the final sample mean, and why does the positive-\(m\) variance formula not describe it?
For \(n=10\), \(m=5\), \(\sigma=4\), and interim posterior variance \(\tau_n^2=1\), calculate the predictive variance of the final sample mean.
Explain the two components of \(m\sigma^2/N^2+m^2\tau_n^2/N^2\).
Why can predictive success decrease when the fixed number of remaining participants increases?
What additional uncertainty is ignored by the formula if the endpoint variance is not known?

Solutions. (1) The final estimate is already observed, so its predictive distribution is a point mass at \(\bar y\); there are no future outcomes. (2) Here \(N=15\), so the variance is \( (5(16)+25(1))/15^2=105/225=0.467 \) squared endpoint units. (3) The first component is sampling variation in future outcomes conditional on \(\theta\); the second is variation in the predictive mean because \(\theta\) remains uncertain under the interim posterior. (4) A different \(m\) changes the future-data weight, predictive spread, and final decision cutoff. If the current evidence is favourable, future outcomes can still shift the final estimate below its cutoff, so the probability of meeting the final rule need not rise. (5) It ignores uncertainty about \(\sigma^2\); integrating over that uncertainty generally changes the predictive distribution and requires a model that includes it.