Deriving Stochastic Curtailment Probabilities

Why Derive the Probability of Eventual Rejection?

At an interim analysis, a trial may stop before its planned sample size. Stochastic curtailment uses the chance of eventual rejection, conditional on the data observed so far, to help decide whether continuing is worthwhile. That chance is not obtained by treating the current test statistic as if it were final. It is calculated from the distribution of the score that will be available at the future analysis, given the current data and a specified model for future outcomes.

This tutorial derives that distribution for a normal-mean test. The derivation makes clear which quantities are observed, which assumptions describe future data, and how the final rejection rule translates into a conditional probability. The probability is conditional on a specified value of the mean; it is not a posterior probability that a hypothesis is true.

Model, Score, and Assumptions

Suppose observations $X_1,\ldots,X_N$ are independent and identically distributed as $N(\mu,\sigma^2)$, where $\sigma$ is known. At an interim analysis, $n_1$ observations have been collected, with $0<n_1<N$, and their sample mean $\bar{x}_1$ is observed. The final sample size $N$ is fixed. We test $H_0:\mu=\mu_0$ against the one-sided alternative $H_a:\mu>\mu_0$, rejecting at the final analysis if the standardized score is at least $c$.

Definition: The final score is $Z_N=\sqrt{N}(\bar{X}_N-\mu_0)/\sigma$, where $\bar{X}_N$ is the mean of all $N$ observations. The interim score is $z_1=\sqrt{n_1}(\bar{x}_1-\mu_0)/\sigma$. The conditional probability of eventual rejection at the final analysis, under a specified mean $\mu$, is $P_\mu(Z_N\ge c\mid X_1,\ldots,X_{n_1})$.

The normal model and known variance give an exact normal score distribution. Independence ensures that the observations still to come are independent of the observed interim data. The fixed final sample size and fixed boundary ensure that the future rejection event is defined without needing to model additional design changes. If the variance is estimated, outcomes are dependent, or the design changes in response to interim data, this calculation may not apply as written.

Write $m=N-n_1$ for the number of future observations. Under a specified future mean $\mu$, their sample mean $\bar{X}_{\mathrm{future}}$ has distribution $N(\mu,\sigma^2/m)$, independently of the interim observations. The final mean is the weighted average of the interim and future means:

\bar{X}_N=\frac{n_1\bar{x}_1+m\bar{X}_{\mathrm{future}}}{N}.

Deriving the Future-Score Distribution

Substitute the weighted mean into the definition of $Z_N$. Once the interim data are fixed, the only random part is the future sample mean. Its conditional expectation and variance give the conditional distribution of the final score.

Z_N =\frac{n_1(\bar{x}_1-\mu_0)+m(\bar{X}_{\mathrm{future}}-\mu_0)} {\sigma\sqrt{N}}.

The first term is fixed given the interim data. The second term is normal because the future sample mean is normal. Its conditional mean and variance are therefore:

\begin{aligned} E_\mu(Z_N\mid X_1,\ldots,X_{n_1}) &=\frac{n_1(\bar{x}_1-\mu_0)+m(\mu-\mu_0)} {\sigma\sqrt{N}}\\ &=\sqrt{\frac{n_1}{N}}\,z_1+ \frac{m(\mu-\mu_0)}{\sigma\sqrt{N}},\\[4pt] \operatorname{Var}_\mu(Z_N\mid X_1,\ldots,X_{n_1}) &=\frac{m^2(\sigma^2/m)}{\sigma^2 N} =\frac{m}{N}. \end{aligned}

Thus, if $t=n_1/N$ is the fraction of the final sample already observed, the conditional distribution is:

Z_N\mid X_1,\ldots,X_{n_1} \sim N\left( \sqrt{t}\,z_1+(1-t)\frac{\sqrt{N}(\mu-\mu_0)}{\sigma}, \,1-t \right).

The second argument is the variance. The term $\sqrt{t}\,z_1$ is the contribution of the observed data; the other mean term describes the expected contribution from future observations under the specified $\mu$. The remaining variance $1-t$ is the uncertainty contributed by the unobserved observations.

For a one-sided final boundary $c$, standardizing this conditional normal distribution gives the rejection probability:

Result: Under the stated model and a specified mean $\mu$, the conditional probability of rejecting at the final analysis is $$ P_\mu(Z_N\ge c\mid X_1,\ldots,X_{n_1}) =\Phi\left( \frac{\sqrt{t}\,z_1+(1-t)\sqrt{N}(\mu-\mu_0)/\sigma-c} {\sqrt{1-t}} \right), $$ where $\Phi$ is the standard normal cumulative distribution function.

Worked Example: One Future Final Analysis

Suppose $n_1=40$, $N=100$, $\sigma=10$ units, $\mu_0=0$ units, and the observed interim mean is $\bar{x}_1=2$ units. For this hypothetical test, the final rejection rule is $Z_N\ge 1.96$. Calculate the conditional rejection probability under the specified future mean $\mu=0.5$ units.

First calculate the interim score and information fraction:

z_1=\frac{\sqrt{40}(2-0)}{10}=1.2649, \qquad t=\frac{40}{100}=0.4.

The conditional mean of the final score is $\sqrt{0.4}(1.2649)+0.6\{\sqrt{100}(0.5-0)/10\}=0.8+0.3=1.1$. Its conditional standard deviation is $\sqrt{0.6}=0.7746$. Therefore:

P_{\mu=0.5}(Z_N\ge 1.96\mid\text{interim data}) =\Phi\left(\frac{1.1-1.96}{0.7746}\right) =\Phi(-1.110) \approx 0.133.

Under the specified future mean, the chance of crossing the final boundary is about 13.3%. The calculation conditions on the observed interim data and assumes future observations have mean 0.5 units; it does not state the probability that $\mu=0.5$ or that $H_0$ is true.

Effect Assumptions and Stochastic Curtailment

The conditional distribution requires a mean for future observations. That mean is not learned from the interim data by the formula itself; it is an input. Common choices include a prespecified design alternative or a plug-in value equal to the interim estimate. These choices answer different conditional questions and can produce very different probabilities. The introductory material on conditional power distinguishes the interpretation of these choices; here, the key point is that each choice determines the drift term in the future-score distribution.

Worked Example: Changing the Assumed Future Mean

Use the same hypothetical interim data and final boundary as above, but now assume that future observations have mean equal to the interim estimate, $\mu=2$ units. The observed contribution to the conditional mean remains $0.8$. The future contribution is $0.6\{\sqrt{100}(2-0)/10\}=1.2$, so the conditional mean of $Z_N$ is $2.0$, and its standard deviation remains $0.7746$.

P_{\mu=2}(Z_N\ge1.96\mid\text{interim data}) =\Phi\left(\frac{2.0-1.96}{0.7746}\right) =\Phi(0.05164) \approx 0.521.

The plug-in assumption gives a conditional rejection probability of about 52.1%, compared with about 13.3% under the prespecified mean of 0.5 units. Neither number is an assumption-free forecast: each is conditional on its stated mean model.

A stochastic-curtailment rule can compare this probability with a prespecified threshold $\gamma$. For example, the trial may stop for futility when the conditional probability of eventual rejection is below $\gamma$. This is a rule based on a model-based conditional probability, not a conclusion that rejection is impossible. Its operating characteristics depend on the chosen future mean, threshold, monitoring schedule, and stopping rules.

Worked Example: Applying a Curtailment Threshold

Suppose the same hypothetical design uses $n_1=40$, $N=100$, $\sigma=10$ units, $\mu_0=0$ units, and boundary $c=1.96$, but the interim score is $z_1=-0.5$. Under the prespecified future mean $\mu=0.5$ units, the conditional mean is $\sqrt{0.4}(-0.5)+0.6(0.5)=-0.01623$, and the conditional standard deviation is $0.7746$.

P_{\mu=0.5}(Z_N\ge1.96\mid\text{interim data}) =\Phi\left(\frac{-0.01623-1.96}{0.7746}\right) =\Phi(-2.551) \approx 0.0054.

If the prespecified futility threshold is $\gamma=0.01$, this probability is below the threshold, so the rule calls for stopping for futility. The interpretation is that, under the specified model, the chance of final rejection is about 0.54%; it is not zero, and the decision does not prove that the treatment has no effect.

Multiple Future Looks and Other Rejection Rules

If rejection can occur at several future analyses, the relevant event is crossing a boundary at any of them. The final-score calculation alone is insufficient: a path might cross an earlier boundary and later finish below the final boundary. Independent increments provide the transition distributions needed to calculate the probability of crossing at least once.

For clarity, let $I_j$ denote cumulative information at look $j$, let $I_K$ be final information, and define $t_j=I_j/I_K$. Under the known-variance normal model, information is proportional to sample size. Write $\lambda=(\mu-\mu_0)\sqrt{I_K}$ when the score is scaled so that its variance at final information is one. Given score $z$ at information fraction $t_j$, the next score has conditional distribution:

Z_{j+1}\mid Z_j=z \sim N\left( \sqrt{\frac{t_j}{t_{j+1}}}\,z+ \frac{(t_{j+1}-t_j)\lambda}{\sqrt{t_{j+1}}}, \frac{t_{j+1}-t_j}{t_{j+1}} \right).

This follows by writing the cumulative score as the sum of the observed score and an independent increment. For upper efficacy boundaries $c_j$, eventual rejection is the sum, over future looks, of the probabilities of first crossing at each look. Each term integrates the product of these transition densities over paths that stayed below earlier boundaries and cross the current boundary. This is a multivariate normal probability in general. The crossing probabilities at separate looks must not simply be added without excluding paths already counted.

For a two-sided final rejection rule $|Z_N|\ge c$, both tails are included. If the conditional mean is $a$ and the conditional standard deviation is $s$, the probability is $\Phi((-c-a)/s)+1-\Phi((c-a)/s)$. The lower-tail term is not a second one-sided calculation with the same direction; it represents rejection for sufficiently negative scores.

Worked Example: Two-Sided Final Rejection

Return to the first hypothetical example, where the conditional final-score distribution has mean $a=1.1$ and standard deviation $s=0.7746$. Suppose instead that the final rule rejects when $|Z_N|\ge1.96$. The upper-tail probability is $\Phi((1.1-1.96)/0.7746)=\Phi(-1.110)\approx0.133$. The lower-tail probability is $\Phi((-1.96-1.1)/0.7746)=\Phi(-3.950)\approx0.00004$.

Adding the disjoint tails gives a total conditional rejection probability of approximately $0.133+0.00004=0.133$, or about 13.3% to three significant figures. The lower tail is small here because the conditional mean is positive, but it is part of the two-sided rejection event.

Diagnostics, Comparisons, and Limitations

Before using a computed probability in a curtailment rule, check that the conditional distribution matches the design. The interim and future observations must be independent under the model; the future sample size and analysis rule must be correctly represented; the assumed future mean and variance must be stated; and the boundary must match the actual rejection criterion. A useful numerical check is that the conditional variance $1-t$ is positive when observations remain and tends to zero as $t$ approaches one. Also check that increasing the assumed mean in an upper-tail test increases the calculated rejection probability.

The final-only formula is simpler than calculating crossing probability across several future looks, but it answers a different question when interim efficacy boundaries can reject early. A multiple-look calculation must account for all opportunities to reject and the dependence among scores. Conditional probabilities under a fixed mean also differ from predictive probabilities that average over uncertainty in the mean. Plugging in an interim estimate does not perform that averaging.

The calculation can be sensitive to effect assumptions, especially when little information remains or the interim estimate is unstable. It can also be inaccurate if the normal known-variance model is a poor representation, if variance estimation materially changes the score distribution, or if unplanned design adaptations alter future sampling or boundaries. The probability is a decision aid under stated assumptions, not a guarantee of success or a substitute for prespecifying the stopping rule.

Common mistake: Treating $\sqrt{t}\,z_1$ as the conditional mean of the final score and ignoring future observations. Correction: Add the expected future contribution under the specified $\mu$, and retain the remaining variance $1-t$.
Common mistake: Calling a plug-in conditional rejection probability the probability that the null hypothesis is false. Correction: It is a probability of a future rejection conditional on observed data and a fixed assumed mean.
Common mistake: Multiplying or adding separate look-specific rejection probabilities as if looks were independent. Correction: Use the joint distribution generated by independent score increments and account for paths that have already crossed.

Key takeaway: For a normal-mean test with independent future observations, the conditional final-score distribution is normal with variance equal to the remaining information fraction. Applying the rejection boundary to this distribution gives the conditional probability used by a stochastic-curtailment rule, conditional on the specified future mean.

Check Your Understanding

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

Why is the conditional variance of the final score $1-t$, rather than 1?
For $t=0.25$, $z_1=1$, $\sqrt{N}(\mu-\mu_0)/\sigma=0.5$, and $c=1.96$, find the one-sided conditional rejection probability.
What does it mean if a conditional rejection probability is below a stochastic-curtailment threshold?
Why can a final-only probability fail to represent eventual rejection when there are earlier efficacy looks?

Solutions. (1) The observed part of the score is fixed after conditioning on the interim data. Only the unobserved portion contributes conditional variance, and its fraction is $1-t$. (2) The conditional mean is $\sqrt{0.25}(1)+0.75(0.5)=0.875$, and the standard deviation is $\sqrt{0.75}=0.8660$. Thus the probability is $\Phi((0.875-1.96)/0.8660)=\Phi(-1.253)\approx0.105$. (3) Under the specified mean and model, the chance of eventual rejection is below the prespecified cutoff; the event remains possible, and the result does not establish that the null is true. (4) Rejection may occur at an earlier look even if the final score would not cross its boundary. The probability across looks must account for the dependent sequence of scores and the paths that crossed earlier.