Why Efficacy Stopping Changes an Interval
When a trial stops early because an interim result crosses a prespecified efficacy boundary, the observed estimate is not simply an estimate from a fixed sample of the same size. It is an estimate observed under a particular selection event: it crossed the boundary. An ordinary fixed-sample interval uses the sampling distribution before accounting for that event. A conditional interval instead calibrates its limits using the distribution of the estimate given that efficacy stopping occurred.
The distinction matters because conditioning changes tail probabilities. Among estimates that cross a high boundary, large observed values are less surprising than they would be in an unrestricted sample. Thus the ordinary interval can be too optimistic about the evidence against the null, and its coverage among trials that stop for efficacy need not equal its nominal level. The construction below treats one prespecified interim look with a known standard error. It does not construct repeated intervals across several looks.
Model and Conditional Interval
Let \(\hat{\theta}\) estimate a treatment effect \(\theta\), measured in outcome units. Assume the normal model \(\hat{\theta}\sim N(\theta,s^2)\), where the standard error \(s\) is known and fixed. The null value is zero, and the trial stops for efficacy at the interim look if \(Z=\hat{\theta}/s\ge c\), for a prespecified boundary \(c\). Let \(E\) denote this stopping event and let \(y\) be the observed estimate after stopping. Thus \(y\ge cs\).
These assumptions describe a single-look setting: the analysis and boundary were specified in advance, the estimator has the stated normal distribution, and stopping depends on the observed interim statistic through the stated rule. In particular, the standard error and boundary are not selected after examining the data. If stopping also depends on earlier looks, changing variance estimates, or other adaptation, the conditional distribution must account for those features rather than using this single-threshold formula unchanged.
A central \(100(1-\alpha)\%\) conditional interval is obtained by retaining the parameter values for which the observed conditional upper-tail probability is between \(\alpha/2\) and \(1-\alpha/2\). Its endpoints solve \(q_\theta(y)=\alpha/2\) and \(q_\theta(y)=1-\alpha/2\), respectively. The lower endpoint solves the smaller-tail equation; the upper endpoint solves the larger-tail equation.
Why Inverting the Conditional Tail Works
Conditional on \(E\), the estimate has a continuous distribution truncated below at \(cs\). Its conditional cumulative distribution function at \(y\) is \(F_\theta(y\mid E)=P_\theta(\hat{\theta}\le y\mid \hat{\theta}\ge cs)\). For \(y\ge cs\), the complementary probability is exactly \(q_\theta(y)\). Under the true \(\theta\), evaluating a continuous conditional distribution function at a draw from that conditional distribution gives a Uniform(0,1) random variable. Therefore \(q_\theta(\hat{\theta})\), conditional on \(E\), is also Uniform(0,1). The probability that it lies between \(\alpha/2\) and \(1-\alpha/2\) is \(1-\alpha\).
To turn that probability statement into interval endpoints, the conditional tail must move monotonically with \(\theta\). For \(\theta_2>\theta_1\), the ratio of the two truncated normal densities at \(x\ge cs\) is a positive constant times \(\exp((\theta_2-\theta_1)x/s^2)\), an increasing function of \(x\). The conditional distribution therefore shifts toward larger values as \(\theta\) increases, so \(q_\theta(y)\) increases with \(\theta\). The two tail equations consequently identify the lower and upper limits in the order stated. This also supplies a useful numerical check: the tail ratio should increase as a trial value of \(\theta\) increases.
Worked Example: Inverting the Conditional Tail
Suppose a hypothetical interim analysis has \(s=1\) outcome unit, an efficacy boundary \(c=2\), and observed estimate \(y=3\) outcome units. Calculate a central 90% conditional interval. The observed statistic is \(Z=3\), so the stopping event \(Z\ge2\) has occurred.
Here \(cs=2\) and the conditional tail is \(q_\theta(3)=Q(3-\theta)/Q(2-\theta)\). For a 90% interval, \(\alpha=0.10\), so the endpoint equations are \(q_\theta(3)=0.05\) and \(q_\theta(3)=0.95\). Numerical inversion gives the lower endpoint from \(Q(3-\theta)/Q(2-\theta)=0.05\), with \(\theta\approx-0.190\), and the upper endpoint from \(Q(3-\theta)/Q(2-\theta)=0.95\), with \(\theta\approx4.604\).
Thus the conditional 90% interval is approximately \([-0.190,\ 4.604]\) outcome units. For comparison, the ordinary fixed-sample normal interval is \(y\pm z_{0.95}s=3\pm1.645(1)\), or \([1.355,\ 4.645]\) outcome units. The ordinary interval excludes zero, whereas the conditional interval includes it. This is not a contradiction: the two intervals calibrate coverage under different sampling conditions.
Calculation on the Treatment-Effect Scale
The boundary and estimate can be reported in the original clinical units without changing the procedure. Standardizing by the known standard error reduces the computation to the same conditional tail ratio. The result is then expressed in the units of the treatment effect.
Worked Example: Rescaling to Clinical Units
Consider a hypothetical continuous-outcome trial with a treatment-effect estimate of \(1.50\) points and known standard error \(0.50\) points. Its prespecified efficacy boundary is \(Z\ge2\). The corresponding boundary on the estimate scale is \(2(0.50)=1.00\) point, and the observed statistic is \(1.50/0.50=3\), so efficacy stopping has occurred.
In standardized units this is the preceding calculation: \(y/s=3\) and \(cs/s=2\). The conditional 90% limits in standardized units are approximately \(-0.190\) and \(4.604\). Multiplying each by \(s=0.50\) point gives limits of \(-0.0950\) and \(2.302\) points. The conditional interval is therefore approximately \([-0.0950,\ 2.302]\) points.
The ordinary fixed-sample interval is \(1.50\pm1.645(0.50)\), giving \([0.6775,\ 2.3225]\) points. Both procedures use the same observed estimate and standard error, but only the conditional interval accounts for observing an estimate beyond the efficacy boundary.
Conditional Tests and the Meaning of the Difference
The adjustment can also be understood through a test of the null value \(\theta=0\). After efficacy stopping, a one-sided conditional p-value is the probability, under the null and conditional on crossing the boundary, of an estimate at least as large as the observed one. It is \(q_0(y)\), not the unrestricted tail probability \(P_0(\hat{\theta}\ge y)\). This gives a direct check on whether the null belongs to a two-sided equal-tailed conditional interval.
Worked Example: The Conditional p-Value at the Null
Use \(y=3\), \(cs=2\), and \(s=1\) outcome unit from the first example. Under \(\theta=0\), the conditional one-sided p-value is \(Q(3)/Q(2)\). Standard normal tail values are \(Q(3)\approx0.001350\) and \(Q(2)\approx0.02275\), so \(q_0(3)\approx0.001350/0.02275=0.05934\).
At a one-sided 5% level, this conditional test does not reject \(H_0:\theta=0\), because \(0.05934>0.05\). The ordinary fixed-sample tail probability is \(0.001350\), which is much smaller, but it does not condition on having crossed the boundary. Consistently, zero lies inside the 90% conditional interval from the first example: its conditional tail probability lies between 0.05 and 0.95.
This comparison is specific to the inferential question. The introductory material on unconditional intervals averaged over adaptation concerns coverage averaged across possible adaptations. A conditional interval here instead targets coverage within the subset of trials that stop at this efficacy look. Neither target automatically substitutes for the other.
Diagnostics, Alternatives, and Limitations
A reliable calculation begins by writing down the actual stopping event. Check that the observed estimate crossed the stated boundary, that the same boundary and standard error appear in the conditioning denominator, and that all quantities are on a consistent scale. During numerical inversion, evaluate both endpoint equations at the reported roots. For the first example, the upper root is about \(4.604\), which rounds to \(4.60\) at two decimal places; \(4.61\) is not an equally accurate two-decimal report. Also check that the tail ratio increases with \(\theta\), that the lower limit corresponds to \(\alpha/2\), and that the upper limit corresponds to \(1-\alpha/2\).
The main alternative is the ordinary fixed-sample interval, which is simple and appropriate when the sample size and analysis are fixed independently of the observed estimate. After efficacy stopping, however, its nominal fixed-sample coverage does not imply nominal coverage conditional on stopping. Conditional inference is more directly aligned with the question, “What values remain compatible with this estimate, given that the boundary was crossed?” The trade-off is that the resulting interval may be wide or include the null even when the ordinary interval does not.
The formula also has limits. It assumes a known standard error and a normal estimate, and it conditions on one specified threshold event. Estimated variance, discrete outcomes, multiple looks, or a stopping rule based on a more complicated history require a conditional distribution appropriate to those features. If the boundary was chosen after inspecting the data, treating it as prespecified does not repair the resulting selection. Finally, numerical tail probabilities can underflow for extreme values; stable survival-function or log-tail routines should be used rather than subtracting nearly equal probabilities.
- Pitfall: interpreting the ordinary interval as conditionally valid. The reasoning “the usual interval has 90% coverage, so it has 90% coverage among efficacy-stopped trials” ignores the conditioning event. The correction is to derive or calculate the distribution conditional on crossing.
- Pitfall: conditioning on the wrong event. Dividing by the probability of exceeding the observed value, rather than the probability of crossing the boundary, gives a different quantity. The denominator must be \(P_\theta(E)\).
- Pitfall: reversing endpoints. Because \(q_\theta(y)\) increases with \(\theta\), its smaller quantile equation gives the lower limit and its larger quantile equation gives the upper limit.
Check Your Understanding
Answer each question, then compare with the solutions below.
- In the one-look normal model, what event belongs in the denominator of the conditional tail probability?
- For a central 90% interval, which conditional tail probabilities define the lower and upper limits?
- In the example with \(y=3\), \(cs=2\), and \(s=1\), why does the conditional 5% one-sided test of zero differ from the ordinary test?
- Name two assumptions that must be reconsidered before applying the displayed formula to a different trial design.
Solutions. (1) The denominator conditions on \(E=\{\hat{\theta}\ge cs\}\), the prespecified efficacy-stopping event. (2) The lower limit solves \(q_\theta(y)=0.05\), and the upper limit solves \(q_\theta(y)=0.95\). (3) The conditional test divides the null tail probability beyond the observation by the null probability of crossing the boundary: \(Q(3)/Q(2)\approx0.05934\), rather than using \(Q(3)\approx0.001350\) alone. (4) Examples include the known standard error, normal sampling distribution, and single prespecified threshold; estimated variance or multiple looks require an appropriate conditional model.