Conditional Confidence Intervals Given the Stopping Stage

Why Condition on the Stopping Stage?

A group sequential trial produces both an estimate and a stopping stage. If a trial stops early after crossing an efficacy boundary, the reported statistic comes from a selected part of its sampling distribution. An interval calibrated under the fixed-sample distribution may therefore fail to have its stated coverage among trials that stop at that stage. One response is to construct an interval whose coverage is calibrated conditional on that stopping event.

The target is different from coverage averaged across every possible stopping stage. A conditional interval asks how often it covers the true parameter among repetitions that stop at a specified look. Its conditional coverage can be nominal even when the interval is wide, and it need not resemble an interval calibrated for the full collection of possible stopping stages.

Definition: Let $E_j$ denote the event that the design stops at look $j$, and let $C_j$ be an interval computed from the data available when it stops there. The interval has $1-\alpha$ confidence coverage conditional on stopping at look $j$ if, for each parameter value $\theta$ for which $P_\theta(E_j)>0$, $$ P_\theta\{\theta\in C_j\mid E_j\}=1-\alpha. $$ This is a repeated-sampling property conditional on $E_j$, not a probability statement about the parameter after observing the data.

To construct such an interval, the sampling distribution must reflect the stopping rule. In general, let $Y$ contain the observed data path that determines $E_j$, and let $f_\theta(y)$ be its density under parameter $\theta$. Then the density conditional on stopping at look $j$ is

f_\theta(y\mid E_j) = \frac{f_\theta(y)\mathbf{1}\{y\in E_j\}}{P_\theta(E_j)}.

The path matters: an endpoint statistic alone may not determine whether the trial reached look $j$. If $Z_j$ is the endpoint statistic and the stopping event also depends on earlier looks, its conditional density is instead

f_\theta(z\mid E_j) = \frac{f_\theta(z)P_\theta(E_j\mid Z_j=z)}{P_\theta(E_j)},

when the conditional probability in the numerator is defined. The factor $P_\theta(E_j\mid Z_j=z)$ accounts for the unobserved parts of the path relevant to the stopping decision. Omitting it generally gives the wrong distribution.

Inverting the Conditional Distribution

Write $F_\theta(y\mid E_j)$ for the conditional cumulative distribution function of a continuous statistic $Y$. For an observed value $y$, an equal-tailed conditional confidence set is obtained by retaining the parameter values satisfying

\frac{\alpha}{2} \leq F_\theta(y\mid E_j) \leq 1-\frac{\alpha}{2}.

When this set is an interval, its endpoints are found by solving the two boundary equations. The direction in which the endpoints correspond to the two tail probabilities depends on how $F_\theta(y\mid E_j)$ changes with $\theta$. It should be checked, not assumed.

The justification is the probability integral transform. For a fixed true $\theta$, if the conditional distribution is continuous, then $F_\theta(Y\mid E_j)$, conditional on $E_j$, is uniform on $[0,1]$. Thus the probability it falls between $\alpha/2$ and $1-\alpha/2$ is $1-\alpha$. This proves the conditional coverage claim when the interval is obtained by inverting that conditional distribution. With discrete data, the transform need not be uniform; exact intervals may then be conservative or require a specified randomization convention.

Worked Example: An Efficacy Stop at the First Look

Suppose a standardized statistic $Z_1\sim N(\theta,1)$, where $\theta$ is the treatment effect in standard-error units. A hypothetical design stops for efficacy at the first look when $Z_1\geq 2.00$. The observed value is $z_1=2.40$. Construct the equal-tailed 95% interval conditional on stopping at this look.

The event is $E_1=\{Z_1\geq2.00\}$, and the observed statistic satisfies it. For any $\theta$, its conditional CDF at the observation is

F_\theta(2.40\mid E_1) = \frac{\Phi(2.40-\theta)-\Phi(2.00-\theta)} {1-\Phi(2.00-\theta)} = 1-\frac{1-\Phi(2.40-\theta)}{1-\Phi(2.00-\theta)}.

For a 95% interval, set this quantity first to $0.975$ and then to $0.025$. Numerical root-finding gives approximately $\theta=-6.90$ for the first equation and $\theta=4.15$ for the second. The conditional CDF decreases with $\theta$ here, so the parameter values satisfying $0.025\leq F_\theta(2.40\mid E_1)\leq0.975$ lie between those roots. The interval is therefore approximately $(-6.90,\,4.15)$ standard-error units.

The width is striking, but it is consistent with the conditioning. A statistic of 2.40 is only 0.40 above the boundary. Under many parameter values, trials reaching this stage are selected to have crossed the boundary; the conditional observation alone can consequently provide limited precision about $\theta$. This interval is not the ordinary untruncated normal interval centered at 2.40.

When Earlier Looks Affect the Conditional Law

At a later stopping stage, the event includes the decisions at earlier looks. For illustration, use the canonical two-look normal model with information levels $I_1=1$ and $I_2=4$. Let $Z_1$ and $Z_2$ have means $\theta\sqrt{I_1}$ and $\theta\sqrt{I_2}$, unit variances, and correlation $\sqrt{I_1/I_2}=0.5$. A hypothetical design stops at look 1 if $Z_1\geq2$, and otherwise proceeds to look 2, where it stops. Thus $E_2=\{Z_1<2\}$. This model assumes the stated joint normal law and a fully specified boundary rule.

Conditionally on $Z_2=z$, the normal conditional distribution gives $Z_1\mid Z_2=z\sim N(0.5z,\,0.75)$. The parameter cancels from this conditional mean because the two statistics have the canonical means and correlation specified above. Consequently,

P_\theta(E_2\mid Z_2=z) = \Phi\left(\frac{2-0.5z}{\sqrt{0.75}}\right), \qquad P_\theta(E_2)=\Phi(2-\theta).

Worked Example: The Endpoint Statistic Does Not Determine the Path

In the two-look design just defined, suppose the observed endpoint statistic is $z_2=1.50$. Find the conditional density at this value under $\theta=0$, and identify the ingredients needed to build a stage-2 interval.

The marginal density of $Z_2$ at $1.50$, when $\theta=0$, is $\phi(1.50)=0.1295$. The probability of having avoided the first-look efficacy boundary given this endpoint is

P_0(E_2\mid Z_2=1.50) = \Phi\left(\frac{2-0.75}{\sqrt{0.75}}\right) = \Phi(1.443) \approx 0.9255.

The unconditional probability of reaching look 2 is $P_0(E_2)=\Phi(2)=0.9772$. Therefore the conditional density is

f_0(1.50\mid E_2) = \frac{\phi(1.50)(0.9255)}{0.9772} \approx \frac{0.1295(0.9255)}{0.9772} \approx 0.1227.

This is a density value, not a probability of observing exactly 1.50. To construct an interval, use the full conditional density $f_\theta(z\mid E_2)$ as a function of $z$ and $\theta$, integrate it to obtain $F_\theta(z\mid E_2)$, and invert the equal-tail inequalities at the observed $z_2$. Using only the indicator $1\{z\in E_2\}$ would be incorrect: $Z_2=1.50$ does not reveal whether the earlier $Z_1$ crossed its boundary.

Conditional Coverage and Coverage Across Stages

Let $C_j$ be the interval used when stopping at stage $j$. The overall coverage is a probability over all stopping stages, and the law of total probability gives

P_\theta\{\theta\in C_J\} = \sum_j P_\theta(E_j) P_\theta\{\theta\in C_j\mid E_j\},

where $J$ is the stopping stage. If every stage-specific interval has conditional coverage $1-\alpha$, the weighted average is also $1-\alpha$, provided the stopping events partition the possible outcomes and have their stated probabilities. Conversely, an interval calibrated to achieve a target after averaging across stages need not achieve that target within each stage.

Worked Example: Averaging Conditional Coverage

At a fixed hypothetical parameter value, suppose a two-stage design has $P_\theta(E_1)=0.30$ and $P_\theta(E_2)=0.70$. Consider intervals whose conditional coverage is 0.95 at stage 1 and 0.90 at stage 2. Their overall coverage is

$$ 0.30(0.95)+0.70(0.90) = 0.285+0.630 = 0.915. $$

Thus 91.5% overall coverage does not mean 91.5% coverage conditional on either stage. If instead both conditional coverages were 0.95, the overall coverage would be $0.30(0.95)+0.70(0.95)=0.95$. These calculations distinguish the targets; they do not imply that an interval calibrated across stages must have a particular stage-specific coverage.

Practical Checks and Limitations

Conditional intervals are useful when the scientific or operational question explicitly concerns trials that stopped at a particular look. Their construction depends on the design being known: the information times, joint sampling model, stopping boundaries, and any other rules that determine $E_j$ must be represented correctly. If an unrecorded adaptation or a binding futility rule changes which paths reach a stage, the conditional distribution used for inversion may be wrong.

A useful computational check is to verify that the conditional density integrates to one and that the resulting endpoints satisfy the intended tail equations. For a proposed interval, simulation under selected parameter values can check the proportion of repetitions that both stop at look $j$ and cover, divided by the proportion that stop at look $j$. Such checks are especially important when the event probability $P_\theta(E_j)$ is small: dividing by a small probability can make conditional calculations numerically unstable and conditional coverage difficult to assess by simulation.

The main alternatives answer different questions. A fixed-sample interval ignores selection by the stopping rule and can have distorted coverage after stopping. A stopping-aware interval based on the stopped sampling law accounts for the rule without necessarily conditioning on one particular stage. A stage-conditional interval targets coverage within the specified stage, which may come at a substantial cost in width or depend strongly on the model. The interval should therefore be reported with its conditioning event and parameter scale stated explicitly.

Common mistake: Treating the endpoint statistic $Z_j$ as if it alone determined the event of stopping at look $j$, then truncating its marginal density with an indicator. If earlier decisions depend on other statistics in the path, that indicator formula is not the conditional density of $Z_j$. Use the full path density with the event indicator, or weight the endpoint density by $P_\theta(E_j\mid Z_j=z)$, and divide by $P_\theta(E_j)$.

Key takeaway: A conditional confidence interval is calibrated under the sampling distribution given the observed stopping stage. Its guarantee is stage-specific; it is not interchangeable with an interval whose coverage is calibrated over all possible stages.

Check Your Understanding

Answer each question, then compare with the solutions below.

In a continuous model, what conditional probability defines a $1-\alpha$ interval at stage $j$?
A stopping event depends on both $Z_1$ and $Z_2$. Why is an indicator based only on the observed $Z_2$ generally insufficient for its conditional density?
For the two-look example, calculate $P_0(E_2\mid Z_2=2)$, using the given model.
If stage probabilities are 0.20 and 0.80 and conditional coverages are 0.90 and 0.95, respectively, what is overall coverage?

Solutions. (1) The required property is $P_\theta\{\theta\in C_j\mid E_j\}=1-\alpha$ for parameter values with $P_\theta(E_j)>0$. (2) $Z_2$ does not reveal the earlier statistic, so it does not determine whether the path met the stage-$j$ event; the density needs the factor $P_\theta(E_j\mid Z_2=z)$. (3) The conditional probability is $\Phi((2-0.5(2))/\sqrt{0.75})=\Phi(1.155)\approx0.876$. (4) Overall coverage is $0.20(0.90)+0.80(0.95)=0.94$, or 94%.