Calibrating A Continuation Boundary

What Boundary Calibration Does

After an interim analysis, a trial may continue under a revised design. A continuation boundary specifies which future test-statistic values lead to rejection. Calibrating that boundary means choosing its value so that the revised continuation rule uses a specified amount of conditional type I error. The target is often the conditional error from the original design at the observed interim history.

The calibration is an inversion problem: evaluate the conditional rejection probability as a function of the proposed boundary, then find a boundary that gives the target probability. That inversion is straightforward only when the function behaves suitably. In particular, a numerical root finder cannot create a solution if the target is unattainable, and a single returned solution does not establish uniqueness.

Definition: Let $H_I=h$ denote the interim information being conditioned on, and let $T$ be a future statistic with conditional null distribution given $H_I=h$. For an upper-tail continuation rule that rejects when $T\ge b$, its conditional error function is $$ C(b\mid h)=P_0(T\ge b\mid H_I=h). $$ To calibrate to a target conditional error $c(h)$, find a boundary $b$ satisfying $C(b\mid h)=c(h)$. Here $P_0$ denotes probability under the null hypothesis.

The target $c(h)$ may be the original design's conditional error at $h$. The candidate rule need not use the original statistic or sample size, but its conditional rejection probability must be evaluated under the null distribution appropriate to the proposed continuation design. A target is a probability, so it lies between 0 and 1; not every value in that range must be attainable by a particular boundary family.

When Inversion Is Unique

For an upper-tail rule, increasing $b$ can only remove outcomes from the rejection region. If $b_1<b_2$, then $\{T\ge b_2\}\subseteq\{T\ge b_1\}$, and therefore $C(b_2\mid h)\le C(b_1\mid h)$. This is the basic monotonicity that makes one-dimensional calibration possible.

Result: If the conditional distribution function $F_h(t)=P_0(T\le t\mid H_I=h)$ is continuous and strictly increasing over the relevant range, then $C(b\mid h)=1-F_h(b)$. For any target $c(h)$ strictly between 0 and 1, the unique calibrated boundary is $$ b=F_h^{-1}\bigl(1-c(h)\bigr). $$ If the conditional distribution has atoms, or the candidate rule is not a nested family of rejection regions, continuity, strict monotonicity, or even monotonicity may fail.

The justification is direct. Under continuity, $P_0(T\ge b\mid h)=1-F_h(b)$. Setting this equal to $c(h)$ gives $F_h(b)=1-c(h)$; strict increase makes the corresponding quantile unique. If $F_h$ has flat portions, more than one boundary can yield the same probability. If it has jumps, some target probabilities can be skipped.

Worked Example: Calibrating a Normal Continuation Boundary

Suppose the original one-sided design rejects when a final statistic exceeds $1.96$. At the interim, the information fraction is $t=0.50$ and the observed interim statistic is $z=1.00$. Under the null, let the final statistic be $Z=\sqrt{t}\,z+\sqrt{1-t}\,Z_2$, where $Z_2\sim N(0,1)$ is the standardized future increment. Find a revised boundary on a future statistic $V\sim N(0,1.2^2)$ that matches the original conditional error.

The conditional error of the original rule is the probability that $Z_2$ is large enough to make $Z\ge1.96$: $$ c=1-\Phi\left(\frac{1.96-\sqrt{0.50}(1.00)}{\sqrt{0.50}}\right) =1-\Phi(1.772)\approx0.0382. $$

For the revised rule, reject when $V\ge b$. Under its stated conditional null distribution, $$ C(b)=P_0(V\ge b)=1-\Phi\left(\frac{b}{1.2}\right). $$ Equating this to the target gives $$ b=1.2\,\Phi^{-1}(1-0.0382) \approx1.2(1.772)=2.126. $$

Thus the revised future-statistic boundary is approximately 2.126 units of $V$. Substitution gives $P_0(V\ge2.126)\approx0.0382$, as required. The normal distribution is continuous and its density is positive everywhere, so this upper-tail function is continuous and strictly decreasing; the calibrated boundary is unique.

Numerical Calibration and Checks

When the conditional null distribution does not have a convenient closed form, numerical integration or simulation can be used to evaluate $C(b\mid h)$. A practical calibration procedure is to establish the function's attainable range, check its shape, and then solve for the boundary. If the function is continuous and monotone, a bracketed root-finding method is preferable to an unconstrained search: the bracket makes clear that the target is crossed and helps avoid convergence to a spurious value.

Specify the target.
Compute or otherwise define the conditional error $c(h)$ to be matched, and state the interim history on which it is conditional.

Define the candidate rule.
Write the rejection region as a function of the boundary parameter and specify the conditional null distribution used to calculate its probability.

Inspect the error function.
Check its range, monotonicity, continuity, and any flat intervals or jumps. Do not assume that every target has a solution.

Solve and verify.
Find a boundary with $C(b\mid h)=c(h)$, then recompute the conditional error at that boundary and report any numerical tolerance.

In numerical work, a small residual $|C(b\mid h)-c(h)|$ supports accuracy but does not prove uniqueness. Check the function on the relevant boundary range, especially if its shape is not known analytically. Simulation adds Monte Carlo error to the estimated conditional error; a solver's apparent convergence should not be reported as exact calibration if that uncertainty is material.

Worked Example: A Discrete Boundary Cannot Match Every Target

Suppose the future statistic is $X\sim\operatorname{Binom}(5,0.5)$ under the null, and a deterministic continuation rule rejects when $X\ge k$, for integer $k$. The target conditional error is $c=0.10$. Determine whether a boundary can match it exactly.

The relevant upper-tail probabilities are $$ P_0(X\ge4)=\frac{\binom54+\binom55}{2^5} =\frac{5+1}{32}=0.1875, \qquad P_0(X\ge5)=\frac{1}{32}=0.03125. $$ For $k\le0$, the error is 1; for $k\ge6$, it is 0. The tail therefore jumps from 0.1875 to 0.03125 as the integer cutoff moves from 4 to 5. No deterministic integer cutoff has conditional error 0.10.

If randomized rejection is permitted at $X=4$, one could always reject for $X=5$ and reject with probability $\gamma$ when $X=4$. Solving $$ \frac{1}{32}+\gamma\frac{5}{32}=0.10 $$ gives $\gamma=0.44$. This achieves the target, but it is a different, randomized rule. Whether it is acceptable depends on the trial design; it should not be presented as a deterministic boundary solution.

Flat Regions and Nonmonotone Functions

A nonunique solution is possible even when a function is monotone. For a discrete statistic, moving a cutoff between adjacent support points may leave the rejection set unchanged, so every such cutoff gives the same conditional error. More generally, a flat section of $C$ gives multiple calibrated boundaries whenever the target equals the value on that section. Report the range of solutions, or apply a prespecified tie-breaking rule, rather than implying that a numerical solver found the only answer.

Monotonicity requires nested rejection regions. It need not hold if changing one parameter moves multiple boundaries in opposing ways, as can happen in a two-sided rule. In that case, a target may be crossed more than once or not crossed at all. A monotone root-finding algorithm is not justified until nesting or monotonicity has been established.

Worked Example: A Moving Two-Sided Rule Is Not Monotone

Let $Y$ be equally likely to be 0, 1, 2, or 3 under the null. Consider the boundary-indexed rule that rejects when $Y\le b$ or $Y\ge b+2$. Evaluate its conditional error at three proposed boundaries.

At $b=0.5$, rejection occurs for $Y=0$ or $Y=3$, so $C(0.5)=2/4=0.50$. At $b=1$, rejection occurs for $Y=0,1,3$, so $C(1)=3/4=0.75$. At $b=1.5$, rejection occurs for $Y=0,1$, so $C(1.5)=2/4=0.50$.

The function first increases and then decreases; the rejection regions are not nested as $b$ changes. A target of 0.60 is unattainable for this example: as $b$ varies, the rejection probability is determined by which of the four support points meets either inequality, and takes only the values 0, 0.25, 0.50, 0.75, or 1. A target of 0.50, by contrast, is achieved at multiple boundary values. Calibration must therefore examine the rule's regions and all relevant boundary ranges, rather than rely on a monotone solver.

Why Conditional Calibration Matters

Conditional error links an interim history to the remaining type I error available under the null. If a revised continuation rule has conditional rejection probability no greater than the original conditional error at every relevant interim history, then its overall null rejection probability is no greater than that of the original design. This follows by averaging the conditional probabilities: $$ P_0(\text{reject}) =E_0\!\left[P_0(\text{reject}\mid H_I)\right] \le E_0\!\left[c(H_I)\right]. $$ Thus exact matching preserves the original error contribution pointwise, while using a smaller conditional error is conservative for that history. This argument depends on comparing probabilities conditional on the same relevant interim information and on evaluating them under the null.

Calibration is useful only when the conditional error function corresponds to the proposed rule and its conditional null distribution. A boundary computed under the wrong distribution, or for a different rejection region, does not establish the intended error control. If an exact match is unavailable, one option is a deterministic rule satisfying $C(b\mid h)\le c(h)$, if such a rule exists. That is conservative rather than exact calibration and should be described as such.

Common Pitfalls and Limitations

Assuming every target is attainable. Discrete statistics can skip probabilities. Check the function's range before solving.
Assuming a unique root. Flat regions can yield many solutions. Verify uniqueness from the function's behavior, not from a solver's output.
Using bisection without a monotonicity check. Bisection requires a valid bracket and a sign change; nonmonotone functions can have several crossings.
Ignoring numerical error. Integration or simulation may approximate $C$; report the achieved error and its precision, not just the chosen boundary.
Equating conditional matching with unconditional matching at one history. The overall type I error argument averages conditional probabilities over interim histories; a single calibrated value alone is not that argument.

The main alternatives are a deterministic conservative cutoff, if its error is below the target, or a randomized boundary when randomization is allowed and can attain the target. Another option is to change the candidate boundary family, but that changes the continuation rule and requires recalculating its conditional error. These choices involve design constraints as well as mathematics; an unattainable target should be stated plainly rather than hidden by rounding.

Check Your Understanding

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

Under the null, $V\sim N(0,1.5^2)$. Find the upper-tail boundary that gives conditional error 0.05.
For $X\sim\operatorname{Binom}(5,0.5)$, can a deterministic rule $X\ge k$ attain conditional error 0.125?
Why can a function $C(b\mid h)$ for an upper-tail boundary be nonincreasing but still fail to have a unique solution?
State the condition that lets pointwise conditional error control imply overall type I error control, and give the probability identity used.

Solutions. (1) Since $P(V\ge b)=0.05$, $b=1.5\Phi^{-1}(0.95)\approx1.5(1.6449)=2.467$. The normal distribution is continuous and strictly increasing, so the solution is unique. (2) No. The neighboring tail probabilities are $P(X\ge4)=0.1875$ and $P(X\ge5)=0.03125$; the deterministic cutoff skips 0.125. (3) A flat interval can occur when moving the boundary does not change the rejection set, for example when the statistic has no support points in that interval. Every boundary there has the same conditional error. (4) The revised conditional rejection probability must be no greater than the original conditional error at every relevant interim history under the null. Averaging gives $P_0(\text{reject})=E_0[P_0(\text{reject}\mid H_I)]\le E_0[c(H_I)]$.