Modifying A Critical Boundary

Why a Changed Boundary Needs a Conditional Check

After an interim analysis, a design may be modified by changing its final critical boundary. A lower boundary makes rejection easier; a higher boundary makes it harder. The relevant question is not just how the new boundary affects the final test in general, but how it changes the probability of rejection conditional on the interim information already observed. That probability is the conditional error for the specified continuation rule.

This tutorial uses an upper-tail, two-stage normal combination test to make the calculation explicit. It focuses on changing the final boundary while keeping the conditioning information and the final statistic fixed. A boundary that preserves the original conditional error at one interim result need not preserve it at other interim results; a design-wide guarantee requires checking the full adaptation rule.

Model and assumptions: Under the null hypothesis, let \(Z_1\) be the first-stage statistic and \(Z_2\) the independent second-stage statistic, with \(Z_1,Z_2\sim N(0,1)\) and independent under \(H_0\). The combined statistic is \(T=(Z_1+Z_2)/\sqrt{2}\). The test rejects when \(T>c\), where \(c\) is its final critical boundary. The interim information consists of the observed value \(Z_1=z\). These assumptions describe an equal-weight normal combination test; other statistics or dependence structures require their own conditional distribution.

Conditional Error as a Function of the Boundary

Once \(Z_1=z\) is observed, rejection occurs precisely when the second-stage statistic exceeds a cutoff determined by \(z\) and \(c\):

\frac{z+Z_2}{\sqrt{2}}>c \quad\Longleftrightarrow\quad Z_2>\sqrt{2}c-z.

Because \(Z_2\) is standard normal under the stated null model, the conditional rejection probability is

\operatorname{CE}(z;c) =P_0(T>c\mid Z_1=z) =1-\Phi(\sqrt{2}c-z),

where \(\Phi\) is the standard normal cumulative distribution function. This is a probability under \(H_0\), conditional on the specified interim result. It is not the probability that \(H_0\) is true.

For fixed \(z\), increasing \(c\) increases the second-stage cutoff \(\sqrt{2}c-z\), so it decreases conditional error. More formally, the derivative is

\frac{\partial}{\partial c}\operatorname{CE}(z;c) =-\sqrt{2}\,\phi(\sqrt{2}c-z)<0,

where \(\phi\) is the standard normal density. Thus, conditional error is strictly decreasing in the boundary. A decrease in \(c\) has the opposite effect.

Worked Example: Comparing Three Final Boundaries

Suppose the hypothetical interim result is \(z=2.000\). Compare boundaries \(c=2.400\), \(2.500\), and \(2.600\) under the model above.

For \(c=2.400\), the second-stage cutoff is \(\sqrt{2}(2.400)-2.000=1.3941\). Therefore,

\operatorname{CE}(2.000;2.400) =1-\Phi(1.3941) \approx 0.0816.

For \(c=2.500\), the cutoff is \(\sqrt{2}(2.500)-2.000=1.5355\), giving

\operatorname{CE}(2.000;2.500) =1-\Phi(1.5355) \approx 0.0623.

For \(c=2.600\), the cutoff is \(\sqrt{2}(2.600)-2.000=1.6770\), so

\operatorname{CE}(2.000;2.600) =1-\Phi(1.6770) \approx 0.0468.

At this interim result, lowering the boundary from 2.500 to 2.400 raises conditional error from about 6.23% to 8.16%. Raising it to 2.600 lowers conditional error to about 4.68%. The calculations concern conditional probabilities under the null, not probabilities of an alternative being true.

Solving for a Boundary That Meets an Error Limit

Suppose the permitted conditional error at interim value \(z\) is \(e\), where \(0<e<1\). We require \(1-\Phi(\sqrt{2}c-z)\le e\). Since \(\Phi\) is increasing, this is equivalent to \(\Phi(\sqrt{2}c-z)\ge 1-e\). Inverting \(\Phi\) gives the boundary condition:

c\ge \frac{z+\Phi^{-1}(1-e)}{\sqrt{2}}.

This is the minimum boundary meeting the specified cap, with equality giving conditional error exactly \(e\). For \(e=0\), no finite boundary achieves zero conditional error in this continuous normal model; for \(e=1\), every boundary has conditional error at most one.

Worked Example: Enforcing a 10% Conditional-Error Limit

Suppose the hypothetical interim value is \(z=2.000\), and the continuation rule must have conditional error no greater than \(e=0.1000\). Use \(\Phi^{-1}(0.9000)\approx1.281552\). The minimum boundary is

c_{\min} =\frac{2.000+\Phi^{-1}(0.9000)}{\sqrt{2}} =\frac{2.000+1.281552}{1.414214} \approx 2.32041.

The inequality is \(c\ge 2.32041\) (approximately). To report a boundary to three decimal places without falling below the computed minimum, choose \(c=2.321\) or a larger value. For example, at \(c=2.321\), the cutoff is

\sqrt{2}(2.321)-2.000\approx 1.28239, \qquad \operatorname{CE}(2.000;2.321) =1-\Phi(1.28239)\approx 0.0999.

By contrast, \(c=2.320\) gives a cutoff of approximately \(1.28098\) and conditional error of about \(0.1001\), which exceeds the limit. Rounding the minimum boundary down to three decimal places would therefore fail the stated requirement.

Preserving the Original Conditional Error

Let \(c_0\) be the original boundary and \(c_1\) a proposed replacement. Under the same statistic and conditioning information, preserving or reducing the original conditional error at a given interim value requires

\operatorname{CE}(z;c_1)\le \operatorname{CE}(z;c_0).

Since conditional error is strictly decreasing in \(c\), this inequality is equivalent to \(c_1\ge c_0\). This simple equivalence depends on keeping the model, statistic, and conditioning information fixed. If any of those change, the conditional probability must be recalculated under the new continuation design rather than inferred from boundary ordering alone.

Worked Example: Checking a Replacement Against the Original Error

Suppose an original design has boundary \(c_0=2.500\) and the observed hypothetical interim value is \(z=2.000\). From the earlier calculation, its conditional error is about \(0.0623\). A proposed replacement boundary is \(c_1=2.450\). Its second-stage cutoff is

\sqrt{2}(2.450)-2.000\approx1.46482.

Thus the new conditional error is \(1-\Phi(1.46482)\approx0.0715\). It exceeds the original conditional error of about \(0.0623\), so this replacement does not preserve the original conditional-error limit at \(z=2.000\). The boundary is lower than the original one, consistent with the increase in rejection probability.

If the intended cap were instead \(0.0500\), the required boundary would be

c\ge \frac{2.000+\Phi^{-1}(0.9500)}{\sqrt{2}} = \frac{2.000+1.644854}{1.414214} \approx 2.577.

For that stricter cap, a boundary of \(2.578\) is a conservative three-decimal choice; a boundary of \(2.577\) should be checked using unrounded values before being treated as sufficient.

Pointwise Preservation and Trial-Wide Error Control

A calculation at one observed \(z\) answers a local question: what is the conditional error at this interim result? An adaptive rule may choose a boundary \(c_1(z)\) as a function of the interim data. To preserve the original conditional error for every possible interim result, require the pointwise inequality

\operatorname{CE}(z;c_1(z)) \le \operatorname{CE}(z;c_0) \quad\text{for every }z.

In the current fixed-statistic model, this is equivalent to \(c_1(z)\ge c_0\) for every \(z\). Checking the inequality only at the observed value does not establish it for other possible interim outcomes. A rule that sometimes selects a lower boundary may spend more conditional error in those regions, even if its boundary is higher elsewhere.

The link to unconditional type I error follows from conditional expectation. If the original design has null rejection probability at most \(\alpha\), and the new conditional error is no larger at every interim history, then

P_0(\text{reject under new rule}) =E_0\!\left[\operatorname{CE}_{\mathrm{new}}\right] \le E_0\!\left[\operatorname{CE}_{\mathrm{old}}\right] =P_0(\text{reject under original rule}) \le\alpha.

This argument assumes that both conditional errors are defined with respect to the same relevant interim information and null model, and that the new rule’s conditional rejection probability is correctly calculated. It explains why pointwise conditional-error control can protect the overall type I error; it does not make a single-point check sufficient.

Key distinction: A boundary can be assessed against an explicit conditional-error cap \(e\), or against the original conditional error at each interim history. Under this tutorial’s unchanged normal combination test, the latter condition reduces to not lowering the original boundary. If the test or its assumptions change, calculate the new conditional error directly.

Diagnostics, Alternatives, and Limitations

Several checks help detect a faulty calculation. First, verify the direction of the inequality: a larger upper-tail boundary must produce a smaller conditional error. Second, confirm the conditioning value is used in the correct place: the cutoff is \(\sqrt{2}c-z\), not \(\sqrt{2}(c-z)\). Third, ensure the assumed distribution of the second-stage statistic is appropriate conditional on the information being held fixed. Dependence between stages, a different combination statistic, or additional conditioning information can change the formula.

When reporting a minimum boundary, retain adequate precision during the calculation and round in the safe direction. If the requirement is \(c\ge c_{\min}\), rounding \(c_{\min}\) downward can violate it, as the 10% example shows. For a boundary reported to a fixed number of decimal places, evaluate the conditional error at the reported value as a final check.

The main alternative to conditional-error preservation is to specify a new unconditional design and calibrate its boundary to meet a chosen overall type I error rate. That approach addresses the average null rejection probability across interim outcomes, whereas conditional-error preservation constrains rejection probability given interim information. The two targets are not interchangeable: a new design could meet an unconditional limit while exceeding the original conditional error in some regions.

The calculations here are specific to a one-sided upper-tail test with independent standard normal stage statistics and the stated equal-weight combination statistic. They do not directly cover two-sided rejection, discrete statistics, nuisance-parameter complications, or changed dependence structures. For those cases, the conditional distribution and rejection event must be specified and evaluated for that design. In particular, a numerical boundary formula should not be carried over merely because another test also reports a statistic called \(Z\).

Common Mistakes

Reversing the boundary effect. A lower upper-tail boundary makes rejection easier and increases conditional error, not decreases it.
Using a marginal probability. Once \(Z_1=z\) is observed, calculate \(P_0(\text{reject}\mid Z_1=z)\), not the unconditional rejection probability.
Rounding a minimum downward. A rounded value below \(c_{\min}\) may fail the error constraint. Check the final reported boundary directly.
Checking only one interim value for an adaptive rule. A design-wide guarantee requires the conditional-error inequality at every relevant interim history.
Changing the statistic but reusing the old formula. The formula depends on the joint null distribution and rejection rule, not just on the numerical boundary.

Check Your Understanding

Use the stated two-stage normal combination model. Show the conditional cutoff and the calculation supporting each answer.

At \(z=1.000\), calculate the conditional error for \(c=2.000\).
At \(z=1.500\), find the minimum boundary for a conditional-error cap of \(0.0500\).
At \(z=2.000\), does changing the original boundary from \(2.500\) to \(2.600\) preserve or increase conditional error? Quantify the new conditional error.
A rule checks its conditional-error constraint at one observed interim value only. What additional check is needed for a guarantee over the adaptive rule?

Solutions. (1) The cutoff is \(\sqrt{2}(2.000)-1.000=1.82843\), so the conditional error is \(1-\Phi(1.82843)\approx0.0337\). (2) The minimum is \([1.500+\Phi^{-1}(0.9500)]/\sqrt{2}=(1.500+1.644854)/1.414214\approx2.224\); use a boundary at least this large, checking the unrounded result before final rounding. (3) The cutoff at \(c=2.600\) is \(1.67696\), giving \(1-\Phi(1.67696)\approx0.0468\). This is lower than the original conditional error, about \(0.0623\), so the increase in boundary preserves the limit and reduces conditional error. (4) Check the conditional-error inequality for every interim history where the adaptive rule can select a boundary, not only the observed value.

What You'll Learn