Why Reverse the Sign of a Statistic Path?
A two-sided group sequential rule has an upper direction and a lower direction. A rule may treat those directions differently by design, for example by using different efficacy limits. But a rule can also look different simply because the statistic has been coded with the opposite sign while the boundaries have not been translated to match. Sign reversal provides a direct way to separate these cases.
The diagnostic compares a complete interim path with its sign-reversed partner. For a path \(z=(z_1,\ldots,z_K)\), the partner is \(-z=(-z_1,\ldots,-z_K)\). The comparison asks whether a stop in one direction for the first path corresponds to a stop at the same look in the opposite direction for the partner. This is a check on the rule itself, not a method for choosing boundaries.
The centered-coordinate condition matters. If the null value is not zero on the original effect scale, first express the statistic relative to that null value. The sign reversal concerns deviation from the null, not necessarily the raw parameter estimate. Also, reverse the whole path: changing only the final statistic does not test whether the sequential decisions are sign-symmetric.
The Boundary Criterion and Its Justification
At a given look, reflecting a statistic value across zero sends the upper crossing condition \(Z_j\geq u_j\) to the lower crossing condition \(-Z_j\leq -u_j\). For this to be exactly the lower crossing condition of the same rule, its lower boundary must be \(l_j=-u_j\). Reflecting the lower crossing condition in turn gives the corresponding upper condition. Thus the rule is sign-symmetric precisely when the boundary pair mirrors at every look:
Equivalently, each pair has midpoint zero. If boundaries are expressed by a center \(c_j=(u_j+l_j)/2\) and half-width \(h_j=(u_j-l_j)/2\), symmetry requires \(c_j=0\). Unequal boundary magnitudes at even one look are enough to make the rule structurally asymmetric, although a particular path may not encounter that difference.
The criterion can also be justified by tracking the first crossing. If a path first crosses the upper boundary at look \(m\), then for every earlier look \(j<m\) it lies between the boundaries, and at look \(m\) it is at least \(u_m\). Under mirrored boundaries, the reversed path remains between the boundaries at each earlier look, then satisfies \(-Z_m\leq -u_m=l_m\). It therefore first crosses the lower boundary at the same look \(m\). The same argument applies with upper and lower interchanged. This proves the pathwise correspondence, including the stopping time.
This is a structural test and does not require a probability model for the path. To interpret directional null error probabilities, however, a model is needed. Under the usual centered Gaussian group sequential null model, the vector of interim statistics has mean zero and a joint normal distribution; for canonical information fractions \(t_j\), its covariance entries are \(\operatorname{Cov}(Z_i,Z_j)=\sqrt{t_i/t_j}\) for \(i\leq j\). The looks are dependent, but the joint null distribution is invariant under \(Z\mapsto -Z\). With mirrored boundaries, the upper-stop event maps to the lower-stop event, so their null probabilities are equal. Without mirrored boundaries, that equality is not guaranteed. An accidental equality of probabilities for a particular model would not establish that the rule itself is symmetric.
Worked Example: A Symmetric Rule at Two Looks
Consider two hypothetical interim looks with boundaries \(l_1=l_2=-2.0\) and \(u_1=u_2=2.0\). The observed statistic path is \(z=(0.8,2.25)\). Compare it with its sign-reversed path.
At look 1, \( -2.0<0.8<2.0\), so the original path continues. At look 2, \(2.25\geq2.0\), so it stops in the upper direction. The stopping look is 2.
The paired path is \(-z=(-0.8,-2.25)\). At look 1, \(-2.0<-0.8<2.0\), so it also continues. At look 2, \(-2.25\leq-2.0\), so it stops in the lower direction at look 2. The stop direction changes, but the stopping look does not. The boundary criterion confirms this result because \(l_j=-u_j\) at both looks.
Using Paired Paths to Expose Asymmetry
A path pair is especially useful when a rule is presented in software output, a design table, or a verbal description whose sign convention is unclear. Apply the stated boundaries to both paths, record the first crossing for each, and compare the look and direction. A difference can reveal a genuine asymmetry, but first check that the boundaries and statistic have been expressed in the same coordinate convention.
Worked Example: An Unequal Boundary Produces Different Decisions
Suppose a hypothetical two-look rule has first-look boundaries \(l_1=-2.4\) and \(u_1=2.0\), and second-look boundaries \(l_2=-2.2\) and \(u_2=2.3\). Consider \(z=(2.2,0.5)\) and its partner \(-z=(-2.2,-0.5)\).
For \(z\), look 1 gives \(2.2\geq2.0\), so the rule stops in the upper direction at look 1. For \(-z\), look 1 gives \(-2.2>-2.4\), so there is no lower crossing and the path continues. At look 2, \(-2.2<-0.5<2.3\), so the reversed path remains in the continuation region. It has no efficacy crossing by the last look.
The paired paths do not have corresponding opposite-direction stops at the same look. The boundary check identifies why: at look 1, \(l_1=-2.4\) is not \(-u_1=-2.0\). The difference is a property of the rule, not evidence that the sign reversal was performed incorrectly. This single path pair witnesses a failure of sign symmetry; it does not quantify how often either outcome occurs.
For a systematic diagnostic, inspect each boundary pair first, then use path pairs as a check on implementation and interpretation. Testing only a few paths cannot prove symmetry: paths may never reach a look where the boundaries differ. Conversely, a single paired path with non-matching decisions can disprove pathwise symmetry. In numerical work, retain the same equality conventions as the design: if a rule uses strict rather than inclusive crossing at a boundary, use that convention for both directions.
Sign Coding Is Not the Same as Design Asymmetry
A change in sign convention must transform both the statistic and the boundaries. If the original statistic is \(Z_j\) with boundaries \(l_j,u_j\), define the recoded statistic \(Z_j^*=-Z_j\). The same stopping rule in the recoded coordinates has boundaries
The order remains valid because \(l_j<u_j\) implies \(l_j^*<u_j^*\). Keeping the old numerical boundary labels after reversing the statistic generally defines a different rule. That mistake can make a symmetric design appear asymmetric, or make a genuinely asymmetric design appear to have changed in an unexplained way.
Worked Example: Correctly Translating a Sign Convention
A hypothetical one-look rule uses \(l=-2.0\) and \(u=2.5\). A path value \(z=2.2\) does not cross either boundary, since \(-2.0<2.2<2.5\). Now reverse the statistic: \(z^*=-2.2\).
For the same rule in the new convention, the boundaries must be \(l^*=-u=-2.5\) and \(u^*=-l=2.0\). Since \(-2.5<-2.2<2.0\), the recoded path also continues. The decision is unchanged, as it must be when only the coding convention changes.
If the old boundaries \((-2.0,2.5)\) were incorrectly retained, then \(-2.2\leq-2.0\) would signal a lower stop. That apparent discrepancy is created by failing to transform the boundaries; it is not evidence that the correctly recoded rule has different behavior.
What the Diagnostic Does and Does Not Establish
Sign reversal is a compact diagnostic because it separates three questions that are easy to conflate: whether the boundaries are symmetric around zero, whether a particular pair of paths receives corresponding decisions, and whether upper and lower null error probabilities are equal. Mirrored boundaries answer the first question; paired-path application checks the second; the third additionally depends on the null distribution.
Under a sign-invariant null distribution, mirrored boundaries yield equal upper and lower null stopping probabilities. This conclusion does not mean the rule has equal power for positive and negative alternatives, nor does a failure of boundary symmetry by itself determine the size of either error probability. The distribution of the path and the full sequence of boundaries both matter. Likewise, unequal boundaries can sometimes yield equal directional error probabilities by coincidence, so equality of probabilities is not proof of structural symmetry.
A practical comparison with alternatives is useful. Inspecting boundary values is the quickest test of the formal symmetry condition. Paired paths make the consequences of a failure concrete and are useful for debugging. Simulation or Gaussian integration is needed when the question concerns error probabilities under a specified path distribution; neither one observed path nor a boundary comparison supplies those probabilities. The sign-reversal check does not replace calibration, and it does not choose a preferred allocation between directions.
The most reliable workflow is to state the statistic’s sign convention and null-centered coordinate, list both boundaries at every look, and then compare first-crossing decisions for \(z\) and \(-z\). If results do not match in the expected way, determine whether the cause is a true boundary mismatch, an untransformed coding convention, or a mistake in applying the stopping rule.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- A rule has boundaries \(l_1=-1.9,u_1=1.9\) and \(l_2=-2.3,u_2=2.1\). Is it sign-symmetric? Identify the look that determines your answer.
- At one look, \(l=-2.4,u=2.4\). A path value is \(2.6\). What decision does the paired value \(-2.6\) receive?
- A rule has \(l=-1.8,u=2.2\). After recoding \(Z^*=-Z\), what boundaries represent the same rule?
- Under a sign-invariant null path distribution, does equality of upper- and lower-stop probabilities prove that the boundaries are symmetric? Explain.
Solutions. (1) No. At look 2, \(-u_2=-2.1\), which differs from \(l_2=-2.3\); the boundaries do not mirror at that look. (2) The original value crosses the upper boundary because \(2.6\geq2.4\). Its paired value satisfies \(-2.6\leq-2.4\), so it crosses the lower boundary at the same look. (3) The recoded lower boundary is \(l^*=-u=-2.2\), and the recoded upper boundary is \(u^*=-l=1.8\). (4) No. Equal probabilities can occur by coincidence for a particular path distribution and rule. Mirrored boundary values establish structural symmetry; equal probabilities alone do not.