Mirror Symmetry Under A Zero-Effect Null

Why Mirror Symmetry Matters

A two-sided sequential design monitors evidence in both directions. Under a zero-effect null, it is tempting to assume that upper- and lower-tail decisions are equally likely. That equality is not guaranteed merely because the null effect is zero or because the test is called two-sided. It follows when the null distribution of the full sequence of statistics is unchanged by reversing every sign, and when the decision rules reverse in the same way.

This distinction matters when interpreting directional error probabilities. If the assumptions for mirror symmetry hold, the probability of an upper-boundary decision equals the probability of a lower-boundary decision. If the model, boundaries, or stopping rules are not symmetric, that conclusion may fail even when the parameter being tested is zero.

Assumptions and Definitions

Consider a design with \(K\) planned looks. Let \(Z_k\) be the signed test statistic at look \(k\), oriented so that positive values favor one direction and negative values favor the other. Write the full sequence as \(\mathbf{Z}=(Z_1,\ldots,Z_K)\). The null considered here is a zero-effect null, under which the statistic is centered at zero.

The key distributional condition concerns the whole sequence, not just each look separately. Reversing every sign must leave the joint null distribution unchanged. For a centered multivariate normal sequence this condition holds, whatever its covariance matrix, because a centered multivariate normal vector and its negative have the same distribution. More generally, the required property is central symmetry of the joint null distribution.

Definition: A sequential statistic has sign symmetry under the zero-effect null if \(\mathbf{Z}\) and \(-\mathbf{Z}=(-Z_1,\ldots,-Z_K)\) have the same joint distribution under \(H_0\). Boundary and stopping rules are mirror images if replacing every \(Z_k\) by \(-Z_k\) swaps the upper- and lower-direction decisions while preserving continuation and any other corresponding decisions.

For example, let \(u_k>0\) be the upper efficacy boundary at look \(k\), and let the lower boundary be \(-u_k\). The boundary rules are mirror images at that look: \(Z_k\geq u_k\) is the upper crossing condition, and \(Z_k\leq -u_k\) is the lower crossing condition. The thresholds may vary across looks; symmetry requires the lower threshold at each look to be the negative of the corresponding upper threshold.

In a sequential design, a crossing usually means a directional decision made at the first look at which an efficacy boundary is reached. The monitoring rule must apply the same first-crossing convention in both directions. If there are additional stopping rules, such as futility rules, they must also be paired under sign reversal for the complete decision process to be mirror-symmetric.

Why the Crossing Probabilities Are Equal

Let \(U\) be the event that the design makes an upper-direction crossing, and \(L\) the event that it makes a lower-direction crossing. Under the mirror-rule condition, applying the transformation \(\mathbf{Z}\mapsto-\mathbf{Z}\) to any path that produces \(U\) produces a path that produces \(L\). The same mapping in reverse takes paths in \(L\) to paths in \(U\).

\mathbf{Z}\in U \quad\Longleftrightarrow\quad -\mathbf{Z}\in L

Since \(\mathbf{Z}\) and \(-\mathbf{Z}\) have the same null distribution, events related by this transformation have equal probability. Therefore,

P_{H_0}(U) = P_{H_0}(-\mathbf{Z}\in L) = P_{H_0}(\mathbf{Z}\in L).

This is a pathwise argument: it pairs complete trajectories, rather than comparing one-look marginal tail probabilities. Dependence between looks does not invalidate the argument. In particular, the statistics at different looks need not be independent. Their joint distribution must satisfy sign symmetry, and the decision rules must be mirror images.

When crossing decisions are absorbing and mutually exclusive, the overall probability of either directional decision is \(P_{H_0}(U)+P_{H_0}(L)\). Under the conditions above, the two terms are equal. This result establishes equality between the two directional probabilities; it does not by itself calculate their common value. That value still depends on the sequential boundaries and the joint distribution across looks.

Worked Example: A One-Look Symmetric Test

Suppose a hypothetical study has one planned analysis. Under \(H_0\), assume its standardized statistic satisfies \(Z_1\sim N(0,1)\). The design declares an upper crossing if \(Z_1\geq1.96\), and a lower crossing if \(Z_1\leq-1.96\). These boundaries are mirror images, and the standard normal distribution is symmetric about zero.

The upper crossing probability is

P_{H_0}(U)=P(Z_1\geq1.96)=1-\Phi(1.96)=1-0.9750=0.0250.

By symmetry, the lower crossing probability is

P_{H_0}(L)=P(Z_1\leq-1.96)=\Phi(-1.96)=0.0250.

Thus each direction has a 0.0250 crossing probability in this one-look example. The conclusion follows from the zero-centered symmetric null distribution and mirrored thresholds; it is not a general consequence of simply using a two-sided test.

Checking the Argument With Complete Paths

At multiple looks, sign reversal applies to all values on a path at once. The next finite example makes that pairing visible. It uses a deliberately simple discrete null distribution rather than a normal model, so that the crossing probabilities can be counted exactly.

Worked Example: Pairing Two-Look Paths

Suppose a hypothetical two-look null distribution assigns equal probability to each of the following four paths and its sign-reversed partner:

(1.0,1.2),\quad (1.7,1.0),\quad (0.4,2.3),\quad (2.2,2.4),

and

(-1.0,-1.2),\quad (-1.7,-1.0),\quad (-0.4,-2.3),\quad (-2.2,-2.4).

Each of these eight paths has probability \(1/8\). At both looks, use upper and lower efficacy boundaries \(1.5\) and \(-1.5\), respectively, and stop at the first efficacy crossing.

Among the four positive representatives, \((1.7,1.0)\), \((0.4,2.3)\), and \((2.2,2.4)\) reach the upper boundary. The first reaches it at look 1; the other two reach it no later than look 2. Their three negative partners reach the lower boundary instead. The path \((1.0,1.2)\) and its negative partner cross neither boundary.

Therefore,

P_{H_0}(U)=\frac{3}{8}=0.375, \qquad P_{H_0}(L)=\frac{3}{8}=0.375.

The common probability is high because these illustrative boundaries are modest relative to the path values. The equality, however, depends on the one-to-one pairing of every path with its negative and the corresponding mirrored stopping rules.

Simulation Check Under a Sequential Null Model

A simulation can check whether a proposed implementation appears to preserve the symmetry. It does not replace the argument: simulation introduces Monte Carlo variability, and an unpaired simulation may show slightly different upper and lower counts even when the exact probabilities are equal.

Worked Example: Sign-Paired Gaussian Simulation

Consider a hypothetical two-look design with information fractions \(t_1=0.5\) and \(t_2=1\). Under a canonical joint normal zero-effect null, assume \(Z_1\) and \(Z_2\) each have mean zero and variance one, with correlation \(\sqrt{t_1/t_2}=\sqrt{0.5}=0.7071\). Use boundaries \(+2.0\) and \(-2.0\) at both looks, and stop at the first efficacy crossing. The joint normal model is centrally symmetric, and the thresholds and stopping rules are mirrored.

Suppose a hypothetical simulation generates 5,000 independent null paths. The results for those paths are 184 upper decisions, 167 lower decisions, and 4,649 paths with neither decision. Now append the sign-reversed version of every simulated path. Each original upper decision becomes a lower decision in its partner, and each original lower decision becomes an upper decision.

In the combined set of 10,000 paths, the upper and lower counts are therefore both \(184+167=351\). The empirical probabilities are

\widehat P(U)=\frac{351}{10{,}000}=0.0351, \qquad \widehat P(L)=\frac{351}{10{,}000}=0.0351.

The equality in these paired results is exact for the combined simulation by construction. The value 0.0351 is only a Monte Carlo estimate of the common null probability; a different set of simulated paths would generally give a different estimate. Also, the 10,000 paths are paired, so they should not be treated as 10,000 independent draws when quantifying simulation uncertainty.

Diagnostics, Alternatives, and Limitations

When checking an analysis or its implementation, look for symmetry in the entire sequential decision process—not only in the final test statistic. Useful checks include:

Confirm the null model. Check that the joint distribution of the signed statistics is centered and sign-symmetric under the zero-effect null. Symmetric one-look marginals alone do not establish joint sign symmetry.
Compare corresponding boundaries. At every look, verify that the lower boundary is the negative of the upper boundary, using the same scale and sign convention.
Audit all stopping rules. Confirm that sign reversal swaps directional decisions and leaves corresponding continuation or other stopping decisions unchanged.
Check the implementation. In a sign-paired simulation, every path’s reversed partner should exchange upper and lower outcomes. A discrepancy points to a coding, sign, boundary, or stopping-rule mismatch.

An unpaired Monte Carlo run is a practical alternative when sign-paired paths are unavailable, but small differences between the directional estimates are expected from simulation error. Pairing paths makes the symmetry check more direct. Neither simulation approach proves that the model assumptions or the implemented rules are correct.

The equality can fail when the joint null distribution is not sign-symmetric, when upper and lower thresholds differ in magnitude, or when other rules treat the directions differently. A zero value of the null effect alone is not enough to rule out those possibilities. Further, equality of the two directional probabilities does not supply their numerical value: calculating that value still requires the joint null behavior over the planned looks. The result here concerns symmetry under a zero-effect null, not general claims about nonzero effects or asymmetric designs.

Common pitfall: “The null effect is zero, so the upper and lower crossing probabilities must be equal.” This skips essential conditions. The correction is to verify joint sign symmetry of the sequential statistics and mirror symmetry of every relevant boundary and stopping rule; only then does sign reversal pair upper and lower outcomes.

Key takeaway: Under a sign-symmetric zero-effect null and mirror-image sequential rules, reversing every statistic on a path swaps upper and lower decisions. This one-to-one probability-preserving mapping proves that the two directional crossing probabilities are equal.

Check Your Understanding

Answer each question using the pathwise symmetry argument.

A three-look null statistic has a centrally symmetric joint distribution. Upper boundaries are \(2.1, 2.0, 1.9\), and lower boundaries are \(-2.1, -2.0, -1.9\). The stopping rules are otherwise mirrored. What can be concluded about the two directional crossing probabilities?
At every look, the upper and lower marginal distributions are symmetric, but the joint distribution of the full statistic vector is not known to be sign-symmetric. Is that enough to establish equal sequential crossing probabilities?
A two-look design has sign-symmetric null paths and mirrored efficacy boundaries, but an additional stopping rule treats positive and negative statistics differently. Does the equality result necessarily apply to the complete decision process?
A sign-paired simulation produces equal upper and lower counts. Does that establish the exact common null crossing probability? Explain.

Solutions. (1) The directional crossing probabilities are equal: the joint null distribution is sign-symmetric, and each lower boundary is the negative of the corresponding upper boundary, with mirrored stopping rules. (2) No. Equal one-look marginals do not by themselves establish sign symmetry of the joint sequence, which is needed to map complete paths. (3) No. The additional rule breaks mirror symmetry of the full decision process, so the pathwise mapping need not swap its final decisions. (4) No. Pairing makes the simulated counts equal by construction, but the common simulated frequency is an estimate. The exact probability requires the model and decision rules; simulation alone does not establish it.