How The Wang-Tsiatis Shape Parameter Works

What the Shape Parameter Changes

A group sequential design has several planned analyses, and its efficacy boundary specifies how strong the evidence must be to stop early for benefit. The Wang-Tsiatis family separates the shape of that boundary from its overall scale. The shape parameter changes how stringent the early cutoffs are relative to the later ones; it does not, by itself, set the design's overall type I error.

This distinction matters in practice. Two designs can share the same target type I error and information schedule but place different amounts of protection against early stopping. A shape comparison is therefore not a valid comparison of calibrated designs if it holds the scale constant without checking the resulting crossing probability.

Definition: At look $k$, let $t_k$ be the information fraction, with $0<t_1<\cdots<t_K=1$. For an upper, one-sided efficacy boundary, the Wang-Tsiatis cutoff is $$ b_k=a\,t_k^{\Delta-\frac12}, $$ where $a>0$ is a scale constant and $\Delta$ is the shape parameter. The design stops for efficacy at the first look $k$ for which $Z_k\ge b_k$. The constant $a$ is chosen to attain the prespecified overall type I error $\alpha$ under $H_0$, for the specified look schedule and statistical model.

Here $Z_k$ is the standardized test statistic at look $k$. The boundary is a sequence of cutoffs, not a sequence of nominal p-values. Its interpretation depends on the information fractions and on the dependence between statistics at different looks.

How a Change in Shape Redistributes the Cutoffs

For a fixed scale $a$, the final cutoff is $b_K=a$, because $t_K=1$. At any earlier look, $t_k<1$, so changing $\Delta$ has a direct, predictable effect. If $\Delta$ increases by $d>0$, the cutoff at look $k$ is multiplied by $t_k^d<1$, while the final cutoff is unchanged when $a$ is held fixed.

\frac{b_i}{b_j} =\left(\frac{t_i}{t_j}\right)^{\Delta-\frac12}, \qquad \frac{b_i(\Delta+d)}{b_i(\Delta)} =t_i^d \quad (t_i<1,\ a\text{ fixed}).

Thus, for a fixed $a$, increasing $\Delta$ lowers earlier cutoffs relative to later ones. Decreasing $\Delta$ makes early evidence harder to attain relative to the final cutoff. The ratio formula is often the clearest way to describe the shape: it compares two looks without confusing that comparison with the overall scale.

In a calibrated design, however, $a$ is not meant to stay fixed when $\Delta$ changes. Lowering interim cutoffs makes it more likely that the process crosses a boundary under $H_0$. To maintain the same overall type I error, the scale constant must be recalculated. Consequently, the final cutoff, which equals $a$, will generally change as well.

Worked Example: Comparing Shapes at a Fixed Scale

Consider four planned looks at information fractions $t=(0.25,0.50,0.75,1)$, and temporarily set $a=2.20$ for each shape. This calculation isolates the geometry; it does not claim that all three designs have the same type I error.

For $\Delta=0$, the exponent is $-0.5$. The cutoffs are $$ b_k=2.20t_k^{-0.5} =(4.400,\ 3.111,\ 2.540,\ 2.200). $$

For $\Delta=0.25$, the exponent is $-0.25$: $$ b_k=2.20t_k^{-0.25} =(3.111,\ 2.616,\ 2.364,\ 2.200). $$

For $\Delta=0.50$, the exponent is zero, so all four cutoffs equal $2.200$. As $\Delta$ increases, the early cutoffs fall relative to the final cutoff. For example, at the first look the cutoff falls from $4.400$ to $3.111$ to $2.200$. Because $a$ was deliberately held fixed, these numbers illustrate shape alone, not an error-rate-matched comparison.

Calibration and Its Justification

For error-rate calibration, assume a one-sided efficacy test and a prespecified schedule $t_1,\ldots,t_K$. Under $H_0$, use the canonical joint-normal model: each $Z_k$ has a standard normal marginal distribution, and for $i<j$, $$ \operatorname{Corr}(Z_i,Z_j)=\sqrt{\frac{t_i}{t_j}}. $$ This model is appropriate for the usual information-based standardized statistics under the null when information accumulates as planned and the underlying score or increment process has the assumed independent-increment structure. The overall type I error is the probability of crossing at least one upper boundary: $$ \alpha_{\mathrm{overall}} =P_{H_0}\left(\bigcup_{k=1}^{K}\{Z_k\ge b_k\}\right). $$

The target $\alpha$ applies to this union event, not separately to each look. The calibration chooses $a$ so that $\alpha_{\mathrm{overall}}=\alpha$. In general, this probability is evaluated from the joint distribution of the sequential statistics, usually by numerical integration or a validated boundary-calibration algorithm.

Specify the design.
Fix the one-sided or two-sided testing convention, the target $\alpha$, and all planned information fractions.

Choose a shape.
Set $\Delta$, then calculate each relative cutoff $t_k^{\Delta-\frac12}$.

Calibrate the scale.
Find $a$ such that the joint-null probability of crossing any boundary equals the target $\alpha$.

Report the calibrated sequence.
Use $b_k=a t_k^{\Delta-\frac12}$, and identify the schedule, model, and error-rate convention used.

There is a simple justification for why calibration must change with shape. For fixed $a$, increasing $\Delta$ lowers each interim cutoff, since $t_k^d<1$ for $t_k<1$, and leaves the final cutoff unchanged. The event of crossing any boundary can therefore only grow; under a nonsingular joint-normal model with at least one interim look, it grows with positive probability. The null crossing probability increases. Increasing $a$, in contrast, raises every cutoff, and the crossing probability decreases continuously. Therefore, to restore the original target after increasing $\Delta$, calibration must increase $a$. This argument explains the direction of the adjustment, but does not give its numerical size.

Worked Example: Calibrating a Two-Look Design

Suppose a hypothetical one-sided design has two looks at $t_1=0.5$ and $t_2=1$, with target overall type I error $\alpha=0.025$. Under $H_0$, the canonical model gives $\operatorname{Corr}(Z_1,Z_2)=\sqrt{0.5}$. Write $\Phi_2(x,y;\rho)=P(Z_1\le x,Z_2\le y)$ for the bivariate standard normal distribution function with correlation $\rho$. For cutoffs $b_1$ and $b_2$, calibration requires $$ 1-\Phi_2(b_1,b_2;\sqrt{0.5})=0.025. $$

For $\Delta=0.25$, $b_1=a(0.5)^{-0.25}$ and $b_2=a$. Numerical solution of the displayed equation gives approximately $a=2.04$. Substitution gives $$ b_1=2.04(0.5)^{-0.25} =2.04(1.1892) \approx2.426, \qquad b_2=2.04. $$ Using the unrounded numerical solution in the bivariate-normal calculation gives a crossing probability of $0.025$; the displayed cutoffs are rounded.

For comparison, numerical calibration for the same schedule and target gives approximately $a=1.98$ at $\Delta=0$, and approximately $a=2.18$ at $\Delta=0.50$. The corresponding interim and final cutoffs, rounded, are:

$\Delta$	Calibrated $a=b_2$	Interim cutoff $b_1$
0	1.98	$1.98(0.5)^{-0.5}\approx2.800$
0.25	2.04	$2.04(0.5)^{-0.25}\approx2.426$
0.50	2.18	$2.18(0.5)^0=2.180$

These are approximate numerical calibration results; small differences in the last digit can arise from rounding. They show the redistribution clearly: with the same target error rate, increasing $\Delta$ makes the interim cutoff lower, while calibration raises the final cutoff. For instance, $a=2.05$ is not the appropriate rounded calibration for $\Delta=0.25$ in this example; using it gives a crossing probability of about $0.0243$, rather than the stated target $0.025$.

Reading the Shape in a Boundary Decision

Boundary shape affects when a trial can stop, not just the appearance of a curve. The next example compares decisions along the same hypothetical sequence of test statistics under three separately calibrated designs. These values illustrate how the rule changes; they are not data from a real study.

Worked Example: Same Evidence, Different First-Crossing Look

Use the two-look schedule and the calibrated cutoffs from the preceding example. Suppose the observed statistics are $Z_1=2.30$ at the interim analysis and $Z_2=2.10$ at the final analysis.

At $\Delta=0$, the interim cutoff is about $2.800$, so $2.30<2.800$ and there is no interim stop. At the final look, $2.10>1.98$, so the design crosses its final efficacy boundary.

At $\Delta=0.25$, $2.30<2.426$, so there is again no interim stop; at the final look, $2.10>2.04$, so the final boundary is crossed. At $\Delta=0.50$, $2.30>2.180$, so this design crosses its interim boundary and stops then. The same evidence path can therefore lead to different stopping times even though each design was calibrated to the same overall null error rate.

The comparison does not imply that one shape is universally preferable. It describes the way each calibrated rule allocates opportunities to stop early versus continue to the planned final analysis.

Assumptions, Diagnostics, and Limitations

The calibration is meaningful only for the design actually analyzed. Check that the information fractions used to calculate the boundary are the planned fractions, that the test statistic has the assumed null distribution and dependence, and that the direction and number of tails match the stated error-rate target. In particular, a one-sided upper-boundary calculation does not automatically control a two-sided type I error of the same numerical size.

Warning signs include analyses conducted at unplanned information times, an incorrect covariance structure, a change in the test statistic or variance model, and use of rounded cutoffs as though they were exact. If the schedule or model changes, the original numerical calibration may no longer control the intended overall error rate. A nominal per-look p-value alone is not a diagnostic for the probability of crossing at least one boundary.

The Wang-Tsiatis family offers a compact way to vary boundary geometry, but it does not select a shape for a particular scientific or operational priority. Nor does fixing $\Delta$ determine the stopping probability or expected sample size under an alternative; those depend on the effect, information schedule, and calibrated boundary. The introductory material on Pocock and O'Brien-Fleming boundaries provides the familiar boundary patterns; the key point here is that changing the shape requires recalibrating the scale before making an error-rate-matched comparison.

Common Pitfall

Wrong reasoning: “The target overall type I error is still $0.025$, so I can change $\Delta$ and keep $a$ unchanged.” The target is a requirement, not a guarantee supplied by the formula. With fixed $a$, increasing $\Delta$ lowers interim cutoffs and increases the null probability of crossing at least one boundary.

Correction: Recalculate $a$ for every shape using the same information schedule, null model, and error-rate convention. Then compare the calibrated cutoffs. Also distinguish a cutoff's geometric movement from its calibrated movement: at a fixed scale the final cutoff stays at $a$, but after recalibration the final cutoff changes because $a$ changes.

Key takeaway: The shape parameter controls the relative stringency of early and late cutoffs through $t_k^{\Delta-\frac12}$. A change in shape alters the null crossing probability when the scale is held fixed, so the scale must be recalibrated to preserve the target overall type I error.

Check Your Understanding

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

For a fixed $a$, what happens to the interim cutoff when $\Delta$ increases by $0.10$ at information fraction $t=0.25$?
At $t=0.50$, calculate the ratio $b_1/b_2$ for $\Delta=0.25$, with the final look at $t_2=1$.
Why must $a$ be recalibrated after increasing $\Delta$ if the target overall type I error is to remain unchanged?
For the two-look example with $\Delta=0.25$ and calibrated $a\approx2.04$, calculate the interim cutoff.

Solutions. (1) The cutoff is multiplied by $0.25^{0.10}\approx0.8706$, so it falls by about $12.9\%$ at fixed $a$. (2) The ratio is $0.5^{0.25-0.5}=0.5^{-0.25}\approx1.1892$, so the interim cutoff is about $1.1892$ times the final cutoff. (3) Increasing $\Delta$ lowers interim cutoffs at fixed $a$, enlarging the event of crossing at least one boundary under $H_0$; increasing $a$ is needed to bring that probability back to the target. (4) $b_1=2.04(0.5)^{-0.25}=2.04(1.1892)\approx2.426$.

\(\Delta\)	Calibrated \(a=b_2\)	Interim cutoff \(b_1\)
0	1.98	\(1.98(0.5)^{-0.5}\approx2.800\)
0.25	2.04	\(2.04(0.5)^{-0.25}\approx2.426\)
0.50	2.18	\(2.18(0.5)^0=2.180\)