Pocock And O'Brien-Fleming Within Wang-Tsiatis

One Boundary Family, Two Familiar Shapes

Pocock and O’Brien-Fleming boundaries are often introduced as different ways to distribute stringency across interim analyses. The Wang-Tsiatis family provides a compact way to relate their shapes: a single exponent determines how the critical value changes as information accumulates. This relationship is useful for understanding what the named boundaries have in common without confusing their shapes with the separate calculation that sets their overall type I error.

Consider a group sequential efficacy design with $K$ planned analyses. Let $I_k$ be the cumulative information at look $k$, and let $t_k=I_k/I_K$ be its information fraction, where $0<t_1<\cdots<t_K=1$. Let $Z_k$ be the standardized test statistic at look $k$. We use the conventional one-sided upper-boundary parameterization below; a lower efficacy boundary for a treatment benefit coded in the opposite direction can be obtained by reversing the sign of the statistic.

Definition: A conventional Wang-Tsiatis upper efficacy boundary has critical value $$ b_k=\frac{a}{t_k^{\Delta}}, \qquad k=1,\ldots,K, $$ and stops for efficacy at look $k$ if $Z_k\ge b_k$. Here $a>0$ is a constant determined by calibration, $t_k$ is the information fraction, and $\Delta$ is the Wang-Tsiatis shape parameter. The same $a$ is used across looks in a given design.

In the conventional range $0\le \Delta\le 1/2$, the endpoints of the parameter range give two familiar boundary shapes. The names refer to the pattern of critical values across information fractions. They do not, by themselves, specify a complete design: the number and timing of looks, the statistic’s joint distribution, and the target type I error are also needed to calibrate $a$.

The Assumptions Behind the Shape Comparison

The formula defines a pattern of cutoffs, but a valid group sequential design also requires a model for the statistics across looks. For the standard canonical model under the null hypothesis $H_0$, the vector of test statistics is jointly normal, with mean zero and covariance determined by the information fractions:

(Z_1,\ldots,Z_K)\sim N(\mathbf{0},\boldsymbol{\Sigma}), \qquad \operatorname{Cov}(Z_i,Z_j)=\sqrt{\frac{t_i}{t_j}} \quad\text{for }i\le j.

This model is appropriate when the test statistics are formed from accumulating information and have the usual independent-increments structure; the standard normal approximation must also be adequate for the data and test. The analysis looks and information fractions are planned, and the test statistic’s direction and scaling are consistent across looks. Under these assumptions, calibration chooses $a$ so that the probability, under $H_0$, of crossing the upper boundary at any planned look equals the prespecified one-sided familywise level $\alpha$:

P_{H_0}\!\left(\bigcup_{k=1}^{K}\{Z_k\ge a/t_k^\Delta\}\right)=\alpha.

The covariance matters because the $K$ test statistics are not independent. Consequently, a critical value cannot generally be selected by treating each look as an isolated test and assigning it the full $\alpha$. Here the focus is the relation between $\Delta$ and the boundary’s shape; the equation above explains why the constant $a$ must be calibrated for the actual design rather than inferred from the exponent alone.

Why the Endpoint Choices Give Pocock and O’Brien-Fleming Shapes

At $\Delta=0$, every positive information fraction raised to the zero power is one. The Wang-Tsiatis formula therefore becomes a constant cutoff:

b_k=\frac{a}{t_k^0}=a.

This is the defining geometric feature of the Pocock boundary: the same critical value is used at each analysis. Its value is not simply the single-look critical value $z_{1-\alpha}$; it is calibrated to control the probability of crossing at any of the planned looks.

At $\Delta=1/2$, the cutoff is inversely proportional to the square root of the information fraction:

b_k=\frac{a}{\sqrt{t_k}}.

This is the conventional O’Brien-Fleming information-scaled shape. Early in the study, when $t_k$ is small, the cutoff is high. As information accumulates, the cutoff declines, reaching $a$ at the final analysis because $t_K=1$. The familiar shape therefore follows directly from the Wang-Tsiatis formula, rather than requiring a separate exponent at every look.

Key result: Under the conventional Wang-Tsiatis parameterization $b_k=a/t_k^\Delta$, setting $\Delta=0$ gives the constant-cutoff Pocock shape, and setting $\Delta=1/2$ gives the inverse-square-root-information O’Brien-Fleming shape. In both cases, the design-specific constant $a$ still requires calibration.

A Short Derivation Using Relative Cutoffs

The relationship can be checked without knowing the calibration constant. At the final analysis $t_K=1$, so $b_K=a$. Dividing the cutoff at any look by this final cutoff gives:

\frac{b_k}{b_K} = \frac{a/t_k^\Delta}{a/1^\Delta} = t_k^{-\Delta}.

Thus, the relative shape is determined entirely by $t_k$ and $\Delta$. When $\Delta=0$, $t_k^{-\Delta}=1$ at every look, so all cutoffs equal the final cutoff. When $\Delta=1/2$, the ratio is $1/\sqrt{t_k}$, which exceeds one for every interim look with $t_k<1$. This proves that the interim cutoff is higher than the final cutoff for the O’Brien-Fleming-shaped member.

This ratio argument also shows why shape comparisons should separate the exponent from the calibration constant. If two designs use different values of $\Delta$, their calibrated values of $a$ will generally differ, even if they use the same planned looks and target $\alpha$. Comparing their raw cutoffs while holding $a$ fixed isolates shape, but does not in general compare two fully calibrated designs.

Worked Example: The Pocock Shape at Three Looks

Suppose a hypothetical design has planned information fractions $t_1=0.25$, $t_2=0.50$, and $t_3=1.00$. For illustrating the shape only, take $a=2.40$ and set $\Delta=0$. This value of $a$ is illustrative, not a claim that it calibrates any particular type I error rate.

The cutoff at each look is $b_k=a/t_k^0$. Since $t_k^0=1$, the calculations are:

b_1=\frac{2.40}{0.25^0}=2.40,\qquad b_2=\frac{2.40}{0.50^0}=2.40,\qquad b_3=\frac{2.40}{1.00^0}=2.40.

The boundary is flat: the hypothetical design would stop for upper-boundary efficacy at a look if that look’s $Z_k$ were at least $2.40$. These calculations demonstrate the Pocock shape only. To use the boundary as a design controlling a specified $\alpha$, the constant must be calibrated under the joint null model for the specified schedule and number of looks.

Worked Example: The O’Brien-Fleming Shape at Three Looks

Use the same hypothetical information fractions, and now set $\Delta=1/2$. For a shape calculation, let $a=2.00$, again without claiming that this constant is calibrated to a particular $\alpha$.

The cutoffs are $b_k=2.00/\sqrt{t_k}$:

b_1=\frac{2.00}{\sqrt{0.25}}=4.000,\qquad b_2=\frac{2.00}{\sqrt{0.50}}\approx2.828,\qquad b_3=\frac{2.00}{\sqrt{1.00}}=2.000.

The interim thresholds are more stringent than the final threshold, and the first-look cutoff is twice the final cutoff. For example, if the hypothetical first-look statistic were $Z_1=3.20$, it would not cross the $4.000$ boundary; a statistic of $4.10$ would cross it. The numbers show the information-scaled shape, not a calibrated design’s promised error rate.

Intermediate Values and the Role of Calibration

Values of $\Delta$ strictly between zero and one-half generate shapes between these endpoints in relative-cutoff terms. At any fixed interim information fraction $t_k<1$, increasing $\Delta$ increases $t_k^{-\Delta}$, and therefore increases the cutoff relative to the final cutoff. This follows because $\log(t_k)<0$, so $t_k^{-\Delta}=\exp(-\Delta\log(t_k))$ increases with $\Delta$. This is a statement about relative shape; after calibration, the constant $a$ may change too.

Worked Example: An Intermediate Wang-Tsiatis Shape

Consider the same three information fractions and choose $\Delta=0.25$. Hold $a=2.40$ fixed solely to compare shapes on a common scale. The formula gives:

b_1=\frac{2.40}{0.25^{0.25}} =\frac{2.40}{\sqrt{\sqrt{0.25}}} =\frac{2.40}{0.7071} \approx3.394,

b_2=\frac{2.40}{0.50^{0.25}} \approx\frac{2.40}{0.8409} \approx2.854,\qquad b_3=\frac{2.40}{1.00^{0.25}}=2.400.

The first-look cutoff is above the final cutoff but below the corresponding Pocock cutoff of $2.40$ only if the constants are held equal? In fact, with the same $a=2.40$, it is $3.394$, which is higher than the flat Pocock cutoff $2.40$, and lower than the O’Brien-Fleming cutoff $2.40/\sqrt{0.25}=4.800$. At the second look it is also higher than $2.40$ and lower than $2.40/\sqrt{0.50}\approx3.394$. This confirms the intermediate relative shape. A real comparison of calibrated boundaries would use a separately calibrated $a$ for each value of $\Delta$.

Shape Is Not the Same as an Entire Boundary Procedure

The O’Brien-Fleming name is also used for designs constructed by alpha-spending methods. Such a method specifies how cumulative type I error is spent as information accrues and derives critical values accordingly. The Wang-Tsiatis member at $\Delta=1/2$ has the O’Brien-Fleming information-scaled shape, but that fact alone does not establish that its calibrated cutoffs exactly match those from every O’Brien-Fleming spending procedure. The calibration approach and design details matter.

Similarly, Pocock’s constant-cutoff shape does not mean that each look is an independent test at level $\alpha$. The relevant error control is the probability of crossing at least one planned boundary under the global null. For an efficacy design with a different number of looks or a different information schedule, the calibrated constant can change even though $\Delta=0$ still gives a flat boundary.

The canonical model and planned information fractions are assumptions, not decoration. If the statistic does not have the assumed null distribution, the information fractions are misspecified, or the analysis schedule changes, the calibrated constant may no longer control the intended error rate. Shape recognition cannot repair a mismatch between the design and the model used for calibration.

Diagnostics and Common Pitfalls

Check the final-look identity. Since $t_K=1$, the formula must give $b_K=a$. A different final cutoff signals an arithmetic or parameterization error.
Check monotonicity for the endpoint shapes. At $\Delta=0$, every cutoff must be equal. At $\Delta=1/2$, earlier looks with smaller $t_k$ must have larger cutoffs.
Do not confuse $a$ with $\alpha$. The constant is a critical-value scale, not the type I error probability. It is chosen by evaluating the crossing probability under the joint null distribution.
Do not assume nominal per-look testing controls overall error. Repeated looks create multiple opportunities to cross. The design’s familywise error concerns crossing at any planned look.
Do not label every spending-function boundary with the same exponent as identical. The $\Delta=1/2$ Wang-Tsiatis choice identifies an O’Brien-Fleming-shaped member, but another construction’s exact cutoffs must be checked rather than assumed equal.

The main alternatives differ in what they specify directly. A Pocock design emphasizes a common cutoff across looks; an O’Brien-Fleming design emphasizes very stringent early evidence with a lower threshold near the final analysis. The Wang-Tsiatis family expresses a range of relative shapes with one parameter, while calibration supplies the constant required for the chosen design. This description identifies the geometry; it does not decide which shape should be selected for a particular trial.

Check Your Understanding

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

For $b_k=a/t_k^\Delta$, what value of $\Delta$ gives the constant Pocock shape, and why?
A design has $t_1=0.36$, $t_2=1$, $\Delta=1/2$, and illustrative $a=1.80$. Calculate both cutoffs and their ratio.
At $t=0.25$, compare the relative cutoff $b_k/b_K$ for $\Delta=0$ and $\Delta=1/2$.
Why does knowing $\Delta=1/2$ not by itself give a calibrated O’Brien-Fleming design at a specified $\alpha$?
True or false: if $a$ is held constant, increasing $\Delta$ raises the interim cutoff whenever $0<t_k<1$. Explain.

Solutions. (1) $\Delta=0$, because $t_k^0=1$ and hence $b_k=a$ at every look. (2) $b_1=1.80/\sqrt{0.36}=3.000$, and $b_2=1.80/\sqrt{1}=1.800$; their ratio is $3.000/1.800=1.667=1/\sqrt{0.36}$. (3) The ratio is $t^{-\Delta}$: it is $0.25^0=1$ for $\Delta=0$, and $0.25^{-1/2}=2$ for $\Delta=1/2$. (4) The exponent gives the relative shape, but $a$ must be calibrated using the planned looks, information fractions, joint null distribution, and target $\alpha$. (5) True. For $0<t_k<1$, $\log(t_k)<0$, so $t_k^{-\Delta}=\exp(-\Delta\log(t_k))$ increases with $\Delta$; with $a$ fixed, so does $b_k$.