A Three-Look Design With Unequal Spacing

Why Unequal Spacing Matters

A three-look design is not defined just by its total significance level. Its planned information fractions determine how strongly the test statistics at different looks are related, and therefore how much probability of crossing a later boundary is already accounted for by earlier looks. When the looks are unequally spaced, the increments of information differ; equal increments of alpha spending do not imply equal increments of information or equal efficacy cutoffs.

This tutorial calibrates an upper-boundary design with three planned information fractions. It uses a one-sided test and a specified spending function. The method applies to a canonical group sequential test under the null: information increases over time, and the standardized score at each look has the joint distribution stated below. No futility boundaries are included.

Definition: Let $0<t_1<t_2<t_3=1$ be the information fractions at three planned looks. Under $H_0$, the standardized test statistics $Z_1,Z_2,Z_3$ have mean zero, variance one, and covariance $$ \operatorname{Corr}(Z_i,Z_j)=\sqrt{\frac{t_i}{t_j}},\qquad i<j. $$ An upper efficacy boundary is a sequence $c_1,c_2,c_3$ such that the trial rejects $H_0$ at the first look $i$ for which $Z_i\ge c_i$.

This canonical model assumes the test statistic at each look is standardized using the information available at that look, the information fractions are correctly specified, and the score process has independent increments on the information scale. Under these assumptions, $Z_i$ conditional on $Z_{i-1}=z$ is normal with mean $\sqrt{t_{i-1}/t_i}\,z$ and variance $1-t_{i-1}/t_i$. Thus the sequential probability calculation can be performed recursively. If the test statistics or analysis plan do not follow this model, the resulting boundaries need not control the intended error rate.

Spending Targets and Recursive Calibration

A cumulative alpha-spending function $A(t)$ specifies how much of the one-sided Type I error budget is intended to have been spent by information fraction $t$. For the design below, use linear spending, $A(t)=\alpha t$, with $\alpha=0.025$. The targets at the three looks are $A(t_1), A(t_2), A(t_3)$. These are cumulative crossing probabilities under $H_0$, not the probabilities of crossing at each look alone.

To calibrate a boundary, calculate the probability of avoiding all boundaries through that look. At look 1, $Z_1\sim N(0,1)$, so $c_1$ is the upper-tail normal quantile that gives crossing probability $A(t_1)$. At later looks, the calculation must exclude paths that crossed earlier. The conditional normal distribution supplies the transition from one look to the next.

q_1(z)=\phi(z)\mathbf{1}(z<c_1),\qquad q_i(z)=\mathbf{1}(z<c_i)\int_{-\infty}^{c_{i-1}}q_{i-1}(x) \frac{1}{s_i}\phi\left(\frac{z-r_i x}{s_i}\right)\,dx, $$ $$ r_i=\sqrt{\frac{t_{i-1}}{t_i}},\qquad s_i=\sqrt{1-\frac{t_{i-1}}{t_i}},\qquad \int_{-\infty}^{c_i}q_i(z)\,dz=1-A(t_i).

Here $\phi$ is the standard normal density, $q_i(z)$ is the null density at look $i$ among paths that have not crossed any boundary through that look, and $\mathbf{1}(\cdot)$ is an indicator. The last equation is the calibration equation: choose $c_i$ so the probability of surviving through look $i$ is $1-A(t_i)$. Numerical integration and a one-dimensional root finder are typically used at looks 2 and 3.

Justification: At each look, the paths that have already crossed are excluded from the surviving density. The probability of surviving through look $i$ is therefore $1-A(t_i)$ when the cutoff solves the displayed equation. Its complement is the cumulative probability of at least one crossing, $A(t_i)$. Subtracting consecutive cumulative targets gives the incremental crossing probabilities under $H_0$, so the final cumulative probability is $A(1)=\alpha$.

Worked Example: Calibrating Three Unequally Spaced Looks

Consider a hypothetical one-sided design with information fractions $t_1=0.20$, $t_2=0.55$, and $t_3=1.00$, using $\alpha=0.025$ and linear spending $A(t)=0.025t$. The information fractions are strictly increasing and end at 1, as required. The canonical null model is assumed to hold.

Step 1: Calculate cumulative and incremental spending. The cumulative targets are $A(0.20)=0.025(0.20)=0.00500$, $A(0.55)=0.025(0.55)=0.01375$, and $A(1)=0.025$. Thus the increments are $0.00500$, $0.01375-0.00500=0.00875$, and $0.025-0.01375=0.01125$.

Step 2: Calibrate the first cutoff. At the first look, the probability of crossing is $P(Z_1\ge c_1)=0.00500$. Therefore $$ c_1=\Phi^{-1}(1-0.00500)=\Phi^{-1}(0.99500)\approx 2.576. $$

Step 3: Calibrate the second cutoff. The correlation between the first two statistics is $\sqrt{0.20/0.55}\approx0.603$, and the conditional standard deviation at look 2 is $\sqrt{1-0.20/0.55}\approx0.798$. Solve the recursive survival equation so the probability of surviving both looks is $1-0.01375=0.98625$. Numerical Gaussian integration gives $c_2$ approximately $2.33$. The incremental probability of first crossing at look 2 is $0.00875$, not the marginal probability $P(Z_2\ge c_2)$.

Step 4: Calibrate the final cutoff. The conditional correlation parameter from look 2 to look 3 is $\sqrt{0.55/1}\approx0.742$, with conditional standard deviation $\sqrt{1-0.55}\approx0.671$. Solving for survival probability $1-0.025=0.975$ gives $c_3$ approximately $2.16$. The final incremental crossing probability is $0.01125$.

The approximate cutoffs are rounded; numerical calibration uses their unrounded values. Under $H_0$, the cumulative rejection probabilities by looks 1, 2, and 3 are $0.00500$, $0.01375$, and $0.0250$, respectively. Their increments sum to $0.0250$, the specified one-sided Type I error rate.

What the Recursion Is Correcting

A common shortcut is to choose each cutoff from its marginal normal tail, as though each look were a separate test. That ignores dependence between looks and counts some paths more than once. The second cutoff in the example illustrates the correction: the calibration equation concerns paths with $Z_1<c_1$, not every possible value of $Z_2$. The same exclusion must be applied at look 3 for both earlier boundaries.

Worked Example: Why the Second Cutoff Is Not a Marginal Quantile

Continue with the design above. Suppose, incorrectly, that the second cutoff is selected so that the marginal tail probability is the incremental target $0.00875$. This gives $c_2=\Phi^{-1}(1-0.00875)$, approximately $2.38$. This is not the correct calibration equation.

Under the canonical model, $Z_2\mid Z_1=x\sim N(0.603x,0.798^2)$. The probability of crossing at look 2 without an earlier crossing is $$ \int_{-\infty}^{2.576}\phi(x)\left[1-\Phi\left(\frac{2.38-0.603x}{0.798}\right)\right]\,dx. $$

This integral is less than the marginal tail $P(Z_2\ge2.38)=0.00875$, because it excludes paths with $Z_1\ge2.576$. The first-look crossing probability $0.00500$ must also be included in the cumulative probability through look 2. Consequently, the incorrect marginal-quantile method does not generally produce the cumulative target $0.01375$; the cutoff must instead be found by solving the no-prior-crossing recursion. In the calibrated design, the second-look increment is $0.00875$ after earlier crossings have been excluded.

Unequal spacing changes the conditional distributions used in this calculation. For example, the correlation from look 1 to look 2 is about $0.603$, while that from look 2 to look 3 is about $0.742$. The conditional variability is also different at the two transitions. A boundary calibrated for one set of information fractions cannot simply be reused after changing those fractions, even if the total alpha and spending function remain the same.

Rejection Probabilities Under an Alternative

The calibrated null crossing probabilities control Type I error, but they do not describe the probability of rejecting under a treatment effect. Under a simple canonical alternative with standardized drift $\delta$, assume $Z_i\sim N(\delta\sqrt{t_i},1)$, with the same covariance structure as under $H_0$. The probability of rejecting by the final look is the probability of crossing at least one of the calibrated upper boundaries. It can be computed by the same recursion, using the alternative means in the transition distributions.

Worked Example: Rejection Probability Under a Specified Alternative

For illustration, retain the hypothetical design and take $\delta=1.5$. This value is a standardized drift per square root of information, not an effect measured in clinical units. The assumed alternative means at the three looks are $1.5\sqrt{0.20}\approx0.671$, $1.5\sqrt{0.55}\approx1.112$, and $1.5\sqrt{1}=1.500$. The previously calibrated cutoffs remain approximately $2.576$, $2.33$, and $2.16$.

At the first look, the marginal crossing probability is $P(Z_1\ge2.576)=1-\Phi(2.576-0.671)\approx0.0284$. At later looks, marginal tails cannot simply be added to this probability: they overlap with paths already rejected. Applying the no-crossing recursion under the alternative, with the means above and the same transition variances, gives a total rejection probability of roughly $0.28$ by the end of the trial. This is the power under the specified canonical alternative, subject to the model assumptions; it is not the Type I error rate.

Diagnostics, Alternatives, and Limitations

A reliable calibration should be checked at every look, not only at the final one. Recalculate the cumulative probability of any crossing under $H_0$ using the final, unrounded boundaries. It should match each spending target within the numerical tolerance of the integration and root-finding procedure. The final value should be $\alpha$. Also check that the boundaries are finite, that the computed surviving probabilities decrease as looks are added, and that each incremental crossing probability is nonnegative.

If the cumulative crossing probability overshoots a target, likely causes include using a marginal tail rather than the survival recursion, mixing up cumulative and incremental spending, or using the wrong information fractions in a conditional correlation. If a root finder reports failure, inspect the objective at plausible low and high cutoffs and verify that the numerical integration is stable. The earlier material on numerical integration and recursive boundary calibration provides the general computational framework; in this three-look calculation, both transition steps must use the specified unequal fractions.

The main alternative is to specify a boundary family directly, rather than begin with a spending function and solve for cutoffs. Direct boundary choices may be useful when a particular timing profile is desired, but their cumulative Type I error must still be evaluated. Spending-function calibration instead begins with the desired cumulative error allocation and derives boundaries consistent with the planned timing. Neither approach removes the need to specify the test direction, information model, and analysis schedule.

This calculation addresses efficacy boundaries in a planned three-look design. It does not account for unplanned analysis times, changes to the design after observing data, or departures from the canonical joint distribution. Rounding cutoffs before use can also cause small deviations from the nominal error rate. Report full-precision boundaries for implementation and rounded values only for exposition.

Key takeaway: In an unequally spaced three-look design, each later cutoff is calibrated against the distribution of paths that have not crossed earlier boundaries. Cumulative spending determines the target rejection probability at each look; recursive integration converts those targets into efficacy cutoffs.

Check Your Understanding

Try the questions, then check the solutions below.

With $\alpha=0.025$ and linear spending, what are the cumulative targets at information fractions $0.20$, $0.55$, and $1.00$?
For the first two fractions in the design, what is $\operatorname{Corr}(Z_1,Z_2)$, to three significant figures?
Why is the second-look incremental crossing probability not generally $P(Z_2\ge c_2)$?
If cumulative targets at three looks are $0.004$, $0.012$, and $0.025$, what are the incremental crossing probabilities?

Solutions. (1) The targets are $0.025(0.20)=0.00500$, $0.025(0.55)=0.01375$, and $0.025(1)=0.0250$. (2) $\sqrt{0.20/0.55}\approx0.603$. (3) The marginal tail includes paths that crossed at look 1, whereas the incremental probability counts only paths that avoided the first boundary; the recursion integrates only over $Z_1<c_1$. (4) The increments are $0.004$, $0.012-0.004=0.008$, and $0.025-0.012=0.013$; they sum to $0.025$.