Why the Alternative Changes the Calculation
A group sequential boundary is crossed by a sequence of correlated statistics, not by a single final statistic. Under the null hypothesis \(H_0\), a Gaussian transition model describes how the statistic moves between planned looks. To calculate power for a specified treatment effect, the transition variances and dependence structure remain the same under the canonical model, but the transition means change. Leaving the null transition means unchanged would calculate null crossing probabilities, not power.
This tutorial uses a fixed treatment effect \(\delta\), measured in the units of the treatment-effect parameter, and cumulative information \(I_k\) at look \(k\). The direction of the effect is chosen so that a positive \(\delta\) favors efficacy, which is assessed by crossing an upper boundary. Information is assumed to be known at each look.
This model assumes a stable treatment effect across the information range, a correctly specified information scale, and Gaussian score increments with independent increments over disjoint intervals. The resulting joint distribution of the \(Z_k\) values is not independent: earlier observations contribute to later statistics. The recursion accounts for that dependence through the conditional transition distribution.
Deriving the Alternative-Drift Transition
Let \(I_{k-1}<I_k\), with \(Z_{k-1}=z\). The score increment \(S_k-S_{k-1}\) is independent of the previous score and has mean \(\delta(I_k-I_{k-1})\) and variance \(I_k-I_{k-1}\). Since \(S_{k-1}=z\sqrt{I_{k-1}}\), adding the increment and standardizing by \(\sqrt{I_k}\) gives the transition.
The first term in the conditional mean carries forward the previous statistic. The second is the mean contribution from information newly added at look \(k\). The conditional variance is the variance of that new score increment after standardization. Under \(H_0\), set \(\delta=0\); under the alternative, retain the specified treatment effect in every transition.
If information is expressed as fractions \(t_k=I_k/I_K\) of final information, it is convenient to write \(\theta=\delta\sqrt{I_K}\). Then the mean of \(Z_k\) is \(\theta\sqrt{t_k}\), and the conditional transition becomes
Here \(\theta\) is the effect measured in standard-error units at final information; it is not the treatment effect in its original measurement units. The formula also shows why simply adding \(\delta\sqrt{I_k-I_{k-1}}\) to a standardized statistic is incorrect: the new score contribution must be divided by \(\sqrt{I_k}\), the denominator for the statistic at the current look.
Recursing the Probability of Efficacy Crossing
Let \(u_k\) be the upper efficacy boundary at look \(k\). If there is also a lower stopping boundary \(l_k\), the continuation region is \(C_k=(l_k,u_k)\); with an upper efficacy boundary only, take \(C_k=(-\infty,u_k)\). The survivor density \(f_k(z)\) is an unnormalized density: its integral is the probability of having reached look \(k\) without having crossed a stopping boundary. It is not a density conditional on survival.
At the first look, the process has its alternative marginal distribution, restricted to the continuation region. At each later look, the survivor density is propagated through the alternative transition and restricted again to the new continuation region. Probability entering the upper stopping region is counted as efficacy crossing. With a competing lower stopping boundary, probability entering that lower region is not counted as efficacy crossing.
Here \(\phi(x;m,v)\) is the normal density with mean \(m\) and variance \(v\), \(m_k(z)\) and \(v_k\) are the conditional mean and variance in the transition formula, and \(\mathbf{1}\{\cdot\}\) is an indicator. The upper-crossing mass at the first look is the probability under the alternative that \(Z_1\ge u_1\). At later looks, the crossing mass is obtained by integrating the propagated density over \(y\ge u_k\). The cumulative power is the sum of these upper-crossing masses over the planned looks. If there is no competing lower stopping boundary and the final look stops everyone remaining, power can equivalently be calculated as one minus the final survivor mass.
Worked Calculations
Worked Example: One-Look Power Under a Specified Effect
Suppose a hypothetical trial has one planned look with information \(I_1=64\) in units reciprocal to the squared treatment-effect units. The specified beneficial effect is \(\delta=0.25\) treatment-effect units, and the upper efficacy boundary is \(u_1=2.10\). Assume the canonical Gaussian model applies and there is no competing stopping boundary.
The statistic has mean \(\delta\sqrt{I_1}=0.25\sqrt{64}=2.00\) and variance 1. Therefore,
The calculated probability of crossing the efficacy boundary by this look is about 46.0% under the specified effect. As a check on the direction of the calculation, under \(H_0\) the corresponding crossing probability would be \(1-\Phi(2.10)\approx 0.0179\), which is smaller because the alternative shifts the statistic toward the upper boundary.
Worked Example: Two-Look Recursive Power
Consider a hypothetical two-look trial with information fractions \(t_1=0.50\) and \(t_2=1.00\), specified final standardized effect \(\theta=1.50\), and upper boundaries \(u_1=2.50\) and \(u_2=2.00\). There is no lower stopping boundary. Assume independent Gaussian score increments and the stated information fractions.
At the first look, the mean of \(Z_1\) is \(1.50\sqrt{0.50}=1.061\). The first-look efficacy-crossing probability is
For the second look, the conditional mean and variance are
Only paths with \(Z_1<2.50\) remain available to cross at the second look. Their contribution is the integral of the first-look density times the conditional probability of crossing \(2.00\). A one-dimensional numerical integration gives approximately \(0.243\) for this remaining-path contribution. Thus the cumulative power is
The probability of crossing an efficacy boundary by either look is therefore about 31.8% under this specified effect. The integral matters: using the final marginal probability alone would include paths that already crossed the first boundary and would not represent the trial's stopping rule.
Worked Example: Checking a Conditional Transition
Use the two-look information fractions and effect from the preceding calculation. Given \(Z_1=1.20\), calculate the second-look probability of crossing \(u_2=2.00\), and compare it with the null transition probability. This conditional comparison checks whether the alternative drift has been included in the right part of the calculation.
Under the alternative, the conditional mean is \(0.7071(1.20)+0.7500=1.5985\), and the standard deviation is \(\sqrt{0.5000}=0.7071\). Hence
Under \(H_0\), the conditional mean is only \(0.7071(1.20)=0.8485\). The corresponding probability is \(1-\Phi((2.00-0.8485)/0.7071)\approx 0.0518\). The larger alternative probability is consistent with a beneficial effect shifting the statistic toward the upper boundary. This is a conditional probability for one state, not the cumulative trial power.
Numerical Checks, Alternatives, and Limitations
In practice, the integrals in the recursion are evaluated numerically, for example by quadrature or a discretized state-space method. The earlier material on Gaussian transition integration and survivor-density recursion supplies the computational framework; the essential change here is to use the alternative conditional mean. At each look, verify that the transition variance is nonnegative, that the mean contains the increment-specific drift, and that the survivor mass does not exceed the mass entering the look.
- Check the information scale. Confirm that \(I_k\) is cumulative and increasing. The transition uses \(I_k-I_{k-1}\), not \(I_k\), for the new score increment.
- Check the drift direction. A beneficial effect should move the mean toward the efficacy boundary under the chosen sign convention. A negative effect may reduce upper-boundary power.
- Check probability accounting. Upper-crossing mass plus lower-crossing mass plus continuing survivor mass should account for the probability that has reached each look, up to numerical integration error.
- Check numerical stability. Refine the grid or quadrature and confirm that cumulative power changes negligibly at the precision being reported. A result outside \([0,1]\), or a materially changing result under modest refinement, signals a computational or implementation problem.
A common mistake is to keep the null transition mean, \(\sqrt{I_{k-1}/I_k}\,z\), and change only the final marginal distribution or boundary. That omits the alternative drift from each new score increment, so the recursive path probabilities are wrong. The correction is to add \(\delta(I_k-I_{k-1})/\sqrt{I_k}\) to every conditional mean, then integrate only over paths that have not already stopped.
This recursive approach respects the sequential stopping rule and can accommodate unequal information increments and look-specific boundaries. A fixed-time calculation using only the final marginal distribution is simpler, but it does not in general give power for a group sequential design. Direct multivariate-normal probability calculations can provide an alternative way to compute crossing probabilities, while simulation can be useful as a rough independent check; neither removes the need to specify the alternative and stopping regions correctly. The recursion also has limitations: it relies on the canonical Gaussian model and specified information times, and its numerical accuracy depends on the integration method. If the treatment effect varies over time, information is misspecified, or the test statistic has materially non-Gaussian behavior, the resulting power may not describe the trial well.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- If the final standardized effect is \(\theta=1.80\) and the information fraction is \(t=0.40\), what is the mean of the standardized statistic at that look?
- For \(t_{k-1}=0.40\), \(t_k=0.70\), \(\theta=1.80\), and \(Z_{k-1}=1.00\), calculate the conditional mean and standard deviation at look \(k\).
- Why is the probability of crossing the final boundary under the final marginal normal distribution not generally the trial's cumulative power?
- Name two checks that can reveal an error in the recursive calculation.
Solutions: (1) The mean is \(\theta\sqrt{t}=1.80\sqrt{0.40}=1.138\). (2) The conditional mean is \(\sqrt{0.40/0.70}(1.00)+1.80(0.30)/\sqrt{0.70}=0.7559+0.6455=1.401\). The conditional standard deviation is \(\sqrt{0.30/0.70}=0.6547\). (3) The final marginal probability includes paths that may already have crossed an earlier boundary and stopped; cumulative power must follow the stopping rule. (4) Suitable checks include verifying the information increments and drift direction, confirming probability-mass accounting, and checking stability under grid or quadrature refinement.