Power Spending Family

What the Power Spending Family Controls

A group sequential design permits formal analyses before the planned final analysis. Repeatedly testing at the nominal significance level can inflate the probability of at least one false rejection of $H_0$. An alpha spending function instead specifies how much of the overall type I error is available cumulatively by each information time. The power spending family offers a compact way to control the timing of that expenditure with one shape parameter.

Let $t$ denote the information fraction at an analysis, with $0\leq t\leq 1$. It is the information available by that analysis divided by the planned information at the final analysis. Thus, $t=1$ at the final analysis. It need not equal the fraction of calendar time elapsed or the fraction of participants enrolled.

Definition: For a one-sided design with total type I error level $\alpha$, the power spending function is $$ A(t)=\alpha t^\rho,\qquad 0\leq t\leq 1,\quad \rho>0. $$ Here $A(t)$ is the cumulative type I error expenditure by information fraction $t$, and $\rho$ is the shape parameter. For a two-sided design, $A(t)$ can be defined as the cumulative total error expenditure across both tails.

The condition $\rho>0$ ensures that $A(0)=0$, that $A(t)$ is nondecreasing, and that $A(1)=\alpha$. In other words, the whole error budget is available by the final analysis, but none is spent at information time zero. The formula describes cumulative spending, not the significance level to use independently at each analysis.

How the Shape Parameter Changes Timing

For any interim time $0<t<1$, raising a number below one to a larger positive power makes it smaller. Consequently, if $\rho>1$, then $A(t)<\alpha t$: cumulative spending is below the linear schedule. If $0<\rho<1$, then $A(t)>\alpha t$: spending is above the linear schedule. When $\rho=1$, spending is linear in information time.

The derivative makes this difference in shape explicit:

\frac{dA(t)}{dt}=\alpha\rho t^{\rho-1},\qquad 0<t<1.

For $\rho>1$, the rate of expenditure increases with $t$; for $0<\rho<1$, it decreases. This derivative describes the slope of the cumulative spending curve, not a separate per-look test level. In particular, when $0<\rho<1$, the slope grows without bound as $t$ approaches zero, even though $A(t)$ itself approaches zero continuously.

Worked Example: Comparing Three Spending Shapes

Suppose a one-sided design has $\alpha=0.05$. Compare cumulative spending at information fractions $t=0.25$ and $t=0.50$ for $\rho=0.5$, $\rho=1$, and $\rho=2$.

At $t=0.25$, the formula is $A(0.25)=0.05(0.25^\rho)$:

For $\rho=0.5$, $A(0.25)=0.05\sqrt{0.25}=0.05(0.5)=0.0250$.
For $\rho=1$, $A(0.25)=0.05(0.25)=0.0125$.
For $\rho=2$, $A(0.25)=0.05(0.25^2)=0.05(0.0625)=0.003125$.

At $t=0.50$:

For $\rho=0.5$, $A(0.50)=0.05\sqrt{0.50}\approx0.03536$.
For $\rho=1$, $A(0.50)=0.05(0.50)=0.0250$.
For $\rho=2$, $A(0.50)=0.05(0.50^2)=0.0125$.

The $\rho=0.5$ curve spends more than half the total alpha by $t=0.50$; the $\rho=2$ curve spends only one quarter. All three spend exactly $0.05$ by $t=1$. These calculations compare cumulative budgets, not the probability of rejection at an individual look.

Choosing a Shape From a Cumulative Target

The parameter can be set by specifying what fraction of the total error budget should have been spent by a particular information fraction. If the desired fraction is $q$, where $0<q<1$, at time $t_0$, then $A(t_0)=q\alpha$. Dividing by $\alpha$ and taking logarithms gives:

t_0^\rho=q \quad\Longrightarrow\quad \rho=\frac{\log(q)}{\log(t_0)}.

Because both logarithms are negative for values between zero and one, this gives a positive $\rho$. The result is a shape parameter that matches one specified point on the curve; it does not imply that this timing target is optimal for power or for any particular clinical objective.

Worked Example: Setting rho From a Target at One-Quarter Information

Consider a hypothetical one-sided design with $\alpha=0.025$. The design team wants 10% of the total alpha spent by $t=0.25$. Thus $q=0.10$, and

\rho=\frac{\log(0.10)}{\log(0.25)} =\frac{-2.302585}{-1.386294} \approx1.660964.

At $t=0.25$, the cumulative expenditure is $A(0.25)=0.025(0.25^{1.660964})=0.025(0.10)=0.002500$. At $t=0.50$, use $0.50^{1.660964}=\sqrt{0.25^{1.660964}}=\sqrt{0.10}\approx0.3162278$, so

A(0.50)=0.025(0.3162278)\approx0.0079057.

The additional cumulative expenditure between $t=0.25$ and $t=0.50$ is $A(0.50)-A(0.25)\approx0.0079057-0.002500=0.0054057$. The values are probabilities on the type I error scale. The increment is the increase in the cumulative budget over that interval; it is not automatically the unconditional probability of crossing a boundary during only that interval.

From Spending to Sequential Boundaries

A spending function gives cumulative targets; boundary calibration translates those targets into critical values for the sequence of test statistics. To make that translation, the design needs a specified joint null distribution for the interim statistics. In the canonical group sequential model, at information fractions $t_1<\cdots<t_K=1$, each standardized statistic has a standard normal marginal distribution under $H_0$, and the correlation between statistics at $s\leq t$ is $\sqrt{s/t}$. This model and the planned information times must reasonably represent the design. The relevant normal approximation must also be adequate for the chosen test statistic.

For a one-sided upper-boundary design, let $Z(t_j)$ be the standardized statistic at look $j$, and let $b_j$ be its upper efficacy boundary. Calibration chooses the boundaries so that, under $H_0$, the probability of having crossed at least one upper boundary by look $i$ is $A(t_i)$. For a symmetric two-sided design, the corresponding event is crossing either boundary, and the cumulative target is the total error across both tails. The previous cumulative probability is accounted for when solving each later boundary; the increment $A(t_i)-A(t_{i-1})$ is not generally a standalone critical probability at look $i$.

Worked Example: Calibrating a First-Look Boundary

Suppose a hypothetical one-sided design has $\alpha=0.04$, uses $\rho=2$, and conducts its first analysis at $t_1=0.50$. Under the canonical null model, $Z(t_1)$ has a standard normal distribution. The cumulative target at the first analysis is

$$ A(0.50)=0.04(0.50^2)=0.0100. $$

At the first look there have been no earlier opportunities to cross a boundary. Therefore choose $b_1$ to satisfy $P(Z(t_1)\geq b_1\mid H_0)=0.0100$. If $\Phi$ is the standard normal cumulative distribution function, then

b_1=\Phi^{-1}(1-0.0100)=\Phi^{-1}(0.9900)\approx2.3263.

Thus the first-look rule rejects $H_0$ for efficacy if the observed $Z(t_1)$ is at least 2.3263. Under the stated null model, the probability of crossing at this first look is 0.0100. A later boundary cannot generally be found by treating the remaining $0.0300$ as an independent per-look level: it must be calibrated jointly with the first-look statistic to attain the cumulative target at the later time.

Comparison, Diagnostics, and Limitations

The earlier material on the Hwang-Shih-DeCani spending family introduces a different way to shape cumulative expenditure. The power family is especially transparent for comparing a specified information fraction with the total budget: $A(t)/\alpha=t^\rho$, and a single target point gives a direct logarithmic solution for $\rho$. The Hwang-Shih-DeCani family uses a normalized exponential shape parameter instead. Neither family is universally preferable; the choice expresses a design preference and must be followed by valid boundary calibration. A fixed per-look significance threshold is different: without a sequential adjustment, it does not in general control the overall type I error at $\alpha$.

Several checks help identify a misapplied power spending design:

Confirm $0< t_1<\cdots<t_K=1$ and that the values are information fractions, not automatically enrollment or calendar-time fractions.
Check that $\rho>0$, $A(0)=0$, and $A(1)=\alpha$. A nonpositive $\rho$ fails the intended zero-at-start behavior or produces an undefined value at zero.
For $\rho>1$, verify that interim cumulative expenditure is below the linear schedule; for $0<\rho<1$, verify that it is above it.
Check that boundary calibration uses the joint null distribution and the actual planned look times. A table of $A(t_i)-A(t_{i-1})$ alone is not a set of independent per-look testing levels.

The family’s simplicity is also a limitation. One parameter cannot independently set the curve at several information times, and a target cumulative expenditure does not itself establish desirable power under alternatives. The name can mislead: power spending is a family for spending type I error, not a formula for statistical power. If timing, test direction, or information differs from the design assumptions, the planned calibration may no longer deliver the intended error control; the design assumptions and analysis plan must be checked rather than interpreting the curve in isolation.

Deriving the Shape Comparisons

For completeness, compare any two positive shape parameters $\rho_2>\rho_1$. For $0<t<1$, $\log(t)<0$, so $(\rho_2-\rho_1)\log(t)<0$. Exponentiating gives $t^{\rho_2-\rho_1}<1$, and hence $t^{\rho_2}<t^{\rho_1}$. Multiplying by the positive $\alpha$ shows that the larger parameter spends less cumulatively at every strictly interim time. Both curves still meet at $t=1$, since $1^\rho=1$. This proves the timing comparison without relying on a plotted curve.

The derivative supplies a second check. If $\rho>1$, then $\alpha\rho t^{\rho-1}$ increases with $t$, so the cumulative curve is convex and its slope grows. If $0<\rho<1$, the derivative decreases with $t$, so the cumulative curve is concave and its slope diminishes. For $\rho=1$, the derivative is the constant $\alpha$. These curvature statements concern expenditure against information fraction; they do not state how the trial’s power changes.

Key takeaway: $A(t)=\alpha t^\rho$ spends exactly $\alpha$ by the final information time. Values $\rho>1$ delay cumulative expenditure relative to a linear schedule, values $0<\rho<1$ spend more early, and $\rho=1$ gives linear spending. Boundaries must still be calibrated using the sequential null distribution.

Check Your Understanding

Try each question, then check the solutions below.

For $\alpha=0.05$, $\rho=2$, and $t=0.40$, calculate $A(t)$.
What value of $\rho$ spends 25% of the total alpha by $t=0.50$?
At $t=0.36$, compare cumulative spending for $\alpha=0.05$, $\rho=0.5$, and $\rho=1$.
Why is $A(t_i)-A(t_{i-1})$ not generally the unconditional probability of crossing the boundary at look $i$?

Solutions. (1) $A(0.40)=0.05(0.40^2)=0.05(0.16)=0.00800$. (2) Solve $0.50^\rho=0.25$, giving $\rho=\log(0.25)/\log(0.50)=2$. (3) For $\rho=0.5$, $A(0.36)=0.05\sqrt{0.36}=0.05(0.60)=0.0300$; for $\rho=1$, $A(0.36)=0.05(0.36)=0.0180$. The first shape has spent more by that time. (4) The statistics across looks are dependent, and $A(t_i)$ is the probability of crossing at least one boundary by look $i$. The increase in that cumulative probability is not generally the unconditional probability of crossing specifically at look $i$; the boundary must be calibrated with the joint null distribution.

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