Hwang-Shih-DeCani Spending Family

Why Use a Spending Family?

A spending function describes how a trial allocates its type I error budget over accumulating information. The Hwang-Shih-DeCani (HSD) family is a one-parameter family of cumulative spending curves. Its shape parameter changes when the error budget is spent: a curve can allocate relatively more of the budget early or hold more for later. The formula specifies cumulative targets; it does not, by itself, specify the test statistic boundaries used to attain those targets.

Let $t$ denote the information fraction, with $0 \leq t \leq 1$, and let $\alpha$ be the planned total type I error budget for the spending scheme. In a one-sided efficacy design, $\alpha$ is the one-sided budget. In a two-sided design, the convention for applying the cumulative spending function and allocating its total across tails must be specified as part of the design.

Definition: For shape parameter $\gamma$, the Hwang-Shih-DeCani cumulative spending function is $$ A(t;\gamma)= \begin{cases} \alpha\displaystyle\frac{1-\exp(-\gamma t)}{1-\exp(-\gamma)}, & \gamma \ne 0,\\[1.2ex] \alpha t, & \gamma=0, \end{cases} \qquad 0\leq t\leq 1. $$ Here $A(t;\gamma)$ is the cumulative type I error spending target by information fraction $t$, not the significance level for a single look.

The function is used with an information-based group sequential design. Under the canonical normal model under $H_0$, the standardized statistics $Z_1,\ldots,Z_K$ at planned information fractions $0<t_1<\cdots<t_K=1$ have mean zero, variance one, and covariance $\operatorname{Cov}(Z_i,Z_j)=\sqrt{t_i/t_j}$ for $i\leq j$. The spending curve supplies cumulative null-crossing probability targets at the looks. Boundary construction then uses the joint null distribution of the statistics to match those targets. This interpretation assumes the information fractions and null model used for calibration are appropriate; changing them can change the actual crossing probabilities.

Reading the Shape Parameter

The sign of $\gamma$ determines whether cumulative spending is relatively front-loaded or back-loaded compared with linear spending. If $\gamma>0$, the cumulative curve is concave and lies above $\alpha t$ for interior $t$: more of the total budget is allocated early. If $\gamma<0$, it is convex and lies below $\alpha t$: less is allocated early and relatively more remains for later. At $\gamma=0$, spending is linear in information fraction.

This can be seen from the slope and curvature. For $\gamma\ne 0$,

\frac{\partial A(t;\gamma)}{\partial t} =\frac{\alpha\gamma\exp(-\gamma t)}{1-\exp(-\gamma)}, \qquad \frac{\partial^2 A(t;\gamma)}{\partial t^2} =-\frac{\alpha\gamma^2\exp(-\gamma t)}{1-\exp(-\gamma)}.

For $\alpha>0$, the first derivative is positive for either sign of $\gamma$, so spending increases with information. When $\gamma>0$, the denominator in the second derivative is positive, making the curve concave. When $\gamma<0$, that denominator is negative, making the curve convex. The endpoints are $A(0;\gamma)=0$ and $A(1;\gamma)=\alpha$, so the full budget is reached at the final information fraction.

An equivalent representation makes the shape easier to interpret:

\frac{A(t;\gamma)}{\alpha} = \frac{\int_0^t \exp(-\gamma u)\,du} {\int_0^1 \exp(-\gamma u)\,du}.

For positive $\gamma$, the weighting function $\exp(-\gamma u)$ is larger at early $u$, so the accumulated share by an interior $t$ exceeds the fraction $t$. For negative $\gamma$, it is larger later, so the accumulated share is below $t$. This is a comparison with linear spending, not a claim that a particular interim boundary is higher or lower: boundary values also depend on the joint distribution of the sequential statistics.

Worked Example: Evaluating an Early Target

Suppose a one-sided design has $\alpha=0.05$, $\gamma=2$, and an interim analysis at $t=0.25$. Calculate the cumulative spending target.

Because $\gamma\ne 0$, substitute into the HSD formula:

A(0.25;2) =0.05\frac{1-\exp(-2(0.25))}{1-\exp(-2)} =0.05\frac{1-\exp(-0.5)}{1-\exp(-2)}.

Using $\exp(-0.5)\approx 0.60653$ and $\exp(-2)\approx 0.13534$, the ratio is approximately $0.39347/0.86466\approx 0.45505$. Therefore,

A(0.25;2)\approx 0.05(0.45505)=0.02275.

The cumulative target is about $0.02275$, or 2.275 percentage points, by the first look. It is not a per-look $p$-value cutoff. A sequential boundary must be calibrated so that the probability of having crossed by this look under $H_0$ matches the target.

The Limit at Zero

The nonzero-$\gamma$ expression appears to have a zero denominator as $\gamma$ approaches zero. Its numerator also approaches zero, however, and the ratio has a finite limit. Expanding the exponential around $\gamma=0$ gives

1-\exp(-\gamma t)=\gamma t-\frac{\gamma^2t^2}{2}+O(\gamma^3), \qquad 1-\exp(-\gamma)=\gamma-\frac{\gamma^2}{2}+O(\gamma^3).

For $\gamma\ne 0$, divide numerator and denominator by $\gamma$. The resulting ratio tends to $t$, so

\lim_{\gamma\to 0} A(t;\gamma) =\alpha\lim_{\gamma\to 0} \frac{t-\gamma t^2/2+O(\gamma^2)} {1-\gamma/2+O(\gamma^2)} =\alpha t.

Thus the definition at $\gamma=0$ is not a separate, incompatible rule. It is the continuous extension of the nonzero-parameter formula. The first-order expansion also shows how the curve departs from linear spending near zero:

A(t;\gamma)=\alpha t\left[1+\frac{\gamma(1-t)}{2}+O(\gamma^2)\right].

For interior $t$, a small positive $\gamma$ raises cumulative spending slightly above $\alpha t$, and a small negative $\gamma$ lowers it slightly. At $t=0$ and $t=1$, the endpoints remain fixed.

Worked Example: The Linear Limit

Let $\alpha=0.04$ and consider $t=0.30$. At $\gamma=0$, the spending is directly

A(0.30;0)=\alpha t=0.04(0.30)=0.012.

To check the limiting interpretation, take $\gamma=0.01$, which is close to zero. Then

A(0.30;0.01) =0.04\frac{1-\exp(-0.003)}{1-\exp(-0.01)} \approx 0.04\frac{0.0029955}{0.0099502} \approx 0.01204.

This is close to $0.012$, the linear-spending value, and is slightly larger because the parameter is positive. As $\gamma$ approaches zero from either side, the value approaches $0.012$.

Comparing Parameter Values and Look Increments

A useful comparison holds $\alpha$ and $t$ fixed while changing $\gamma$. For example, at the midpoint $t=0.5$, positive and negative values of equal magnitude illustrate the opposite shapes. Once cumulative targets are known, the target allocated between successive looks is their difference. If the planned looks occur at $t_1,\ldots,t_K$, the incremental target at look $k$ is $A(t_k;\gamma)-A(t_{k-1};\gamma)$, with $A(0;\gamma)=0$. These increments are bookkeeping for cumulative targets; they are not independent per-look tests.

Worked Example: Positive Versus Negative Shape

Suppose $\alpha=0.05$ and compare $\gamma=2$ with $\gamma=-2$ at $t=0.5$. For $\gamma=2$,

A(0.5;2) =0.05\frac{1-\exp(-1)}{1-\exp(-2)} \approx 0.05\frac{0.63212}{0.86466} \approx 0.03655.

For $\gamma=-2$, rewrite numerator and denominator to avoid subtracting large negative quantities:

A(0.5;-2) =0.05\frac{\exp(1)-1}{\exp(2)-1} \approx 0.05\frac{1.71828}{6.38906} \approx 0.01345.

Linear spending at $t=0.5$ would be $0.05(0.5)=0.025$. The positive parameter has spent more than this by halfway, while the negative parameter has spent less. Both targets still reach $0.05$ at $t=1$.

Using and Checking the Family

The HSD family is useful when a design needs a smooth, monotone cumulative allocation that can be adjusted with one shape parameter. Unlike a fixed-boundary plan tied to a particular set of analysis times, the spending-function approach specifies targets in terms of information fraction; the boundaries still need calibration for the planned analysis structure. The parameter should be chosen as part of design, not adjusted after observing interim results without a separately justified adaptive procedure.

Several checks help detect mistakes. Confirm that $A(0;\gamma)=0$, $A(1;\gamma)=\alpha$, and $A(t;\gamma)$ is nondecreasing and stays between zero and $\alpha$. For interior $t$, positive $\gamma$ should give more cumulative spending than $\alpha t$, and negative $\gamma$ should give less. If a computed curve violates these checks, first inspect the sign in the exponential and the denominator. When $\gamma$ is close to zero, direct evaluation can lose numerical precision because both $1-\exp(-\gamma t)$ and $1-\exp(-\gamma)$ are small. Use the limit $\alpha t$, a stable exponential calculation, or the series expansion instead.

A spending curve does not guarantee that the design’s realized type I error matches the target if the assumed information fractions or joint null distribution are wrong. Nor does a particular $\gamma$ automatically produce the best power or operating characteristics for every trial. The curve defines an allocation preference; boundary calibration and design assessment remain necessary.

Common pitfall: Treating $A(t;\gamma)$ as the nominal significance level at the look $t$. It is cumulative spending through that look. The increment since the preceding look is $A(t_k;\gamma)-A(t_{k-1};\gamma)$, and the test boundary is calibrated using the sequential null distribution. Neither quantity should be confused with a stand-alone $p$-value cutoff.

Check Your Understanding

Answer each question, then compare with the solutions below.

For $\alpha=0.04$, what is $A(0.30;0)$?
With $\alpha=0.05$ and $\gamma=2$, the cumulative target at $t=0.5$ is about $0.03655$. Is this above or below linear spending, and by how many percentage points approximately?
For $\alpha=0.05$, $\gamma=-2$, and $t=0.5$, approximately what is the cumulative target?
In one sentence, explain why the HSD formula at $\gamma=0$ is consistent with the formula for nonzero $\gamma$.
Why is $A(t_k;\gamma)$ not generally the per-look significance level at analysis $k$?

Solutions. (1) $A(0.30;0)=0.04(0.30)=0.012$. (2) Linear spending is $0.05(0.5)=0.025$, so the HSD target is above it by $0.03655-0.025=0.01155$, or about 1.155 percentage points. (3) $A(0.5;-2)=0.05(\exp(1)-1)/(\exp(2)-1)\approx 0.05(0.26894)=0.01345$. (4) Expanding the exponentials near zero shows that their ratio tends to $t$, so $A(t;\gamma)$ tends to $\alpha t$. (5) It is cumulative through look $k$; the increment since the preceding look is a difference of cumulative targets, and the sequential boundary is calibrated using the joint null distribution.