Choosing Fixed Stage Weights From Information

Why Choose Weights From Information?

An inverse-normal combination test turns stagewise evidence into one statistic by weighting the normal scores from each stage. The weights determine how much each stage contributes. If the stages are planned to provide different amounts of information, equal weights give them equal influence despite that difference. Information-proportional weights instead assign greater influence to stages planned to be more informative.

The weights are part of the test specification, not a result of the observed data. This tutorial considers weights fixed in advance from planned information increments. That distinction matters: the null reference distribution described below depends on the weights being fixed and on the stagewise scores satisfying the stated independence assumptions.

Definition: For $K$ stages, let $I_k$ denote the planned information increment contributed by stage $k$, with $I_k>0$. The information-proportional inverse-normal weight for stage $k$ is $$ w_k=\sqrt{\frac{I_k}{\sum_{j=1}^{K} I_j}}. $$ These weights are positive, fixed before observing the stagewise results, and satisfy $\sum_{k=1}^{K}w_k^2=1$. Equal weights are $w_k=1/\sqrt{K}$ for every stage.

Information is a measure of how much precision a stage is expected to contribute to estimating the effect of interest. Its precise calculation depends on the statistical model and endpoint. The formula above needs the planned information increments on a common scale; it does not require that the increments be equal. For two stages, the first contributes $I_1$ and the second contributes $I_2$, so the total planned information is $I_1+I_2$.

Be careful when a design reports information at interim and final analyses cumulatively. If the first analysis has cumulative information $I(1)$, and the final analysis has cumulative information $I(F)$, then the stage-two increment is $I(F)-I(1)$, not $I(F)$. The weights use increments from nonoverlapping stages, not repeated cumulative totals.

Reference Distribution and Rationale

For a one-sided alternative in the direction of benefit, let $p_k$ be the valid one-sided p-value from stage $k$, and define its upper-tail normal score by $Z_k=\Phi^{-1}(1-p_k)$, where $\Phi$ is the standard normal cumulative distribution function. The inverse-normal combination statistic is $Z_{\mathrm{comb}}=\sum_{k=1}^{K}w_kZ_k$. The earlier material on inverse-normal combination tests defines this construction; here the key question is how to select its fixed weights.

The null calibration requires that, under $H_0$, each $Z_k$ has a standard normal distribution and the stagewise scores are independent. These assumptions are standard when each stage supplies a valid continuous stagewise p-value and the evidence comes from independent stage data. The weights must be specified independently of the observed stagewise scores. If these conditions hold and the weights are fixed with squared sum one, the combination statistic has a standard normal null distribution.

Result: If $Z_1,\ldots,Z_K$ are independent $N(0,1)$ random variables under $H_0$, and fixed weights satisfy $\sum_{k=1}^{K}w_k^2=1$, then $$ Z_{\mathrm{comb}}=\sum_{k=1}^{K}w_kZ_k\sim N(0,1)\quad\text{under }H_0. $$ For a one-sided test at significance level $\alpha$, reject $H_0$ when $Z_{\mathrm{comb}}\geq \Phi^{-1}(1-\alpha)$. The one-sided combined p-value is $1-\Phi(Z_{\mathrm{comb}})$.

The justification follows from the distribution of a linear combination of independent normal random variables. Its mean under $H_0$ is zero, and its variance is the sum of the squared coefficients multiplied by the component variances:

E(Z_{\mathrm{comb}}\mid H_0)=\sum_{k=1}^{K}w_kE(Z_k\mid H_0)=0, \qquad \operatorname{Var}(Z_{\mathrm{comb}}\mid H_0) =\sum_{k=1}^{K}w_k^2\operatorname{Var}(Z_k\mid H_0) =\sum_{k=1}^{K}w_k^2=1.

A linear combination of independent normal variables is normal, so its zero mean and unit variance establish the stated reference distribution. This also explains why arbitrary unnormalized coefficients cannot simply be used with the standard normal cutoff: if their squared sum is not one, the combination will not have unit variance.

There is also a power rationale for information-proportional weights under a common-effect model. Suppose, as a normal approximation under a particular alternative, $E(Z_k)=\delta\sqrt{I_k}$, with the same positive standardized effect $\delta$ in every stage. For any nonnegative weights with $\sum w_k^2=1$, the combined mean is $\delta\sum w_k\sqrt{I_k}$. By the Cauchy–Schwarz inequality, this is maximized by choosing weights proportional to $\sqrt{I_k}$, which gives the information-proportional formula. This is a rationale under that model, not a claim that these weights maximize power under every alternative or endpoint.

\sum_{k=1}^{K}w_k\sqrt{I_k} \leq \sqrt{\sum_{k=1}^{K}w_k^2}\sqrt{\sum_{k=1}^{K}I_k} = \sqrt{\sum_{k=1}^{K}I_k}, \qquad w_k=\sqrt{\frac{I_k}{\sum_{j=1}^{K}I_j}} \text{ attains equality.}

Calculating and Comparing Weights

The following examples use hypothetical data and a prespecified beneficial direction. In each, the displayed stagewise scores are treated as independent standard normal scores under $H_0$, and the weights are fixed from the planned information before those scores are observed. Those assumptions make the standard normal calibration applicable.

Worked Example: Two Stages With Unequal Planned Information

Suppose a hypothetical two-stage trial plans information increments $I_1=40$ and $I_2=60$ units. At the end of the stages, the respective one-sided normal scores are $Z_1=0.8000$ and $Z_2=2.1000$. Calculate the information-proportional weights and combined result, then compare with equal weights.

Weights. The planned total is $40+60=100$ information units. Thus $$ w_1=\sqrt{\frac{40}{100}}=0.6325,\qquad w_2=\sqrt{\frac{60}{100}}=0.7746. $$ Their squared sum is $0.4+0.6=1$, as required.

Information-proportional combination. Using the unrounded square roots for the calculation, $$ Z_{\mathrm{comb}} =\sqrt{0.4}(0.8000)+\sqrt{0.6}(2.1000) \approx0.5060+1.6267 =2.1326. $$ The one-sided p-value is $1-\Phi(2.1326)\approx0.0165$. At $\alpha=0.05$, the critical value is $1.6449$; since $2.1326>1.6449$, reject $H_0$. The combined evidence is statistically significant at the 5% one-sided level.

Equal-weight comparison. With two stages, each equal weight is $1/\sqrt{2}\approx0.7071$, giving $$ Z_{\mathrm{comb,eq}} =\frac{0.8000+2.1000}{\sqrt{2}} \approx2.0506. $$ Its one-sided p-value is approximately $0.0202$, also below $0.05$. The information-proportional result gives the larger statistic here because the higher score came from the stage with the larger planned information. The difference is modest in this example; the two choices do not generally produce the same combined statistic.

When a final information total is reported cumulatively, first convert it to increments. Applying the square-root rule to cumulative totals as though they were separate stage contributions counts information from the first stage twice. The next example makes the conversion explicit.

Worked Example: Converting Cumulative Information to Stage Increments

A hypothetical design plans cumulative information of $120$ units at its first analysis and $300$ units at its final analysis. Suppose the stagewise scores are $Z_1=1.4000$ and $Z_2=1.6000$. Find the weights and combined statistic.

Find the increments. The first-stage increment is $I_1=120$. The second-stage increment is the additional information after the first analysis: $$ I_2=300-120=180. $$ The total increment is $120+180=300$, matching the planned final cumulative information.

Calculate weights. $$ w_1=\sqrt{\frac{120}{300}}=\sqrt{0.4}\approx0.6325,\qquad w_2=\sqrt{\frac{180}{300}}=\sqrt{0.6}\approx0.7746. $$ Then, retaining full precision for the arithmetic, $$ Z_{\mathrm{comb}} =\sqrt{0.4}(1.4000)+\sqrt{0.6}(1.6000) \approx0.8854+1.2394 =2.1248. $$ The one-sided p-value is approximately $0.0168$. At the 5% one-sided level, $2.1248>1.6449$, so the combined test rejects $H_0$.

If equal weights were used, $Z_{\mathrm{comb,eq}}=(1.4000+1.6000)/\sqrt{2}\approx2.1213$, with one-sided p-value about $0.0169$. The results happen to be close because the stage scores are similar. That closeness does not mean the weights are interchangeable; it reflects these particular scores and planned increments.

When Equal Weights Change the Conclusion

Equal weights are a legitimate fixed choice if specified in advance and normalized to have squared sum one. They give each stage the same coefficient regardless of planned precision. Information-proportional weights encode a different design judgment: a stage expected to contribute more information receives more influence. Neither choice can be selected after inspecting which gives the more favorable result while continuing to use the same nominal standard normal calibration.

Worked Example: A Small Stage With a High Score

Consider a hypothetical design with planned information increments $I_1=25$ and $I_2=75$, and observed scores $Z_1=2.0000$ and $Z_2=1.0000$. The first stage has less planned information, but its score is higher.

Information-proportional weights and result. $$ w_1=\sqrt{\frac{25}{100}}=0.5000,\qquad w_2=\sqrt{\frac{75}{100}}\approx0.8660, $$ so $$ Z_{\mathrm{comb}} =0.5000(2.0000)+\sqrt{0.75}(1.0000) \approx1.0000+0.8660 =1.8660. $$ The one-sided p-value is approximately $0.0310$, which is below $0.05$.

Equal-weight result. $$ Z_{\mathrm{comb,eq}} =\frac{2.0000+1.0000}{\sqrt{2}} \approx2.1213. $$ The one-sided p-value is approximately $0.0169$, also below $0.05$. Equal weights give the high first-stage score more influence than the information-proportional choice does, so the equal-weight statistic is larger in this example. This illustrates why a weight choice affects power and the observed statistic, even though both fixed, normalized choices have the same standard normal null reference distribution under the assumptions stated above.

Diagnostics, Comparisons, and Limitations

Before using a fixed information-based weighting rule, check that the information values actually represent stage increments and are on a common scale. Verify the normalization numerically, $\sum w_k^2=1$. Also confirm that the stagewise p-values are valid for the specified hypotheses and that the stagewise scores are independent under the null. If stages share observations or their scores are otherwise dependent, the variance of the weighted sum includes covariance terms; the unit-variance result above cannot be assumed without addressing that dependence.

The main alternatives differ in how they allocate influence. Equal weights are simple and do not require planned information values, but can give a relatively small stage the same coefficient as a much larger one. Information-proportional weights use the planned information structure and, under a common-effect normal approximation, maximize the combined mean among nonnegative normalized weights. A different fixed allocation may be appropriate for other design goals, but its weights and reference calibration must be specified coherently.

Information-proportional weighting has limitations. Planned information may be uncertain, and the common-effect power rationale may not describe settings where effects differ across stages. The rule does not itself guarantee high power, protect against bias, or repair invalid stagewise tests. It is also not a license to recalculate weights from observed interim results: choosing weights in response to data raises additional validity questions that require separate treatment.

Using cumulative information as increments. Substituting $I(F)$ for the second-stage increment double-counts the first stage. Subtract the first cumulative total from the final total.
Using information fractions as the weights. The weights are square roots of the information fractions. For example, information shares $0.4$ and $0.6$ produce weights $\sqrt{0.4}$ and $\sqrt{0.6}$, not $0.4$ and $0.6$.
Forgetting normalization. The standard normal cutoff applies when the weights’ squared sum is one. Check this before calculating a p-value.
Claiming one weighting rule always wins. The information-based choice has a specific rationale under common effects, but realized combined statistics depend on the stage scores. As the third example shows, equal weights can yield a larger statistic for a particular set of results.

Key takeaway: For fixed information-proportional inverse-normal weights, use the square root of each planned stage information fraction. Keep stage increments distinct from cumulative information, verify the squared weights sum to one, and apply the standard normal reference only when the fixed-weight independence assumptions hold.

Check Your Understanding

Answer each question, then compare with the solutions below.

A two-stage design plans information increments of $16$ and $84$ units. What are its information-proportional weights?
Cumulative information is $50$ units at the first analysis and $125$ units at the final analysis. What are the two stage increments and their information-proportional weights?
Why must the weights’ squared sum equal one to use the standard normal reference distribution for the weighted score?
For two independent standard normal stage scores under $H_0$, compare the null reference distribution for equal weights with that for information-proportional weights.

Solutions. (1) The total is $100$, so $w_1=\sqrt{16/100}=0.4000$ and $w_2=\sqrt{84/100}\approx0.9165$; their squared sum is $0.16+0.84=1$. (2) The increments are $I_1=50$ and $I_2=125-50=75$, giving $w_1=\sqrt{50/125}\approx0.6325$ and $w_2=\sqrt{75/125}\approx0.7746$. (3) Independence and unit component variances give $\operatorname{Var}(Z_{\mathrm{comb}}\mid H_0)=\sum w_k^2$; setting this to one yields a standard normal statistic rather than a normal statistic with another variance. (4) Both choices have squared weights summing to one, so both give $Z_{\mathrm{comb}}\sim N(0,1)$ under the stated null assumptions. Equal weights affect the contribution of each stage differently, not the null reference distribution.