Deriving Weights for Multiple Stages

Why Derive the Weights?

A multi-stage inverse normal test combines stagewise evidence on the normal scale. Its weights determine how strongly each stage contributes to the combined statistic. For a \(K\)-stage test, the central requirement is that the prespecified weights have squares summing to one. This normalization makes the combined statistic have variance one under the null when the stagewise statistics are independent standard normal variables.

It is often easier to plan the allocation of squared weight across stages than to choose the weights themselves. The derivation is direct, but the distinction matters: an allocation \(q_j\) of squared weight produces a weight \(\sqrt{q_j}\), not \(q_j\). The allocation describes each stage's contribution to the null variance, not necessarily its information or its influence under every alternative.

Definition: For \(K\) stages, let \(Z_j\) be the stagewise statistic at stage \(j\), oriented so that larger values indicate evidence in the prespecified direction. A fixed-weight inverse normal combination statistic is \(Z_{\mathrm{comb}}=\sum_{j=1}^{K} w_j Z_j\). A squared-weight allocation is a set of prespecified numbers \(q_1,\ldots,q_K\), with \(q_j\geq 0\) and \(\sum_{j=1}^{K}q_j=1\). The corresponding nonnegative weights are \(w_j=\sqrt{q_j}\).

Assumptions Behind the Normalization

The exact unit-variance and standard-normal result uses specific assumptions. Under the null hypothesis, each \(Z_j\) must have mean zero and variance one, and the stagewise statistics must be mutually independent. In the standard inverse normal construction, a one-sided stagewise \(p\)-value \(p_j\) is transformed as \(Z_j=\Phi^{-1}(1-p_j)\), where \(\Phi\) is the standard normal cumulative distribution function. If the stagewise \(p\)-value is continuous and exactly uniform under the null, this transform is standard normal.

Independence is typically justified when stages use independent, non-overlapping data and each stage statistic is formed according to its prespecified analysis. The derivation below does not establish independence merely because stages are labelled separately; the data and the way the statistics are constructed must support it. The weights must also be fixed independently of the observed stage results. If stagewise tests are only approximately normal, the resulting null reference distribution is approximate as well.

The direction of evidence must be consistent across stages: positive \(Z_j\) values should represent evidence in the same prespecified direction. The derivation concerns one-sided, directionally aligned stagewise statistics. It does not make a two-sided \(p\)-value into a signed statistic.

Deriving Weights from an Allocation

Suppose the design assigns stage \(j\) a fraction \(q_j\) of the total squared weight. The weight is the nonnegative square root of that fraction. If the planned amounts are given as nonnegative numbers \(c_j\) that do not yet sum to one, first convert them to fractions by dividing each by their total.

w_j=\sqrt{q_j} \qquad\text{when}\qquad \sum_{j=1}^{K}q_j=1.

\text{If }c_j\geq 0\text{ and }\sum_{j=1}^{K}c_j>0, \qquad q_j=\frac{c_j}{\sum_{\ell=1}^{K}c_\ell}, \qquad w_j=\sqrt{\frac{c_j}{\sum_{\ell=1}^{K}c_\ell}}.

The normalization has a simple variance justification. For independent stagewise statistics with variance one, the variance of their weighted sum is the sum of the squared weights. Thus, allocating squared weight fractions that add to one gives the combined statistic variance one.

\begin{aligned} \operatorname{Var}(Z_{\mathrm{comb}}) &=\operatorname{Var}\left(\sum_{j=1}^{K}w_jZ_j\right)\\ &=\sum_{j=1}^{K}w_j^2\operatorname{Var}(Z_j) +2\sum_{j<\ell}w_jw_\ell\operatorname{Cov}(Z_j,Z_\ell)\\ &=\sum_{j=1}^{K}w_j^2 =\sum_{j=1}^{K}q_j =1. \end{aligned}

This variance calculation is not, by itself, a proof of a standard normal distribution. For that conclusion, the \(Z_j\) must also be jointly normal; independent standard normal stage statistics satisfy this condition. Their weighted sum is then normal, with mean zero and variance one, so \(Z_{\mathrm{comb}}\sim N(0,1)\) under the null. If the stage statistics are independent, standard normal, and the weights are fixed, this gives the usual standard normal null reference for the combined statistic.

Result: If \(Z_1,\ldots,Z_K\) are independent standard normal variables under the null, and the fixed weights satisfy \(\sum_{j=1}^{K}w_j^2=1\), then \(Z_{\mathrm{comb}}=\sum_{j=1}^{K}w_jZ_j\sim N(0,1)\). Choosing \(w_j=\sqrt{q_j}\) from a squared-weight allocation with \(\sum_jq_j=1\) satisfies this condition.

Worked Weight Derivations

Worked Example: Three Stages with Prespecified Fractions

Suppose a hypothetical three-stage design allocates 50% of the squared weight to stage 1, 30% to stage 2, and 20% to stage 3. These fractions are fixed before any stage results are observed. Find the weights and calculate the combined statistic for stagewise values \(Z_1=0.40\), \(Z_2=1.20\), and \(Z_3=1.50\).

The allocations sum to one: \(0.50+0.30+0.20=1.00\). Taking square roots gives \(w_1=\sqrt{0.50}=0.7071\), \(w_2=\sqrt{0.30}=0.5477\), and \(w_3=\sqrt{0.20}=0.4472\). Their squared weights sum to \(0.5000+0.3000+0.2000=1.000\).

The combined statistic is \(Z_{\mathrm{comb}}=(0.7071)(0.40)+(0.5477)(1.20)+(0.4472)(1.50)\). The stage contributions are \(0.2828\), \(0.6572\), and \(0.6708\), respectively, so \(Z_{\mathrm{comb}}=1.6108\), approximately \(1.611\). Under the stated independence and normality assumptions, its null distribution is standard normal. The allocations describe the stages' shares of null variance; the observed contributions need not be in the same proportions.

Worked Example: Normalizing Relative Allocation Amounts

Consider a hypothetical four-stage design whose planned squared-weight amounts are in the ratio \(2:3:5:0\). Find the normalized allocations and weights. The zero assigned to stage 4 means that stage receives no weight in this combination statistic.

The total amount is \(2+3+5+0=10\). Therefore the allocation fractions are \(q_1=2/10=0.20\), \(q_2=3/10=0.30\), \(q_3=5/10=0.50\), and \(q_4=0/10=0\). The corresponding weights are \(w_1=\sqrt{0.20}=0.4472\), \(w_2=\sqrt{0.30}=0.5477\), \(w_3=\sqrt{0.50}=0.7071\), and \(w_4=0\).

For hypothetical stagewise statistics \(Z_1=0.60\), \(Z_2=-0.20\), \(Z_3=1.00\), and \(Z_4=2.00\), the fourth statistic contributes nothing because its weight is zero. The calculation is \(Z_{\mathrm{comb}}=(0.4472)(0.60)+(0.5477)(-0.20)+(0.7071)(1.00)+(0)(2.00)\) \(=0.2683-0.1095+0.7071=0.8659\). The squared weights sum to \(0.20+0.30+0.50+0=1\). Under the assumptions already stated, this weighted sum has a standard normal null distribution. The zero weight is mathematically allowed, but it means stage 4 supplies no evidence to this combination test.

Worked Example: Correcting a Squared-Weight Mistake

A hypothetical three-stage plan specifies relative squared-weight amounts \(2:1:1\). A common error is to use the normalized fractions themselves as the weights. Compare that incorrect calculation with the correct weights, using \(Z_1=1.00\), \(Z_2=0.50\), and \(Z_3=-0.20\).

The normalized squared-weight fractions are \(q_1=2/4=0.50\), \(q_2=1/4=0.25\), and \(q_3=1/4=0.25\). The correct weights are their square roots: \(w_1=0.7071\), \(w_2=0.5000\), and \(w_3=0.5000\). Their squares sum to one, as required.

The correct combined statistic is \(Z_{\mathrm{comb}}=(0.7071)(1.00)+(0.5000)(0.50)+(0.5000)(-0.20)=0.7071+0.2500-0.1000=0.8571\). If the fractions \(0.50,0.25,0.25\) are mistakenly used as weights, the resulting sum is \(0.50(1.00)+0.25(0.50)+0.25(-0.20)=0.5750\). Its variance under independent unit-variance stages is \(0.50^2+0.25^2+0.25^2=0.3750\), not one. The error is using squared-weight allocations as the weights without taking square roots.

Checks, Alternatives, and Limitations

Before using the standard normal reference, check both the algebra and the assumptions. The arithmetic check is \(\sum_jw_j^2=1\). The design checks are that weights were prespecified, that stagewise statistics are independent under the null, and that each stage uses the same direction of evidence. A weight allocation cannot repair dependence between stages or make an incorrectly oriented statistic valid.

A useful comparison is equal weighting: choosing \(q_j=1/K\) for every stage gives \(w_j=1/\sqrt{K}\). More generally, multiplying all preliminary coefficients by the same positive constant does not change the normalized weighting scheme if they are subsequently divided by the square root of their summed squares. This is equivalent to specifying relative squared-weight amounts and normalizing them.

Fisher combination is a different approach: it combines stagewise \(p\)-values through a logarithmic transformation and uses a different null reference distribution. The inverse normal construction instead transforms one-sided \(p\)-values to normal scores and forms a weighted sum. Its squared-weight allocation provides a direct way to set the stages' contributions to the null variance; it does not imply that inverse normal combination is uniformly more powerful or that a chosen allocation is optimal.

The exact standard normal result has limits. If stagewise \(p\)-values are discrete, their normal-score transforms need not be exactly standard normal; if stage statistics are dependent, covariance terms affect the combined variance. Likewise, variance one alone does not establish a normal distribution. These conditions should be checked rather than inferred from the fact that the formula contains weights.

Common mistake: “The first stage receives 40% of the squared-weight allocation, so its weight is 0.40.” The correction is \(w_1=\sqrt{0.40}\), approximately \(0.6325\). Then \(w_1^2=0.40\), as intended. Confusing the allocation with the weight changes the total variance and can invalidate the standard normal reference.

Check Your Understanding

Answer each question, then compare with the solutions below.

A two-stage design allocates 25% and 75% of squared weight. What are the weights, and what is their squared sum?
Three relative squared-weight amounts are \(1:1:2\). Find the normalized allocations and weights.
If independent stage statistics each have variance one, but the chosen weights are \(0.6\) and \(0.6\), what is the combined variance?
Does a weighted sum have a standard normal distribution just because its variance is one? State the additional conditions used in this tutorial.

Solutions. (1) The weights are \(\sqrt{0.25}=0.5000\) and \(\sqrt{0.75}=0.8660\); their squared sum is \(0.25+0.75=1\). (2) The total amount is 4, so the allocations are \(0.25,0.25,0.50\), and the weights are \(0.5000,0.5000,0.7071\). Their squared sum is one. (3) The variance is \(0.6^2+0.6^2=0.72\), so it is not one. (4) No. For the exact result here, the stage statistics must be independent standard normal variables under the null, and the weights must be fixed with squared sum one; then their weighted sum is standard normal.