Deriving the Inverse Normal Critical Value

From a Combined Statistic to a Final-Stage Threshold

A critical value gives a rejection rule for a test statistic. In an adaptive trial, it can also be useful to express that rule in terms of the evidence still to come: once the first-stage p-value is observed, what is the largest second-stage p-value that will lead to rejection? For a weighted inverse normal test, this threshold follows by rearranging the statistic’s rejection inequality.

The threshold is not a new significance level chosen independently for stage two. It is determined by the overall one-sided significance level, the weights, and the observed first-stage evidence. The derivation depends on the null distribution of the weighted statistic, so the assumptions behind that distribution must be clear.

Definition: For two stages with one-sided p-values $p_1$ and $p_2$, define the inverse normal scores $Z_i=\Phi^{-1}(1-p_i)$, where $\Phi$ is the standard normal cumulative distribution function. For positive weights $w_1,w_2$ satisfying $w_1^2+w_2^2=1$, the weighted inverse normal statistic is $$ Z_{\mathrm{comb}}=w_1Z_1+w_2Z_2. $$

This direction convention makes smaller one-sided p-values correspond to larger normal scores and hence stronger evidence against $H_0$. The derivation below assumes a prespecified one-sided alternative in that direction. It also assumes independent stagewise p-values that are uniform on $(0,1)$ under $H_0$. In an adaptive design, the stage-two p-value must remain valid under the null conditional on the stage-one information used to adapt the trial. These conditions make each $Z_i$ standard normal under the null and independent across stages. For fixed positive weights, the normalization $w_1^2+w_2^2=1$ then gives $Z_{\mathrm{comb}}$ a standard normal null distribution.

Deriving the Critical Value

A one-sided test at significance level $\alpha$ rejects for sufficiently large values of $Z_{\mathrm{comb}}$. Define $c_\alpha$ to be the upper-tail standard normal critical value, so that an observation at the boundary has upper-tail probability $\alpha$.

P(Z_{\mathrm{comb}}\geq c_\alpha\mid H_0)=\alpha, \qquad c_\alpha=\Phi^{-1}(1-\alpha).

Here is the justification for using that critical value. Under the null assumptions, $Z_1$ and $Z_2$ are independent standard normal random variables. A linear combination of independent normal variables is normal; its mean is the weighted sum of their means, and its variance is the sum of their variances multiplied by the squared weights.

E(Z_{\mathrm{comb}}\mid H_0)=w_1(0)+w_2(0)=0, \qquad \operatorname{Var}(Z_{\mathrm{comb}}\mid H_0) =w_1^2+w_2^2=1.

Thus $Z_{\mathrm{comb}}\sim N(0,1)$ under $H_0$, and setting its upper-tail probability equal to $\alpha$ gives $c_\alpha=\Phi^{-1}(1-\alpha)$. In particular, for $\alpha=0.05$, $c_\alpha=\Phi^{-1}(0.95)\approx1.6449$. The rejection rule is $Z_{\mathrm{comb}}\geq1.6449$, with equality conventionally included; under a continuous null distribution, the boundary itself has probability zero.

Theorem: Under independent uniform stagewise p-values under $H_0$, with fixed positive weights satisfying $w_1^2+w_2^2=1$, the one-sided weighted inverse normal test rejects at level $\alpha$ when $$ w_1\Phi^{-1}(1-p_1)+w_2\Phi^{-1}(1-p_2)\geq\Phi^{-1}(1-\alpha). $$

Solving for the Final-Stage P-Value

Suppose the first-stage p-value has been observed. With $w_2>0$, the rejection inequality can be rearranged to isolate the stage-two normal score, then translated back to the stage-two p-value. Because $\Phi$ is increasing, each rearrangement preserves the direction of the inequality.

\begin{aligned} w_1\Phi^{-1}(1-p_1)+w_2\Phi^{-1}(1-p_2)&\geq c_\alpha,\\ \Phi^{-1}(1-p_2)&\geq \frac{c_\alpha-w_1\Phi^{-1}(1-p_1)}{w_2},\\ 1-p_2&\geq \Phi\!\left( \frac{c_\alpha-w_1\Phi^{-1}(1-p_1)}{w_2} \right),\\ p_2&\leq q(p_1) = 1-\Phi\!\left( \frac{c_\alpha-w_1\Phi^{-1}(1-p_1)}{w_2} \right). \end{aligned}

The quantity $q(p_1)$ is the final-stage p-value threshold for the observed first-stage p-value. A stage-two p-value at or below this threshold leads to rejection; one above it does not. If the first-stage evidence is stronger, $p_1$ is smaller and $\Phi^{-1}(1-p_1)$ is larger. That leaves less of the critical value for stage two to supply, so $q(p_1)$ generally increases. This does not change the overall type I error rate: the threshold is the algebraic form of the same combined-statistic rule, whose null rejection probability is $\alpha$ under the stated assumptions.

Worked Threshold Calculations

Worked Example: A Strong First Stage and a Failing Final P-Value

Suppose a hypothetical two-stage trial uses $\alpha=0.05$, $w_1=0.5$, and $w_2=\sqrt{0.75}\approx0.8660$, so that $w_1^2+w_2^2=1$. The observed first-stage p-value is $p_1=0.01$. Find the final-stage threshold and decide whether a hypothetical $p_2=0.30$ rejects.

The first-stage score is $Z_1=\Phi^{-1}(0.99)\approx2.3263$, and $c_{0.05}\approx1.6449$. Substituting in the threshold formula gives

\begin{aligned} q(0.01) &=1-\Phi\!\left( \frac{1.6449-(0.5)(2.3263)}{0.8660} \right)\\ &=1-\Phi\!\left(\frac{0.48175}{0.8660}\right) \approx1-\Phi(0.5563) \approx0.2890. \end{aligned}

Thus rejection requires $p_2\leq0.2890$, and $p_2=0.30$ does not meet that condition. Directly checking the combined statistic gives the same result. Here $Z_2=\Phi^{-1}(0.700)\approx0.5244$. Using the displayed rounded inputs,

Z_{\mathrm{comb}} =(0.5)(2.3263)+(0.8660)(0.5244) \approx1.1632+0.4541 =1.6173<1.6449.

The combined statistic is below the critical value, consistent with the threshold decision. The p-value cutoff is not $0.05$: a sufficiently strong first-stage result can permit a larger stage-two p-value while retaining the overall one-sided test rule.

Worked Example: A Weaker First Stage Requires More Stage-Two Evidence

Consider a second hypothetical analysis with $\alpha=0.05$, weights $w_1=0.5$ and $w_2=0.8660$, and $p_1=0.20$. Calculate the final-stage threshold and assess $p_2=0.05$.

Since $Z_1=\Phi^{-1}(0.80)\approx0.8416$, the threshold is

\begin{aligned} q(0.20) &=1-\Phi\!\left( \frac{1.6449-(0.5)(0.8416)}{0.8660} \right)\\ &=1-\Phi\!\left(\frac{1.2241}{0.8660}\right) \approx1-\Phi(1.4135) \approx0.0788. \end{aligned}

A stage-two p-value of $0.05$ is below $0.0788$, so the test rejects. As a direct check, $Z_2=\Phi^{-1}(0.95)\approx1.6449$, and

Z_{\mathrm{comb}} \approx(0.5)(0.8416)+(0.8660)(1.6449) \approx0.4208+1.4245 =1.8453>1.6449.

The weaker first-stage evidence gives a smaller final-stage cutoff than in the previous example. The decision is nevertheless based on the same critical value and weighted statistic.

Worked Example: Translating a Different Weighting Scheme

For a hypothetical trial, let $\alpha=0.05$, $w_1=0.6$, $w_2=0.8$, and $p_1=0.04$. These weights satisfy $0.6^2+0.8^2=1$. Calculate the final-stage cutoff and verify the decision for $p_2=0.20$.

The first-stage score is $Z_1=\Phi^{-1}(0.96)\approx1.7507$. The required stage-two score and corresponding p-value cutoff are

\begin{aligned} Z_2&\geq\frac{1.6449-(0.6)(1.7507)}{0.8} =\frac{0.5945}{0.8}\approx0.7431,\\ p_2&\leq1-\Phi(0.7431)\approx0.2287. \end{aligned}

Because $0.20<0.2287$, the test rejects. In the statistic form, $Z_2=\Phi^{-1}(0.80)\approx0.8416$, and

Z_{\mathrm{comb}} \approx(0.6)(1.7507)+(0.8)(0.8416) =1.0504+0.6733 =1.7237>1.6449.

This check confirms that solving for the p-value cutoff and evaluating the combined statistic are two equivalent ways to apply the rejection rule.

Assumptions, Checks, and Alternatives

The standard normal critical value is exact when the stagewise p-values are independent and uniform under the null, the weights are positive and fixed for the test, and their squared sum is one. In an adaptive trial, validity of $p_2$ must account for the stage-one information that guided adaptation. If stage-two choices are made in a way that invalidates its conditional null calibration, treating the transformed scores as independent standard normals is not justified.

Several practical checks catch common errors:

Check the direction. With $Z_i=\Phi^{-1}(1-p_i)$, smaller one-sided p-values produce larger scores. Reversing the transform without changing the rejection direction reverses the evidence.
Check the weights. Confirm that $w_1^2+w_2^2=1$, and use the intended positive weights. If the weights do not have unit squared sum, the combined null variance is not one, so the usual standard normal critical value does not apply without adjustment.
Check the tail. The one-sided upper-tail critical value is $\Phi^{-1}(1-\alpha)$, not $\Phi^{-1}(\alpha)$. For $\alpha=0.05$, it is approximately $1.6449$.
Check the arithmetic twice. Keep extra digits until the final comparison. A small rounding error near the critical value can change a reported decision.
Check p-value reporting. A p-value rounded to zero cannot be transformed to a finite normal score. Use an unrounded p-value or an appropriate prespecified numerical method rather than substituting zero.

If stagewise p-values are conservative rather than exactly uniform under the null, the transformed scores need not have an exact standard normal distribution. The standard critical value then need not yield exact equality at level $\alpha$; calibration must respect the p-value distributions and dependence structure. Likewise, if $w_2=0$, the final-stage p-value has no effect on the statistic and no finite stage-two threshold can be obtained by division by $w_2$. The displayed threshold formula is intended for $w_2>0$.

The threshold can be compared with a fixed allocation of $\alpha$ between stages. A fixed allocation sets separate stagewise cutoffs in advance; the inverse normal threshold instead expresses the same overall combined-statistic boundary as a cutoff that depends on the observed first-stage p-value. It should not be described as a newly selected stage-two level that replaces the overall design. A pooled-data analysis is another possible approach, but it requires an appropriate model and assumptions about the combined data; its rejection rule is not automatically the inverse normal rule. The combination test is useful when valid evidence from separate stages is to be combined, but it does not remove the need for valid stagewise p-values or make different choices of weights interchangeable.

A Common Misinterpretation

Common mistake: “The final-stage cutoff is $0.05$ because the overall test uses $\alpha=0.05$.” This confuses the overall type I error rate with a fixed stage-two p-value cutoff. The correct threshold is $q(p_1)$, calculated from the critical value, weights, and observed first-stage p-value. Its dependence on $p_1$ is exactly what makes the p-value cutoff equivalent to the combined-statistic rule.

For example, in the first worked calculation, the cutoff was about $0.2890$, not $0.05$, because the observed first-stage score was strong. That does not mean the design’s overall type I error rate has become $0.2890$. It is a conditional boundary for stage-two evidence given the observed first-stage result; the unconditional null rejection probability is governed by the full combination test.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

For $\alpha=0.025$, what is the one-sided critical value in terms of $\Phi$?
With $w_1=0.8$ and $w_2=0.6$, write the formula for the stage-two p-value threshold after observing $p_1$.
Why does a stronger first-stage result generally allow a larger stage-two p-value while keeping the same combined-statistic boundary?
What key null-distribution condition is needed to treat the weighted statistic as standard normal?
Explain why a stage-two cutoff calculated from $p_1$ is not itself the overall significance level.

Solutions. (1) The critical value is $c_{0.025}=\Phi^{-1}(0.975)\approx1.9600$. (2) Since $0.8^2+0.6^2=1$, the threshold is $q(p_1)=1-\Phi\!\left((c_\alpha-0.8\Phi^{-1}(1-p_1))/0.6\right)$. (3) A stronger first-stage result gives a larger $Z_1$, so $w_1Z_1$ supplies more of the score needed to reach $c_\alpha$; the required $Z_2$ falls, permitting a larger $p_2$. (4) The stagewise p-values must be independent and uniform under the null for their transformed scores to be independent standard normals; in an adaptive trial, stage-two validity must hold conditional on the adaptation information. (5) The cutoff is a conditional boundary for the observed first-stage result. Overall type I error is the null probability of rejection across the complete test, not the stage-two cutoff considered alone.