Worked Conditional-Error Update After One Arm Is Dropped

Why Recalibrate After an Arm Is Dropped?

An interim analysis can change which treatment arms continue, while leaving the trial’s original familywise error rate (FWER) guarantee in force. A dropped arm is no longer tested for efficacy, but its removal does not by itself authorize a surviving arm to use a more liberal significance threshold. The relevant question is how much type I error the original design still allowed after the observed interim data.

Conditional error answers that question. It measures, under a null hypothesis, the probability that the original procedure would make a future false rejection, conditional on the information available at the interim analysis. A new continuation rule can preserve the original guarantee if its conditional probability of false rejection is no larger, for every relevant interim history, than the original procedure’s conditional error.

The calculation below uses a simple normal model to make the update explicit. The same logic applies more generally, but the conditional distribution and the set of null hypotheses that must be protected have to match the design.

Definition and Guarantee

Definition: Let $\mathcal{F}_1$ denote all information available at the interim analysis, and let $R_i^{\mathrm{old}}$ be the event that the original procedure rejects hypothesis $H_i$ after that analysis. Under $H_i$, the conditional error for $H_i$ at interim history $D_1$ is $$ CE_i(D_1)=P_{H_i}(R_i^{\mathrm{old}}\mid \mathcal{F}_1=D_1). $$ A revised procedure controls conditional error for $H_i$ if its conditional probability of rejecting $H_i$, given every relevant interim history, is no greater than $CE_i(D_1)$.

This is a conditional, path-by-path requirement: it is not enough for the revised procedure to have a small average error under one simulated configuration. The conditional error must be evaluated under the null being protected, and the revised rule must respect it after the data-dependent decision to drop or retain arms.

For a simple strong-FWER construction, suppose the original design assigns each arm $i$ a type I error allocation $\alpha_i$, with $\sum_i \alpha_i\leq\alpha$, and tests each arm so that its null rejection probability is at most $\alpha_i$. If the revised procedure has conditional rejection probability for each true $H_i$ no greater than the original $CE_i(D_1)$, then

P_{H_i}(\text{revised procedure rejects }H_i) =E_{H_i}\!\left[P_{H_i}(\text{revised rejection of }H_i\mid\mathcal{F}_1)\right] \leq E_{H_i}[CE_i(\mathcal{F}_1)] =P_{H_i}(R_i^{\mathrm{old}}) \leq\alpha_i.

The first equality is the law of total probability. The final equality follows because averaging the original procedure’s conditional rejection probability over interim histories recovers its original null rejection probability. For any configuration of true null hypotheses $I_0$, the union bound then gives

P(\text{at least one false rejection}) \leq \sum_{i\in I_0}P(\text{reject }H_i) \leq \sum_{i\in I_0}\alpha_i \leq \alpha.

This argument is one transparent route to strong FWER control. For a procedure based on intersection tests or another strong-control construction, the conditional error requirement must protect the relevant intersection hypotheses as well; checking only the global null is not generally sufficient.

Normal Independent-Increments Calculation

Suppose one-sided efficacy is assessed for an arm using a standardized statistic at two information times. Let $Z_1$ be the interim statistic and $Z_2$ the final statistic. Under the arm’s null hypothesis, assume the canonical normal independent-increments model

Z_2=\sqrt{t}\,Z_1+\sqrt{1-t}\,Z_{\mathrm{inc}}, \qquad Z_{\mathrm{inc}}\sim N(0,1),

where $t$ is the interim information fraction, $0<t<1$, and the future increment $Z_{\mathrm{inc}}$ is independent of the interim information. Assume the original rule makes no efficacy rejection at interim and rejects at the final look if $Z_2\geq c$, where $c$ is its original one-sided critical value. Conditional on $Z_1=z_1$, the conditional error is therefore

CE(z_1) =P_{H_0}(Z_2\geq c\mid Z_1=z_1) =1-\Phi\!\left(\frac{c-\sqrt{t}\,z_1}{\sqrt{1-t}}\right).

The corresponding critical value for a rule based directly on the future increment is

Z_{\mathrm{inc}}\geq \frac{c-\sqrt{t}\,z_1}{\sqrt{1-t}}.

This threshold gives exactly the original conditional error under the stated model. A more conservative threshold gives a conditional error no greater than the original budget. The calculation requires the original cutoff, information fraction, interim statistic, and correct null distribution of the future increment. If any of these changes, the conditional-error calculation may change too.

Worked Example: Updating the Surviving Arm’s Boundary

Consider a hypothetical two-arm trial with one-sided FWER target $\alpha=0.05$. The original design uses Bonferroni allocations $\alpha_A=\alpha_B=0.025$, and no arm can be rejected for efficacy at interim. For each arm the final cutoff is $c=1.95996$, the standard normal cutoff for a one-sided test at level $0.025$. The interim information fraction is $t=0.50$. At interim, arm B is dropped for futility and the standardized result for arm A is $z_1=0.80$. Assume the normal independent-increments model holds for arm A under $H_A$.

The conditional error available for a future rejection of $H_A$ is

\begin{aligned} CE_A(0.80) &=1-\Phi\!\left( \frac{1.95996-\sqrt{0.50}(0.80)}{\sqrt{0.50}} \right)\\ &=1-\Phi(1.97180)\\ &\approx 0.0243. \end{aligned}

The arithmetic uses $\sqrt{0.50}(0.80)=0.565685$, so the numerator is $1.394275$, and $1.394275/\sqrt{0.50}=1.97180$ to the displayed precision. Under the null, the future increment for arm A is standard normal. Thus a later-stage rule that rejects when

Z_{\mathrm{inc},A}\geq 1.97180

uses approximately $0.0243$ conditional type I error at this interim history. It reproduces the original procedure’s conditional error for arm A; it does not give arm A the full $0.05$ FWER allowance. The calculation assumes the planned information fraction and future variance structure are unchanged for arm A.

Applying the Updated Boundary

The conditional-error calculation produces a critical value, but a decision still requires an observed later-stage statistic. Under the example’s assumptions, that statistic is the independent future increment. The conditional p-value for an observed increment $u$ is $1-\Phi(u)$; comparing it with $CE_A(0.80)$ is equivalent to comparing $u$ with the conditional-error critical value.

Worked Example: A Later Increment Above the Boundary

Continue the hypothetical trial above. Suppose the observed future increment for arm A is $u=2.05$. The conditional one-sided p-value under $H_A$ is

1-\Phi(2.05)\approx 0.0202.

Because $0.0202<0.0243$, equivalently because $2.05>1.97180$, the updated rule rejects $H_A$. This is a decision under the conditional-error rule for this interim history. It does not mean that the null hypothesis has a $2.02\%$ probability of being true; the p-value is a tail probability calculated assuming $H_A$.

Why Dropping an Arm Does Not Automatically Free Its Budget

A tempting but incorrect argument is that dropping arm B frees its $\alpha_B=0.025$, so arm A may now be tested at the original familywise level $0.05$. The problem is that strong FWER control must hold under configurations in which $H_A$ is true and $H_B$ is false. In that configuration, a false rejection of B is not an error, so B’s nominal allocation is not automatically a conditional-error allowance for rejecting A.

The conditional error for $H_A$ is defined under $H_A$, and it is determined by the original probability of rejecting A given the interim history. In the preceding example, that amount is about $0.0243$. Testing the future increment at the one-sided $0.05$ cutoff $1.64485$ would give conditional error $0.05$, which exceeds $0.0243$. It would therefore fail the required pathwise comparison, even though one arm has been removed.

Worked Example: Checking a Proposed Liberal Threshold

At the same interim result $z_1=0.80$, suppose a proposed update uses $Z_{\mathrm{inc},A}\geq1.64485$, arguing that the dropped arm’s allocation can be reassigned to A. Under $H_A$, the future increment is standard normal, so the conditional rejection probability is

P_{H_A}(Z_{\mathrm{inc},A}\geq1.64485\mid\mathcal{F}_1) =1-\Phi(1.64485)\approx0.0500.

This exceeds the available conditional error, $0.0500>0.0243$. The proposal does not preserve the original arm-specific conditional error at this history. The fact that B was dropped does not resolve that violation; a different allocation would need a valid argument establishing the conditional bound for every relevant true-null configuration.

Practical Checks and Limitations

Before using a conditional-error update, verify the ingredients that make the conditional probability valid. In particular, check that the interim information fraction and test statistic correspond to the planned model, and that the future increment has the assumed null distribution conditional on the interim data. The independence and variance decomposition in the normal model are substantive assumptions, not algebraic conveniences.

Check the decision history. Include any interim efficacy rejection, futility rule, or other path restriction in the original rejection event. The displayed formula applies to the stated design with no interim efficacy rejection.
Check the information structure. If sample sizes, allocations, or follow-up change, the future increment’s conditional distribution may no longer be the one used in the formula.
Check all true-null configurations. A global-null calculation alone does not establish strong FWER control. With a Bonferroni construction, verify the bound for each true elementary hypothesis; with closed or intersection testing, protect the relevant intersections.
Do not confuse unconditional and conditional error. An original unconditional level of $0.025$ does not imply that the conditional error equals $0.025$ at every interim result. Here it is approximately $0.0243$ at $z_1=0.80$.

Conditional-error methods are flexible because they permit a new continuation rule without requiring that the original future test be copied exactly. Their limitation is equally important: flexibility is constrained by the conditional error left by the original design. If the model for future data is misspecified, or if the relevant null configurations are omitted, the calculated budget may not deliver the intended familywise guarantee.

Key takeaway: After an arm is dropped, calculate the surviving arm’s conditional error under its null using the original design and the observed interim history. Choose a later rule whose conditional false-rejection probability does not exceed that amount. Dropping an arm alone is not a license to increase another arm’s error budget.

Check Your Understanding

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

In the normal independent-increments model, write the formula for the conditional error when the original final boundary is $c$, the information fraction is $t$, and the observed interim statistic is $z_1$.
If the original conditional error at an interim history is $0.0243$, what constraint should a revised future test satisfy under the corresponding null?
Why does dropping one of two arms not automatically allow the remaining arm to use the full familywise level $0.05$?
At an interim history, the conditional-error critical value for a standard normal future increment is $1.97180$. Does an observed increment $1.90$ reject under this rule?

Solutions. (1) $CE(z_1)=1-\Phi((c-\sqrt{t}\,z_1)/\sqrt{1-t})$, assuming the stated normal independent-increments model and no earlier efficacy rejection. (2) Its conditional probability of rejecting that null, given the interim history, must be no greater than $0.0243$. (3) Under a configuration where the remaining arm’s null is true and the dropped arm’s null is false, a false rejection of the remaining arm is still an error; the dropped arm’s allocation is not automatically conditional error for that null. (4) No. Since $1.90<1.97180$, it does not cross the boundary; its one-sided tail probability is about $0.0287$, greater than $0.0243$.