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Conditional Error Functions · Tutorial 234 of 799

Proof By Conditional Expectation

See how a bound on rejection probability conditional on interim information yields control of the modified test’s overall Type I error.

Advanced 10 min read

What You'll Learn

  • Define the original and modified tests’ conditional rejection probabilities at the adaptation time.
  • State the assumptions that make conditional-error preservation valid under a null hypothesis.
  • Derive the modified test’s marginal rejection probability using iterated expectation.
  • Apply the argument to a two-stage normal test and a finite interim-outcome example.
  • Diagnose a modified rule that exceeds the original conditional error on some interim outcomes.
  • Distinguish conditional-error control from unconditional recalibration.

Why a Conditional Argument Controls Overall Error

An adaptive trial may use interim information to choose how to continue. That choice changes the rule applied to future data, so the original test’s overall Type I error cannot simply be assumed to carry over. A useful way to justify a modification is to compare the two tests at each possible interim history: under the null, the modified rule must have no greater conditional probability of rejection than the original rule had at that history.

The key step is then to average those conditional probabilities over the interim histories that could occur. The law of iterated expectation says that this average is the test’s overall rejection probability. This tutorial develops that argument and makes explicit which conditional probability is being compared.

Definition: Let \(\mathcal{F}\) denote the information available at the adaptation time, including the interim data used to make the adaptation. For a test with rejection indicator \(\phi\), its conditional rejection probability under parameter value \(\theta\) is \(E_\theta(\phi\mid\mathcal{F})\). It is a random variable determined by the interim information; at a particular interim history, it gives the probability of rejection over the remaining data, conditional on that history.

A rejection indicator equals 1 when the test rejects and 0 otherwise. The conditional expectation notation also accommodates randomized tests: if \(\phi\) is a conditional rejection probability between 0 and 1, the same argument applies. In either case, the conditioning must use the information actually available when the adaptation is made. The conditional distribution of future data is taken under the null parameter value being considered.

The Conditional-Expectation Argument

Write \(\phi_0\) for the original test’s rejection indicator and \(\phi_1\) for the modified test’s rejection indicator. For a fixed null parameter value \(\theta\), define the original conditional error by \(A_\theta=E_\theta(\phi_0\mid\mathcal{F})\). The modified test’s conditional rejection probability at the adaptation time is \(E_\theta(\phi_1\mid\mathcal{F})\). A sufficient condition for preserving the original test’s error bound is that the latter does not exceed the former, almost surely under that null parameter value.

Theorem: Suppose that for every \(\theta\in H_0\), the original test satisfies \(E_\theta(\phi_0)\leq\alpha\), and the modified test satisfies \(E_\theta(\phi_1\mid\mathcal{F})\leq E_\theta(\phi_0\mid\mathcal{F})\) almost surely. Then \(E_\theta(\phi_1)\leq\alpha\) for every \(\theta\in H_0\). Thus the modified test controls Type I error at level \(\alpha\).

The almost-sure qualification is important: the conditional inequality must hold except possibly on interim histories having probability zero under the null parameter value in question. For a composite null, the argument is applied separately at each \(\theta\in H_0\). The conditional distribution used to compute each expectation is the one under that same \(\theta\).

$$ \begin{aligned} E_\theta(\phi_1) &=E_\theta\!\left[E_\theta(\phi_1\mid\mathcal{F})\right]\\ &\leq E_\theta\!\left[E_\theta(\phi_0\mid\mathcal{F})\right]\\ &=E_\theta(\phi_0)\\ &\leq \alpha. \end{aligned} $$

The first and third equalities are applications of the law of iterated expectation: averaging a conditional expectation over the conditioning information recovers the original expectation. The inequality in the middle follows from the assumed conditional comparison. This proves the result without requiring the modified test to have the same form as the original test.

Worked Example: Averaging Conditional Error Across Interim Outcomes

Consider a hypothetical trial with two possible interim categories, \(G_1\) and \(G_2\). Under a fixed null hypothesis, \(P(G_1)=0.40\) and \(P(G_2)=0.60\). Suppose the original test’s conditional rejection probabilities are 0.020 in \(G_1\) and 0.030 in \(G_2\). A proposed modification has conditional rejection probabilities 0.015 and 0.025, respectively. Check the conditional comparison and calculate both overall rejection probabilities.

In \(G_1\), the modified probability, 0.015, is no greater than the original probability, 0.020. In \(G_2\), 0.025 is no greater than 0.030. The conditional inequality therefore holds in both possible interim categories.

The original overall rejection probability is the probability-weighted average of its conditional probabilities:

$$ P(\text{original rejection}) =(0.40)(0.020)+(0.60)(0.030) =0.008+0.018 =0.026. $$

For the modified rule, the same calculation gives

$$ P(\text{modified rejection}) =(0.40)(0.015)+(0.60)(0.025) =0.006+0.015 =0.021. $$

The modified overall rejection probability, 0.021, is below the original value, 0.026. If the original test was known to have Type I error at most \(\alpha=0.026\) at this null parameter value, this calculation verifies the bound for the modified rule at that value as well. The protection comes from checking each interim category, not merely from observing that the final weighted average happens to be small.

Applying the Argument to a Two-Stage Normal Test

A normal model makes the conditional rejection probability especially clear. Suppose a planned test uses independent stage statistics \(Z_1\) and \(Z_2\), each standard normal under a fixed null hypothesis, and rejects when the weighted combination exceeds a critical value. Let the information fraction be \(t=0.5\), and let the rejection rule be \(\sqrt{t}Z_1+\sqrt{1-t}Z_2>1.96\). Under the null, the combined statistic is standard normal, so the original test’s rejection probability is \(P(Z>1.96)\), approximately 0.0250. The assumptions used here are independence of the two stage statistics under this null and their standard normal distributions.

At the adaptation time, \(Z_1=z_1\) is observed. Conditional on that value, only \(Z_2\) remains random. Rearranging the original rejection rule gives its conditional error:

$$ \begin{aligned} A(z_1) &=P\!\left(\sqrt{0.5}\,z_1+\sqrt{0.5}\,Z_2>1.96\mid Z_1=z_1\right)\\ &=1-\Phi\!\left(\frac{1.96-\sqrt{0.5}\,z_1}{\sqrt{0.5}}\right), \end{aligned} $$

where \(\Phi\) is the standard normal cumulative distribution function. The second equality uses the stated independence: conditional on \(Z_1=z_1\), \(Z_2\) is still standard normal.

Worked Example: Preserving the Conditional Error at an Interim Value

For the hypothetical two-stage test above, suppose the observed interim statistic is \(z_1=0.800\). Calculate the original conditional error, then set a second-stage cutoff that gives the modified test the same conditional rejection probability.

The original test rejects when \(\sqrt{0.5}(0.800)+\sqrt{0.5}Z_2>1.96\). Solving for \(Z_2\), the cutoff is

$$ Z_2>\frac{1.96-\sqrt{0.5}(0.800)}{\sqrt{0.5}} =\frac{1.3943}{0.7071} \approx 1.972. $$

Thus its conditional rejection probability is \(1-\Phi(1.972)\), approximately 0.0243. A modified rule that, at this interim value, rejects when \(Z_2>1.972\) has the same conditional rejection probability. To apply the preservation argument across all possible interim values, this cutoff must be defined as a prespecified function of \(z_1\), not chosen after inspecting which cutoff gives a desirable result.

In this example the matching rule uses the original conditional-error function at every interim value. Consequently, its conditional rejection probability equals the original one, and iterated expectation gives the same overall rejection probability as the original test: approximately 0.0250 under the null. The calculation at \(z_1=0.800\) illustrates one history; the overall conclusion follows from the rule’s conditional relationship across histories.

Checking the Condition and Recognizing Failure

For a proposed adaptation, the proof reduces to a specific diagnostic: for each relevant interim history, calculate or bound the modified test’s conditional probability of eventual rejection and compare it with the original test’s conditional error at that same history. A marginal calculation alone does not establish this pointwise condition. Nor does it suffice to compare two conditional probabilities evaluated under different null models or different conditioning information.

Worked Example: A Rule That Exceeds Conditional Error

Suppose a hypothetical null model gives interim categories \(G_1\) and \(G_2\) equal probabilities of 0.50. The original test has conditional rejection probabilities 0.010 in \(G_1\) and 0.040 in \(G_2\). A proposed modified rule has probabilities 0.020 and 0.060. Determine whether the conditional-error argument protects the original level.

The original overall rejection probability is \((0.50)(0.010)+(0.50)(0.040)=0.025\). But in each category the modified probability exceeds the original one: \(0.020>0.010\) and \(0.060>0.040\). The required conditional inequality fails.

The modified overall rejection probability is \((0.50)(0.020)+(0.50)(0.060)=0.040\), which exceeds 0.025. In this example the modified test does not preserve the original 0.025 level. More generally, once the conditional inequality fails, this theorem no longer supplies a guarantee; the actual overall error would need a separate valid argument.

A common mistake is to say, “The original test has level \(\alpha\), so a modified test that uses the same interim data also has level \(\alpha\).” Sharing interim data does not imply that the modified test has a controlled conditional rejection probability. The correction is to write down both conditional rejection probabilities under the null, verify the inequality at the adaptation time, and only then average using iterated expectation.

This conditional proof differs from direct unconditional recalibration. An unconditional argument calculates the modified rule’s rejection probability after averaging over interim outcomes; it can be valid if that distribution and the post-adaptation rule are correctly accounted for. The conditional-error argument instead checks a bound at each interim history, then obtains the unconditional guarantee by averaging. Its assumptions are explicit, but calculating the conditional probabilities can be difficult when the future-data distribution is complex or when nuisance parameters affect it.

Key takeaway: Conditional-error preservation is an application of iterated expectation. If, under each null parameter value, the modified test’s rejection probability conditional on the adaptation information is no greater than the original test’s conditional rejection probability, then its overall Type I error is no greater than the original test’s.

Check Your Understanding

Answer each question using the conditional-expectation argument. Solutions follow the questions.

  1. Under a fixed null, interim categories \(A\) and \(B\) have probabilities 0.25 and 0.75. The original conditional rejection probabilities are 0.020 and 0.040; the modified probabilities are 0.018 and 0.035. Calculate both overall rejection probabilities.
  2. State the conditional inequality needed to apply the theorem, using \(\phi_0\), \(\phi_1\), and \(\mathcal{F}\).
  3. In the two-stage normal example, explain why \(Z_2\) is standard normal conditional on the observed \(Z_1=z_1\).
  4. If the conditional inequality fails for one interim category with positive probability, does this proof establish that the modified test violates the level? Explain.

Solutions. (1) The original probability is \((0.25)(0.020)+(0.75)(0.040)=0.005+0.030=0.035\). The modified probability is \((0.25)(0.018)+(0.75)(0.035)=0.0045+0.02625=0.03075\). The modified probability is smaller. (2) For each null value \(\theta\), require \(E_\theta(\phi_1\mid\mathcal{F})\leq E_\theta(\phi_0\mid\mathcal{F})\) almost surely. (3) The stage statistics are independent under the stated null model, so observing \(Z_1=z_1\) does not change the distribution of \(Z_2\); by assumption \(Z_2\sim N(0,1)\). (4) No. Failure means this conditional-error proof gives no guarantee; it does not by itself show the modified test’s marginal rejection probability exceeds the level. A separate calculation or valid argument would be needed to determine that probability.

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