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Type I Error Under Data-Dependent Changes · Tutorial 152 of 799

Conditioning on a Selection Event

You will derive and apply conditional null distributions to selected statistics, distinguishing selection on the statistic itself from selection independent of a later test.

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What You'll Learn

  • Define a selection event and the null distribution conditional on that event.
  • Derive a conditional upper-tail p-value for a normally distributed statistic screened by a threshold.
  • Calculate the selected maximum’s conditional tail probability under an independent-normal model.
  • Explain why an independent confirmatory statistic is not changed by screening on separate data.
  • Distinguish conditional error among selected cases from an unconditional error probability.
  • Diagnose when an ordinary reference distribution is inappropriate after selection.

Why Selection Changes the Reference Distribution

In an adaptive trial, interim data can determine which population, treatment, endpoint, or hypothesis proceeds to the final analysis. Once a statistic has helped determine what will be analyzed, its distribution among the analyses that proceed need not be its original null distribution. The relevant question is not only, “What values can this statistic take under the null?” It is also, “What values can it take under the null given that the selection rule chose it?”

This distinction matters for interpretation. A p-value calculated from an unselected reference distribution can appear small because the statistic was large enough to pass a screen, even when such a value is common among selected statistics under the null. The distribution after selection depends on the rule that selected the statistic and on how that statistic is related to the data used by the rule.

Definition: Let \(S\) denote a selection event, such as “the interim statistic exceeds a threshold” or “this candidate has the largest interim statistic.” For a test statistic \(T\), its null distribution conditional on selection is the distribution of \(T\) under \(H_0\), restricted to cases where \(S\) occurs. For example, its conditional upper-tail probability at an observed value \(t\) is \(P(T\ge t\mid S,H_0)\). This is distinct from the marginal null probability \(P(T\ge t\mid H_0)\).

The conditional distribution is defined only when \(P(S\mid H_0)>0\). In a continuous model, a useful general expression for the conditional density is \(f_{T\mid S,H_0}(t)=f_{T\mid H_0}(t)P(S\mid T=t,H_0)/P(S\mid H_0)\). The factor \(P(S\mid T=t,H_0)\) describes how selection weights different values of the statistic. If selection is determined by \(T\) itself, this factor can sharply favor large values. If selection depends only on independent data, it does not change the distribution of \(T\).

A Threshold Screen on the Same Statistic

Suppose a one-sided test statistic \(Z\) has the standard normal distribution under the null, \(Z\sim N(0,1)\), and a candidate proceeds only when \(Z>c\). Assume \(c\) is fixed in advance, the null model is correct, and the statistic is continuous. The selection event is \(S=\{Z>c\}\). For a selected observed value \(z\ge c\), the conditional upper-tail probability is

$$ P(Z\ge z\mid Z>c,H_0) =\frac{P(Z\ge z\mid H_0)}{P(Z>c\mid H_0)} =\frac{1-\Phi(z)}{1-\Phi(c)}. $$

The numerator is the probability of a value at least as large as the observation; because \(z\ge c\), every such value also passes the screen. The denominator is the probability of passing the screen at all. Dividing by it expresses the tail probability on the selected subset. By contrast, the unselected upper-tail p-value is \(1-\Phi(z)\). The two answer different questions.

Worked Example: A Normally Distributed Statistic Passes a Screen

Suppose, hypothetically, a trial screens a candidate using a one-sided interim statistic \(Z\), and proceeds only if \(Z>1.645\). Under the point null, assume \(Z\sim N(0,1)\). The observed selected value is \(z=2.000\). Calculate the ordinary and selection-conditional upper-tail probabilities.

The ordinary upper-tail probability is \(1-\Phi(2.000)\approx 0.0227501\). It is the null probability of a statistic at least this large before conditioning on the screen.

The probability of passing the screen under the null is \(1-\Phi(1.645)\approx 0.0499849\). Thus,

$$ P(Z\ge 2.000\mid Z>1.645,H_0) =\frac{0.0227501}{0.0499849} \approx 0.45514 \approx 0.455. $$

Among null cases that pass this screen, about 45.5% have a statistic at least 2.000. The ordinary p-value, about 0.0228, is not the tail probability conditional on selection. This calculation does not say that the null is true or that it has a 45.5% probability; it describes a probability under the specified null model.

Deriving the Conditional Distribution

The threshold result follows directly from the definition of conditional probability. For any \(z\ge c\), the event \(\{Z\ge z\}\) is contained in \(\{Z>c\}\), apart from a boundary of probability zero. Therefore their intersection has probability \(P(Z\ge z\mid H_0)\). Dividing by the probability of selection gives the displayed conditional tail formula.

The same argument gives the conditional cumulative distribution function. For \(z\le c\), a selected statistic cannot be at or below \(z\), so the conditional CDF is zero. For \(z>c\), subtract the probability of exceeding \(z\) from the probability of passing the screen:

$$ P(Z\le z\mid Z>c,H_0) =\frac{\Phi(z)-\Phi(c)}{1-\Phi(c)},\qquad z>c. $$

This is a truncated normal distribution: values at or below the threshold have been removed and the remaining probabilities rescaled to sum to one. It is not a standard normal distribution. In particular, treating the selected \(Z\) as though it were an unselected standard normal statistic uses the wrong reference distribution for a conditional calculation.

Selection of the Largest Candidate

Selection can also favor a statistic because it is the largest among several candidates. Consider \(K\) independent statistics \(Z_1,\ldots,Z_K\), each standard normal under a joint null, and let \(M=\max_j Z_j\). Independence is important for the following formula. For any \(m\),

$$ P(M\le m\mid H_0) =P(Z_1\le m,\ldots,Z_K\le m\mid H_0) =\Phi(m)^K. $$

Consequently, \(P(M\ge m\mid H_0)=1-\Phi(m)^K\). If selection means that the maximum exceeds a threshold \(a\), then, for \(m\ge a\), \(P(M\ge m\mid M>a,H_0)=[1-\Phi(m)^K]/[1-\Phi(a)^K]\). The selected maximum is not distributed like one prespecified \(Z_j\). The selection rule, including the number of candidates, is part of the null distribution.

Worked Example: The Maximum of Two Candidates Passes a Screen

Suppose, hypothetically, two candidate treatment populations produce independent interim statistics \(Z_1,Z_2\), each standard normal under the joint null. The trial selects the larger value \(M=\max(Z_1,Z_2)\), and proceeds only if \(M>1.960\). The selected maximum is \(m=2.200\). Find its conditional upper-tail probability.

For two independent statistics, \(P(M\le x)=\Phi(x)^2\), so \(P(M\ge x)=1-\Phi(x)^2\). Using \(\Phi(2.200)\approx0.9860966\) and \(\Phi(1.960)\approx0.9750021\),

$$ \begin{aligned} P(M\ge2.200\mid M>1.960,H_0) &=\frac{1-\Phi(2.200)^2}{1-\Phi(1.960)^2}\\ &\approx\frac{1-(0.9860966)^2}{1-(0.9750021)^2}\\ &\approx\frac{0.0276136}{0.0493709}\\ &\approx0.5593. \end{aligned} $$

Thus, under this model, about 55.9% of selected maxima are at least 2.200. This differs from the ordinary upper-tail probability for one standard normal statistic, \(1-\Phi(2.200)\approx0.01390\), and from the marginal tail probability for the maximum, about 0.02761. Each probability uses a different reference distribution or conditioning event. The calculation assumes independence; correlated candidate statistics require their joint null distribution rather than \(\Phi(m)^2\).

When Selection Does Not Change the Test Statistic

Selection does not automatically distort every later statistic. If the screen uses data independent of the data used to calculate the final statistic, the final statistic’s distribution can remain unchanged under the null. Formally, if \(T\) and the selection event \(S\) are independent under \(H_0\), then \(P(T\ge t\mid S,H_0)=P(T\ge t\mid H_0)\). This is a conditional probability identity, not a claim that independence is always plausible in a trial.

Worked Example: An Independent Confirmation Statistic

Suppose, hypothetically, an interim statistic \(Z_1\) is used to select a candidate when \(Z_1>1.645\). A later statistic \(Z_2\) is calculated from new, independent observations. Under the null, assume \(Z_1\) and \(Z_2\) are independent and each has a standard normal distribution. For an observed \(Z_2=2.000\), calculate the upper-tail probability conditional on selection.

By independence, \(P(Z_2\ge2.000\mid Z_1>1.645,H_0)=P(Z_2\ge2.000\mid H_0)\). Therefore,

$$ P(Z_2\ge2.000\mid Z_1>1.645,H_0) =1-\Phi(2.000) \approx0.02275. $$

Here the ordinary standard normal reference distribution remains appropriate for \(Z_2\), because the screen used independent data. The conclusion depends on the stated independence and correct null distribution. Reusing interim observations in \(Z_2\), or selecting based on information associated with \(Z_2\), would require a new distributional argument.

Diagnostics, Alternatives, and Limits

Before interpreting a p-value after an adaptive choice, specify the selection event precisely. Ask which hypotheses, populations, endpoints, or statistics could have been selected; what data determined the choice; whether the final statistic reuses those data; and whether multiple candidates were compared. Then identify the null distribution jointly with that rule. A derivation that conditions on a threshold but omits the fact that the largest of several candidates was chosen is not conditioning on the actual selection process.

A useful alternative is to base inference on independent confirmatory data, when that independence and the analysis plan are defensible. Another approach is to calculate probabilities using the joint null distribution and the selection rule, as in the maximum example. A prespecified analysis that accounts for the full data-dependent procedure can also be used. These approaches are not interchangeable by default: each relies on its own model and design assumptions.

Conditioning changes the question, not necessarily the trial’s unconditional error probability. For example, the conditional probability of rejection among selected null cases can differ greatly from the probability of rejection across all null cases. They are related by \(P(\text{reject},S\mid H_0)=P(\text{reject}\mid S,H_0)P(S\mid H_0)\). A conditional probability alone therefore does not determine the unconditional probability. Conversely, a marginal error calculation does not automatically establish a desired conditional guarantee.

The calculations above also have limits. The standard-normal formulas rely on the specified null model, continuity, and (for the maximum formula) independence. With correlated statistics, the joint dependence must be included. With discrete statistics, conditional tail probabilities must be computed from the actual probability mass function; ties can affect the attainable levels. In practice, model misspecification, unrecorded selection steps, or reuse of data can invalidate a claimed conditional distribution.

A common mistake is: “The observed statistic is 2.000, and its ordinary standard normal p-value is 0.0228, so it is equally unusual among the cases selected by a rule requiring \(Z>1.645\).” The correction is that 0.0228 is an unselected tail probability. Under the stated screen, the conditional tail probability is about 0.455. The ordinary value is not intrinsically wrong; it answers a different question and is inappropriate if the intended reference distribution is conditional on passing that screen.

Key takeaway: A data-dependent selection rule can reshape a statistic’s null distribution. Use the unselected distribution only when the statistic is independent of selection under the null or when the analysis otherwise justifies that reference distribution; otherwise, account for the selection event explicitly.

Check Your Understanding

Answer each question using the stated model. Brief solutions follow the questions.

  1. Under \(H_0\), \(Z\sim N(0,1)\), and selection requires \(Z>1.645\). For an observed selected value \(z=2.000\), what is the conditional upper-tail probability, approximately?
  2. Two independent standard normal statistics are observed and the larger is selected. Write the null CDF of their maximum \(M\).
  3. A screen uses \(Z_1\), while a final statistic \(Z_2\) is independent of \(Z_1\) and standard normal under \(H_0\). Does conditioning on the screen alter the null distribution of \(Z_2\)? State why.
  4. Explain the difference between \(P(\text{reject}\mid S,H_0)\) and \(P(\text{reject},S\mid H_0)\).

Solutions. (1) The value is \([1-\Phi(2.000)]/[1-\Phi(1.645)]\approx0.0227501/0.0499849\approx0.455\). (2) Independence gives \(P(M\le m\mid H_0)=\Phi(m)^2\). (3) No. Independence gives \(P(Z_2\ge t\mid S,H_0)=P(Z_2\ge t\mid H_0)\), so its standard normal null distribution is unchanged. (4) The first is the rejection probability among selected null cases; the second is the probability of both selection and rejection across all null cases. They satisfy \(P(\text{reject},S\mid H_0)=P(\text{reject}\mid S,H_0)P(S\mid H_0)\).

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