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Deriving Selection Bias After Significance Thresholding

Learn how a significance cutoff alters expected estimates, and why signed and absolute selection produce different conditional effects.

Advanced 12 min read

What You'll Learn

  • State the normal sampling model and conditions for the selection-bias formulas.
  • Derive the expected estimate conditional on crossing a positive one-sided threshold.
  • Derive the signed and absolute conditional expectations after two-sided selection.
  • Distinguish directional selection from selection by an absolute test statistic.
  • Calculate selection probabilities and conditional effects in hypothetical examples.
  • Recognize when the simple threshold model does not describe the actual selection process.

Why a Significance Threshold Changes the Estimate

A genetic association estimate varies from sample to sample, even when the underlying effect is fixed. If researchers report an estimate only when its test statistic crosses a significance threshold, the reported estimates are drawn from a restricted part of their sampling distribution. Estimates near the true effect that happen to be unusually large are more likely to pass the filter than estimates that happen to be small. The conditional mean among selected results can therefore differ from the underlying effect.

This tutorial derives that difference under a normal sampling model. The key distinction is what was selected: a result exceeding a threshold in a prespecified positive direction, or a result exceeding a threshold in either direction. The latter selects on the absolute test statistic. These events are not interchangeable: under a zero true effect, two-sided selection leaves the mean signed estimate at zero by symmetry, even though the selected absolute estimates are large on average.

Model, Assumptions, and Definitions

Let \(\hat{\beta}\) estimate a genetic association effect \(\beta\), measured in the study's effect units. Assume its standard error \(SE\) is fixed and known for this calculation and that \(\hat{\beta}\sim N(\beta,SE^2)\). Define the standardized estimate \(Z=\hat{\beta}/SE\) and the standardized true effect \(\delta=\beta/SE\). Then \(Z\sim N(\delta,1)\). Let \(\phi\) and \(\Phi\) denote the standard normal density and cumulative distribution functions, and let \(Q(x)=1-\Phi(x)\).

Definition: For a threshold \(c>0\), positive-direction selection is the event \(Z>c\). Two-sided selection is the event \(|Z|>c\), which includes both \(Z>c\) and \(Z<-c\). Conditional selection bias on the signed scale is \(E[\hat{\beta}\mid S]-\beta\), where \(S\) is the specified selection event. When the selected estimate is aligned to its observed direction, the relevant magnitude is \(E[|\hat{\beta}|\mid |Z|>c]\).

The formulas below require the normal model, a fixed \(SE\), and a fixed threshold \(c\). They describe selection for one estimate under that threshold. If the standard error is estimated with substantial uncertainty, the test statistic is not well approximated by this fixed-variance normal model. Likewise, selecting the largest result among many correlated tests involves an additional selection rule not captured solely by conditioning one statistic on \(|Z|>c\).

Positive-Direction Selection

For a prespecified positive-direction test, the probability of selection is \(P(Z>c)=Q(c-\delta)\). To obtain the conditional mean, write \(Z=\delta+\varepsilon\), where \(\varepsilon\sim N(0,1)\). The normal tail identity \(E[\varepsilon\,1(\varepsilon>a)]=\phi(a)\) follows by integrating \(u\phi(u)\) from \(a\) to infinity, since \(\phi'(u)=-u\phi(u)\). With \(a=c-\delta\), this gives

$$ E[Z\mid Z>c] =\delta+\frac{\phi(c-\delta)}{Q(c-\delta)}. $$

Thus the conditional mean exceeds \(\delta\): the added term is positive. Returning to the effect scale, \(\hat{\beta}=SE\cdot Z\) and \(\beta=SE\cdot\delta\), so

$$ E[\hat{\beta}\mid Z>c] =\beta+SE\frac{\phi(c-\delta)}{Q(c-\delta)}. $$

The amount of conditional inflation depends on both the threshold and the true standardized effect. It is not a fixed correction that can be applied to every selected estimate.

Worked Example: Selection in a Prespecified Positive Direction

Suppose a hypothetical association has true effect \(\beta=0.20\) units and fixed \(SE=0.20\) units. Then \(\delta=0.20/0.20=1\). Select results when \(Z>2\), so \(c=2\). The selection probability is \(Q(2-1)=Q(1)=0.1587\).

Using \(\phi(1)=0.2420\) and \(Q(1)=0.1587\), the conditional mean standardized estimate is \(1+0.2420/0.1587=1+1.525=2.525\).

On the effect scale, the expected selected estimate is \(0.20+(0.20)(0.2420/0.1587)=0.20+0.3050=0.5050\) units. Thus, under this model, the average estimate among results passing the positive threshold is about 0.505 units, although the true effect is 0.20 units. This is a conditional expectation over repeated samples, not a claim that every selected estimate is inflated by that amount.

Two-Sided Selection: Signed Estimates and Absolute Magnitudes

For selection by \(|Z|>c\), both tails contribute. The selection probability is \(Q(c-\delta)+\Phi(-c-\delta)\). The positive-tail contribution to the unnormalized signed expectation is \(E[Z\,1(Z>c)]=\delta Q(c-\delta)+\phi(c-\delta)\). The negative-tail contribution is \(E[Z\,1(Z<-c)]=\delta\Phi(-c-\delta)-\phi(c+\delta)\). Adding these contributions and dividing by the selection probability yields the signed conditional mean:

$$ E[Z\mid |Z|>c] = \frac{ \delta\{Q(c-\delta)+\Phi(-c-\delta)\} +\phi(c-\delta)-\phi(c+\delta) }{ Q(c-\delta)+\Phi(-c-\delta) }. $$

For the absolute magnitude, the negative-tail contribution changes sign because \(|Z|=-Z\) there. Therefore,

$$ E[|Z|\mid |Z|>c] = \frac{ \delta\{Q(c-\delta)-\Phi(-c-\delta)\} +\phi(c-\delta)+\phi(c+\delta) }{ Q(c-\delta)+\Phi(-c-\delta) }. $$

Multiplying either conditional expectation by \(SE\) gives the corresponding estimate in effect units. Under a zero true effect, \(\delta=0\), the signed formula is zero: the selected positive and negative estimates cancel. The absolute formula instead simplifies to

$$ E[|Z|\mid |Z|>c,\delta=0] = \frac{2\phi(c)}{2Q(c)} = \frac{\phi(c)}{Q(c)}. $$

Worked Example: A Two-Sided Threshold Under the Null

Consider a hypothetical null association, so \(\beta=0\) and \(\delta=0\), with fixed \(SE=0.10\) effect units. Suppose the two-sided threshold is \(c=1.96\). The probability of selection is \(2Q(1.96)\), approximately \(2(0.0250)=0.0500\).

Since \(\phi(1.96)\approx0.05844\), the conditional mean absolute standardized estimate is \(\{2(0.05844)\}/\{2(0.0250)\}=0.1169/0.0500\approx2.338\).

Consequently, the expected absolute selected estimate is \((0.10)(2.338)=0.2338\) effect units. But the expected signed selected estimate is zero: under the null, positive and negative selected estimates are equally likely and symmetric. If estimates are reported after orienting each one to its observed direction, the reported magnitudes have mean 0.2338 units in this hypothetical model.

Worked Example: Signed and Absolute Means Under Two-Sided Selection

Return to a hypothetical association with \(\delta=1\), \(\beta=0.20\) units, and \(SE=0.20\) units. Use the two-sided threshold \(c=2\). The positive and negative tail probabilities are \(Q(1)=0.158655\) and \(\Phi(-3)=Q(3)=0.001350\), so the total selection probability is \(0.160005\). Also, \(\phi(1)=0.241971\) and \(\phi(3)=0.004432\).

The numerator for the signed conditional mean is \(1(0.160005)+0.241971-0.004432=0.397544\). Therefore, \(E[Z\mid |Z|>2]=0.397544/0.160005=2.485\) (approximately). The expected signed estimate after selection is \((0.20)(2.485)=0.4970\) units.

For the absolute conditional mean, the numerator is \(1(0.158655-0.001350)+0.241971+0.004432=0.403708\). Thus, \(E[|Z|\mid |Z|>2]=0.403708/0.160005=2.523\), and the expected selected magnitude is \((0.20)(2.523)=0.5046\) units. The absolute mean is larger because the selected negative estimates contribute positively to magnitude but negatively to the signed mean.

Why the Distinction Matters

A two-sided \(p\)-value threshold selects on \(|Z|\), not on \(Z\) alone. If the effect direction was specified before looking at the data, positive-direction selection can be represented by \(Z>c\) (or by the corresponding negative tail for a prespecified negative direction). If the direction is chosen after examining the estimate, the selected magnitude is relevant, and treating the event as a prespecified positive-direction test gives the wrong conditional expectation.

The difference is especially clear under the null. Two-sided significance filtering does not make the mean signed estimate positive when both signs are retained; it makes the mean absolute estimate positive and large relative to the threshold. If a report flips signs so every selected estimate points in the observed direction, it has changed the quantity being summarized from a signed estimate to a magnitude.

In a genomic analysis, the threshold may be very stringent, and the number of tested variants may be large. These formulas explain the conditional effect of crossing a specified threshold for a given statistic. They do not by themselves account for choosing a lead variant among many correlated candidates, changing the analysis after seeing results, or uncertainty in the estimated standard error. Those features require a model of the actual selection process.

Diagnostics, Alternatives, and Limitations

Before using a conditional expectation, write down the selection event in the same terms as the statistic. Check whether the direction was prespecified, whether the cutoff is one- or two-sided, and whether the reported estimate retains its original sign. Confirm that \(\hat{\beta}\) and \(SE\) are on the same scale and that \(Z=\hat{\beta}/SE\) is the statistic being thresholded. A mismatch between the reported selection rule and the event in the calculation invalidates the result.

Useful warning signs include interpreting \(E[|Z|\mid |Z|>c]\) as a signed mean, treating a two-sided \(p\)-value filter as positive-tail selection, or applying the one-test formula to a lead estimate selected as the largest among many loci. The normal model is also a poor approximation if the estimator distribution is strongly skewed or if \(SE\) is highly uncertain. In those cases, the selection mechanism and sampling distribution need to be represented more fully.

Selection or summaryConditional quantityKey implication
Prespecified positive tail, \(Z>c\)\(E[Z\mid Z>c]\)Exceeds \(\delta\) under the normal model.
Two tails, \(|Z|>c\), signs retained\(E[Z\mid |Z|>c]\)Can be zero under the null by symmetry.
Two tails, direction aligned\(E[|Z|\mid |Z|>c]\)Positive even under the null; describes magnitude.

The formulas here derive and quantify selection bias; they are not correction procedures. In particular, a conditional mean is not enough to recover an individual locus's true effect. A correction must also specify how the selection event and the information in the observed estimate are used; that is a separate estimation problem.

Common Mistake

Common mistake: “A two-sided test selected a significant result, so its expected signed \(Z\) must be greater than the cutoff.” This reasoning silently replaces \(|Z|>c\) with \(Z>c\). Under a zero effect and symmetric normal sampling, \(E[Z\mid |Z|>c]=0\), whereas \(E[|Z|\mid |Z|>c]=\phi(c)/Q(c)>c\). Keep the sign if the question is about signed estimates; use absolute values only when the reported quantity is direction-aligned magnitude.

Check Your Understanding

Answer each question using the selection event stated. Solutions are included below for checking the reasoning.

  1. Under a null normal model, what is \(E[Z\mid |Z|>c]\), and why?
  2. For \(\delta=0\) and \(c=1.96\), compute the conditional mean of \(|Z|\), using \(\phi(1.96)=0.05844\) and \(Q(1.96)=0.0250\).
  3. With \(\delta=1\), \(SE=0.20\) units, and positive-tail selection \(Z>2\), express the conditional expected estimate using \(\phi(1)=0.241971\) and \(Q(1)=0.158655\), then calculate it.
  4. Why can the expected signed estimate under two-sided selection differ from the expected absolute estimate?

Solutions. (1) It is zero: the selected positive and negative tails have equal probability and cancel in the signed mean. (2) \(E[|Z|\mid |Z|>1.96]=0.11688/0.0500\approx2.338\). (3) \(\beta+SE\phi(1)/Q(1)=0.20+0.20(0.241971/0.158655)\approx0.5050\) units. (4) Negative selected estimates subtract from the signed mean but add to the absolute magnitude.

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