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Selection and Bias in Prediction · Tutorial 776 of 989

Winner’s Curse After GWAS Significance Selection

Learn to calculate the expected effect estimate among significant variants and distinguish average selection-induced inflation from a guarantee about any one estimate.

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What You'll Learn

  • Set up the normal sampling model for a GWAS effect estimate and its test statistic.
  • Derive the conditional expectation after one-sided or two-sided significance selection.
  • Calculate selection probabilities and conditional effect estimates from concrete values.
  • Distinguish the mean signed estimate from the mean absolute estimate among selected variants.
  • Explain why winner’s curse describes an average over repeated discovery samples, not certain inflation in every result.

Why Significance Selection Inflates Estimates

A genetic association study tests many variants, then commonly highlights those whose test statistics pass a significance threshold. Even when each effect estimate is unbiased before selection, estimates that happen to be unusually far from zero are more likely to pass. Among selected variants, sampling errors therefore tend to push estimates away from zero. This selection-induced difference between the average reported estimate and the underlying effect is called the winner’s curse.

The effect is about the estimate, not a change in the true genetic effect. A particular estimate may be close to, above, or below its true value. The key result is an expectation: across repeated studies of the same underlying effect, what is the mean estimate among studies in which the variant is significant?

Sampling Model and Selection Rule

Definition: For a variant with true genetic effect \(\beta\), suppose its estimated effect \(\hat{\beta}\), measured in the outcome’s units per effect allele, has sampling distribution \(\hat{\beta}\mid\beta\sim N(\beta,SE^2)\), with a fixed standard error \(SE\). Define \(Z=\hat{\beta}/SE\) and \(\delta=\beta/SE\). Then \(Z\mid\beta\sim N(\delta,1)\). A two-sided significance rule selects the variant when \(|Z|>c\); a one-sided rule selects it when \(Z>c\).

Here \(c\) is the critical value for the chosen significance level. For a two-sided test with level \(\alpha\), \(c\) satisfies \(P(|Z|>c\mid\beta=0)=\alpha\). The standard normal density and cumulative distribution function are denoted by \(\phi\) and \(\Phi\), respectively. The model assumes an unbiased estimate before selection, a normal sampling distribution, and a standard error treated as fixed. These are useful approximations for deriving the effect; departures from them can change the amount of inflation.

The quantity \(\delta\) is the true effect in standard-error units. It determines how likely the variant is to pass the threshold: a larger positive \(\delta\) makes positive-significance results more likely. Since \(\hat{\beta}=SE\,Z\), the conditional expectation of the estimate follows directly from the conditional expectation of \(Z\).

Deriving the Conditional Expectation

For a two-sided test, selection includes both tails: unusually large positive and unusually large negative test statistics. Under \(Z\sim N(\delta,1)\), the selection probability is

$$ P(|Z|>c\mid\beta) =\Phi(\delta-c)+\Phi(-c-\delta). $$

To find the mean among selected results, use the truncated-normal tail identities. If \(Z\sim N(\delta,1)\), then \(E[Z\,\mathbf{1}(Z>c)]=\delta\Phi(\delta-c)+\phi(c-\delta)\), and \(E[Z\,\mathbf{1}(Z<-c)]=\delta\Phi(-c-\delta)-\phi(c+\delta)\). These follow by writing \(Z=\delta+U\), where \(U\sim N(0,1)\), and integrating the standard normal density; in particular, \(\int_a^\infty u\phi(u)\,du=\phi(a)\).

Adding the contributions from the two selected tails gives the total selected first moment. Dividing by the probability of selection gives the conditional mean:

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

Multiplying by \(SE\) and using \(\beta=SE\,\delta\) yields the result for the estimated effect:

Result: Under the stated normal model and a two-sided threshold, the mean signed effect estimate among selected variants is
$$ E[\hat{\beta}\mid |Z|>c,\beta] =\beta+SE\, \frac{\phi(c-\delta)-\phi(c+\delta)} {\Phi(\delta-c)+\Phi(-c-\delta)}. $$
For \(\beta>0\), this conditional mean exceeds \(\beta\); for \(\beta<0\), it is below \(\beta\). In both cases, the signed mean is shifted away from zero.

For a one-sided rule \(Z>c\), only the upper tail contributes. The corresponding result is

$$ E[\hat{\beta}\mid Z>c,\beta] =\beta+SE\,\frac{\phi(c-\delta)}{1-\Phi(c-\delta)}. $$

The extra term is positive because selection requires an unusually large positive estimate. This directional formula is not interchangeable with the two-sided formula: a two-sided rule can also select a negative estimate.

Worked Example: A One-Sided Selection Rule

Suppose a hypothetical variant has true effect \(\beta=0.020\) outcome units per allele and \(SE=0.020\) units per allele. A one-sided test selects when \(Z>1.96\). Calculate the selection probability and the mean estimate among selected studies.

The standardized true effect is \(\delta=\beta/SE=0.020/0.020=1\). The selection probability is \(1-\Phi(1.96-1)=1-\Phi(0.96)\approx0.1685\). Thus, under this model, about 16.85% of repeated studies would pass this one-sided threshold.

For the conditional mean, \(\phi(0.96)\approx0.2516\), so the one-sided formula gives

$$ E[\hat{\beta}\mid Z>1.96,\beta] =0.020+0.020\left(\frac{0.2516}{0.1685}\right) \approx0.0499. $$

The mean selected estimate is about \(0.0499\) outcome units per allele, approximately 2.49 times the true effect. This is an average over studies meeting the rule, not a prediction that every selected estimate will equal \(0.0499\).

Genome-Wide Thresholds and Effect-Size Inflation

For a conventional two-sided genome-wide significance level near \(5\times10^{-8}\), the standard normal critical value is approximately \(c=5.45\). The formula shows that inflation depends not only on this threshold but also on the true signal-to-noise ratio \(\delta\). A variant with a large \(\delta\) is likely to pass even without an unusually favorable sampling error, so the conditional shift can be modest. A weaker variant passes mainly in studies with unusually large estimates, so its selected estimate can be much more inflated relative to its true effect.

Worked Example: Selection at a Genome-Wide Threshold

Consider a hypothetical variant with true effect \(\beta=0.090\) outcome units per allele and \(SE=0.020\) units per allele. Use the two-sided threshold \(c=5.45\) to calculate its probability of selection and mean signed estimate among selected studies.

Here, \(\delta=0.090/0.020=4.5\). The selection probability is

$$ P(|Z|>5.45\mid\beta) =\Phi(-0.95)+\Phi(-9.95) \approx0.1711. $$

The lower-tail probability at \(-9.95\) is negligible at the displayed precision. Also, \(\phi(0.95)\approx0.2541\) and \(\phi(9.95)\) is negligible. Therefore,

$$ E[\hat{\beta}\mid |Z|>5.45,\beta] =0.090+0.020\left(\frac{0.2541}{0.1711}\right) \approx0.1197. $$

The selected mean is about \(0.1197\) outcome units per allele, or \(0.1197/0.090\approx1.33\) times the true effect. In this hypothetical setting, about 17.11% of studies pass the threshold. The example illustrates that a stringent threshold does not determine inflation by itself: the true effect relative to its standard error matters as well.

Signed Estimates and Reported Magnitudes

The two-sided formula above concerns the signed estimate in the allele coding used for the analysis. When the true effect is positive but small, some selected studies can still have significant negative estimates. If results are instead summarized by the magnitude of the estimated effect, the relevant quantity is \(E[|\hat{\beta}|\mid |Z|>c,\beta]\), which is not generally the absolute value of the conditional signed mean.

The two selected tails contribute positively to the absolute first moment. For \(Z\sim N(\delta,1)\), their combined contribution is \(\delta\{\Phi(\delta-c)-\Phi(-c-\delta)\}+\phi(c-\delta)+\phi(c+\delta)\). Dividing by the selection probability gives

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

This expression is useful when discussing effect sizes by absolute magnitude or when the estimated sign is used to orient the reported effect. In particular, even at a null effect, the selected signed estimates average to zero under symmetric two-sided selection, while their selected absolute magnitudes are positive.

Worked Example: A Weak Effect and Absolute Inflation

Suppose a hypothetical variant has \(\beta=0.010\) outcome units per allele and \(SE=0.020\) units per allele. Select studies with \(|Z|>2\), so \(\delta=0.5\). Compare the conditional signed mean with the conditional mean absolute estimate.

The selection probability is \(\Phi(-1.5)+\Phi(-2.5)\approx0.06681+0.006210=0.07302\). For the signed mean, \(\phi(1.5)\approx0.1295\) and \(\phi(2.5)\approx0.01753\). Thus,

$$ E[\hat{\beta}\mid |Z|>2,\beta] =0.010+0.020\left(\frac{0.1295-0.01753}{0.07302}\right) \approx0.0407. $$

The signed conditional mean is about \(0.0407\) outcome units per allele, around 4.07 times the true positive effect.

For absolute estimates, the numerator on the standardized scale is

$$ 0.5(0.06681-0.006210)+0.1295+0.01753 \approx0.1773. $$

Therefore, \(E[|\hat{\beta}|\mid |Z|>2,\beta]\approx0.020(0.1773/0.07302)=0.0486\) outcome units per allele. This is about \(0.0486/0.010=4.86\) times the true absolute effect. The signed and absolute summaries differ because a small number of selected studies have significant estimates in the negative tail.

Interpretation and Common Pitfall

The winner’s curse matters because significance selection affects how discovery estimates should be interpreted: the estimate reported for a selected variant is not an unbiased summary of its true effect when conditioning on having passed the threshold. The formulas quantify this conditional expectation under a specified model and threshold. They do not imply that every selected estimate is too large, nor do they provide a correction for an individual variant without additional information.

  • Wrong reasoning: “The estimate was unbiased before testing, so it remains unbiased among significant variants.” Correction: Unbiasedness before selection means \(E[\hat{\beta}\mid\beta]=\beta\) across all studies. Selection changes the distribution being averaged, giving \(E[\hat{\beta}\mid |Z|>c,\beta]\), which generally differs from \(\beta\).
  • Wrong reasoning: “A significant estimate must be inflated.” Correction: Inflation describes a conditional average over repeated studies. A particular selected estimate can be below or above that average, and it can even be closer to the true effect than the conditional mean suggests.
  • Wrong reasoning: “The two-sided conditional signed mean is the mean reported effect magnitude.” Correction: Signed and absolute estimates differ when selected results occur in both tails. Use the absolute-value expectation when the question concerns magnitudes.

The calculation is conditional on a fixed true effect and standard error. Across variants with different true effects or precisions, the average among all selected variants also depends on how those effects and standard errors are distributed. Thus, the result for a single variant should not be mistaken for a universal inflation factor across a collection of GWAS findings.

Check Your Understanding

Use the normal sampling model and distinguish the selection rule from the quantity being summarized.

  1. In the stated model, how are \(\delta\), \(\beta\), and \(SE\) related?
  2. For a one-sided rule \(Z>c\), why is the conditional mean estimate shifted in the positive direction?
  3. What two tail probabilities make up the selection probability for a two-sided threshold?
  4. Why can the conditional mean signed estimate and conditional mean absolute estimate differ?
  5. Does the winner’s curse imply that every selected estimate exceeds its true effect in magnitude? Explain.
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