Shrinkage Corrections for Selected Genetic Effects

Why Correct a Selected Effect Estimate?

A genetic association estimate that passes a significance threshold is not an ordinary draw from all estimates. Even if the underlying effect is modest, random sampling error can push some estimates over the threshold. Among the selected estimates, sampling errors therefore tend to point away from zero. This is the winner’s curse: a selected discovery estimate can exaggerate the magnitude of the underlying effect.

A correction does not recover the true effect with certainty. It uses a model for the sampling process and the selection rule to reduce the bias expected under that model. Two approaches are useful here. A truncated-normal likelihood models the estimate conditional on having passed the selection threshold. Empirical-Bayes shrinkage combines an estimate with a distribution of plausible effects, whose parameters are learned across many genetic effects. They address selection in different ways and need not give identical answers.

Definition: Let $\hat{\beta}$ be an estimated genetic effect with known standard error $s$. Under the normal sampling model, $\hat{\beta}\mid\beta\sim N(\beta,s^2)$, where $\beta$ is the underlying effect. For a prespecified positive-direction selection rule $\hat{\beta}/s>c$, the selection-adjusted likelihood is the density of $\hat{\beta}$ conditional on passing that rule: $$ L_{\mathrm{sel}}(\beta;\hat{\beta})= \frac{\frac{1}{s}\phi\left(\frac{\hat{\beta}-\beta}{s}\right)} {P(\hat{\beta}/s>c\mid\beta)} = \frac{\frac{1}{s}\phi\left(\frac{\hat{\beta}-\beta}{s}\right)} {\Phi\left(\frac{\beta}{s}-c\right)},\qquad \hat{\beta}/s>c. $$ Here $\phi$ and $\Phi$ are the standard normal density and cumulative distribution function. The selection-adjusted estimate is the value of $\beta$ that maximizes this conditional likelihood.

This definition uses a one-sided threshold because it describes positive effects selected in a prespecified direction. For negative selected estimates, reverse the sign of the estimate and effect before applying the same calculation. A two-sided selection rule requires conditioning on the event $|\hat{\beta}/s|>c$, whose probability is $1-\Phi(c-\beta/s)+\Phi(-c-\beta/s)$. The selection event used in the likelihood must match the rule that actually produced the reported estimate.

Truncated-Normal Likelihood Correction

The ordinary normal likelihood uses the density of the observed estimate. A selected estimate instead has a truncated density: the ordinary density divided by the probability of selection at the proposed value of $\beta$. That denominator matters because effects with lower probability of passing the threshold are less likely to appear in the selected set.

Taking the log of the likelihood for a positive selected estimate gives $\ell(\beta)= -\frac{1}{2}((\hat{\beta}-\beta)/s)^2-\log(s)-\log(\sqrt{2\pi})-\log\Phi(\beta/s-c)$. Differentiating with respect to $\beta$ produces a score equation. Define the inverse Mills ratio $m(a)=\phi(a)/\Phi(a)$. Then:

\frac{d\ell(\beta)}{d\beta} = \frac{\hat{\beta}-\beta}{s^2} - \frac{1}{s}m\left(\frac{\beta}{s}-c\right).

At an interior maximum, the score is zero, so the estimate satisfies

\hat{\beta}_{\mathrm{sel}} = \hat{\beta}-s\,m\left(\frac{\hat{\beta}_{\mathrm{sel}}}{s}-c\right).

The subtracted term is positive: under this model, the observed estimate exceeds the selection-adjusted estimate for a positive selected result. This is an implicit equation because the inverse Mills ratio depends on the unknown adjusted estimate. It is solved numerically, for example by iterating the right-hand side or finding a root of the score equation. The formula is not a universal fixed subtraction: the correction depends on the standard error, threshold, and observed estimate.

Worked Example: Correcting a Positive Selected Estimate

Suppose a hypothetical study estimates a per-allele genetic effect as $\hat{\beta}=0.30$ log-odds units, with known standard error $s=0.10$ log-odds units. Positive results are selected when $Z=\hat{\beta}/s>1.96$. Here $Z=3.00$, so the estimate passes the threshold. Find the conditional-likelihood estimate.

The score equation in standardized units is $b=3.00-m(b-1.96)$, where $b=\hat{\beta}_{\mathrm{sel}}/s$. Numerical solution gives approximately $b=2.52$. At that value, $b-1.96=0.56$, and $m(0.56)=\phi(0.56)/\Phi(0.56)\approx0.341/0.712=0.479$. Thus $3.00-0.479=2.521$, consistent with the root to the displayed precision.

Converting back to the effect scale gives $\hat{\beta}_{\mathrm{sel}}=0.10(2.52)=0.252$ log-odds units per allele. The correction reduces the discovery estimate from 0.30 to about 0.252 under the stated sampling and selection model. It does not prove that the underlying effect equals 0.252.

Empirical-Bayes Shrinkage Across Effects

A single selected estimate provides limited information about its own selection bias. Empirical-Bayes shrinkage uses a collection of genetic effects to estimate a distribution of underlying effects, then combines that distribution with each estimate. A simple model assumes that effects are drawn from a zero-centered normal distribution and that their sampling errors are normal.

Model: For effects $i=1,\ldots,M$, assume $\beta_i\sim N(0,\tau^2)$ and $\hat{\beta}_i\mid\beta_i\sim N(\beta_i,s_i^2)$. The effects are independent under this model, and their standard errors $s_i$ are treated as known. The prior variance $\tau^2$ describes the across-effect spread. Given $\tau^2$, the posterior distribution for an effect is normal, with mean and variance $$ E(\beta_i\mid\hat{\beta}_i)= \frac{\tau^2}{\tau^2+s_i^2}\hat{\beta}_i,\qquad \operatorname{Var}(\beta_i\mid\hat{\beta}_i)= \frac{\tau^2s_i^2}{\tau^2+s_i^2}. $$

The posterior mean is a weighted estimate: it multiplies the observed effect by a factor between zero and one. Estimates with larger sampling error relative to the prior spread are pulled more strongly toward zero. In empirical Bayes, $\tau^2$ is estimated from the ensemble of effects rather than fixed in advance. If those effects were selected, estimation of $\tau^2$ should account for the selection rule; otherwise the estimated distribution can be too dispersed because the observed effects are disproportionately large.

For example, under a common standard error $s$ and positive selection threshold $c$, the marginal estimate has distribution $\hat{\beta}_i\sim N(0,\tau^2+s^2)$ before selection. Its density among selected positive estimates is that normal density divided by $P(\hat{\beta}_i>cs)$. A selection-aware estimate of $\tau^2$ can maximize the product of these conditional densities across selected effects. This uses the threshold information when learning the prior spread.

Once $\tau^2$ is estimated, the posterior mean given the observed $\hat{\beta}_i$ is still the normal-model shrinkage formula above. Indeed, selection is determined by the observed estimate: if its exact value is known, knowing additionally that it passed the threshold supplies no new information about $\beta_i$. The selection correction is needed when fitting the distribution from selected estimates, not as an extra multiplier in the posterior for an already observed value.

Worked Example: Applying an Empirical-Bayes Shrinkage Factor

Suppose a hypothetical selection-aware fit across a discovery panel estimates the across-effect standard deviation as $\tau=0.50$ log-odds units. For one selected variant, $\hat{\beta}=0.80$ and $s=0.20$, both in log-odds units per allele. Treat the fitted $\tau$ as given and calculate the posterior mean and standard deviation.

The shrinkage factor is $\tau^2/(\tau^2+s^2)=0.50^2/(0.50^2+0.20^2)=0.25/0.29=0.8621$. Therefore the posterior mean is $0.8621(0.80)=0.6897$ log-odds units per allele.

The posterior variance is $\tau^2s^2/(\tau^2+s^2)=(0.25)(0.04)/0.29=0.0100/0.29=0.03448$, so the posterior standard deviation is $\sqrt{0.03448}=0.1857$ log-odds units per allele. This uncertainty is conditional on the normal model and the fitted prior variance; it does not include uncertainty in estimating $\tau$.

Comparing Corrections With Independent Replication

An independent replication estimate is informative in a different way from either model-based correction. If the replication sample was not used to select the discovery result, its estimate is not subject to that discovery threshold’s direct selection bias. It can still be noisy, and it may differ because of sampling variation or differences in the populations, phenotype, or analysis. Thus, neither a corrected discovery estimate nor a replication estimate should automatically be treated as the true effect.

Worked Example: Comparing Discovery, Shrinkage, and Replication

Continue the hypothetical example with a discovery estimate of 0.30 and standard error 0.10 log-odds units per allele, selected at $Z>1.96$. Its conditional-likelihood estimate was 0.252. Suppose the selection-aware empirical-Bayes fit gives $\tau=0.50$, and an independent, unselected replication sample estimates the effect as 0.24 with standard error 0.12 log-odds units per allele.

The empirical-Bayes estimate for the discovery result is $(0.50^2/(0.50^2+0.10^2))(0.30)=(0.25/0.26)(0.30)=0.2885$ log-odds units per allele.

The four point estimates are therefore: unadjusted discovery, 0.300; selection-adjusted conditional likelihood, 0.252; empirical-Bayes posterior mean, 0.2885; and independent replication, 0.240 log-odds units per allele. The replication standard error is 0.12, so its approximate 95% normal interval is $0.24\pm1.96(0.12)=0.24\pm0.2352$, or 0.0048 to 0.4752 log-odds units per allele.

These estimates are reasonably close in this hypothetical example, but their meanings differ. The conditional-likelihood estimate conditions on the positive discovery threshold. The empirical-Bayes mean uses the fitted distribution of effects. The replication estimate comes from a separate sample and has its own sampling uncertainty. Agreement is reassuring, but does not establish that their shared value is the exact underlying effect.

Assumptions, Interpretation, and Pitfalls

Both corrections rely on an accurate account of the data-generating and selection processes. The truncated-normal method assumes a suitable normal approximation for the estimate, a known or adequately estimated standard error, and a correctly specified threshold and direction of selection. The empirical-Bayes model additionally assumes an appropriate distribution of true effects; a zero-centered normal prior may be unsuitable if effects are asymmetric, have a heavy-tailed distribution, or combine many near-zero effects with a small number of large effects. Model misspecification can leave residual bias or over-shrink genuine large effects.

The corrections also have different scopes. A conditional likelihood for one estimate uses its selection event but does not borrow information across variants. Empirical Bayes borrows strength across effects, but its answer depends on how the prior variance is estimated. If discovery reporting uses several filters—for example, a significance threshold followed by selective reporting of only some loci—the fitted selection process must represent those filters to the extent relevant. A correction for one known threshold cannot automatically remove biases from an unmodeled selection step.

Do not interpret an adjusted estimate as a guaranteed true effect. It is a model-based estimate under stated assumptions.
Do not fit an ordinary effect distribution to selected estimates without considering selection. Selected estimates can make the distribution appear more variable than it is.
Do not confuse shrinkage with replication. Shrinkage is a model-based combination; replication is new evidence from another sample.
Do not apply the wrong threshold or direction. The conditional likelihood must reflect the actual rule that selected the result.

Key takeaway: Selection-aware likelihoods correct an estimate by conditioning on the event that selected it. Empirical Bayes shrinks estimates toward a learned distribution of effects, and selection must be considered when that distribution is fitted. Independent replication provides a separate comparison, not a guarantee of the true effect.

Check Your Understanding

Answer each question using the assumptions and distinctions developed above.

Why does the truncated-normal likelihood divide the ordinary estimate density by a probability of selection?
For a positive selected estimate, what direction does the conditional-likelihood correction generally move the estimate, and why?
Under the normal empirical-Bayes model, how does the shrinkage factor change when the standard error increases while $\tau$ stays fixed?
Why should selection be accounted for when estimating $\tau^2$ from a collection of significant effects?
Give one reason an independent replication estimate may differ from a selection-adjusted discovery estimate even when both analyses are correctly conducted.