Why the Largest Observed Effect Is Usually Too Large
When several treatments are estimated, selecting the one with the largest observed effect also selects partly on random error. Even if every estimate is unbiased before selection, the selected estimate tends to have a positive error: a treatment can win because its true effect is large, because its estimate happened to be high, or both. The resulting exaggeration depends on how the winner is chosen and on the variability and dependence of the estimates.
The calculation below isolates this selection effect for two estimates. It does not say that every selected estimate is exaggerated, nor that the winner’s true effect must exceed the other treatment’s true effect. It describes an expectation over repeated hypothetical trials conducted under the stated model.
Model and Assumptions
For the main result, suppose the estimates are jointly normal and unbiased before selection:
The errors may be correlated. Their variances and covariance are fixed, finite, and known for the calculation; write \(v_i^2=\operatorname{Var}(\varepsilon_i)\) and \(c=\operatorname{Cov}(\varepsilon_1,\varepsilon_2)\). For selection by the larger raw estimate, assume ties have probability zero, which holds when the difference of the estimates has positive variance. If its variance is zero, the estimates’ difference is deterministic and there is no random ranking to analyze.
Write \(\delta=\theta_1-\theta_2\), \(D=\hat{\theta}_1-\hat{\theta}_2\), and \(\tau^2=v_1^2+v_2^2-2c\). Then \(D\sim N(\delta,\tau^2)\). Selecting the largest raw estimate means \(J=1\) if \(D\geq 0\), and \(J=2\) otherwise. Normality is used for the exact formulas; unbiasedness alone does not generally determine the amount of selection bias.
Expected Maximum and Winner’s Noise Bias
The identity \(\max(\hat{\theta}_1,\hat{\theta}_2)=(\hat{\theta}_1+\hat{\theta}_2+|D|)/2\) gives the expected maximum. For \(a=\delta/\tau\), the absolute-value expectation for a normal variable is \(E|D|=2\tau\phi(a)+\delta\{2\Phi(a)-1\}\), where \(\phi\) and \(\Phi\) are the standard normal density and cumulative distribution functions. Therefore:
This is the expected reported effect when the largest raw estimate is reported. The selected treatment’s true effect is also random across repetitions because the selected index can change. Since \(P(J=1)=\Phi(a)\), its expected true effect is \(\theta_2+\delta\Phi(a)\). Subtracting this from the expected maximum gives a particularly direct expression for winner’s noise bias:
A short derivation shows why dependence enters through \(\tau\). Set \(V=\varepsilon_1-\varepsilon_2\), so \(D=\delta+V\), and \(J=1\) when \(V\geq-\delta\). For jointly normal errors, the conditional expectation of an error given \(V\) is linear in \(V\). With \(a=\delta/\tau\), the standard normal tail identity \(E[Z\,1(Z\geq-a)]=\phi(a)\) yields
Adding the two contributions gives the expected selected error. Since \(\operatorname{Cov}(\varepsilon_1,V)-\operatorname{Cov}(\varepsilon_2,V)=\operatorname{Var}(V)=\tau^2\), their sum is \(\tau\phi(a)\). Thus, positive dependence can reduce the noise in the ranking difference and hence reduce selection bias; the relevant variability is not either estimate’s standard error considered alone.
Worked Example: Equal True Effects and Independent Estimates
Suppose two hypothetical treatment-effect estimates are independent, each has true effect \(0.20\) percentage points, and each has standard error \(0.30\) percentage points. The reported treatment is the one with the larger raw estimate. The estimates are unbiased and jointly normal, and the independence assumption makes \(c=0\).
Here \(\delta=0\) and \(\tau=\sqrt{0.30^2+0.30^2}=0.4243\) percentage points. With \(a=0\) and \(\phi(0)=0.3989\), winner’s expected noise is
The expected maximum is consequently \(0.20+0.1693=0.3693\) percentage points. Across repetitions, the estimate reported for the selected treatment averages \(0.1693\) percentage points above that selected treatment’s true effect. This is an expectation, not a guarantee that the winning estimate in any particular trial is too high.
Worked Example: Unequal True Effects
Consider two hypothetical independent, normal estimates with standard errors \(0.40\) percentage points and true effects \(\theta_1=0.50\) and \(\theta_2=0.20\) percentage points. Select the treatment with the larger raw estimate. Independence and the stated standard errors imply \(\tau=\sqrt{0.40^2+0.40^2}=0.5657\) percentage points. Thus \(a=0.30/0.5657=0.5303\), \(\phi(a)\approx0.3466\), and \(\Phi(a)\approx0.7021\).
The probability that treatment 1 is selected is about \(0.7021\). The expected true effect of the selected treatment is therefore
The expected winner’s noise is \(0.5657(0.3466)=0.1961\) percentage points, so the expected largest reported estimate is \(0.4106+0.1961=0.6067\) percentage points. In contrast, its excess over the best fixed true effect is \(0.6067-0.50=0.1067\) percentage points. These two excesses are not interchangeable: the first compares the estimate with the true effect of whichever treatment wins, while the second compares it with the best true effect, treatment 1.
The Selection Rule Changes the Bias
The raw-estimate result applies only when selection is based on \(\hat{\theta}_1-\hat{\theta}_2\). Suppose instead the rule selects the larger standardized score, \(\hat{\theta}_i/SE_i\), while the reported quantity remains the selected raw effect estimate. If the standard errors are fixed, define the score difference \(L=\hat{\theta}_1/SE_1-\hat{\theta}_2/SE_2\), with mean \(m=\theta_1/SE_1-\theta_2/SE_2\), and let \(U=L-m\) have standard deviation \(\omega>0\). Joint normality gives the corresponding winner’s error:
This follows by writing the selected error as \(\varepsilon_1\) when \(L\geq0\) and \(\varepsilon_2\) otherwise, then applying the same normal tail identity used above. Unlike the raw-selection formula, the covariance in this expression need not equal \(\omega^2\): the score used to choose the winner is not generally the difference between the raw estimates. If standard errors are equal, the two rules rank the estimates identically. Otherwise they can choose different winners and produce different expected exaggeration.
Worked Example: Selecting by Standardized Score
Suppose two hypothetical independent normal estimates have fixed standard errors \(SE_1=0.20\) and \(SE_2=0.40\) percentage points, with true effects \(\theta_1=0.40\) and \(\theta_2=0.20\) percentage points. Select by the larger standardized score, but report the selected raw estimate. The score-difference mean is \(m=0.40/0.20-0.20/0.40=1.50\), and its error standard deviation is \(\omega=\sqrt{1^2+1^2}=\sqrt{2}\). Independence gives \(\operatorname{Cov}(\varepsilon_1-\varepsilon_2,U)=SE_1+SE_2=0.60\) percentage points.
Thus \(m/\omega=1.0607\), \(\phi(1.0607)\approx0.2273\), and the expected error in the reported winner is
This value is for selection by standardized score, not by the largest raw estimate. It illustrates why the selection rule must be stated before quantifying the winner’s exaggeration.
Assumptions, Diagnostics, and Limits
The formulas are useful when two estimates have a defensible joint normal approximation, are unbiased before selection, and their error variances and covariance are adequately specified. The selection rule must be applied to the same estimates represented in the model. If the rule uses additional outcomes, constraints, or a more complicated ranking, these two-estimate expressions may not describe the resulting bias.
In an adaptive trial, dependence can arise when estimates share participants or data. A practical diagnostic is to check whether the assumed standard errors and covariance reflect that shared information and the actual analysis plan. If \(\tau\) is nearly zero for raw selection, the estimates move almost identically and their ranking is nearly determined by the true-effect difference; little bias is expected from noisy ranking. If the estimates are strongly dependent but that dependence is ignored, using \(\sqrt{v_1^2+v_2^2}\) instead of \(\sqrt{v_1^2+v_2^2-2c}\) can misstate the bias.
Normality can be doubtful with very small samples, highly skewed estimators, or boundary-constrained treatment-effect estimates. In those settings, the expected maximum may still be studied, but it should be calculated under a more appropriate joint sampling distribution, for example by simulation that reproduces both estimation and selection. With more than two candidate treatments, the maximum depends on the joint distribution of all estimates; applying a two-estimate formula to a large candidate set does not account for the additional selection.
A prespecified single treatment estimate avoids this particular selection step, though it does not remove other sources of bias or uncertainty. An independent confirmation estimate, not used to choose the winner, is another useful comparison: conditional on the selection data, an unbiased independent confirmation estimate does not inherit the same selection error. Neither alternative changes the fact that the selected estimate can be useful for choosing a treatment; it changes how cautiously its magnitude should be interpreted.
Common Mistake
A related mistake is to call \(\tau\phi(\delta/\tau)\) the expected excess over the best true effect. It is the expected error relative to the true effect of the selected treatment. The best-effect comparison also depends on the probability of selecting each treatment, as the unequal-effects example demonstrates.
Check Your Understanding
Use the assumptions stated in each question. The solutions follow so that each calculation can be checked.
- Two independent normal estimates have equal true effects and equal standard errors \(0.20\) percentage points. What is the expected winner’s noise when selecting the larger raw estimate?
- For two jointly normal estimates selected by the larger raw estimate, what aspect of their joint error distribution determines the scale of selection bias?
- If the estimates have unequal true effects, why is the winner’s noise bias not generally the same as the expected maximum minus the best true effect?
- When might selection by standardized score differ from selection by raw estimate, and what additional quantity is needed to calculate the resulting reported-estimate bias?
Solutions. (1) Here \(\delta=0\) and \(\tau=\sqrt{0.20^2+0.20^2}=0.2828\) percentage points, so the bias is \(0.2828\phi(0)=0.2828(0.3989)=0.1128\) percentage points. (2) The relevant scale is \(\tau=\sqrt{\operatorname{Var}(\varepsilon_1-\varepsilon_2)}\), which includes the covariance between errors. (3) Winner’s noise compares the reported estimate with the true effect of whichever treatment is selected; the best-effect comparison instead uses the larger fixed true effect, so it also depends on selection probabilities. (4) With unequal standard errors, standardizing can change the ranking. The calculation requires the covariance between the error contrast \(\varepsilon_1-\varepsilon_2\) and the centered score difference \(U\), as well as the standard deviation \(\omega\) and score mean \(m\): the bias is \(\operatorname{Cov}(\varepsilon_1-\varepsilon_2,U)\phi(m/\omega)/\omega\).