Why a One-Look Correction Is Needed
At a single efficacy look, an estimate is reported only if it crosses a prespecified threshold. Among trials that cross, estimates tend to be larger than the underlying effect: values that fall below the threshold are not part of the selected set. The correction in this tutorial quantifies that selection effect in a normal model and uses the resulting conditional expectation to adjust an observed estimate.
The target is specifically the mean estimate among trials that cross the efficacy boundary. This is a conditional target, not the average over all trials under a stopping rule that might report something different when efficacy is not declared. The earlier material on conditional and unconditional bias targets distinguishes these two averages; here the calculation is for the conditional target at one efficacy look.
The model assumes a normal sampling distribution, a known standard error, and a threshold fixed in advance on the \(X\) scale. These assumptions matter: for example, a boundary on a standardized statistic with an estimated standard error need not correspond to a fixed \(c\) in this model.
The Truncated-Normal Expectation
Write \(\phi\) and \(\Phi\) for the standard normal density and cumulative distribution functions. Define the upper inverse Mills ratio by \(\lambda(a)=\phi(a)/[1-\Phi(a)]\). Under a candidate value of \(\theta\), the threshold in standard-normal units is \(a=(c-\theta)/\sigma\), and the probability of crossing is \(P(X\geq c\mid\theta)=1-\Phi(a)\).
To derive the conditional expectation, standardize \(X\) as \(Z=(X-\theta)/\sigma\). Then \(Z\sim N(0,1)\), and crossing \(c\) is equivalent to \(Z\geq a\). Since \(\phi'(z)=-z\phi(z)\),
It follows that
The term \(\sigma\lambda((c-\theta)/\sigma)\) is the conditional selection bias: the difference between the mean reported estimate among efficacy-crossing trials and \(\theta\). It is positive for every finite threshold and finite \(\theta\). Its size depends on the standardized distance from the mean to the boundary, not just on \(\sigma\).
Inverting the Conditional Mean
Because \(\theta\) appears both directly and inside \(\lambda\), the correction is generally found by solving a one-dimensional equation numerically. This inversion is more than subtracting a bias evaluated at the observed estimate: the unknown parameter also determines how strongly the threshold selects the reported values.
A short monotonicity argument establishes uniqueness. Put \(m(\theta)=\theta+\sigma\lambda(a)\), where \(a=(c-\theta)/\sigma\). Differentiation gives \(\lambda'(a)=\lambda(a)(\lambda(a)-a)\). For a standard normal variable truncated to \(Z\geq a\), its conditional variance is \(1+a\lambda(a)-\lambda(a)^2\). Therefore,
Thus \(m\) is strictly increasing. As \(\theta\) tends to positive infinity, \(a\) tends to negative infinity, the crossing probability tends to one, and \(m(\theta)\) tends to infinity. As \(\theta\) tends to negative infinity, \(a\) tends to positive infinity. The upper-tail approximation \(\lambda(a)\sim a+1/a\) then gives \(m(\theta)\to c\) from above. Hence the range of \(m\) is \((c,\infty)\), which proves existence and uniqueness for every \(x>c\). Also, \(\lambda(a)>0\), so \(m(\theta)>\theta\); at the solution, this means \(\tilde{\theta}(x)<x\).
Worked Example: Calculating the Selection Bias
Suppose a hypothetical treatment-effect estimator has \(X\sim N(\theta,1^2)\), with effects measured in percentage points, and efficacy requires \(X\geq 1\) percentage point. Evaluate the conditional mean when \(\theta=0\).
Here \(a=(1-0)/1=1\). Using \(\phi(1)\approx0.2420\) and \(1-\Phi(1)\approx0.1587\),
Among hypothetical trials whose estimates cross the one-percentage-point threshold, the mean estimate is about 1.525 percentage points even though the underlying effect is zero. The conditional selection bias is therefore about 1.525 percentage points in this setting. This calculation describes the selected group, not the unconditional average estimate across all trials.
Solving the Correction in Practice
For computation, one can solve the equation directly in \(\theta\), or solve for \(a\) first. Since \(\theta=c-\sigma a\), the equation is equivalent to \(x-c=\sigma\{\lambda(a)-a\}\). Numerical root-finding can use a bracketed method or Newton’s method. The uniqueness result supplies a check on the root, but numerical output should still be substituted back into the original equation.
Worked Example: Correcting an Estimate at an Efficacy Stop
Suppose a hypothetical one-look trial uses \(c=0.8\) percentage points and has known \(\sigma=0.4\) percentage points. Its observed estimate is \(x=1.0\) percentage point, so the efficacy threshold was crossed. Find the conditional-mean correction.
The equation in percentage-point units is
Set \(a=(0.8-\theta)/0.4\), so \(\theta=0.8-0.4a\). Substitution reduces the equation to \(\lambda(a)-a=(1.0-0.8)/0.4=0.5\). Numerical solution gives approximately \(a=1.131\), for which \(\lambda(a)\approx1.631\). Thus
Check by substituting into the original equation: \((0.8-0.348)/0.4=1.13\), and \(0.348+0.4(1.631)\approx0.348+0.6524=1.0004\), or approximately 1.000 percentage points. The corrected value is below the observed estimate, as expected under upward selection. The calculation does not say there is a probability that the true effect equals this corrected value; it is the solution to a conditional-mean matching equation.
Relation to Likelihood and Simpler Adjustments
For this particular model and selection event, the same equation also gives the conditional maximum likelihood estimate. Given that efficacy was declared, the density of the observed \(x\) is the normal density divided by the crossing probability:
Differentiating its log with respect to \(\theta\) gives \((x-\theta)/\sigma^2-\lambda((c-\theta)/\sigma)/\sigma\). Setting this score to zero yields the same conditional-mean equation. This agreement is a feature of the one-parameter normal location model with a single fixed truncation boundary; it does not imply that mean matching and conditional maximum likelihood are always identical.
A tempting shortcut is to estimate the bias by substituting the observed \(x\) for the unknown \(\theta\) in the bias term, then subtracting once. In the preceding hypothetical example, this plug-in calculation would use \(a=(0.8-1.0)/0.4=-0.5\). Since \(\lambda(-0.5)\approx0.5092\), it gives \(1.0-0.4(0.5092)\approx0.7963\) percentage points. That is not the solution to the conditional-mean equation: the corrected value changes the standardized distance to the boundary, so the bias term must be recomputed at the corrected value. The exact inversion above gives about 0.348 percentage points instead.
Worked Example: Checking a Candidate by Substitution
Use the same hypothetical threshold and standard error, and suppose a numerical routine returns \(\theta=0.726\) percentage points as the correction for \(x=1.0\). Check the proposed value rather than accepting the routine’s output.
It implies \(a=(0.8-0.726)/0.4=0.185\). At this value, \(\lambda(0.185)\approx0.919\), so the conditional mean predicted by the candidate is
This does not equal the observed 1.0 percentage point, so 0.726 is not a solution. Solving \(\lambda(a)-a=0.5\) instead gives \(a\approx1.131\), hence \(\theta\approx0.348\); substitution gives a conditional mean of approximately 1.000 percentage point. This check illustrates why every numerical correction should be verified against the defining equation.
An unconditional bias correction has a different target: it averages over the full sampling and reporting rule, including what happens when efficacy is not declared. A conditional likelihood estimate uses the sampling distribution given the observed selection event. In this one-look normal setting, the latter coincides with the conditional-mean inversion above. Neither should be confused with a median-unbiased estimate obtained by inverting a stopping-rule probability statement; that procedure targets a different property.
Diagnostics, Pitfalls, and Limitations
- Check that selection occurred. The equation requires \(x>c\). An observation at or below \(c\) is inconsistent with the stipulated rule that only crossing estimates are reported, apart from boundary or data-recording conventions.
- Check the model scale. \(X\), \(c\), and \(\sigma\) must use compatible units. A threshold on a \(Z\)-statistic can be rewritten as a fixed \(X\) threshold only when the standard error is fixed and known.
- Check the root numerically. Recalculate \(a=(c-\tilde{\theta})/\sigma\), evaluate \(\lambda(a)\), and verify that \(\tilde{\theta}+\sigma\lambda(a)\) agrees with \(x\) to the reported precision.
- Do not call the result an unbiased estimator. The equation matches the observed value to a model-based conditional expectation. It does not establish \(E[\tilde{\theta}(X)\mid X\geq c]=\theta\).
- Do not use this correction unchanged with estimated standard errors or multiple looks. Those settings alter the selection distribution or stopping event and require methods that account for those features.
The correction is model-dependent, and it is not itself a confidence interval. It relies on a normal location model with known variance and a fixed, one-sided threshold. If the normal approximation is poor, if the boundary depends on additional random quantities, or if estimates are reported after different stopping outcomes, this formula may not describe the relevant selection bias. In practice, report the assumptions and the threshold alongside the corrected estimate, and treat disagreement between the model and the trial’s actual design as a reason to use a more appropriate stopping-aware analysis.
Check Your Understanding
Answer each question, then compare with the solutions below.
- In a normal one-look model, why does conditioning on \(X\geq c\) shift the expected estimate upward?
- For \(X\sim N(\theta,1^2)\) and \(c=1\), calculate \(E[X\mid X\geq c]\) when \(\theta=0\), using \(\lambda(1)\approx1.525\).
- For a threshold \(c=0.8\), known \(\sigma=0.4\), and observed \(x=1.0\), what equation defines the correction, and why is a candidate \(\theta=0.726\) invalid?
- State two assumptions under which the one-look correction is derived, and explain why an unconditional bias target is different.
Solutions. (1) The event retains only values in the upper tail. In standardized units, \(E[Z\mid Z\geq a]=\lambda(a)>0\), so \(E[X\mid X\geq c]=\theta+\sigma\lambda(a)>\theta\). (2) Here \(a=(1-0)/1=1\), so the conditional mean is \(0+1(1.525)=1.525\). (3) Solve \(1.0=\theta+0.4\lambda((0.8-\theta)/0.4)\). At \(\theta=0.726\), \(a=0.185\) and \(\lambda(a)\approx0.919\), giving a predicted conditional mean \(0.726+0.4(0.919)\approx1.094\), not 1.0. (4) Assumptions include a normal estimator with known, fixed standard error and a fixed efficacy threshold on the estimate scale. The conditional target averages only among trials crossing the boundary; an unconditional target averages over all outcomes under the reporting or stopping rule.