Why Calibrate Across the Complete Stopping Rule?
A sequential trial does not always produce an estimate at the same sample size. Some trials stop early; others continue to later analyses. If the ordinary estimate is systematically pulled away from the treatment effect by this data-dependent stopping, one possible target is to calibrate an estimator so that its repeated-sampling expectation under the entire design equals the effect. This is an unconditional target: it averages over every possible stopping time and every data path that can lead to it.
The calibration must use the full sampling procedure, not just the distribution conditional on the look actually observed. That distinction matters because the stopping decision carries information about the data. A correction that works on average across all paths need not work within any one stopping group.
This definition is a precise target, not a guarantee that a suitable estimator exists or is useful. It also depends on the assumed sampling model and the actual rule: changing the boundaries, analysis times, or treatment of a stopping decision can change the expectation that must be calibrated.
Writing the Calibration Equation
Suppose the possible observed paths are indexed by \(\omega\). A path records the observed data and the look at which the trial stops. If the design and model specify its probability \(q_\theta(\omega)\), then the unconditional calibration requirement is
For continuous data, the sum is replaced by an integral over the possible paths, using their sampling density under \(\theta\). Crucially, paths at different stopping looks are all included. The probabilities of those paths must reflect the stopping rule; it is not enough to calculate an expectation as if every trial had reached the maximum sample size.
A useful way to formulate the task is to start with a familiar estimator \(\hat\theta\) and find a fixed correction \(c(D)\) such that
The quantity on the right is the bias of the ordinary estimator under the stopping rule. The correction must be a function of observable data and known design features, not of the unknown true \(\theta\). Simply subtracting a bias expression evaluated at the true parameter would not define an estimator. A plug-in correction may be useful in some settings, but does not by itself establish exact unconditional unbiasedness.
Worked Example: Calculating Bias Under a Two-Stage Rule
Worked Example: Calculating Bias Under a Two-Stage Rule
Consider a deliberately small hypothetical model to make the path probabilities visible. Let \(X_1\) and, if needed, \(X_2\) be independent Bernoulli observations with success probability \(\theta\), where \(0\leq\theta\leq1\). After \(X_1=1\), stop; after \(X_1=0\), observe \(X_2\). Use the ordinary stopped estimate \(\hat\theta\), equal to the sample proportion at the stopping time. The model is fully specified, the observations are independent with the stated common probability, and the stopping rule is fixed in advance.
There are three observable paths: stop after a first-stage success, continue and then observe a success, or continue and observe a second failure. Their probabilities are, respectively,
The corresponding ordinary estimates are \(1\), \(1/2\), and \(0\). Thus
The bias is therefore
For the hypothetical value \(\theta=0.4\), the ordinary estimator has expectation \(1.5(0.4)-0.5(0.4)^2=0.52\), so its bias is \(0.52-0.4=0.12\). This calculation averages over early stopping and continuation rather than conditioning on either outcome.
Worked Example: Solving for an Unconditionally Calibrated Estimator
Worked Example: Solving for an Unconditionally Calibrated Estimator
Use the same hypothetical design and model. Assign a calibrated estimate \(a\) to the path \((1)\), \(b\) to \((0,1)\), and \(c\) to \((0,0)\). These values must be fixed numbers; they cannot vary with the unknown \(\theta\). Unconditional unbiasedness requires
Expand the left side and match the constant, \(\theta\), and \(\theta^2\) coefficients. The constant coefficient gives \(c=0\). The \(\theta^2\) coefficient gives \(-b+c=0\), hence \(b=0\). The \(\theta\) coefficient then gives \(a+b-2c=1\), so \(a=1\). The calibrated estimator is therefore 1 after the first success and 0 on either continuation path.
Direct verification confirms the result:
This estimator is unbiased throughout the parameter space under the stated model and stopping rule. It is not simply the ordinary stopped proportion with a constant subtracted: the correction is \(1/2\) on the \((0,1)\) path and 0 on the other paths. Its expected correction is \(\frac12\theta(1-\theta)\), exactly the ordinary estimator’s bias derived above.
Worked Example: Checking the Unconditional Target at a Fixed Parameter
Worked Example: Checking the Unconditional Target at a Fixed Parameter
At the hypothetical value \(\theta=0.4\), the probabilities of the three paths are \(0.4\), \(0.4(0.6)=0.24\), and \((0.6)^2=0.36\). The calibrated estimates from the previous example are 1, 0, and 0. Their unconditional mean is
The ordinary stopped estimate instead has mean
The calibrated estimator’s result can look surprising: conditional on continuing, it is always 0 in this example, even though the true parameter is 0.4. That is not a contradiction. The estimator is calibrated across the complete procedure: the probability 0.4 of an early success, where it reports 1, supplies its entire unconditional mean. It is not calibrated within the continuation group.
Justification, Assumptions, and Practical Computation
The central justification is direct. If a fixed correction function satisfies the calibration equation, linearity of expectation gives
This establishes unbiasedness for every parameter value at which the correction equation holds. It does not establish small variance, good performance in a particular stopping group, or robustness to a misspecified model.
In practice, calibration requires a specified sampling model and the complete stopping rule. One can enumerate all paths when the data are discrete and the design is small. For more complex models, calculate the expectation by numerical integration or simulate complete trials under a grid of parameter values. A numerical solution should be checked over the parameter space, not just at one fitted value. Simulation also introduces Monte Carlo error: close agreement in a finite simulation is evidence of approximate calibration, not proof of exact unbiasedness.
For a parametric calibration, one may express \(c(D)\) using a chosen set of observable path features and solve the resulting expectation equations for its coefficients. The equations must hold for the parameter values claimed. Fitting them only at a few grid points may match those points while leaving bias between them. If nuisance parameters affect the path probabilities, the calibration target must specify whether unbiasedness is required at every nuisance-parameter value; that requirement can make the equations harder or impossible to satisfy.
What Unconditional Calibration Does—and Does Not—Guarantee
Unconditional calibration differs from several other common targets. An estimator calibrated conditional on the observed stopping look targets an expectation within that selected group. A median-unbiased estimator targets a probability statement about whether its estimate lies above or below the parameter, rather than matching the mean. Conditional likelihood methods use a likelihood conditioned on the stopping event. These methods answer different questions; none is automatically a substitute for unconditional mean calibration.
Unbiasedness is a statement about repeated sampling under a model, not a claim that any one estimate is close to the unknown effect. An unbiased estimator may have substantial variance or take values outside a scientifically plausible range. In the worked example, assigning 0 to a continuation path is an algebraic consequence of the selected target and design, not evidence that a continuing trial’s effect is zero.
The estimator may also fail to exist in a requested class. For example, restrictions such as nonnegative estimates, a bounded range, or dependence only on a particular summary statistic can conflict with the expectation equations. Even when solutions exist, they may not be unique. The choice among them may require considering variance, interpretability, and performance across parameter values, in addition to unbiasedness.
Diagnostics and Common Pitfalls
- Omitting stopping paths. The expectation must include every possible stopping time and the probability of reaching it. Calculate path probabilities from the complete rule.
- Conditioning without noticing. An expectation given that the trial stopped at a particular look is not the unconditional expectation. Keep the target explicit.
- Using the unknown parameter in the correction. A correction that depends on the true \(\theta\) is not an operational estimator. Solve for fixed functions of observed data and known design quantities.
- Checking only one parameter value. Unbiasedness requires the expectation equation across the parameter space, unless a narrower pointwise target is explicitly intended.
- Assuming unbiased means precise. Unbiasedness controls the mean over repetitions; it does not control variance, interval coverage, or the behavior of estimates after a particular stopping outcome.
- Ignoring design changes or misspecification. Recalculate or reassess calibration if boundaries, analysis timing, or the sampling model differ from those used to derive it.
The key practical question is therefore not just whether a correction removes a calculated bias, but under which model, for which complete stopping rule, and over which parameter values the expectation identity has been verified.
Check Your Understanding
Use the two-stage Bernoulli design and calibration derived above unless a question says otherwise.
- What does the expectation in unconditional unbiasedness average over?
- At \(\theta=0.5\), calculate the expectation and bias of the ordinary stopped estimate.
- Why is the calibrated estimate 0 on the \((0,1)\) path not evidence that the treatment effect is zero?
- Would checking the calibration equation only at \(\theta=0.4\) establish unbiasedness throughout the parameter space? Explain.
Solutions. (1) It averages over the data and all stopping paths generated by the complete rule. (2) The expectation is \(1.5(0.5)-0.5(0.5)^2=0.625\); the bias is \(0.625-0.5=0.125\). (3) The estimator targets the unconditional mean, not the mean conditional on continuing or the true effect on an individual path. (4) No. It verifies the equation at one parameter value only; unbiasedness throughout requires it to hold for every parameter value in the stated space.