Median-Unbiased Estimation by Stopping-Rule Inversion

Why the Stopping Rule Belongs in the Distribution

When a trial stops early, the reported treatment-effect estimate is based on a data-dependent amount of information. Its sampling distribution is therefore not generally the same as the fixed-sample distribution at the observed look. To define an estimate that accounts for this selection, we ask: under what treatment-effect value would the observed stopped estimate be a median of its sampling distribution under the actual stopping rule?

The answer requires the probability of stopping at each possible stage. The distribution must include both the estimate that would be reported at that stage and the probability that the trial reaches and stops there. In particular, using only the distribution conditional on the observed stage answers a different question.

Definition: Let $\theta$ be the parameter of interest, let $T$ be the stopping stage under a prespecified rule, and let $\hat{\theta}_k$ be the estimate that would be reported if the trial stopped at stage $k$. The stopped estimate is $\hat{\theta}_T$. Its exact cumulative distribution function under parameter value $\theta$ is $$ G_\theta(y)=P_\theta(\hat{\theta}_T\le y) =\sum_k P_\theta(T=k,\hat{\theta}_k\le y). $$ A value $\theta$ is compatible with $y$ as a median of the stopped estimate when $$ G_\theta(y-)\le \frac12\le G_\theta(y), $$ where $G_\theta(y-)=P_\theta(\hat{\theta}_T<y)$. The set of such parameter values is the median-unbiased inversion set for the observed estimate $y$.

The two inequalities are important when the stopped estimate has positive probability at $y$. If its distribution is continuous, $G_\theta(y-)=G_\theta(y)$, so inversion usually reduces to solving $G_\theta(y)=1/2$. With a discrete distribution, the inequalities may hold over an interval of parameter values rather than at one value. In that case the exact inversion is a set; reporting a single number requires an additional, stated convention.

Assumptions and the Inversion Procedure

Exactness is relative to a specified statistical model and stopping rule. The model must give the joint distribution of the data across all planned looks, not just a separate marginal distribution at each look. The rule must specify when sampling stops, including any efficacy, futility, or maximum-sample-size decisions that affect the reported estimate. The estimate used at each possible stop must also be defined.

For example, in a canonical group sequential model with independent information increments, the joint distribution of the cumulative statistics follows from the increment distributions. The stopping events are then determined by the prespecified boundaries. Calculating $P_\theta(T=k,\hat{\theta}_k\le y)$ requires accounting for the path of earlier statistics that leads to stage $k$, as well as the estimate at that stage. One cannot generally replace this joint probability with the fixed-sample probability at look $k$.

Specify the model and rule.
State the parameter, the joint data distribution across looks, the stopping rule, and the estimate reported at each possible stopping stage.

Construct the stopped distribution.
For each stage, calculate the joint probability of stopping there and reporting an estimate no greater than $y$. Sum those probabilities over stages.

Invert at the observed estimate.
Find all parameter values for which the probability below the observed estimate is at most one half and the probability at or below it is at least one half.

Check the result.
Verify that stage probabilities are included, that the inversion respects any atoms, and that the reported value or set matches the chosen median convention.

The inversion is exact if the specified model gives the correct sampling distribution and the probabilities are calculated exactly. In practice, probabilities may be evaluated by numerical integration or simulation; those calculations introduce numerical or Monte Carlo error, even when the underlying model-based definition is exact.

A Two-Stage Discrete Example

The following hypothetical score model makes every stopping-stage probability visible. At each stage, an independent observation $X_i$ is either $+1$ or $-1$, with $P_\theta(X_i=+1)=q=(1+\theta)/2$, where $-1<\theta<1$. Thus $E_\theta(X_i)=\theta$. Think of $\theta$ as a treatment-effect score measured in score units. After the first observation, stop if $X_1=+1$; otherwise collect one more observation and stop. Report the mean of the observations collected.

This is a deliberately small sequential model, not a model for every clinical endpoint. Its observations are independent with the stated distribution, and the stopping decision follows the stated rule exactly. Those assumptions let us enumerate the complete stopped sampling distribution.

Worked Example: Constructing the Exact Stopped Distribution

If $X_1=+1$, the trial stops at stage 1 and reports $\hat{\theta}_T=1$. This outcome has probability $q$. If $X_1=-1$, the trial continues. A second $-1$ gives a stage-2 estimate of $-1$, with probability $(1-q)^2$; a second $+1$ gives a stage-2 estimate of $0$, with probability $(1-q)q$. Therefore,

P_\theta(\hat{\theta}_T=-1)=(1-q)^2,\qquad P_\theta(\hat{\theta}_T=0)=(1-q)q,\qquad P_\theta(\hat{\theta}_T=1)=q.

The probabilities sum to one: $(1-q)^2+(1-q)q+q=1$. The stage-1 stopping probability is $q$; the stage-2 stopping probability is $1-q$, split between the two possible reported estimates. For $y=0$, for example, the stopped CDF is

G_\theta(0)=(1-q)^2+(1-q)q=1-q.

This is an unconditional distribution under the full rule. The stage-2 contribution is not its conditional probability given stage 2: its unconditional contribution includes the probability $1-q$ of reaching that stage.

Inverting a Discrete Stopped Distribution

Worked Example: The Median-Unbiased Set When the Estimate Is Zero

Suppose the hypothetical trial stops at stage 2 and reports the observed mean $y=0$. The probability strictly below zero is $(1-q)^2$; the probability at or below zero is $1-q$. The median condition is

(1-q)^2\le \frac12\le 1-q.

Since $0<q<1$, the right inequality gives $q\le 1/2$. The left gives $1-q\le 1/\sqrt{2}$, or $q\ge 1-1/\sqrt{2}$. Because $\theta=2q-1$, the inversion set is

1-\sqrt{2}\le \theta\le 0,

or approximately $-0.4142\le\theta\le 0$ score units. Every value in this interval makes zero a median of the estimate’s exact stopped distribution. The correct discrete result is an interval of compatible parameter values, not a unique root of an equality.

Worked Example: The Median-Unbiased Set When the Estimate Is Minus One

If the observed stopped estimate is $y=-1$, its probability at or below that value is $(1-q)^2$, and its probability strictly below is zero. The median condition becomes

0\le \frac12\le (1-q)^2.

Taking nonnegative square roots gives $1-q\ge 1/\sqrt{2}$, so $q\le 1-1/\sqrt{2}$. Converting to $\theta$,

-1<\theta\le 1-\sqrt{2}\approx -0.4142.

The parameter and the estimate are both measured in score units. The open lower endpoint reflects the model parameter space $-1<\theta<1$; the upper endpoint is included because equality satisfies the median condition. This example also shows why a discrete inversion may extend over a range rather than identify one parameter value.

Why Conditioning on the Observed Stage Changes the Answer

Worked Example: Unconditional and Conditional Inversion for a Stage-2 Result

Suppose the observed path is $X_1=-1, X_2=+1$, so the trial stops at stage 2 and reports $y=0$. Unconditional inversion uses the probabilities from the full stopped distribution and gives $1-\sqrt{2}\le\theta\le0$, as above.

If instead we condition on $T=2$, the first observation is known to be $-1$. The stage-2 estimate is $-1$ when $X_2=-1$, with conditional probability $1-q$, and $0$ when $X_2=+1$, with conditional probability $q$. For $y=0$, the conditional probability strictly below is $1-q$, while the conditional probability at or below is 1. The conditional median condition is therefore

1-q\le \frac12\le 1,

which gives $q\ge1/2$, or $0\le\theta<1$. This differs from the unconditional set. The two calculations are not competing answers to the same inversion: one uses the distribution under the complete stopping rule, and the other uses a distribution conditional on the observed stage.

The distinction is central to stopping-rule inversion. Each stage’s contribution to the unconditional distribution is its joint probability of stopping there and observing an estimate in the specified range. Equivalently, it is the stage’s stopping probability multiplied by the corresponding conditional probability. Omitting the stopping probability changes the distribution being inverted.

Continuous Models, Diagnostics, and Limitations

In a continuous model, there is usually no point mass at the observed estimate. If $G_\theta(y)$ is continuous in $y$, median inversion seeks a parameter value satisfying $G_\theta(y)=1/2$. For a model in which $G_\theta(y)$ varies monotonically with $\theta$ and crosses one half once, this gives a unique median-unbiased estimate. The monotonicity and uniqueness should be checked for the model and stopping rule; they are not guaranteed merely because the procedure is called inversion.

Useful diagnostics include verifying that the probabilities of all stopping stages sum to one, checking that each stage term incorporates the path required to reach that stage, and confirming that the final inversion satisfies the median inequalities. Numerical calculations should be checked for stability near boundaries and against higher-accuracy calculations when a root is close to an endpoint.

Common failures include:

Dropping stage probabilities. Using only the conditional distribution at the observed look omits the chance of reaching that look. The correction is to use the joint stage-and-estimate probabilities in $G_\theta$.
Using a fixed-sample distribution. Treating the observed look as though it had been chosen in advance ignores the selection induced by the stopping rule. The correction is to reproduce the prespecified rule when constructing the distribution.
Forcing a single answer in a discrete model. Setting a CDF equal to one half can miss parameter values that satisfy the correct median inequalities. The correction is to report the full inversion set, or state a separate convention if a scalar summary is required.
Assuming a root exists or is unique. A numerical solver may return one root even when the CDF is nonmonotone or crosses one half more than once. The correction is to examine the function over the parameter space and report all compatible values or explain the failure of uniqueness.
Ignoring model uncertainty. Exactness is conditional on the assumed joint distribution and stopping rule. A misspecified outcome model, dependence structure, or rule can invalidate the calculated distribution.

The ordinary maximum-likelihood estimate, the conditional likelihood estimate, and the stopping-rule median-unbiased estimate target different features of the data distribution. Stopping-rule inversion defines its target through the median of the unconditional distribution of the reported estimate. It does not automatically give an unbiased estimate of the mean, and the median condition alone does not provide a confidence interval or guarantee good performance under model misspecification. These are limitations of the target and assumptions, not reasons to omit the stopping stages from its definition.

Key takeaway: Invert the sampling distribution of the estimate that the trial would actually report under the full stopping rule. Sum the joint probabilities across stopping stages, and use median inequalities—not just an equality—when the distribution has atoms.

Check Your Understanding

Use the two-stage score model above. Let $q=(1+\theta)/2$, and use the exact stopped distribution when answering.

What are the probabilities of stopping at stages 1 and 2? Solution: Stage 1 occurs when $X_1=+1$, so its probability is $q$. Stage 2 occurs when $X_1=-1$, so its probability is $1-q$. They sum to one.
For an observed stopped estimate of $1$, find the median-unbiased inversion set. Solution: The probability strictly below 1 is $1-q$, and the probability at or below 1 is 1. The condition $1-q\le1/2\le1$ gives $q\ge1/2$, hence $0\le\theta<1$.
Why is $(1-q)q$, rather than $q$, the unconditional probability of a stage-2 estimate of zero? Solution: A zero estimate requires both reaching stage 2, with probability $1-q$, and then observing $X_2=+1$, with probability $q$ conditional on reaching stage 2. Independence gives the joint probability $(1-q)q$.
For an observed estimate of zero, state the unconditional median-unbiased set and the conditional-on-stage-2 median set. Solution: Unconditional inversion gives $1-\sqrt{2}\le\theta\le0$. Conditional inversion gives $0\le\theta<1$. They differ because conditioning changes the distribution being inverted.