Median-Unbiased Estimate at an Efficacy Stop

Why Condition on an Efficacy Stop?

An efficacy stop selects data: a trial stops at a look because its statistic has crossed a prespecified upper boundary. Consequently, the distribution of the statistic among trials that stop at that look is not its ordinary, unselected normal distribution. A median-unbiased estimate at an efficacy stop accounts for that selection by matching the observed statistic to the median of this stopping-conditioned distribution.

The earlier material on stopping-rule inversion introduced the general idea of finding a parameter value that makes the observed statistic a median under the relevant sampling law. Here the target event is specific: the trial first crosses its efficacy boundary at a particular look. The conditioning event must therefore include both the crossing at that look and survival without an efficacy stop at every earlier look.

Definition: Suppose an observed statistic \(Z_k=z_k\) occurs when a one-sided group sequential design first stops for efficacy at look \(k\). A median-unbiased estimate \(\tilde{\theta}_k\) is a parameter value satisfying \[ P_{\theta=\tilde{\theta}_k}(Z_k\le z_k\mid T=k)=\frac12, \] where \(T\) is the first efficacy-crossing look. This definition uses the distribution conditional on the observed stopping event, not the unconditional distribution of \(Z_k\).

Model and Stopping Event

Use the canonical normal model with fixed, known information values \(0<I_1<\cdots<I_K\). The parameter \(\theta\) is the treatment effect per unit of the study’s effect scale; the \(Z_j\) statistics are dimensionless. Under the model,

E_\theta(Z_j)=\theta\sqrt{I_j},\qquad \operatorname{Var}_\theta(Z_j)=1,\qquad \operatorname{Corr}_\theta(Z_i,Z_j)=\sqrt{\frac{I_i}{I_j}}\quad(i<j).

Equivalently, the standardized statistics have the joint normal distribution induced by independent normal increments in information time. Assume the information values and one-sided efficacy boundaries \(b_1,\ldots,b_K\) are fixed in advance, the normal model is correctly specified, and there is no futility stopping in the event under study. There are no unknown nuisance parameters or estimated standard errors in this model.

Define \(T=\min\{j:Z_j\ge b_j\}\), taking \(T\) as beyond the final look if no boundary is crossed. For an efficacy stop at look \(k\), the observed event is

E_k=\{Z_1<b_1,\ldots,Z_{k-1}<b_{k-1},\ Z_k\ge b_k\}.

Thus \(T=k\) and \(E_k\) describe the same event. The conditional CDF needed for estimation is the probability of \(Z_k\le z\) given \(E_k\). Its numerator includes all earlier survival conditions and requires \(b_k\le Z_k\le z\); its denominator includes the earlier survival conditions and requires \(Z_k\ge b_k\).

Conditional Distribution and Median Equation

Write \(q_k(u)\) for the probability of surviving all earlier efficacy looks given \(Z_k=u\):

q_k(u)=P(Z_1<b_1,\ldots,Z_{k-1}<b_{k-1}\mid Z_k=u).

In the canonical normal model, this conditional probability does not depend on \(\theta\). To see why, the conditional mean of an earlier statistic is

E_\theta(Z_i\mid Z_k=u) =\theta\sqrt{I_i} +\sqrt{\frac{I_i}{I_k}}\bigl(u-\theta\sqrt{I_k}\bigr) =\sqrt{\frac{I_i}{I_k}}u,

and its conditional covariance matrix is also independent of \(\theta\). Let \(\phi\) and \(\Phi\) denote the standard normal density and CDF. Since the marginal density of \(Z_k\) is \(\phi(u-\theta\sqrt{I_k})\), conditioning on \(E_k\) gives a density proportional to \(q_k(u)\phi(u-\theta\sqrt{I_k})\) for \(u\ge b_k\). Therefore,

P_\theta(Z_k\le z\mid E_k) = \frac{\displaystyle\int_{b_k}^{z}q_k(u)\phi(u-\theta\sqrt{I_k})\,du} {\displaystyle\int_{b_k}^{\infty}q_k(u)\phi(u-\theta\sqrt{I_k})\,du}, \qquad z\ge b_k.

The median-unbiased estimate is a root of this ratio set equal to \(1/2\), with \(z=z_k\). For \(k=1\), there are no earlier survival conditions, so \(q_1(u)=1\) and the integrals reduce to a truncated-normal expression.

Key condition: If \(z_k=b_k\), the numerator integral is zero for every finite \(\theta\). The conditional CDF at the observed statistic is then zero, so the median equation has no solution. A solution is considered only when \(z_k>b_k\). With finite boundaries and a nondegenerate canonical normal design, the conditional endpoint density is positive on intervals above \(b_k\); the monotone-likelihood-ratio property then gives a unique solution for \(z_k>b_k\).

Why the Root Is Unique

The conditional density of \(Z_k=u\) given \(E_k\) has the form

f_\theta(u\mid E_k)\ \propto\ \mathbf{1}(u\ge b_k)\,q_k(u)\,\phi(u-\theta\sqrt{I_k}).

For two parameter values \(\theta_2>\theta_1\), the ratio of their unnormalized densities at \(u\) is a positive constant times \[ \exp\{(\theta_2-\theta_1)\sqrt{I_k}\,u\}. \] This ratio increases strictly with \(u\). The conditional family therefore has a monotone likelihood ratio in \(Z_k\): increasing \(\theta\) shifts the conditional distribution toward larger endpoint statistics. In particular, for fixed \(z_k>b_k\), \(P_\theta(Z_k\le z_k\mid E_k)\) decreases strictly as \(\theta\) increases.

Under the stated nondegeneracy conditions, the endpoint density has positive support above the boundary. As \(\theta\) becomes very negative, conditioning on \(Z_k\ge b_k\) concentrates the endpoint near its lower support, so the conditional CDF at any fixed \(z_k>b_k\) tends to 1. As \(\theta\) becomes very positive, the endpoint shifts above that fixed value and the CDF tends to 0. Continuity and strict decrease then imply exactly one parameter value where the CDF equals \(1/2\). At \(z_k=b_k\), however, the numerator remains zero: the uniqueness argument does not create a solution at the boundary itself.

Worked Example: One-Look Median-Unbiased Estimate

Suppose a hypothetical one-look normal design has information \(I_1=100\), efficacy boundary \(b_1=1.645\), and observed statistic \(z_1=2.645\). The observed statistic is strictly above the boundary, so the median equation can have a solution.

At one look, \(Z_1\sim N(\theta\sqrt{I_1},1)\), and conditioning on efficacy is conditioning on \(Z_1\ge b_1\). Let \(\mu=\theta\sqrt{I_1}\). The conditional median equation is

\frac{\Phi(z_1-\mu)-\Phi(b_1-\mu)} {1-\Phi(b_1-\mu)}=\frac12.

Substituting the observed values and rearranging gives

\Phi(2.645-\mu)=\frac{1+\Phi(1.645-\mu)}{2}.

For an explicit numerical solution, set \(x=1.645-\mu\). Then \(2.645-\mu=x+1\), so the equation becomes \(\Phi(x+1)=[1+\Phi(x)]/2\). Numerical root-finding gives \(x\approx-0.687\), and hence \(\mu=1.645-x\approx2.332\). Since \(\sqrt{I_1}=10\),

\tilde{\theta}_1=\frac{\mu}{\sqrt{I_1}} =\frac{2.332}{10}\approx0.2332.

The estimate is \(0.2332\) treatment-effect units per information-scale unit. It is the parameter value for which the statistic, among hypothetical repetitions that cross the boundary at this look, has conditional median 2.645. It is not a probability statement about the fixed parameter.

Applying the Formula at a Later Look

At a later look, the earlier survival conditions affect the endpoint distribution through \(q_k(u)\). The conditional endpoint \(Z_k=u\) tends to make earlier statistics larger when \(u\) is larger; consequently, the chance that all earlier statistics stayed below their boundaries can vary with \(u\). Omitting \(q_k(u)\) generally gives the wrong conditional distribution and therefore the wrong median equation.

Worked Example: A Two-Look Design With Earlier Survival

Consider a hypothetical two-look design with \(I_1=25\), \(I_2=100\), \(b_1=5.00\), and \(b_2=1.00\). Suppose it did not stop at look 1 and observed \(z_2=2.00\) at look 2. Because \(2.00>1.00\), the relevant conditional median equation has a solution.

For this design, the canonical model gives \(Z_1\mid Z_2=u\sim N(0.5u,0.75)\): the conditional mean is \(\sqrt{25/100}\,u=0.5u\), and the variance is \(1-25/100=0.75\). Thus the earlier-survival weight is

q_2(u)=P(Z_1<5.00\mid Z_2=u) =\Phi\left(\frac{5.00-0.5u}{\sqrt{0.75}}\right).

Substituting the boundaries, observation, and information into the conditional-CDF equation gives

\frac{\displaystyle\int_{1.00}^{2.00} \Phi\left(\frac{5.00-0.5u}{\sqrt{0.75}}\right) \phi(u-10\theta)\,du} {\displaystyle\int_{1.00}^{\infty} \Phi\left(\frac{5.00-0.5u}{\sqrt{0.75}}\right) \phi(u-10\theta)\,du} =\frac12.

Numerical quadrature of the two integrals followed by one-dimensional root-finding gives \(\tilde{\theta}_2\approx0.1687\). The factor \(10\theta\) follows from \(\sqrt{I_2}=10\). In this example, the early boundary is far above the earlier statistic values typically paired with endpoint values near 1 to 2, so the survival weight is close to 1 over much of the relevant range. It is nevertheless included in the calculation; it cannot generally be discarded just because it has little numerical effect in this particular design.

Worked Example: An Observation Exactly on the Boundary

Suppose a hypothetical two-look design has \(I_1=25\), \(I_2=100\), \(b_1=1.50\), and \(b_2=2.00\). At look 2, the observed statistic is exactly \(z_2=2.00\), after \(Z_1<1.50\).

The numerator of the conditional CDF at the observation is

P_\theta(Z_1<1.50,\ 2.00\le Z_2\le 2.00)=0.

The denominator \(P_\theta(Z_1<1.50,\ Z_2\ge2.00)\) is positive for every finite \(\theta\) under the nondegenerate normal model. Thus \(P_\theta(Z_2\le2.00\mid T=2)=0\) for every such parameter value, not \(1/2\). There is no median-unbiased estimate defined by this inversion equation for this boundary-equality observation. The fact that exact equality has probability zero in a continuous model does not justify claiming a root if equality is the reported observed value.

Computation, Diagnostics, and Alternatives

For \(k>1\), compute the multivariate normal survival weight \(q_k(u)\), evaluate the numerator and denominator integrals, and solve the scalar equation. Adaptive quadrature and a bracketed root solver are useful because the conditional CDF is continuous and strictly decreasing in \(\theta\) for an observation above the boundary. A practical check is to evaluate the CDF at the proposed root and verify that it is close to \(0.5\); also check that values at lower and higher parameter values lie on opposite sides of \(0.5\).

Boundary check: Verify \(z_k>b_k\) before root-finding. If \(z_k=b_k\), the equation has no solution.
Event check: Ensure the denominator includes survival at all earlier looks as well as crossing at look \(k\).
Numerical check: The ratio must lie between 0 and 1, and it should decrease as the candidate \(\theta\) increases. Failure of either check suggests an integration, indexing, or conditioning error.
Model check: Confirm that information values and boundaries are treated as fixed and that the canonical normal model is appropriate. Estimated standard errors or additional stopping rules require a different sampling law.

This estimator differs from the ordinary maximum-likelihood estimate \(z_k/\sqrt{I_k}\), which does not in general target the median of the distribution conditional on stopping at look \(k\). It also differs from conditional maximum likelihood, which uses the conditional likelihood rather than the conditional median equation. Unconditional bias-corrected estimators target a different averaging framework; they should not be substituted for this conditional inversion. The median-unbiased estimate is specific to the observed stopping event and does not by itself provide a confidence interval or guarantee small mean bias.

Common mistake: “The observed statistic crossed the boundary, so I can use the ordinary normal CDF and solve \(\Phi(z_k-\theta\sqrt{I_k})=1/2\).” That equation ignores the selection event and reduces to the unadjusted estimate. The correct equation divides the probability of the observed statistic being no larger than \(z_k\), jointly with the observed stopping event, by the probability of that stopping event.

Check Your Understanding

Answer each question, then check the solutions below.

For an efficacy stop at look \(k\), what events must appear in the conditioning event?
Why does the conditional endpoint density at look \(k\) include \(q_k(u)\)?
What happens to the median equation when the observed statistic equals \(b_k\)?
In the two-look example, what is the conditional distribution of \(Z_1\) given \(Z_2=u\)?
Why is the ordinary estimate \(z_k/\sqrt{I_k}\) not generally the median-unbiased estimate after an efficacy stop?

Solutions. (1) It includes no efficacy crossing at looks \(1,\ldots,k-1\) and crossing at look \(k\). (2) Given endpoint value \(u\), \(q_k(u)\) is the probability that all earlier statistics survived their boundaries; weighting by it retains only paths that stop first at \(k\). (3) The numerator is zero for every parameter, so no solution to the equation equaling \(1/2\) exists. (4) It is \(N(0.5u,0.75)\), from conditional mean \(0.5u\) and variance \(1-25/100=0.75\). (5) The ordinary estimate uses the unconditioned normal model, whereas median-unbiased inversion uses the distribution conditional on the observed stopping event.