Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 32 / 3

A group sequential design can stop early when the treatment is insufficiently promising, reducing expected recruitment, patient exposure and cost when the effect is poor. It also allows a planned decision before maximum enrollment.

A disadvantage is that the analysis is selected by the interim results: estimation and uncertainty must account for stopping, and power and operational planning depend on the stopping rule. The following parts quantify the resulting conditional estimation bias. If a design also allows repeated efficacy tests, its rejection boundaries must control overall Type I error. In the present futility-only design, stopping without rejection need not inflate the error rate of a single prespecified final test; it can make that test conservative. Earlier decisions are valuable, but the ordinary fixed-sample analysis is not automatically appropriate after selection.

At stage 1, the maximum-likelihood estimators of the two normal arm means are their sample means. Their difference estimates the treatment effect:Each arm has $n_1$ patients and variance ${\sigma }^2/n_1$, so independence across arms givesThe quantity $I_1$ is the Fisher information for the mean difference when the common variance is known. The PDF's preliminary expression uses $n_i$ for the variance denominator, although $i$ indexes arms and $n_0$ is not defined. The stage-specific sample size is $n_j$; the calculation uses this intended reading, consistent with the specified $n_1,n_2$ per-arm recruitment and the later weighted estimator. Fresh stage-2 patients are independent of stage-1 patients in the usual trial model; the cumulative stage means themselves are not independent across stages.

The Wald statistic has law $Z_1={\hat{\delta }}_1\sqrt{I_1}\sim N(\delta \sqrt{I_1},1)$. Write $C=\{Z_1\ge f_1\}$ for continuation. By the normal cumulative distribution function,The final sample size per arm is $N=n_1+n_2^{\ast }{1}_C$, soIt lies between $n_1$ and $n_2=n_1+n_2^{\ast }$, increasing with the treatment effect: a more promising treatment is more likely to reach full enrollment. Total expected recruitment over both arms is twice this expression. Equality at the futility boundary has probability zero under the normal model.

Conditioning on continuation selects the upper part of the stage-1 estimator's distribution, since $C=\{{\hat{\delta }}_1\ge f_1/\sqrt{I_1}\}$. Thus ${\mathbb{E}}_{\delta }({\hat{\delta }}_1\mid C)>\delta$ for every finite threshold. The fresh stage-2 estimate ${\hat{\delta }}_2^{\ast }$ is independent of $C$ and remains unbiased for $\delta$. The pooled estimator retains a positive weight on the selected stage-1 data, givingConditional on full enrollment, the ordinary pooled MLE overestimates the treatment effect. This is conditional selection bias after futility continuation. Randomization of treatment assignments does not undo selection on the interim outcome.

For $U\sim N(\mu ,s^2)$ and $a=(T-\mu )/s$, substitution $U=\mu +sZ$ givessince ${\varphi }^{\prime }(z)=-z\varphi (z)$. Divide by $1-\Phi (a)$ to obtain the upper-truncated normal mean.

For $U={\hat{\delta }}_1$, set $\mu =\delta$, $s=1/\sqrt{I_1}$ and $T=f_1/\sqrt{I_1}$. Then $a=f_1-\delta \sqrt{I_1}$. Symmetry of the normal density and distribution yieldsThe ratio $r(x)=\varphi (x)/\Phi (x)$ is an Inverse Mills ratio. It is positive, and the pooled estimator's conditional bias is thereforeThis identity supplies both correction procedures below.

Both alternatives can be made explicit. For a conditional bias-corrected normal mean estimate, let $m$ be the observed pooled estimate and $w=n_1/n_2$, with $0<w<1$. Invert its conditional mean:This can be solved by bracketing or by Newton iteration $d_{k+1}=d_k-[g(d_k)-m]/g^{\prime }(d_k)$. Since $r^{\prime }(x)=-r(x)[x+r(x)]$,The variance of a standard normal conditional on exceeding $-x$ is $1-r(x)[x+r(x)]$, strictly between zero and one. Positivity follows from nondegeneracy; the upper bound follows because the conditional mean $r(x)$ exceeds the truncation threshold $-x$. Hence $1-w<g^{\prime }(d)<1$, so the equation has at most one root. As $d\to \infty$, $g(d)\sim d$; as $d\to -\infty$, the truncated stage-1 mean approaches its threshold and $g(d)\sim (1-w)d+w{f}_1/\sqrt{I_1}$, so a root exists for every finite $m$. Equivalently, differentiating the conditional likelihood divides the ordinary likelihood by ${\mathrm{Pr}}_d(C)$ and gives the same score equation. This conditional-likelihood correction is not exactly conditionally unbiased merely because it inverts a mean.

For an illustration, take $I_1=1$, $w=1/2$, $f_1=0$ and $m=0.5$. The equation is $d+\frac{1}{2}\varphi (d)/\Phi (d)=0.5$. Since $g(0)=0.39894$ and $g(0.5)\simeq 0.75458$, the corrected estimate lies between zero and $0.5$; numerical solution gives $\tilde{\delta }\simeq 0.14649$, below the selected ordinary estimate.

For the uniform minimum variance conditionally unbiased estimator, put $I_2=n_2/(2{\sigma }^2)$, $J=I_2-I_1>0$, $A={\hat{\delta }}_1$ and $V={\hat{\delta }}_2^{\ast }$. The fresh estimate $V$ is conditionally unbiased because it is independent of continuation. LetBefore selection, conditional Gaussian calculations give $A\mid T=t\sim N(t,s^2)$. Conditional on $C$ as well, this normal variable is truncated below $c$, soUsing $V=(I_{2}T-I_{1}A)/J$, Rao-Blackwellization therefore givesIts conditional expectation is $\mathbb{E}[V\mid C]=\delta$, and its conditional variance cannot exceed that of $V$. To justify uniform minimum variance, the joint conditional density of $(A,V)$ is a base density on $\{A\ge c\}$ multiplied by $\exp (\delta I_{2}T-{\delta }^2{I}_2/2)/{\mathrm{Pr}}_{\delta }(C)$. Thus $T$ is a complete sufficient statistic in the one-parameter conditional exponential family, whose natural parameter ranges over an open real interval. The Lehmann–Scheffé theorem proves the claim. The orthogonal pooled arm-average statistic is independent of the entire difference process and carries the nuisance common mean. Together with $T$, it gives a complete sufficient statistic in the selected two-parameter normal family, with an open natural-parameter space. Thus allowing that nuisance statistic does not improve the conditional unbiased estimate of the difference. For the same illustration, $I_2=2$, $J=1$ and $s=1/\sqrt{2}$ give ${\hat{\delta }}_{\mathrm{U}}\simeq 0.21102$. It differs from the conditional-likelihood estimate because exact conditional unbiasedness is a different criterion.