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

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.