Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 207 / 2

Put $v_k=p_k(1-p_k)$ and assume $0<p_k<1$. For fixed $n$ and a fixed nonzero difference $\Delta =p_0-p_1$, the approximate statistical power of the Wald test increases as its noncentrality $|\Delta |/\sqrt{V}$ increases, whereDifferentiating with respect to the continuous allocation givesHence the unique minimum and its associated minimum variance areThis is Neyman allocation. Integer allocations use a feasible adjacent integer to the continuous optimum, comparing the resulting variances; $n_0,n_1\ge 1$ must still hold. It is a large-sample normal approximation argument, not an exact finite-sample power theorem for arbitrary response-adaptive randomization. Under the null $p_0=p_1$, the approximate power is the significance level for any allocation, although the same allocation still minimizes the stated variance.

The difference of the Bernoulli distribution variances factors asThe first factor is positive, whereas $p_0>1-p_1$ makes the second negative. Thus $v_0<v_1$ and $R^{\ast }=\sqrt{v_0/v_1}<1$. Therefore Neyman allocation assigns more patients to treatment 1, the inferior treatment. Minimizing the variance of a treatment comparison and maximizing the number of desirable responses are different objectives.

For the plug-in Neyman allocation, the estimated Bernoulli distribution variances are $6/25$ and $3/16$, soThis uses the target ratio directly as the allocation odds; a procedure that actively corrects previous allocation imbalances would need an additional rule.

For the randomized play-the-winner rule RPW$(1,1)$, return the drawn ball and add one ball of the same treatment after a success or of the opposite treatment after a failure. Treatment 0 has three successes and two failures, and treatment 1 has one success and three failures. The urn therefore has $1+3+3=7$ treatment-0 balls and $1+1+2=4$ treatment-1 balls. HenceFor dynamic programming with only the final patient left, there is no future value from learning. Independent uniform prior distributions and Beta-binomial conjugacy give posteriors $p_0\mid \mathcal{D}\sim \mathrm{Beta}(4,3)$ and $p_1\mid \mathcal{D}\sim \mathrm{Beta}(2,4)$. Their posterior predictive probabilities of success are $4/7$ and $2/6=1/3$. The optimal terminal action therefore givesThis is optimal for expected successes with no imposed lower bound on the randomization probabilities.

A second opportunity to reject raises the trial's overall Type I error above the nominal 10% in general. If $A$ and $B$ denote rejection at the first and second analyses, the overall error isThe cumulative-data test statistics are correlated, so neither 20% nor $1-0.9^2=19\%$ is the general answer. The latter requires independent tests, which is not the situation here.

The rejection thresholds and the stopping/continuation rule must be calibrated jointly. A prespecified group sequential design or alpha-spending function can allocate a total error budget of 10% across the analyses, using their joint null distribution. A conservative alternative planned in advance is two tests each at 5%, controlled by the Bonferroni inequality. An unplanned change to a design with a remaining error budget can be justified by the conditional error principle. Here, however, an original single-look test at 10% has no remaining conditional rejection probability after its final nonrejection. Keeping its entire original rejection region and adding a new positive-probability rejection opportunity cannot in general retain a 10% overall error rate. The two-look rule therefore needs prospective joint calibration; an already observed result cannot retrospectively create an unused error budget.