Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 221 / 1

Under complete randomization, every assignment vector with exactly $n_z$ patients receiving treatment $z$ is equally likely. Thus, for $z=(z_1,\dots ,z_n)$ and $N_r(z)={\sum }_i{1}_{\{z_i=r\}}$,The reciprocal probability $n!/(n_0!n_1!n_2!)$ is the multinomial coefficient counting assignments with those arm sizes.

Under independent assignment,The arm-count vector has a multinomial distribution. Conditional on $N_r=n_r$ for $r=0,1,2$, every compatible vector has the same factor ${\prod }_r{\pi }_r^{n_r}$, sofor compatible $z$, and zero otherwise. Conditioning independent assignment on its arm sizes therefore recovers complete randomization.

For a two-by-two contingency table, Fisher's exact test conditions on both row totals and both column totals. Under the null of no association, the upper-left count $K$ then has the hypergeometric distributionwhere $r_1$ and $c_1$ are the first row and column totals. A one-sided p-value sums the appropriate hypergeometric tail; a common two-sided p-value sums the probabilities of all feasible tables no more likely under the null than the observed table. Because the conditional distribution is discrete, the p-value is super-uniform random variable rather than generally exactly uniform.

Cross-classify the binary exposure $A_i=1_{\{Z_i\ge 1\}}$ against the binary outcome $Y_i$. The four cells count placebo survivors or deaths and active-treatment survivors or deaths.

Under the sharp causal null hypothesis, every observed outcome equals the fixed value $Y_i(0)=Y_i(1)=Y_i(2)$, regardless of assignment. With fixed arm sizes, the active set is a uniformly chosen subset of size $n_1+n_2$. Under independent assignment, $A_i$ are independent Bernoulli variables with probability ${\pi }_1+{\pi }_2$, and conditioning on their total again makes the active set uniform. Conditional on the table margins, the active-group outcome count therefore has exactly the hypergeometric distribution used by Fisher's exact test.

The conditional p-value is super-uniform for every set of margins. The law of total probability then givesHence the test is valid under either assignment mechanism.

Within the active set, cross-classify dosage $Z_i\in \{1,2\}$ against $Y_i$. Conditional on the vector $A=(A_1,\dots ,A_n)$, complete randomization assigns $n_1$ of the active patients to low dosage and $n_2$ to high dosage uniformly. Under independent assignment, active patients receive low and high dosage with conditional probabilitiesconditioning further on their low- and high-dose totals again gives the same uniform allocation. Under $H_B$, active patients' outcomes are fixed as their common $Y_i(1)=Y_i(2)$, so the conditional Fisher p-value obeys

Under $H_A$, $P_A$ is a function only of $A$ and the fixed outcomes. Applying the law of iterated expectation under the joint null givesThis conditional argument explains the stated near-independence even though the two tests reuse outcomes.