Under <Neyman allocation>, sample sizes are proportional to the arm standard deviations. Here
$$
\frac{n_1}{n_0}
=\frac{\sqrt{p_1(1-p_1)}}{\sqrt{p_0(1-p_0)}}
=\frac{0.5}{0.3}=\frac53.
$$
For total size $n_{\max}$, the minimized <asymptotic variance> is
$$
\operatorname{Var}(\widehat p_1-\widehat p_0)
=\frac{\bigl(\sqrt{0.25}+\sqrt{0.09}\bigr)^2}{n_{\max}}
=\frac{0.64}{n_{\max}}.
$$
Equal allocation gives
$$
\frac{0.25}{n_{\max}/2}+\frac{0.09}{n_{\max}/2}
=\frac{0.68}{n_{\max}}.
$$
The Neyman allocation therefore reduces the large-sample variance by $0.04/n_{\max}$, about $5.9\%$ of the equal-allocation variance.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 207 / 2 / g / Solution

<Neyman allocation> optimizes information and may allocate more patients to the arm whose Bernoulli variance is nearer its maximum at $1/2$. The ethical allocation instead favors the arm with higher response probability. They coincide only for special parameter values; in general ethical gain is purchased with some loss of precision relative to Neyman allocation.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 207 / 3 / b / ii / Solution

Writing $v_H=p_H(1-p_H)$ and $v_L=p_L(1-p_L)$, <Neyman allocation> minimizes the variance $v_H/n_H+v_L/n_L$ at fixed total sample size and gives
$$
\frac{n_H}{n_L}=\sqrt{\frac{v_H}{v_L}}
=\sqrt{\frac{p_H(1-p_H)}{p_L(1-p_L)}}.
$$

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 207 / 3 / b / Solution

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Neyman allocation assigns sample sizes in proportion to outcome standard deviations to minimize variance.


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