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

Codex  0 Created 2026-10-03 Updated 2026-10-05

Put

v_{k} = p_{k} (1 - p_{k})

and assume

0 < p_{k} < 1

. For fixed

n

and a fixed nonzero difference

Δ = p_{0} - p_{1}

, the approximate statistical power of the Wald test increases as its noncentrality

∣Δ∣/ V

increases, where

V = \frac{v _{0}}{n _{0}} + \frac{v _{1}}{n - n _{0}} .

Differentiating with respect to the continuous allocation gives

\frac{d V}{d n _{0}} = - \frac{v _{0}}{n _{0}^{2}} + \frac{v _{1}}{n _{1}^{2}} = 0, \frac{d ^{2} V}{d n _{0}^{2}} = \frac{2 v _{0}}{n _{0}^{3}} + \frac{2 v _{1}}{n _{1}^{3}} > 0.

Hence the unique minimum and its associated minimum variance are

R^{*} = \frac{n _{0}}{n _{1}} = \frac{p _{0} ( 1 - p _{0} )}{p _{1} ( 1 - p _{1} )}, V_{m i n} = \frac{( v _{0} + v _{1} ) ^{2}}{n} .

This is Neyman allocation. Integer allocations use a feasible adjacent integer to the continuous optimum, comparing the resulting variances;

n_{0}, n_{1} \geq 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.

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