Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 207 / 2 / b

Table of contents
- i b
  - Solution i
- ii b
  - Solution ii

i

Solution

The expected number of failures is

F (n_{0}, n_{1}) = n_{0} (1 - p_{0}) + n_{1} (1 - p_{1}) .

Holding the alternative and test size fixed, constant power is equivalent to fixing

V (n_{0}, n_{1}) = C

. The ethical allocation problem is therefore

min_{n_{0}, n_{1} > 0} F (n_{0}, n_{1}) subject to \frac{a}{n _{0}} + \frac{b}{n _{1}} = C .

The Lagrange multiplier stationary equations determine the ratio; the second-order condition requires the constrained stationary point to be a local minimum.

ii

Solution

For

L = n_{0} (1 - p_{0}) + n_{1} (1 - p_{1}) + η (\frac{a}{n _{0}} + \frac{b}{n _{1}} - C),

the stationary equations are

1 - p_{0} = \frac{η a}{n _{0}^{2}}, 1 - p_{1} = \frac{η b}{n _{1}^{2}} .

Dividing them gives

R^{*} = \frac{n _{0}}{n _{1}} = \frac{a ( 1 - p _{1} )}{b ( 1 - p _{0} )} = \frac{p _{0}}{p _{1}} \frac{1 - p _{1}}{1 - p _{0}} .

The fixed-power constraint then determines the total sample size. Strict convexity after eliminating one variable supplies the second-order minimum condition.

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