Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 224 / 1 / c

Table of contents
- Solution c

Solution

Let

B_{n}

be the region on which the test chooses

P_{n}

. Then

e_{1}^{(n)} (B_{n}) + e_{2}^{(n)} (B_{n}) = P_{n} (B_{n}^{c}) + Q_{n} (B_{n}) = 1 - {P_{n} (B_{n}) - Q_{n} (B_{n})} .

Minimizing over decision regions is therefore equivalent to maximizing the signed difference. Part b gives

min_{B_{n} \subseteq A^{n}} P_{e}^{(n)} (B_{n}) = 1 - sup_{B_{n}} {P_{n} (B_{n}) - Q_{n} (B_{n})} = 1 - \frac{1}{2} ∥ P_{n} - Q_{n} ∥_{TV} .

The minimizing region is

{x : P_{n} (x) \geq Q_{n} (x)}

, the equal-prior Neyman-Pearson decision region.

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