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
$$
\min_{B_n\subseteq A^n}P_e^{(n)}(B_n)
=1-\sup_{B_n}\{P_n(B_n)-Q_n(B_n)\}
=1-\frac12\lVert P_n-Q_n\rVert_{\rm TV}.
$$
The minimizing region is $\{x:P_n(x)\geq Q_n(x)\}$, the equal-prior <Neyman-Pearson decision region>.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-224/1/c/solution