 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-3/18h/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The likelihood ratio of the simple alternative to the simple null is

\frac{f ( x ∣ 1 )}{f ( x ∣ 0 )} = e^{- 1} (\frac{1 + e ^{x}}{1 + e ^{x - 1}})^{2} .

x

. By the Neyman-Pearson lemma, the most powerful test therefore rejects for

X > c

F_{0} (c) = \frac{e ^{c}}{1 + e ^{c}} .

The size condition

P_{0} (X > c) = α

gives

\frac{e ^{c}}{1 + e ^{c}} = 1 - α, c = lo g \frac{1 - α}{α} .

Because the distribution is continuous, no boundary randomization is needed. The most powerful size-

α

test is therefore

reject H_{0} ⟺ X > lo g \frac{1 - α}{α} .

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