 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ib/paper-1/17h/iii/solution

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

Under the null,

R = (I - P_{0}) ε

, so

R / σ = d (I - P_{0}) Z

for

Z \sim N (0, I)

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