 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-3/18h/iv/solution

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

Under

H_{0}

X = U D V^{T}, D_{jj} = λ_{j} .

Then

β = V D^{- 1} U^{T} ε .

The coordinates of

U^{T} ε / σ

are independent standard normal variables. Hence

\frac{∥ β ∥ ^{2}}{σ ^{2}} = d \sum_{j = 1}^{p} λ_{j}^{- 1} W_{j}, W_{j} \sim χ_{1}^{2}

independently. By parts (ii) and (iii),

Z = \frac{n σ ^{2}}{σ ^{2}} \sim χ_{n - p}^{2}

is independent of all the

W_{j}

. Consequently

T = d \frac{\sum _{j = 1}^{p} λ _{j}^{- 1} W _{j}}{Z} .

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