 Solution

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

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

For arbitrary

β

E ∥ β ∥^{2} = ∥ β ∥^{2} + tr (σ^{2} (X^{T} X)^{- 1}) = ∥ β ∥^{2} + σ^{2} \sum_{j = 1}^{p} λ_{j}^{- 1} .

Independence from part (ii) and

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

give

E T = \frac{E ∥ β ∥ ^{2}}{σ ^{2}} E (\frac{1}{Z}) .

Because

n - p > 2

E (Z^{- 1}) = \frac{1}{n - p - 2} .

Thus

E T = \frac{∥ β ∥ ^{2} / σ ^{2} + \sum _{j = 1}^{p} λ _{j}^{- 1}}{n - p - 2} .

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