 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ib/paper-4/19h/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Any unbiased linear estimator has the form considered in part (b), and its error is the linear combination of independent normal random variables

β - β = \sum_{i} a_{i} ε_{i} \sim N (0, σ^{2} \sum_{i} a_{i}^{2}) .

θ

E_{β, σ^{2}} [e^{θ (β - β)}] = exp (\frac{θ ^{2} σ ^{2}}{2} \sum_{i} a_{i}^{2}) .

Since the exponential function is strictly increasing and

θ^{2} σ^{2} /2 > 0

, minimizing this exponential moment is exactly the same as minimizing

\sum_{i} a_{i}^{2}

. Part (b) shows that the unique minimizer is

a_{i} = x_{i} / S_{xx}

. Thus, independently of the sign or magnitude of

θ

β = β = \frac{\sum _{i} x _{i} Y _{i}}{\sum _{i} x _{i}^{2}} .

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