 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-4/17h/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Write the linear mean model as

Y_{i} = θ x_{i} + ε_{i}, E ε_{i} = 0, var (ε_{i}) = θ x_{i} .

The errors are independent but heteroscedastic. Minimizing the unweighted residual sum of squares

Q (θ) = \sum_{i = 1}^{n} (Y_{i} - θ x_{i})^{2}

0 = Q^{'} (θ) = - 2 \sum_{i} x_{i} (Y_{i} - θ x_{i}) .

Thus, provided

\sum_{i} x_{i}^{2} > 0

θ_{L S} = \frac{\sum _{i} x _{i} Y _{i}}{\sum _{i} x _{i}^{2}} .

Since

E Y_{i} = θ x_{i}

E θ_{L S} = \frac{θ \sum _{i} x _{i}^{2}}{\sum _{i} x _{i}^{2}} = θ .

Hence it is an unbiased estimator. This is the least-squares part of the estimators for a Poisson exposure model.

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