 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-30/2/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-07

Write

m_{i} = μ + α_{i}

. Apart from a constant, the log-likelihood is

ℓ = - \frac{I J}{2} lo g σ^{2} - \frac{1}{2 σ ^{2}} \sum_{i = 1}^{I} \sum_{j = 1}^{J} (Y_{ij} - m_{i})^{2} .

For each group,

\sum_{j} (Y_{ij} - m_{i})^{2} = \sum_{j} (Y_{ij} - \overline{Y}_{i})^{2} + J (\overline{Y}_{i} - m_{i})^{2}, \overline{Y}_{i} = J^{- 1} \sum_{j} Y_{ij} .

m_{i} = \overline{Y}_{i}

α_{1} = 0

identifies

m_{1} = μ

μ = \overline{Y}_{1}, α_{i} = \overline{Y}_{i} - \overline{Y}_{1} (i = 2, \dots, I) .

The baseline mean is the first group mean, rather than the grand mean, because of the chosen identifiability constraint.

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