 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-iii/9/a/solution

Use the corrected decomposition

y_{ij} - w_{j} = (y_{ij} - μ_{j}) + (μ_{j} - w_{j}) .

The hint's minus sign before

(μ_{j} - w_{j})

is a typographical error. Squaring, summing, and using

\sum_{i} (y_{ij} - μ_{j}) = 0

\sum_{i, j} (y_{ij} - w_{j})^{2} = \sum_{i, j} (y_{ij} - μ_{j})^{2} + 2 \sum_{j} (μ_{j} - w_{j}) 0 i \sum (y_{ij} - μ_{j}) + m \sum_{j} (μ_{j} - w_{j})^{2} .

Since

m > 0

and

w \neq = μ

, the final term is strictly positive. Hence

i, j \sum (y_{ij} - μ_{j})^{2} < i, j \sum (y_{ij} - w_{j})^{2} .

Thus the centroid uniquely minimizes the sum of squared Euclidean distances to the observations.

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