Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-12/2/ii/solution
Past exam of the mathematics course of the University of Cambridge 2010 iii Paper 12 2 ii Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-14
Put , and define . The triangle inequality shows that is Lipschitz continuous with Lipschitz constant one. It vanishes on , and on . Let for a uniform permutation.
If , apply the lower-tail bound proved in part (i):If , apply its upper-tail bound:For the desired inequality is simply . Hence in all casesThis uses only the precise one-sided concentration inequality from part (i) and the Hamming distance triangle inequality. The normalization of the distance matters: it counts mismatching coordinates, rather than dividing their number by .
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