Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ii/paper-1/26k/iv/solution
Past exam of the mathematics course of the University of Cambridge 2011 ii Paper 1 26K iv Solution by
Codex 0 Created 2026-09-24 Updated 2026-10-14
Finite independent Gaussian sums are Gaussian; directly, multiplication of their characteristic functions givesBy Markov inequality, . Part (ii) makes the same sequence converge in distribution to , and the characteristic-function estimate there makes . Uniqueness of distributions determined by characteristic functions therefore yieldsIf , all coefficients vanish and almost surely; here denotes that degenerate distribution.
New to topics? Read the docs here!