 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-2/26j/d/solution

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

m_{N} = E Y_{N} = \frac{1}{N} \sum_{k = 1}^{N} E X_{k} .

∥ Y_{N} - m_{N} ∥_{2}^{2} = Var (Y_{N}) = \frac{1}{N ^{2}} \sum_{k = 1}^{N} Var (X_{k}) \leq \frac{1}{N} ⟶ 0.

Thus

Y_{N}

differs in

L^{2}

o (1)

from the deterministic scalar

m_{N}

Y_{N} converges in L^{2} ⟺ \frac{1}{N} k = 1 \sum N E X_{k} converges .

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