 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-201/4/d/ii/solution

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

Reversing the finite summations gives

\frac{1}{n ^{3/2}} \sum_{k = 1}^{n} \sum_{j = 1}^{k} X_{j} = \frac{1}{n} \sum_{k = 1}^{n} \frac{S _{k}}{n} .

This is a Riemann sum for the continuous functional

f \mapsto \int_{0}^{1} f (t) d t

evaluated at the interpolated random walk; the interpolation error tends to zero in probability. The continuous mapping theorem and part (b) yield

\frac{1}{n ^{3/2}} k = 1 \sum n j = 1 \sum k X_{j} d \int_{0}^{1} B_{t} d t \sim N (0, 1/3) .

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