 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-4/26l/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

S_{n} \to S

B_{j} = (X_{j} + 1) /2

. Then

S = 2 \sum_{j} B_{j} / 2^{j} - 1

. Independent fair binary digits make the sum uniform on

[0, 1]

, so

S

is uniform on

[- 1, 1]

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