 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-2/11f/ii/solution

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

The preceding representation and the central limit theorem give

\frac{B _{n} - n p}{n pq} d N (0, 1) .

Since

S_{n} - n (p - q) = 2 (B_{n} - n p),

one may take

b_{n} = n (p - q), c_{n} = 2 n pq .

For

0 < p < 1

this gives the required convergence to

Φ (x)

at every real

x

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