 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-2/30l/b/solution

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

E [X_{n + 1} ∣ F_{n}] = X_{n} [p c + (1 - p) c^{- 1}]

. Thus the condition is

p c + (1 - p) / c \leq 1

, or

p c^{2} - c + 1 - p \leq 0

. Since the roots are

(1 - p) / p

and

1

, the range is

(1 - p) / p \leq c \leq 1

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