 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ia/paper-2/3f/b/solution

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

M_{X} (t) = \frac{1}{2 π} \int_{- \infty}^{\infty} e^{t x - x^{2} /2} d x = e^{t^{2} /2} \frac{1}{2 π} \int_{- \infty}^{\infty} e^{- (x - t)^{2} /2} d x = e^{t^{2} /2} .

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