 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-2/9f/b/solution

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

Normalization of the Gaussian distribution gives

A = \frac{1}{2 π} .

t x - \frac{x ^{2}}{2} = - \frac{( x - t ) ^{2}}{2} + \frac{t ^{2}}{2} .

Therefore

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

and hence

M_{X} (t) = e^{t^{2} /2} .

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