 Solution

ID: past-exam-of-the-probability-course-of-the-university-of-oxford/2017/2/a/iii/solution

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} .

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

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