 Solution

ID: past-exam-of-the-probability-course-of-the-university-of-oxford/2017/2/b/i/solution

M_{X} (0) = 1

M_{X}^{'} (0) = 0

and

M_{X}^{''} (0) = 1

. The Taylor expansion therefore gives

M_{X} (t) = 1 + \frac{t ^{2}}{2} + o (t^{2}) .

Since

lo g (1 + u) = u + o (u)

u \to 0

lo g M_{X} (t) = \frac{t ^{2}}{2} + o (t^{2}), α = \frac{1}{2} .

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