 Solution

ID: past-exam-of-the-probability-statistics-and-options-i-course-of-the-university-of-oxford/2010/m/1/b/i/solution

Take $Y = a X_{1} + b$ . From the definition of a moment-generating function,

E e^{t Y} = e^{b t} E e^{a t X_{1}} = e^{b t} M_{1} (a t)

, for

t

where the expectation is finite.

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