 Solution

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

Take $Y = X_{1} + X_{1}^{'}$ , where

X_{1}^{'}

is an independent copy of

X_{1}

. The preceding factorization gives

E e^{t Y} = M_{1} (t)^{2}

. Using

2 X_{1}

would instead give

M_{1} (2 t)

, so an independent copy is required.

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