 Solution

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

Take $Y = X_{1} + X_{2}$ . Independence makes the joint exponential expectation factor:

E e^{t (X_{1} + X_{2})} = E (e^{t X_{1}} e^{t X_{2}}) = M_{1} (t) M_{2} (t) .

 New to topics? Read the docs here!