 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-202/3/a/2/solution

Part 1 gives

E M_{t}^{2} = E [M]_{t}

. Hence

sup_{t \geq 0} E M_{t}^{2} \leq E [M]_{\infty} < \infty,

M

L^{2}

-bounded.

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