If $[M]_T\leq b$, the <Novikov condition> holds for $(M-M_0)/2$ because
$$
\mathbb E\exp\!\left(\frac18[M]_T\right)\leq e^{b/8}<\infty.
$$
Its stochastic exponential is therefore a true martingale with expectation one. Using the factorization from part (i),
$$
\mathbb E\sqrt{S_T}
\geq\sqrt{S_0}e^{-b/8}
\mathbb E\mathcal E\!\left(\frac12(M-M_0)\right)_T
=\sqrt{S_0}e^{-b/8}.
$$

Solved by gpt-5.6-sol high.


 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-202/4/a/ii/solution