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

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 202 / 4 / a / ii / Solution

{c}
{wiki=Novikov's_condition}

If a continuous local martingale $M$ satisfies
$$
\mathbb E\exp\!\left(\frac12[M]_T\right)<\infty,
$$
then its <stochastic exponential> is a true martingale through time $T$.

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