{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 $T$.

Novikov condition

Factor the square root of the <stochastic exponential> as
$$
\sqrt{S_T}
=\sqrt{S_0}\,
\mathcal E\!\left(\frac12(M-M_0)\right)_T
\exp\!\left(-\frac18[M]_T\right).
$$
The stochastic exponential in this expression is a positive local martingale and hence a <supermartingale>, so its expectation is at most one. If $[M]_T\geq a$ almost surely, then
$$
\mathbb E\sqrt{S_T}
\leq\sqrt{S_0}e^{-a/8}
\mathbb E\mathcal E\!\left(\frac12(M-M_0)\right)_T
\leq\sqrt{S_0}e^{-a/8}.
$$

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 / i / Solution

{title2=$\mathcal E(X)$}
{wiki=Doléans-Dade_exponential}

For a continuous semimartingale $X$, its stochastic exponential is
$$
\mathcal E(X)_t=\exp\!\left(X_t-X_0-\frac12[X]_t\right).
$$
It solves $dZ_t=Z_t\,dX_t$ with $Z_0=1$.


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