Apply <Itô formula> to $Z_t^{(A)}=\exp(A X_t-A^2t/2)$. Its <semimartingale decomposition> is
$$
dZ_t^{(A)}=A Z_t^{(A)}\,dX_t
+\frac{A^2}{2}Z_t^{(A)}\bigl(d[X]_t-dt\bigr).
$$
The second term is a continuous <finite-variation process>. Since $Z^{(A)}$ is assumed to be a <local martingale>, uniqueness of the semimartingale decomposition makes this term identically zero. Both $A$ and $Z^{(A)}$ are nonzero, so the <quadratic variation> of $X$ is $[X]_t=t$.

The <Lévy characterization of Brownian motion> now says that $X_t-X_0$ is a <Brownian motion>. Consequently
$$
Z_t^{(a)}=e^{aX_0}
\exp\!\left(a(X_t-X_0)-\frac{a^2t}{2}\right)
$$
is a constant multiple of an <exponential Brownian martingale>. It is therefore a true martingale for every $a\in\mathbb R$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 202 / 1 / a / Solution

A semimartingale has a decomposition $X=X_0+M+A$ into a <local martingale> $M$ and an adapted <finite-variation process> $A$. Under standard normalizations the decomposition is unique.

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