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

Set $K_t=\sqrt{H_t}$ and define the <stochastic integral>
$$
W_t=\int_0^t\frac1{\sqrt{H_s}}\,dX_s.
$$
Strict positivity and predictability of $H$ make the integrand locally admissible. The process $W$ is a continuous local martingale starting from zero, and the <quadratic variation of a stochastic integral> gives
$$
[W]_t=\int_0^t\frac1{H_s}\,d[X]_s
=\int_0^t\frac1{H_s}H_s\,ds=t.
$$
By the <Lévy characterization of Brownian motion>, $W$ is a Brownian motion. The <associativity of stochastic integration> then yields
$$
\int_0^tK_s\,dW_s
=\int_0^t\sqrt{H_s}\frac1{\sqrt{H_s}}\,dX_s
=X_t-X_0,
$$
which is the required representation.

Solved by gpt-5.6-sol high.

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

{c}
{wiki=Lévy's_characterization_of_Brownian_motion}

A continuous local martingale $M$ with $M_0=0$ is standard Brownian motion exactly when its <quadratic variation> satisfies $[M]_t=t$.

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 Incoming links: Lévy characterization of Brownian motion