Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 24 / 3 / b / Solution

There is an initial-value qualification in the printed conclusion. As written, nothing requires $A_0=0$. For example, $M\equiv 0$ and $A\equiv 1$ satisfy the hypotheses, because $M^2-A\equiv -1$ is a local martingale, but $M_t$ is not $N(0,1)$. The correct variance is $A_t-A_0$; the printed formula holds under the usual normalization $A_0=0$.

Indeed, $M^2-[M]$ is a continuous local martingale. Subtracting it from $M^2-A$ shows that $[M]-A$ is a continuous local martingale with finite variation, hence is its initial constant $-A_0$. ThusIn particular, $C$ is deterministic, continuous, nonnegative and increasing, regardless of the apparent allowance of a general finite-variation process in the premise.

For a real parameter $\theta$, the Itô formula gives the complex local martingaleOn every deterministic interval $[0,T]$, its modulus is at most $e^{{\theta }^2{C}_T/2}$, so its real and imaginary parts are bounded local martingales, hence true martingales. Taking expectations and using $Z_0=1$ yieldsBy the uniqueness theorem for characteristic functions,A zero variance means the point mass at zero. This proves the requested normalized version by a direct characteristic function calculation.