Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 25 / 5 / d / Solution

At the terminal time, $M_T=e^{i\theta X_T}$, so the martingale from part (c) has terminal value $e^{-{\theta }^2⟨X{⟩}_T/2}$. Its initial value is $M_0$, since $X_0=⟨X{⟩}_0=0$. Taking expectations givesHere $M_0$ may be a nonconstant ${\mathcal{F}}_0$-measurable variable; the tower property of conditional expectation still gives $\mathbb{E}{M}_0=\mathbb{E}{e}^{i\theta X_T}$. This characteristic function under conditionally symmetric martingale increments identity relates the characteristic function of the terminal martingale to the Laplace transform of a nonnegative random variable given by its quadratic variation.