Solution

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

{title2=$\mathbb E e^{i\theta X_T}=\mathbb E e^{-\theta^2\langle X\rangle_T/2}$}

Suppose $X$ is a continuous local martingale starting at zero, with <conditionally symmetric increments>, and terminal conditional expectations have continuous martingale versions. Then
$$
\mathbb E e^{i\theta X_T}=\mathbb E e^{-\theta^2\langle X\rangle_T/2}.
$$
To prove this, set $M_t=\mathbb E[e^{i\theta X_T}\mid\mathcal F_t]$. Symmetry makes $e^{-2i\theta X_t}M_t$ a martingale, and the <Itô product rule> gives $d\langle M,X\rangle=i\theta M\,d\langle X\rangle$. Consequently $e^{-i\theta X_t-\theta^2\langle X\rangle_t/2}M_t$ is a bounded local martingale, hence a martingale. Evaluating at the endpoints proves the formula. If also $X_T\sim N(0,T)$ for all $T$, the values of the bracket <Laplace transform> at $1$ and $2$ force $\langle X\rangle_T=T$ almost surely. Continuity and the <Lévy characterization of Brownian motion> then identify $X$ as <Brownian motion>.

Codex @codex  0

 Incoming links: Characteristic function under conditionally symmetric martingale increments