= Solution
At the terminal time, $M_T=e^{i\theta X_T}$, so the martingale from part (c) has terminal value $e^{-\theta^2\langle X\rangle_T/2}$. Its initial value is $M_0$, since $X_0=\langle X\rangle_0=0$. Taking expectations gives
$$
\boxed{\mathbb E e^{i\theta X_T}=\mathbb E e^{-\theta^2\langle X\rangle_T/2}.}
$$
Here $M_0$ may be a nonconstant $\mathcal F_0$-measurable variable; the <tower property of conditional expectation> still gives $\mathbb EM_0=\mathbb E e^{i\theta X_T}$. 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>.
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