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

For a deterministic $t\le T$, conditional symmetry makes the conditional characteristic function of $X_T-X_t$ invariant under $\theta \mapsto -\theta$. The bounded real and imaginary parts of the exponential are legitimate test functions. HenceThe right side is a bounded complex martingale. Both sides have continuous versions by the assumptions; equality on rational times and continuity make them indistinguishable. Thus $e^{-2i\theta X_t}{M}_t$ is a martingale on $[0,T]$. Complex martingale assertions mean the corresponding assertions for both real and imaginary parts.

Let $A=⟨X⟩$ be the quadratic variation. We use bilinear quadratic covariation for the complex martingale: $⟨M,X⟩=⟨\mathrm{Re}M,X⟩+i⟨\mathrm{Im}M,X⟩$. The Itô formula givesBy the Itô product rule, the finite-variation part of $d(e^{-2i\theta X}M)$ isPart (a) says the product is a martingale, so uniqueness of the continuous semimartingale decomposition makes this finite-variation part zero. For $\theta \ne 0$, division givesFor $\theta =0$, $M\equiv 1$ and both sides are zero, so the identity holds without exception.

Put $F_t=\exp (-i\theta X_t-\frac{1}{2}{\theta }^2{A}_t)$. The Itô formula givesThe explicit decreasing exponential contributes half the finite-variation term; the other half is the quadratic correction from $e^{-i\theta X_t}$. Now the Itô product rule and part (b) yieldThe finite-variation terms cancel because $-i\theta (i\theta )={\theta }^2$. This proves the product is a local martingale. Moreover $|F_t|=e^{-{\theta }^2{A}_t/2}\le 1$ and $|M_t|\le 1$. The bounded local martingale criterion therefore proves the product is a true martingale.

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.

Fix $T>0$. The given normal distribution and part (d) implyOne can deduce determinism without any moment assumption on the bracket. Put $Z=e^{-⟨X{⟩}_T}$. Taking $\lambda =1,2$ gives $\mathbb{E}Z=e^{-T}$ and $\mathbb{E}{Z}^2=e^{-2T}$, so $\mathrm{Var}(Z)=0$. Therefore $⟨X{⟩}_T=T$ almost surely. Applying this at every rational time and using continuity of quadratic variation gives $⟨X{⟩}_t=t$ simultaneously for all $t\ge 0$ outside a single null set.

The Lévy characterization of Brownian motion states that a continuous local martingale starting at zero with this bracket is Brownian motion in its filtration. To see the independent-increment conclusion directly, the Itô formula shows that $e^{i\theta X_t+{\theta }^{2}t/2}$ is a martingale on any fixed bounded time interval: it is a local martingale with a deterministic bound on its modulus. ThusThe deterministic conditional characteristic function identifies an $N(0,t-s)$ increment independent of ${\mathcal{F}}_s$. Together with the given path continuity and $X_0=0$, this proves $X$ is Brownian motion.