Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 202 / 2

For a deterministic partition of an interval $0=t_0<\cdots <t_m=t$, put $Q_{\pi }={\sum }_j{\Delta }_{j}B{\Delta }_j\tilde{B}$. Independence of the two Brownian motions and their independent increments givesThe defining sums for quadratic covariation therefore converge to zero in the L2 space and in probability. Hence, with the continuous versions of the quadratic covariation,Equality holds simultaneously for all $t$ by checking rational times and using continuity. For continuous local martingales, the angle-bracket notation here agrees with quadratic covariation.

The stochastic process $W$ starts from zero and has continuous paths. Its increments on disjoint intervals are independent, since the corresponding pairs of Brownian motion increments are independent, andThus $W$ is a standard Brownian motion, including $\rho =\pm 1$. Here the intended filtration is the joint natural filtration, or another filtration relative to which the pair is a two-dimensional Brownian motion; independence of their path laws alone would not justify arbitrary extra information in a larger filtration.

By bilinearity of quadratic covariation and part (i),

The Itô formula applied to $F(t,x)=e^{t/2}\cos x$ gives exact cancellation of the drift:On every finite interval $[0,T]$, the Itô integral is square-integrable since $\mathbb{E}{\int }_0^T{e}^s{\sin }^2{B}_{s}ds\le e^T-1$. Therefore $e^{t/2}\cos B_t$ is a true martingale, and hence a local martingale.

For the other stochastic process, $d(B_t-t^2)=d{B}_t-2tdt$. If $B_t-t^2$ were a local martingale, its difference with $B$ would make the continuous finite-variation process $-t^2$ a local martingale, contradicting the fact that a continuous finite-variation local martingale is constant. Also $\mathbb{E}(B_t-t^2)=-t^2$ is not constant. Thus $B_t-t^2$ is neither a local martingale nor a martingale.

On the right-hand side, ${\mathbb{E}}_x$ means that the reference Brownian motion starts at $B_0=x$; on the left it means the stochastic differential equation starts at $X_0=x$. Since $h$ is smooth and compactly supported, $h^{\prime }$ is bounded and globally Lipschitz continuous. Define the stochastic exponentialThe Novikov condition holds on every finite horizon $T$, because $\mathbb{E}\exp (\frac{1}{2}{\int }_0^T{h}^{\prime }(B_s{)}^{2}ds)\le \exp (T\Vert h^{\prime }{\Vert }_{\infty }^2/2)$. Thus $D$ is a true martingale. Under $d\mathbb{Q}=D_{T}d\mathbb{P}$, the Girsanov theorem makes ${\hat{B}}_t=B_t-x-{\int }_0^t{h}^{\prime }(B_s)ds$ a standard Brownian motion for $t\le T$. Consequently $B$ under $\mathbb{Q}$ solves the prescribed stochastic differential equation; the existence and pathwise uniqueness theorem for a stochastic differential equation identifies its law with that of $X$.

The Itô formula gives ${\int }_0^t{h}^{\prime }(B_s)d{B}_s=h(B_t)-h(x)-\frac{1}{2}{\int }_0^t{h}^{\prime \prime }(B_s)ds$. Hence, for any bounded measurable $f$,The finite-horizon argument applies separately to every $t\ge 0$.