Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 36 / 3

For càdlàg semimartingales, the semimartingale integration by parts formula isFor continuous processes the left limits can be replaced by the values. The bracket includes the jump products in the general formula.

For a continuous ${\mathbb{R}}^d$-valued semimartingale $X$ and $f\in C^2({\mathbb{R}}^d)$, the Itô formula isThe formula is localized when the derivatives are unbounded; continuity of $X$ makes them bounded after stopping on compact sets.

To prove the one-dimensional formula for polynomials without assuming that formula, use induction. It is immediate for constants and $X$. SupposeWrite $X=X_0+M+A$ with $M$ a continuous local martingale and $A$ finite variation. The induction hypothesis identifies the martingale part of $X^k$ as $k{X}^{k-1}\cdot M$. The stochastic-integral covariation identity, obtained by polarization from the quadratic-variation identity, givesApply semimartingale integration by parts to $X^{k}X$. Its first-order terms combine to $(k+1)X^{k}dX$, and its bracket terms combine toThis proves the induction step. Taking linear combinations proves the polynomial Itô formula from integration by parts for every polynomial $f$.

Fix a finite horizon $T$ and define ${\tau }_n=\inf \{s\ge 0:|B_s|\ge n\}$. These stopping times increase to infinity almost surely because continuous paths are bounded on each finite interval. Applying the Itô formula to $B_{t\wedge {\tau }_n}^k$, for $k\ge 2$, givesThe first integrand is bounded by $k{n}^{k-1}$, so its integral is square-integrable and has expectation zero. ThusTo justify both limits explicitly, let $S_T={\sup }_{s\le T}|B_s|$ and choose an integer $q>\max \{k,1\}$. The Gaussian density gives $\mathbb{E}|B_T{|}^q<\infty$ by its exponential tail. The Doob Lp maximal inequality statesTherefore $S_T^k$ and $1+S_T^{k-2}$ are integrable. The stopped terminal powers converge almost surely to $B_t^k$ and are dominated in absolute value by $S_T^k$. The integrands in the time integral converge almost surely and are bounded by $1+S_T^{k-2}$, so their integrals are dominated by $T(1+S_T^{k-2})$. Dominated convergence theorem on both sides, and then Fubini justified by the same bound, give the Brownian moment recursionThis uses only finiteness of Gaussian moments, not the closed moment formula requested in the next part.

The initial values are ${\beta }_0(t)=1$ and ${\beta }_1(t)=0$. Repeatedly applying the Brownian moment recursion to odd indices reduces them to ${\beta }_1$, so every odd moment vanishes. For even indices, if the formula holds at $2k-2$, thenThe base case $k=0$ is ${\beta }_0=1$. ThusEquivalently the even moments are $(2k-1)!!t^k$, the familiar Gaussian moment formula with variance $t$.

Interpret the otherwise undefined $b_k(t)$ here as the moment ${\beta }_k(t)$ defined above. For $k=1$, the process is $B_t$, and for $k=2$ it is $B_t^2-t$; both are martingales.

For $k\ge 3$, the Itô formula and the derivative of the Brownian moment recursion yieldThe stochastic integral is a true martingale on finite horizons, since the required $2k-2$ moment is integrable in time. The continuous finite-variation term is not identically zero. If it were, its continuous derivative would vanish at every time, whereas at a fixed $t>0$, the nondegenerate Gaussian $B_t$ does not almost surely satisfy the polynomial equation $B_t^{k-2}={\beta }_{k-2}(t)$. Uniqueness of continuous semimartingale decomposition therefore rules out even the local martingale property. HenceFor example $\mathbb{E}(B_t^3\mid {\mathcal{F}}_s)=B_s^3+3(t-s)B_s$, so merely subtracting its zero expectation cannot make the cubic power a martingale. The centered Brownian powers need not be martingales distinction is resolved by subtracting the random drift compensator $k(k-1){\int }_0^t{B}_s^{k-2}ds/2$ instead of the deterministic mean.