Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 38 / 2

Use the standard global convention that ${\mathcal{M}}_c^2$ consists of L2-bounded continuous martingales, here starting at zero. This matters for arbitrary unbounded stopping times in part (iv); square-integrability at each separate deterministic time alone is weaker. Such martingales have terminal $L^2$ limits and satisfy the optional sampling theorem at arbitrary finite stopping times. Infinite stopping times can also be included by evaluating at the terminal limit. The weakly orthogonal continuous martingales condition compares cross-time expectations. The following four parts prove its equivalence with the equal-time, bracket-mean and forward-stopping tests.

If $M,N$ are weakly orthogonal continuous martingales, taking the two times equal gives $\mathbb{E}[M_s{N}_s]=0$. Conversely suppose this equal-time condition holds. For $s\le t$, the martingale property and conditional expectation giveFor $t\le s$, condition the other factor at time $t$ instead. Products are integrable by the Cauchy-Schwarz inequality. Thus the equal-time condition implies the full cross-time definition, proving (i) equivalent to (ii).

The equal-time condition is the criterion proved equivalent to weak orthogonality in solution. Its equivalence with the quadratic-covariation condition follows from the covariance identity for continuous square-integrable martingales in solution; its equivalence with the forward-stopping condition follows from the optional sampling theorem in solution. These combine to give all four equivalences.

The polarization identity for quadratic covariation, together with $\mathbb{E}[M\pm N{]}_s=\mathbb{E}(M_s\pm N_s{)}^2$, yieldsThe bracket identity for each zero-starting square-integrable martingale is the Itô isometry for integration of the constant integrand against that martingale. Polarizing both identities gives the displayed covariance identity for continuous square-integrable martingales. The bracket is integrable: its total variation is at most $\sqrt{[M{]}_s[N{]}_s}$ by the Kunita-Watanabe inequality, whose expectation is finite by the Cauchy-Schwarz inequality. Therefore condition (iii) is equivalent to (ii).

For a stopping time $T\ge s$, the optional sampling theorem for the L2-bounded continuous martingale $M$ gives $\mathbb{E}[M_T\mid {\mathcal{F}}_s]=M_s$. ConsequentlyIf $N_s$ is unbounded, first multiply by a bounded truncation and then remove it using $L^2$ bounds and the Cauchy-Schwarz inequality. Thus (ii) implies (iv). Taking $T=s$ proves the converse. For $T=\infty$, the same calculation uses the $L^2$ terminal value.

For a stopping time $T\le s$, optional sampling for $N$ gives $\mathbb{E}[N_s\mid {\mathcal{F}}_T]=N_T$, henceThis is justified by truncation and the Cauchy-Schwarz inequality, as in solution. If $MN$ is a martingale, bounded optional sampling gives $\mathbb{E}[M_T{N}_T]=\mathbb{E}[M_0{N}_0]=0$, proving the required condition.

Conversely suppose the required condition holds. Every bounded stopping time $T$ lies below some deterministic $s$, so the displayed identity gives $\mathbb{E}[(MN{)}_T]=0$. The product is continuous and adapted, and its stopped values are integrable: the Doob L2 maximal inequality bounds the product of the two path suprema in $L^1$ by the Cauchy-Schwarz inequality. The characterization of a martingale by bounded continuous-time stopped expectations therefore makes $MN$ a martingale. Thus orthogonality is equivalent to the backward-stopping expectation test. In contrast to weakly orthogonal continuous martingales, this tests the product at random times, not only deterministic times.

Take a standard Brownian motion $B$ and defineThe Itô formula gives $N_t=2{\int }_0^{t\wedge 1}{B}_{s}d{B}_s$, so both are continuous martingales starting at zero. Their second moments are $\mathbb{E}{M}_t^2=t\wedge 1$ and $\mathbb{E}{N}_t^2=2(t\wedge 1{)}^2$, making them L2-bounded continuous martingales. At equal times, the odd moments of a centered normal random variable vanish, soThey are consequently weakly orthogonal continuous martingales by solution.

Their product, for $t<1$, is $B_t^3-t{B}_t$. The Itô formula givesThe continuous finite-variation term $2{\int }_0^t{B}_{s}ds$ is not identically zero: it has variance $4t^3/3$ at every $t>0$. Uniqueness of the continuous semimartingale decomposition therefore shows that $MN$ is not even a local martingale. Equivalently $[M,N{]}_t=2{\int }_0^{t\wedge 1}{B}_{s}ds$ is not zero, so the pair is not orthogonal continuous local martingales.

For the integrable terminal bracket criterion, assume $M_0=0$, or more generally $M_0\in L^2$. This initial-value assumption is necessary if it is not already part of the notation: an integrable but non-square-integrable random constant process has zero quadratic variation and still fails square-integrability.

For the zero-starting case, choose a localizing sequence ${\tau }_n$ bounding both $|M|$ and its quadratic variation. The stopped processes are true square-integrable martingales, and the Itô isometry givesFor each fixed time these variables are uniformly integrable by their bounded second moments. Almost sure convergence as $n\to \infty$ is therefore convergence in $L^1$. Passing to the limit in the stopped conditional-expectation identity gives $\mathbb{E}[M_t\mid {\mathcal{F}}_s]=M_s$: the local martingale is a true martingale. The Fatou lemma also givesso $M$ belongs to the global ${\mathcal{M}}_c^2$ space. For general square-integrable $M_0$, apply this argument to $M-M_0$ and use $|M_t{|}^2\le 2|M_0{|}^2+2|M_t-M_0{|}^2$.