Past exam of the continuous martingales and stochastic calculus course of the University of Oxford / 2023 / 2

Let $0=t_0<t_1<\cdots <t_m<\infty$ and let ${\xi }_j$ be bounded ${\mathcal{F}}_{t_j}$-measurable random variables. For the bounded simple predictable processthe Itô integral against a continuous martingale $M\in {\mathcal{H}}^{2,c}$ is defined by the increment sumThe same definition applies to elementary predictable processes with stopping-time intervals, replacing $t_j$ by ordered stopping times $T_j$ and taking bounded ${\mathcal{F}}_{T_j}$-measurable coefficients. Refinement of the intervals leaves the increment sum unchanged. The initial value of the integral is zero, and its martingale property follows from the conditional expectation identity for subsequent increments of $M$.

For a bounded simple predictable process $H$ and a continuous martingale $M\in {\mathcal{H}}^{2,c}$, the Itô isometry statesHere $⟨M⟩$ is the predictable quadratic variation, equal to $[M]$ for a continuous martingale. On the whole half-line, the terminal form isThe Itô integral starts at zero even when $M_0\ne 0$. For Brownian motion, $d⟨B{⟩}_s=ds$.

For completeness, cross terms of the finite increment sum have expectation zero by conditioning on the earlier endpoint of the later increment. Also $M^2-⟨M⟩$ is a martingale, so the remaining terms satisfyThis proves the displayed identity without a change-of-variables formula. Completing bounded simple predictable processes in the norm $(\mathbb{E}\int H^{2}d⟨M⟩{)}^{1/2}$ defines the Itô integral of every square-integrable stochastic integrand. This completion will justify the unbounded Brownian integrands below.

Fix $t>0$, a deterministic partition of an interval $\pi$, and write $\eta (s)=t_j$ on $[t_j,t_{j+1})$. Its mesh is $|\pi |={\max }_j(t_{j+1}-t_j)$. The Brownian increment variance givesThus the left-endpoint sums ${\sum }_j{B}_{t_j}\Delta B_j$ converge in L2 space to the Itô integral. Although these step coefficients are unbounded, clipping them at finite levels first makes them bounded simple predictable processes; the Itô isometry and their finite second moments remove the clipping.

The elementary identity $B_{t_{j+1}}^2-B_{t_j}^2=2B_{t_j}\Delta B_j+(\Delta B_j{)}^2$ givesThe squared Brownian increments are independent, with means $\Delta t_j$ and variances $2(\Delta t_j{)}^2$, since a centered normal random variable of variance $v$ has fourth moment $3v^2$. ThereforePassing to the convergence in L2 limits yieldsThis holds for each fixed $t$; continuous sample paths of both sides extend equality from rational times to all $t$ on one event of probability one.

To establish the Brownian weighted quadratic variation in L2 for the deterministic partition of an interval, decomposewhere $U_{\pi }={\sum }_j{B}_j((\Delta B_j{)}^2-\Delta t_j)$ and $V_{\pi }={\sum }_j{B}_j\Delta t_j-{\int }_0^t{B}_{s}ds$.

The summands in $U_{\pi }$ are martingale differences: $B_j$ is ${\mathcal{F}}_{t_j}$-measurable, and the following Brownian increment is independent of ${\mathcal{F}}_{t_j}$. Their conditional means are zero, so distinct summands are orthogonal in L2 space. Since $\mathbb{E}{B}_j^2=t_j$ and $\mathbb{E}[((\Delta B_j{)}^2-\Delta t_j{)}^2]=2(\Delta t_j{)}^2$,For the Riemann sum error, the Cauchy-Schwarz inequality givesThus the triangle inequality in L2 space provesThe time integration variable is $s$.

Apply the exact discrete product identityand sum over the deterministic partition of an interval. Since $B_0=0$,The left side converges in L2 space to ${\int }_0^{t}sd{B}_s$ by the definition and Itô isometry, sinceThe right-endpoint Riemann sum converges in L2 space to ${\int }_0^t{B}_{s}ds$: its squared error is at mostConsequently,As in (a)(iii), continuity extends fixed-time equality to equality of the continuous processes.

Summing the exact cubic increment identity givesThe first sum tends to zero in L2 space. Indeed, odd moments of centered normal random variables vanish, a centered normal random variable of variance $v$ has sixth moment $15v^3$, and independence of the Brownian increments givesThe second sum converges in L2 space to $3{\int }_0^t{B}_{s}ds$ by (i).

For the last sum, first note that $\mathbb{E}{\int }_0^t{B}_s^{4}ds={\int }_0^{t}3s^{2}ds=t^3$, so $B_s^2$ is a square-integrable stochastic integrand. If $s=t_j+u$ and $D=B_s-B_j$, then $D$ is independent of $B_j$, with $D\sim N(0,u)$. HenceIt follows thatThe Itô isometry therefore makes ${\sum }_j{B}_j^2\Delta B_j$ converge in L2 space to ${\int }_0^t{B}_s^{2}d{B}_s$, with bounded truncations supplying the initial simple predictable processes if needed. Taking limits in the cubic identity provesThe proof uses only increment algebra, moments of normal random variables and the definition of the Itô integral.

Put $M_t=B_t^3-3t{B}_t$. It is an adapted process and is integrable at every finite time, because normal random variables have finite absolute moments of every order. Fix $0\le s<t$ and let $D=B_t-B_s$. Then $D$ is independent of ${\mathcal{F}}_s$, has mean zero, variance $t-s$, and third moment zero. Expanding the cube inside the conditional expectation givesAlso $\mathbb{E}[B_t\mid {\mathcal{F}}_s]=B_s$. ThereforeThese are exactly the defining conditions for a martingale.

Interpret the ordered Iterated Itô integral recursively as $I_0(t)=1$ andThe diagonal boundary of the ordered region does not affect this recursive definition. For $n=1$, $I_1(t)=B_t$. For $n=2$, part (a)(iii) gives $I_2(t)=\frac{1}{2}(B_t^2-t)$. For $n=3$, use parts (b)(ii) and (b)(iii):Thus the three Probabilists' Hermite polynomials giveThe right sides are interpreted as their polynomial extensions at $t=0$. These space-time Hermite polynomials arise here directly from ordered Iterated Itô integrals.

Part (ii) represents the process as an Iterated Itô integral:For a centered normal random variable $B_s$ of variance $s$,Consequently, for every finite $T$,The Itô integral of this square-integrable stochastic integrand is a true square-integrable martingale on $[0,T]$. Since $T$ is arbitrary,The last identity is the Itô isometry.