Past exam of the stochastic differential equations and Brownian motion in complex analysis course of the University of Oxford / 2008 / A (Stochastic Differential Equations)

For a partition $0=t_0<\cdots <t_n=1$, approximate the integral by $X_{\pi }={\sum }_j{t}_j(B_{t_{j+1}}-B_{t_j})$, using left endpoints. The defining independent Brownian increment of Brownian motion make this a centered Gaussian variable with variance ${\sum }_j{t}_j^2(t_{j+1}-t_j)$. As the mesh tends to zero, that variance tends to ${\int }_0^1{t}^{2}dt=1/3$.

The stochastic integral is the mean-square convergence of these sums. mean-square convergence implies convergence in distribution, while the characteristic function converge to $e^{-u^2/6}$. HenceThe probability is exactly one half because this centered nondegenerate normal distribution is symmetric and has no atom at zero.

Use a filtration with respect to which $B$ is Brownian. The process is adapted and continuous because it is a continuous function of $B_t$ and $t$. The moment-generating function givesFor $0\le s<t$, writeThe increment is independent of ${\mathcal{F}}_s$ and normal with variance $t-s$, so the conditional expectation of the second factor is one. ThusThis verifies the martingale property from the definition and proves square integrability at each finite time, as well as continuity. It does not assert uniform square integrability over an infinite time horizon.

Lévy's characterization of Brownian motion states that a real continuous local martingale $M$ with $M_0=0$ and quadratic variation $⟨M{⟩}_t=t$ is a standard Brownian motion with respect to its filtration. Equivalently, its increments are independent of the past and have centered normal distribution with variances equal to elapsed times.

Let $H_s=\mathrm{sgn}({\sup }_{r\le s}{B}_r-1)$. The running maximum is continuous and adapted, so $H$ is bounded and progressively measurable process. It has a predictable version for integration; changing its value at the single switching time does not change its stochastic integral. Thus $X=\int HdB$ is a continuous local martingale starting at zero. Since $H_s^2=1$,Lévy's characterization of Brownian motion yields $X$ is standard Brownian motion. The sign depends on the entire past through the running maximum, but the unit quadratic-variation rate is exactly what the characterization needs.

The drift coefficient $b(z)=rz$ and diffusion coefficient $a(z)=\sigma z$ are Lipschitz continuity and have linear growth. The standard existence and pathwise uniqueness theorem for stochastic differential equations therefore supplies a unique global strong solution of a stochastic differential equation, adapted to the driving Brownian filtration.

Apply Itô's formula to $x\exp ((r-{\sigma }^2/2)t+\sigma B_t)$. Its time derivative contributes $(r-{\sigma }^2/2)X_t$ and its Itô correction contributes ${\sigma }^2{X}_t/2$, giving the required equation and initial value. HenceIn particular the solution stays positive. For $t>s$, factor out $X_s$ and use independence and the normal law of $B_t-B_s$:The exponential moment exists, so this conditional expectation is well-defined for every finite $s,t$.

Itô's formula identifies the infinitesimal generator on twice continuously differentiable functions:Subtract the drift coefficient integral from that formula to obtainStopping when $|X|$ leaves a bounded interval makes the integrand bounded on each finite time interval. Therefore $M^f$ is a continuous local martingale. In fact $f^{\prime }\in L^{\infty }$ and the lognormal distribution solution has finite second moments on finite intervals, so here it is square integrable on each such interval as well.

The quadratic covariation of two stochastic integrals against the same Brownian motion is the integral of the product of their integrands. ConsequentlyThis is the quadratic covariation, which agrees with the pathwise quadratic covariation for these continuous local martingales.

Taking the logarithm of the explicit solution gives a linear function of the normal random variable $B_t$:ThusFor $z>0$ its density isand it is zero for $z\le 0$. The factor $1/z$ is the Jacobian of the logarithmic change of variable.

Fix a finite horizon $T$ and use the completed natural Brownian filtration. Martingale representation theorem states that every $F\in L^2({\mathcal{F}}_T)$ can be writtenwhere $H$ is predictable and is unique up to equality for $dsd\mathbb{P}$ almost every $(s,\omega )$.

The Martingale representation theorem states that every square-integrable martingale $M$ in this filtration has a continuous version of the formwith the same integrability and uniqueness condition. The completed natural filtration has only null-set randomness at time zero, so $M_0$ is almost surely constant. The first statement gives the second by representing $M_T$ and taking conditional expectations; the second gives the first by taking $M_t=\mathbb{E}(F\mid {\mathcal{F}}_t)$.

There is also the local version: every local martingale in the Brownian filtration has a continuous version representable by a predictable, locally square-integrable integrand. The filtration restriction matters; an enlarged filtration with independent additional martingales need not have representation using $B$ alone.

For $f(z)=\sin z$, $f^{\prime }(z)=\cos z$ and $f^{\prime \prime }(z)=-\sin z$. Itô's formula givesSince $B_0=0$, this is exactlyThe integrand retains $\cos B_t$; replacing it by a fixed sign of $\sqrt{1-X_t^2}$ would lose information about the phase of $B_t$.

Condition on the Brownian path up to $t\le T$. The independent normal increment $B_T-B_t$ has $\mathbb{E}\cos (B_T-B_t)=e^{-(T-t)/2}$ and zero expected sine. HenceApply Itô's formula to this expression. Its time derivative and second spatial derivative cancel, leavingTherefore the required representation isThe bounded integrand is square integrable, and uniqueness follows from the Itô isometry. The drift-containing decomposition in part (b) alone would not be the required representation as a constant plus one stochastic integral.

Apply Itô's formula to $M_t=e^{N_t-⟨N{⟩}_t/2}$. The quadratic-variation correction cancels the finite-variation term:Localization of the integrand proves that $M$ is a nonnegative continuous local martingale.

For completeness, Conditional Fatou lemma says that for nonnegative random variables $Y_n$,Choose a localizing sequence ${\tau }_n\uparrow \infty$ such that each stopped $M$ is a martingale. Fatou first gives $\mathbb{E}{M}_t\le 1$. For $s<t$, its conditional form then givesThus $M$ is a supermartingale.

A sufficient condition for it to be a true martingale on $[0,T]$ is Novikov's conditionUnder this condition the stochastic exponential is a uniformly integrable martingale on that finite interval. Requiring it for every finite $T$ gives the martingale property at all finite times; it does not automatically give uniform integrability as $t\to \infty$.

For $s<t<1$, independence and the Gaussian Brownian transition density yield the Chapman-Kolmogorov equationThe numerator is positive and its expectation is $p(1,0,y)$, soThese statements hold for every $t<1$, in particular on the specified closed interval $[0,T]$.

The function $h(t,z)=p(1-t,z,y)$ solves the backward heat equation $h_t+\frac{1}{2}{h}_{zz}=0$. Itô's formula therefore givesOn each interval ending before one this adapted continuous integrand is locally square integrable. Thusgives $M_t=\exp (N_t-⟨N{⟩}_t/2)$. This also follows by applying Itô's formula to $\log M_t$. The source's notation $t\ge 0$ for this density process is restricted by its defining formula to $t<1$.

A finite-horizon form of Girsanov theorem is as follows. If $\theta$ is predictable, ${\int }_0^T{\theta }_s^{2}ds<\infty$ almost surely, and its stochastic exponential $M$ is a true martingale of mean one, then under the probability $d\mathbb{Q}=M_{T}d\mathbb{P}$ the processis Brownian motion. The same conclusion holds on the completed natural filtration, since a strictly positive density gives equivalent probabilities.

Part (b) establishes precisely the required density martingale and integrand ${\theta }_s={\partial }_x\log p(1-s,B_s,y)$. ThereforeIn particular it is the requested local martingale. Equivalently, under $\mathbb{Q}$, $B$ solves $d{B}_t=(y-B_t)(1-t{)}^{-1}dt+d{X}_t$, the Brownian bridge drift coefficient toward $y$ at time one. Keeping $T<1$ avoids the singular endpoint in this density change.