Past exam of the stochastic differential equations and Brownian motion in complex analysis course of the University of Oxford / 2007 / A (Stochastic Differential Equations)

The raw natural filtration isEquivalently it is the smallest sigma-algebra making every observation $B_u$, $u\le t$, measurable. As $s\le t$ enlarges the set of observations, ${\mathcal{F}}_s^B\subseteq {\mathcal{F}}_t^B$. It is the smallest filtration to which the process is adapted. If the usual conditions for a filtration are imposed, take the completion by the probability-zero sets and its right-continuous augmentation. Brownian independent increments remain valid for this usual augmentation. Adding arbitrary future information is not part of the generated filtration.

Independent increments give $B_s\sim N(0,s)$ independent of $B_t-B_s\sim N(0,t-s)$. Their joint probability density function at $(x,v)$ is the productThe change of variables $(x,v)\mapsto (x,y)=(x,x+v)$ has Jacobian determinant one. Therefore the joint probability density function of $(B_s,B_t)$ isIn particular these coordinates are correlated, with covariance $s$, although the first pair has independent coordinates.

The increment $B_t-B_s$ is independent of ${\mathcal{F}}_s$, with characteristic function $e^{-{\xi }^2(t-s)/2}$. Since $B_s$ is measurable at time $s$,The heat kernel identity in the hint gives $(P_{u}f)(x)=e^{i\xi x}{e}^{-{\xi }^{2}u/2}$, so this is exactly $(P_{t-s}f)(B_s)$. Multiplying by the deterministic factor $e^{{\xi }^{2}t/2}$ givesThe process is adapted, and $|M_t|=e^{{\xi }^{2}t/2}$ is finite at each fixed time. Thus it is a complex martingale, meaning that its real and imaginary parts are integrable martingales.

Apply Itô's lemma to $F(t,x)=e^{i\xi x+{\xi }^{2}t/2}$. Since $F_t={\xi }^{2}F/2$, $F_x=i\xi F$ and $F_{xx}=-{\xi }^{2}F$, the drift terms cancel:This is initially a local martingale. On every finite interval $[0,T]$,so the stochastic integral is a true square-integrable martingale. Conditional expectation gives $\mathbb{E}(M_t\mid {\mathcal{F}}_s)=M_s$; divide by the nonzero ${\mathcal{F}}_s$-measurable factor $e^{i\xi B_s+{\xi }^{2}t/2}$ to obtain

If $X$ is Brownian motion with respect to the given filtration, independent Gaussian increments give the identity immediately.

Conversely, fix $s<t$ and let $\Delta =X_t-X_s$. Taking expectations gives its characteristic function $e^{-{\xi }^2(t-s)/2}$, so $\Delta \sim N(0,t-s)$. To prove independence, for any $A\in {\mathcal{F}}_s$ define the finite measure ${\nu }_A(E)=\mathbb{P}(A\cap \{\Delta \in E\})$. The conditional identity givesUniqueness of Fourier transforms of finite measures implies ${\nu }_A=\mathbb{P}(A)N(0,t-s)$. This is independence of $\Delta$ from ${\mathcal{F}}_s$. Iterating over times proves independent increments. Together with the stated continuity and $X_0=0$, this gives standard Brownian motion.

Here a process with the stated filtration is understood to be adapted, as required when saying it is Brownian with respect to that filtration. If adaptedness is not implicit, it must be added: an ordinary Brownian motion with a trivial filtration satisfies the displayed conditional characteristic function but is not adapted to that filtration.

Lévy characterization of Brownian motion says that a continuous local martingale $M$ with $M_0=0$ is standard Brownian motion with respect to its filtration if its quadratic variation is $⟨M{⟩}_t=t$. Equivalently one may require $M_t^2-t$ to be a local martingale.

Any weak solution has the integral form $Z_t={\int }_0^t\mathrm{sgn}(Z_s-2)d{W}_s$. Its bounded integrand makes it a continuous local martingale starting at zero, andThe specified value of the sign at zero has square one as well. Lévy characterization of Brownian motion therefore proves that every such weak solution is Brownian motion. This is a statement about its law and filtration; it does not assert $Z=W$ or provide a strong solution as a function of a specified driver.

Let $A_t={\int }_0^{t}h(s)d{B}_s$ and $v_t={\int }_0^{t}h(s{)}^{2}ds$. Itô's lemma for the exponential givesTo justify that this local martingale is true, note that a deterministic Itô integral is Gaussian: step-function approximations are centered Gaussian sums with variances tending to $v_t$, and Itô isometry gives their convergence. Thus $A_t\sim N(0,v_t)$ andContinuity of $h$ on the compact interval bounds $h$ and $v_t$, soThe differential above is therefore a square-integrable stochastic integral, proving the martingale property on $[0,T]$, with initial value one.

Integrating the preceding differential gives $M(h_j{)}_T=1+{\int }_0^T{h}_j(s)M(h_j{)}_{s}d{B}_s$. Consequently $\mathbb{E}F={\sum }_j{c}_j$, and the representation isThe process is continuous and adapted, hence progressively measurable, and the finite-sum inequality together with the previous second-moment estimate gives $\mathbb{E}{\int }_0^T|f_s{|}^{2}ds<\infty$.

If $g$ is another square-integrable predictable integrand with the same representation, Itô isometry givesThus uniqueness is up to $ds\otimes d\mathbb{P}$ null sets in the square-integrable integrand class. Pointwise uniqueness is impossible: changing a value at one deterministic time changes no Itô integral. Square integrability is the natural class for this $L^2$ representation and is needed for this uniqueness argument; the mere word adapted does not by itself specify an integrable stochastic integral.

The semimartingale integration by parts givesHere $d⟨X,M{⟩}_t=h(t)M_{t}dt$. Substitution therefore gives cancellation of the drift:The continuous adapted integrand is locally square integrable. Stopping when $|X|+M$ first exceeds an integer produces true stochastic-integral martingales, hence $Y$ is a local martingale. No unproved global integrability of the product is needed.

Put $a(t)=e^{{\int }_0^t\mu (s)ds}$ and $q(t)=a(t{)}^{-1}$. Since $q^{\prime }=-\mu q$ and $q$ has finite variation, the product rule has no quadratic-covariation contribution:Integrating and multiplying by $a(t)$ yieldsThe deterministic integrand is square integrable on every compact interval. This constructs a strong solution. Two solutions driven by the same Brownian motion have difference solving $dD=\mu Ddt$, $D_0=0$, hence their difference vanishes, proving pathwise uniqueness.

The deterministic stochastic integral in the solution is centered Gaussian with varianceTherefore $X(t,x)$ has mean $a(t)x$ and varianceFor $t>0$ this is strictly positive, since $\sigma >0$. Its density at $y$ isThe boundedness assumptions on $\sigma$ and its reciprocal in particular ensure nondegeneracy; compact-interval continuity is already sufficient for the displayed finite-time construction.

With $G$ a standard normal random variable,For $f\in C_b^1$, in the usual sense that $f$ and its continuous derivative are bounded, the mean value theorem bounds the difference quotient by $a(t)\Vert f^{\prime }{\Vert }_{\infty }$. Dominated convergence theorem therefore allows differentiation inside the integral:This also follows pathwise from $X(t,x+h)-X(t,x)=a(t)h$. If the textual description were read as assuming only bounded differentiability, some control of $f^{\prime }$ is needed for the displayed right side; the stated $C_b^1$ hypothesis supplies it. Gaussian smoothing itself still differentiates a bounded measurable $f$, using the derivative of the Normal distribution instead.