Past exam of the stochastic differential equations course of the University of Oxford

Source: Oxford examination paper C8.1, A12217H1, 9 June 2025; parent PDF pages 31–33.

A convenient sufficient condition is that $f\in C^2({\mathbb{R}}^d)$ has a bounded gradient. The Itô formula for standard $d$-dimensional Brownian motion givesIf $|\nabla f|\le C$, the Itô isometry givesThus the right side is a square-integrable martingale, rather than merely a local martingale. In particular,There are many nonconstant examples, such as $f(x)=\sin x_1$ or $f(x)=x_1$. More generally it is enough that $\mathbb{E}{\int }_0^t|\nabla f(B_s){|}^{2}ds<\infty$ for each finite $t$. Smooth test functions with compact support satisfy the bounded-gradient condition and will provide the domain in the next part.

TakePart (i) proves that Brownian motion solves this martingale problem. For the converse, let $X$ be any continuous adapted process satisfying this martingale problem and $X_0=0$.

Set ${\tau }_n=\inf \{t\ge 0:|X_t|\ge n\}$. Choose a smooth cutoff function ${\chi }_n$ equal to one on a neighbourhood of the closed ball of radius $n$. Apply the martingale problem to ${\chi }_n(x)x_i$ and ${\chi }_n(x)x_i{x}_j$, and stop at ${\tau }_n$. Inside that ball, their Laplace operators are respectively $0$ and $2{\delta }_{ij}$. The optional sampling theorem therefore shows thatare martingales. Continuity makes ${\tau }_n\uparrow \infty$ almost surely. Hence every $X^i$ is a continuous local martingale. The Itô product rule and the fact that a continuous finite-variation local martingale is constant identify its quadratic covariations:Here ${\delta }_{ij}$ is the Kronecker delta.

The Lévy characterization of multidimensional Brownian motion now applies. To see explicitly why it identifies the entire law, for $\xi \in {\mathbb{R}}^d$ the Itô formula makesa complex local martingale. Its absolute value on $[0,T]$ is at most $e^{|\xi {|}^{2}T/2}$, so the bounded local martingale criterion makes it a true martingale there. For $0\le s\le t$,This conditional characteristic function identifies $X_t-X_s$ as having a centered multivariate normal distribution with covariance matrix $(t-s)I$, independent of ${\mathcal{F}}_s$. Together with continuity and $X_0=0$, these are exactly the defining properties of standard $d$-dimensional Brownian motion. Thus the martingale problem for Brownian motion has a unique law.

The existence assertion is false under the hypotheses printed in the source. No measurability of $\theta$ is specified. Even requiring $\theta$ to be a Borel measurable function does not suffice: a coefficient can be positive at every point while approaching zero too rapidly near the starting point. This is the obstruction described by pointwise positivity does not ensure weak existence.

Here is a counterexample in dimension one, with $\Theta =1$:Suppose that a solution to the martingale problem with domain $C_c^{\infty }(\mathbb{R})$ existed. The cutoff argument from part (a)(ii), now with $\mathcal{L}=\theta \Delta$, makes $X$ a continuous local martingale withThe criterion for a continuous local martingale with bounded quadratic variation makes it a square-integrable martingale, and the Itô isometry gives $\mathbb{E}{X}_t^2=\mathbb{E}[X{]}_t$.

Apply the occupation-times formula with the indicator of the singleton $\{0\}$. Since that singleton has zero Lebesgue measure,Thus $X$ spends zero ordinary time at zero. Away from zero, $\theta (x)\le x^2$, soThe Gronwall inequality forces $\mathbb{E}{X}_t^2=0$ for every $t$. Continuity then forces $X\equiv 0$ almost surely, contradicting either the zero occupation time just proved or $[X{]}_t=2t$ for that constant path. Consequently

For completeness, the intended construction works if $\theta$ is a Borel measurable function and bounded away from zero on every compact set. In particular, it works for continuous strictly positive $\theta$. Let $B$ be standard $d$-dimensional Brownian motion, and define a clock and its inverse byThe Brownian path on each finite interval has compact range, so the added hypothesis makes $A_u$ finite. The clock is continuous and strictly increasing, and $A_u\ge u/(2\Theta )$, so $A_u\to \infty$. Its inverse is continuous and satisfies ${\tau }_t\le 2\Theta t$. Each ${\tau }_t$ is a stopping time for the Brownian filtration.

For $f\in C_c^{\infty }({\mathbb{R}}^d)$, the optional sampling theorem applied at these bounded random times shows thatis a true martingale in the filtration ${\mathcal{G}}_t={\mathcal{F}}_{{\tau }_t}^B$. The inverse-clock substitution $d{\tau }_t=2\theta (X_t)dt$, valid almost everywhere, rewrites it asThis proves the Brownian time change for a variable diffusion speed:The factor $2$ in the clock is necessary because standard Brownian motion has infinitesimal generator $\Delta /2$.

A sufficient condition is that $x\mapsto \sqrt{2\theta (x)}$ is globally Lipschitz. A more familiar, stronger condition implying this is that $\theta$ is globally Lipschitz and $\theta \ge \epsilon >0$. Both conditions repair the existence issue in part (i).

To connect this to the suggested stochastic differential equation, apply the coordinate and coordinate-product cutoff tests to any solution of the martingale problem. As before they giveDefine, using stochastic integrals against the coordinates of $X$,These integrals exist even without a global positive lower bound: their quadratic variations are exactly $t$. Indeed,The Lévy characterization of multidimensional Brownian motion identifies $W$ as standard $d$-dimensional Brownian motion. By associativity of stochastic integration,The diffusion matrix $\sqrt{2\theta (x)}I$ is globally Lipschitz under the stated condition, and boundedness of $\theta$ gives bounded growth. The global existence theorem for stochastic differential equations with Lipschitz coefficients gives a unique strong solution of a stochastic differential equation and uniqueness in law. Every solution to the martingale problem has just been represented as a weak solution of a stochastic differential equation of this form, so they all have the same law. Conversely, the Itô formula applied to any $f\in C_c^{\infty }$ gives a true martingale, since its stochastic integrand is bounded. Thus the martingale problem is well posed under this sufficient condition.

Write $N_t=M_t-M_0$ and $V_t=⟨M{⟩}_t=[N{]}_t$. The Dambis-Dubins-Schwarz theorem supplies a standard Brownian motion $W$ such thatOn the event $V_{\infty }=\infty$, continuity and monotonicity imply that the range of $t\mapsto V_t$ is all of $[0,\infty )$.

For completeness, Brownian motion hits every positive integer almost surely. The Brownian reflection principle gives, for $a>0$,Intersect over integer $a$. The times of these increasingly large values must escape every compact interval because each Brownian path is continuous there. Therefore ${\mathrm{limsup}}_{u\to \infty }{W}_u=\infty$. The unbounded clock transfers these values to arbitrarily late times for $N$, and adding the finite initial value $M_0$ preserves unboundedness. HenceThis is an instance of infinite quadratic variation forces unbounded oscillations; applying the same argument to $-W$ also gives negative oscillations.

For $t>0$, $B_t$ has the normal distribution $N(0,t)$. Evaluate its Gaussian integral:At $t=0$ the defining random variable is one, consistent with the same formula. Thus

Let $Y_t=-\beta {\int }_0^t{B}_{u}d{B}_u-{\int }_0^t{h}_{u}du$. Its quadratic variation is $[Y{]}_t={\beta }^2{\int }_0^t{B}_u^{2}du$. The Itô formula for $Z=e^Y$ givesChooseThe chosen process is continuous and adapted, hence progressively measurable. Its time integral is finite on compact intervals because Brownian paths are continuous. The resulting $Z$ is the stochastic exponential of $-\beta {\int }_0^{\cdot }{B}_{u}d{B}_u$, and is a positive local martingale.

The Itô formula for $B_t^2$ yields ${\int }_0^t{B}_{u}d{B}_u=(B_t^2-t)/2$. Substituting into the expression for $Z_t$ gives the pathwise identityConsequentlyTaking expected values provesThe random variable on the preceding line is bounded by one for $\alpha \ge 0$, so its integrability is immediate.

The expression obtained in part (a)(iii) gives a direct deterministic bound. For every fixed $T<\infty$ and $0\le s\le T$,Thus the bounded local martingale criterion makes $Z$ a true martingale on every finite time interval. More explicitly, if ${\sigma }_n$ is a localizing sequence, the identities $\mathbb{E}[Z_{t\wedge {\sigma }_n}\mid {\mathcal{F}}_s]=Z_{s\wedge {\sigma }_n}$ pass to the limit by dominated convergence, using the displayed bound for $t\le T$. ThereforeThis establishes exactly the finite-horizon property needed for the subsequent change of measure.

The relation between $B$ and $W$ is the linear equation $d{B}_s=d{W}_s-\beta B_{s}ds$ with $B_0=0$. Multiplication by the deterministic integrating factor $e^{\beta s}$ and the Itô product rule giveHenceThe second equality follows by semimartingale integration by parts. It expresses $B$ as an explicit function of the continuous path of $W$, using only an ordinary time integral. This inversion holds before changing measure; $W$ is already a continuous semimartingale under the original measure.

Fix $t<\infty$. Part (b)(i) makes $d{\mathbb{Q}}_t=Z_{t}d\mathbb{P}$ a probability measure on ${\mathcal{F}}_t$, equivalent to $\mathbb{P}$ since $Z_t>0$. Its density process on $[0,t]$ is $Z_s=\mathbb{E}[Z_t\mid {\mathcal{F}}_s]$. The Girsanov theorem, applied to the stochastic exponential with integrand $-\beta B_s$, says thatis standard Brownian motion under ${\mathbb{Q}}_t$. The required local square-integrability follows from path continuity, and the true martingale property was proved in part (b)(i).

By part (b)(ii), under ${\mathbb{Q}}_t$ the process $B$ is therefore the zero-started Ornstein-Uhlenbeck process:This identifies its full law on continuous path space, rather than just its one-time marginal. It is a centered Gaussian process, since it consists of deterministic Itô integrals against Brownian motion. For $r,s\in [0,t]$, the Itô isometry gives its covariance function:In particular, its Ornstein-Uhlenbeck transition distribution at the terminal time is

Put $c=\beta /2-\alpha$. The change of measure and part (a)(iii) yieldFor a centered normal random variable $G$ of variance $v>0$, a Gaussian integral gives the Gaussian quadratic exponential momentHere $c\le \beta /2$ and $\beta v_t<1/2$ for finite positive $t$, so the required positivity holds, including when $c>0$. ThusExpanding the two exponentials in terms of hyperbolic functions givesAt $t=0$ this is one. As $\beta \downarrow 0$, the denominator tends to $1+2\alpha t$, recovering part (a)(i). At $\alpha =0$ it gives the Laplace transform of the integrated square of Brownian motion. The joint terminal-and-integral transform is the Joint Laplace transform of Brownian square and its time integral.

Write the diffusion and drift coefficients asBoth are continuously differentiable, withThese derivatives are bounded. For example $2|x|\le 1+x^2$ bounds both absolute values by one. The mean value theorem therefore makes both coefficients globally Lipschitz. They are also bounded, so satisfy the linear growth condition for an SDE.

The global existence theorem for stochastic differential equations with Lipschitz coefficients consequently givesIn particular, there is no finite explosion time. The solution has continuous paths, so all subsequent level-hitting and stopping arguments apply.

Apply the Itô formula to $M_t=e^{X_t}$. Since $d[X{]}_t=\sigma (X_t{)}^{2}dt$ and $b=-{\sigma }^2/2$, the two drift terms cancel:The integrand is continuous and hence locally square integrable after stopping at bounded levels of $X$. Thus $M$ is a positive continuous local martingale, with $M_0=1$ and quadratic variationThe cancellation is the reason $e^x$ is the relevant scale function for this diffusion.

Take $L>0$, as needed for a nontrivial interval. Since $X$ is continuous and adapted, for every level $a$ and $t\ge 0$,The infimum is attained on this compact interval and can be computed on the countable dense set $(\mathbb{Q}\cap [0,t])\cup \{t\}$. It is therefore ${\mathcal{F}}_t$-measurable. Thus $T_a$ is a stopping time, and so is $T=T_{-L}\wedge T_L$, because $\{T\le t\}=\{T_{-L}\le t\}\cup \{T_L\le t\}$.

By continuity, $X_{t\wedge T}\in [-L,L]$, even before knowing that $T$ is finite. HenceStopping preserves the local martingale property. The bounded local martingale criterion now shows that $M^T=(M_{t\wedge T}{)}_{t\ge 0}$ is a uniformly integrable martingale; its deterministic bound also gives square integrability. In particular,

The stopped representation from part (a)(ii) and the Itô isometry give the exact second-moment identityThe cross term with the initial value is zero because the stopped stochastic integral has zero mean. This is the useful finite-time computation; evaluating the remaining occupation integral explicitly is unnecessary for proving integrability of $T$.

For $|x|\le L$,Combining the second-moment identity with part (i) yieldsLet $t\to \infty$ and use monotone convergence:In particular $T<\infty$ almost surely, a conclusion obtained without assuming finite exit in the stopping argument.

One can also evaluate the limiting second moment exactly. At exit $M_T$ is either $e^{-L}$ or $e^L$, soBy bounded convergence theorem and part (i), $\mathbb{E}{M}_T=1$. ThereforeThe same bound also holds at every finite time, by applying $(y-e^{-L})(y-e^L)\le 0$ to $y=M_{t\wedge T}$ and taking expected values. It improves the preceding bound to $\mathbb{E}T\le c_L^{-1}(e^L+e^{-L}-2)$ if desired.

Let $S=T_b\wedge T_{-a}$ and $p=\mathbb{P}(T_b<T_{-a})$. Set $R=\max (a,b)$. The exit time $S$ from $(-a,b)$ is at most the exit time from $(-R,R)$, so part (ii) gives $\mathbb{E}S<\infty$. Continuity ensures that exactly one of the two different boundary levels is reached at $S$.

The stopped local martingale $e^{X_{t\wedge S}}$ lies in $[e^{-a},e^b]$, hence is a uniformly integrable martingale. The optional sampling theorem, justified at the terminal time also by bounded convergence theorem, givesThus the boundary hitting probability from a diffusion scale function isFor integer $n\to \infty$, the events $\{T_b<T_{-n}\}$ increase to $\{T_b<\infty \}$. Indeed, on any path with finite $T_b$, continuity makes the path bounded below on $[0,T_b]$, so some lower barrier $-n$ has not yet been hit. The reverse inclusion is immediate. Taking the limit in the displayed probability gives

Since $X_0=0$, its supremum $X^{\star }$ is nonnegative. For $b>0$, continuity givesThe events increase with $n$. Part (iii) therefore yields $\mathbb{P}(X^{\star }>b)={\lim }_n{e}^{-(b+1/n)}=e^{-b}$. Letting $b\downarrow 0$ gives $\mathbb{P}(X^{\star }>0)=1$; letting $b\to \infty$ gives finiteness almost surely. ConsequentlyIts probability density function is $e^{-x}{1}_{\{x>0\}}$, with no atom at zero. Using strict exceedances above also avoids presupposing that the infinite-horizon supremum is attained.

Let $K={\sup }_{x}f(x)<\infty$ and let $C$ be a Lipschitz constant for $f$. Sinceboth coefficients $f$ and $-f^2/2$ are bounded and globally Lipschitz. The global existence theorem for stochastic differential equations with Lipschitz coefficients gives a unique global strong solution of a stochastic differential equation.

The same Itô formula cancellation as before givesFor $S=T_{-a}\wedge T_b$, the stopped exponential is bounded between $e^{-a}$ and $e^b$. Continuity and strict positivity of $f$ giveThe Itô isometry therefore yieldsThe monotone convergence theorem proves $\mathbb{E}S<\infty$. The bounded optional stopping argument is now identical to part (b)(iii), and givesPassing from level hitting to strict exceedances as in part (b)(iv),No global positive lower bound on $f$ was needed: compact-interval positivity suffices for each exit argument. The law is independent of the particular $f$ because the scale function is always $s(x)=e^x$: its infinitesimal generator satisfiesThis is the exponential scale function for a drift tied to diffusivity. Changing $f$ changes how quickly the path is traversed, but leaves these boundary probabilities and the maximum law unchanged.