Past exam of the stochastic differential equations course of the University of Oxford / 2025 / 3

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.