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

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.