Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 202 / 4

A strong solution of a stochastic differential equation is an adapted process $X$ on a prescribed filtered probability space carrying a prescribed Brownian motion $W$, satisfyingalmost surely. A weak solution of a stochastic differential equation consists of a filtered probability space, a Brownian motion, and an adapted process on that space satisfying the same integral equation; the space and driving Brownian motion are part of what may be chosen.

Pathwise uniqueness means that two solutions on the same filtered probability space, driven by the same Brownian motion and having the same initial value almost surely, are indistinguishable. Uniqueness in law means that any two weak solutions with the same initial distribution induce the same probability law on path space.

Let $X$ and $Y$ be two solutions with the same initial value and Brownian motion, and put $Z=X-Y$. Itô formula givesStop when either process or the stochastic integral becomes large. Taking expectations, using the assumed one-sided Lipschitz bound, and then removing the localization givesThe Gronwall inequality yields $\mathbb{E}{Z}_t^2=0$. Thus $X_t=Y_t$ almost surely at every rational time, and path continuity makes the two processes indistinguishable. This proves pathwise uniqueness.

Fix $t$ and apply Itô formula to $M_s=U(t-s,X_s)$ for $0\le s\le t$. Its drift isby the Kolmogorov backward equation. HenceLocalization makes this a martingale, and boundedness of $U$ permits passage to the limit. Conditioning the identity $M_t=U(0,X_t)$ on $X_0$ givesThis is the required special case of the Feynman-Kac formula, proved directly.

The stochastic-integral identity obtained in part (i), evaluated at $s=t$, isSince $d{X}_s-b(X_s)ds=\sigma (X_s)d{W}_s$, this has the requested form with the previsible process

For $T_n=T\wedge n$, Itô formula and the differential equation show thatup to $T$. After localization this is a martingale, and boundedness of $V$ permits optional stopping. Thus, conditionally on $X_0$,On $\{T<\infty \}$, path continuity gives $X_T\in \{\ell ,r\}$ and hence $V(X_T)=1$. On $\{T=\infty \}$, boundedness of $V$ makes $e^{-\lambda n}V(X_n)\to 0$. The dominated convergence theorem therefore yieldson $\{\ell <X_0<r\}$, with the stated convention when $T=\infty$.