Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 25 / 6

Let $a(x)=\sigma (x)\sigma (x{)}^{\mathsf{T}}$. The multidimensional Itô formula gives the second-order diffusion generatorIndeed,The integrand is locally bounded along the continuous path after stopping on compact sets, because the coefficients and derivatives are continuous. Thus the final stochastic integral is a local martingale. No global growth or uniqueness assumption on this already-given weak solution of a stochastic differential equation is needed.

Apply the Itô product rule to the deterministic discount factor and $u(X_t)$:The prescribed differential equation makes the drift vanish. ThereforeThis is the discounted generator-eigenfunction martingale underlying the Feynman-Kac formula.

Continuity and adaptedness make the first boundary hit a stopping time. A continuous path starting in the open domain cannot leave it before meeting its boundary. Thus $X_{t\wedge T}\in \mathcal{D}\cup \partial \mathcal{D}$, with the usual interpretation when $T=\infty$. If $|u|\le K$ on this set, thenThe stopped process is a bounded local martingale; the bounded local martingale criterion makes it a martingale and, in fact, a uniformly integrable martingale. The Martingale convergence theorem gives almost sure and $L^1$ convergence as $t\to \infty$.

Its limit can also be identified pathwise. On $\{T<\infty \}$ the stopped process is eventually constant at $e^{-\lambda T}u(X_T)$. On $\{T=\infty \}$ its absolute value is at most $K{e}^{-\lambda t}$ and hence tends to zero. Thus the limit is $e^{-\lambda T}u(X_T)1_{\{T<\infty \}}$.

On the event $T<\infty$, continuity puts $X_T$ on the boundary, where $u=1$. The limit found in part (c) is therefore $e^{-\lambda T}$, defining this expression to be zero when $T=\infty$. Taking expectations and using the bounded convergence justified in part (c) givesThe conclusion does not require almost-sure finiteness of $T$. It is the discounted boundary-hitting representation for a bounded solution of $\mathcal{L}u=\lambda u$.

For the drifted Brownian motion $Y_t=y+B_t+at$ on $(0,\infty )$, the diffusion generator is $\mathcal{L}=\frac{1}{2}\frac{d^2}{d{z}^2}+a\frac{d}{dz}$. Seek a bounded solution of $\mathcal{L}u=\lambda u$ with $u(0)=1$. The exponential ansatz $u(z)=e^{rz}$ givesBecause $\lambda >0$, the plus root is positive and the minus root is negative. Boundedness on $[0,\infty )$ therefore selectsIt satisfies the boundary condition and all hypotheses of part (d). HenceThis is the first-passage Laplace transform for Brownian motion with drift, with $e^{-\lambda \infty }=0$. At $a=0$ it reduces to $e^{-y\sqrt{2\lambda }}$. As $\lambda \downarrow 0$, it gives $\mathbb{P}(S<\infty )=e^{-2y\max (a,0)}$, agreeing with certainty of hitting when the drift points towards zero and a possible escape when it points away.