Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 202 / 3

For an SDE driven by Brownian motion, a strong solution of a stochastic differential equation is adapted to the completed filtration generated by a prescribed Brownian motion and satisfies the equation on that space. A weak solution of a stochastic differential equation consists of a probability space, filtration, Brownian motion, and adapted solution satisfying the equation. Uniqueness in law means that any two weak solutions with the same initial law have the same law as processes. Pathwise uniqueness means that two solutions on the same filtered space, driven by the same Brownian motion and having the same initial value, are indistinguishable.

Since $B_{T_a}=0$, applying Itô formula to $B_t^3$ after $T_a$ givesBefore $T_a$, both sides vanish. Since $T_a$ is a stopping time determined by $B$, this is a strong solution of a stochastic differential equation.

Taking $a=0$ gives $X_t=B_t^3$, whereas any $a>0$ gives a solution that remains zero until $T_a$; these differ with positive probability while using the same Brownian motion and initial value. Therefore pathwise uniqueness fails.

One form of the Feynman-Kac formula is the following. For bounded sufficiently regular $f,V$ andone hasConversely, a bounded classical solution has this representation. To prove it, fix $t$ and apply Itô formula toThe PDE cancels its drift. The remaining stochastic integral is a martingale, so taking expectations at $s=0,t$ gives the representation; the converse follows by the same calculation and uniqueness for the parabolic boundary-value problem.

Apply the Feynman-Kac formula with $V(x)=\sigma x$ and terminal function $1$. The ansatz $u(t,x)=e^{A(t)x+C(t)}$ givesThus $A(t)=-\sigma t$ and $C(t)={\sigma }^2{t}^3/6$, soEquivalently, the Integral of Brownian motion is Gaussian with mean $xt$ and variance $t^3/3$.