Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 29K (Stochastic Financial Models)

For $t>0$, differentiateThe growth assumptions justify differentiation under the expectation, and Gaussian integration by parts givesSince $u(0)=f(0)$, integration from $0$ to $t$ yields the Brownian transition semigroup identityContinuity at zero covers $t=0$.

A process $(W_t{)}_{t\ge 0}$ is a standard Brownian motion if:

Equivalently, every future increment is independent of the past filtration and has the stated centred normal law.

Let ${\mathcal{F}}_s=\sigma (W_r:0\le r\le s)$. For $0\le s<t$, the increment $W_t-W_s$ is independent of ${\mathcal{F}}_s$ and distributed as $\sqrt{t-s}Z$. Apply part (a) to the translated function $u\mapsto f(W_s+u)$, conditionally on ${\mathcal{F}}_s$:ThereforeThe assumed growth conditions supply integrability, so $(M_t)$ is the Brownian compensator martingale.

Let ${\mathcal{F}}_s=\sigma (W_r:0\le r\le s)$ and fix $c\in \mathbb{R}$. The martingale property between $s$ and $t$ givesFor fixed $s$, write the left-hand conditional expectation at time $t$ as $G(t)$. This integral equation has the unique solutionMultiplication by the ${\mathcal{F}}_s$-measurable factor $e^{-c{W}_s}$ therefore yields the conditional moment-generating functionThis is the moment-generating function of $N(0,t-s)$ and is deterministic. Hence $W_t-W_s$ has that normal law conditionally on ${\mathcal{F}}_s$, and its conditional law does not depend on ${\mathcal{F}}_s$; the increment is therefore independent of the past.

The process starts at zero by assumption and has continuous paths. The independent centred Gaussian increments just obtained complete the definition of Brownian motion. This proves the exponential test-function characterization of Brownian motion.