Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 27 / 5

Strict positivity lets us define the continuous local martingaleThe integrand is locally bounded because a positive continuous path has positive minimum on every compact time interval. The Itô formula for $\log M$ givesThis is the stochastic exponential representation of a positive continuous local martingale.

If $⟨X{⟩}_{\infty }$ were finite on an event of positive probability, the finite-bracket convergence lemma proved in Question 2(a) would make $X_t$ converge to a finite limit there. The exponential would then have a strictly positive limit, contradicting the assumed $M_t\to 0$. ThereforeThis is the divergent logarithmic clock for a positive local martingale tending to zero.

Fix $a>1$. Continuity ensures $M_{T_a}=a$ when $T_a<\infty$, even though the defining inequality is strict. Before that infimum the process is at most $a$. Thus $M^{T_a}$ is a bounded nonnegative local martingale and hence a true martingale, with expectation one. For deterministic $t$,The stopped process converges to $a$ on $\{T_a<\infty \}$ and to zero on its complement. It is bounded by $a$, so dominated convergence gives $1=a\mathbb{P}(T_a<\infty )$. The crossing event is exactly $\{{\sup }_{t\ge 0}{M}_t>a\}$. ThereforeThis is the maximal identity for a continuous nonnegative local martingale tending to zero. It also shows there is no atom at a level greater than one. The tail tends to one as $a\downarrow 1$, so the overall maximum has no atom at its lower endpoint either.

Let $S$ denote this maximum and put $\tau =\tau (-b)$. Brownian motion reaches $-b$ almost surely: the Brownian reflection principle gives crossing probability $2\mathbb{P}(W_t<-b)\to 1$. Continuity gives $W_{\tau }=-b$, with the strict-crossing infimum interpreted as in part (b). The processis a continuous nonnegative local martingale starting at one and tending to zero. Its maximum is $1+S/b$. Apply part (b) at $a=1+x/b$, for $x>0$:Differentiating gives the maximum before a lower Brownian barrier densityThe tail tends to one as $x\downarrow 0$, so there is no atom at zero. The density integrates to one.

Use the positive exponential Brownian martingaleThe strong law for Brownian motion, $W_t/t\to 0$, makes its exponent tend to $-\infty$ and hence $M_t\to 0$. If $S={\sup }_{t\ge 0}(W_t-\lambda t)$, then $\sup M=e^{2\lambda S}$. Part (b) gives, for $x>0$,Thus the maximum is exponentially distributed with rate $2\lambda$:There is no atom at zero, by letting $x\downarrow 0$ in the tail. This is the infinite-horizon crossing probability for Brownian motion with negative drift.