Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 33 / 2 / c / Solution

If $S=1$, the strict inequality in question already fails at time zero, almost surely, since $X_0=0$. Its probability is therefore zero, as required. Assume for the rest that $0<S<1$, so $M_0=S<1$.

Before time one the boundary inequality is equivalent to $M_t<1$. The process $M$ increases continuously between arrivals and has only downward jumps. Its first time $T<1$ of reaching level one must consequently be a continuous crossing, except possibly if an arrival occurs at the very same instant. All possible first crossing times belong to the finite deterministic setA uniform arrival hits none of these times almost surely. Thus $M_T=M_{T-}=1$ on $\{T<1\}$. This finite candidate set also proves the stopping-time property directly: determining whether a crossing has occurred by time $t$ only uses the arrival history up to $t$. The boundary set always contains time one because $X_1=S$, so $T\le 1$.

For a deterministic $r<1$, the martingale is bounded on $[0,r]$ by $S/(1-r)$. The optional stopping theorem for a bounded stopping time therefore givesMoreover $0\le M_{T\wedge r}\le 1$: if $T\le r$ its value is one, and otherwise no crossing has yet occurred. As $r\uparrow 1$, the stopped value tends to one on $\{T<1\}$ and to zero on $\{T=1\}$, because after the last arrival the unstopped process is zero. Dominated convergence now givesThis is the finite-jump boundary crossing identity. It uses uniform integrability of the bounded stopped values, not uniform integrability of the original family, which fails.