= 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 set
$$
\left\{1-S+\sum_{i\in I}\theta_i:I\subseteq\{1,\ldots,n\}\right\}.
$$
A uniform arrival hits none of these times almost surely. Thus \b[$M_T=M_{T-}=1$ 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 $T\leq1$.
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 gives
$$
\mathbb E M_{T\wedge r}=S.
$$
Moreover $0\leq M_{T\wedge r}\leq1$: if $T\leq r$ its value is one, and otherwise no crossing has yet occurred. As $r\uparrow1$, 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 gives
$$
S=\mathbb P(T<1),\qquad\boxed{\mathbb P(1-S+X_t>t\text{ for every }t<1)=\mathbb P(T=1)=1-S.}
$$
This 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.
Back to article page
