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

Write the residual process asFor $s<t<1$, the filtration tells us whether each $U_i$ has arrived by time $s$, and its exact arrival time if it has. On $\{U_i>s\}$, its conditional law is uniform on $(s,1)$. Independence of the original $U_i$ means that the histories of the other arrivals do not alter this law. HenceThis conditional identity can also be checked against products of bounded functions of the censored arrival times and extended to ${\mathcal{F}}_s$ by the Monotone class theorem. Summing the identity gives $\mathbb{E}[M_t\mid {\mathcal{F}}_s]=M_s$.

Each $M_t$ is adapted and integrable, withIts paths are smooth between the finitely many arrival times and have downward jumps of size ${\theta }_i/(1-U_i)$ at arrivals before time one. The indicators use $U_i>t$, so the value at a jump is the post-jump value: paths are right continuous with finite left limits on $[0,1)$. Thus $M$ is a càdlàg martingale, specifically the uniform-arrival residual mass martingale.

Since there are finitely many arrival times and each is strictly below one almost surely, $U_{\ast }:={\max }_i{U}_i<1$ almost surely. For every $t\ge U_{\ast }$ with $t<1$, all weights have arrived and $M_t=0$. Therefore $M_t\to 0$ almost surely as $t\uparrow 1$.

Nevertheless $\mathbb{E}{M}_t=S>0$ for every $t<1$. If the family were uniformly integrable, its almost-sure convergence along any deterministic sequence tending to one would imply convergence in $L^1$, and hence convergence of expectations to zero. This is impossible. Thus the martingale is not uniformly integrable. Its constant expected mass is carried by increasingly rare, late arrivals.

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.