Graph of an accessible stopping time

Question

If we consider an accessible stopping time $\tau$, on an underlying filtered probability space, we know that, by definition, $\exists$ a sequence $(\tau_n)_n$ of predictable stopping times such that $$[\tau]\subseteq \bigcup_n [\tau_n],$$ where $[\tau]$ denotes the graph of $\tau$, i.e., $$[\tau]:=\{(\omega,t)\in\Omega\times\mathbb{R}_+:\, \tau(\omega)=t\}.$$ I would like to know if we can actually construct a sequence of predictable stopping times $(\tilde{\tau_n})_n$ (reasonably dealing with $(\tau_n)_n$) such that $$[\tau]= \bigcup_n [\tilde{\tau_n}].$$ I have been searching in many stochastic analysis books but I have not found an answer to my question. Every answer would be really appreciated!

Thanks,

Alessandro

Answer 1

The answer is No. Consider, for an example, a Brownian motion $(B_t)_{t\ge 0}$ on $[0,\infty)$ with $0$ an absorbing state. Let $\tau_1:=\inf\{t>0:B_t=0\}$. (And for completeness, take $\tau_n=+\infty$ for $n\ge 2$.) Now construct a process $X$ with values in $\{-1\}\cup[0,\infty)$ as follows. Independently of $B$, toss a fair coin. If the coin shows Heads then $X_t=B_t$ for all $t$. But if the coin comes up Tails, then $X_t=B_t$ for $0\le t<\tau_1$ but $X_t=-1$ for $t\ge\tau_1$. Finally, $\tau:=\inf\{t>0:X_t=0\}$, with the stipulation that $\tau=+\infty$ should $X$ never reach state $0$ (that is, should the coin show Tails). Then $\tau_1$ is predictable with respect to the filtration $(\mathcal F^X_t)_{t\ge 0}$ generated by $X$, $\tau$ is a stopping time with respect to this filtration, and $[\tau]\subset[\tau_1]$. Thus $\tau$ is accessible. But the predictable projection of $1_{[\tau]}$ is the process ${1\over 2}1_{[\tau_1]}\not=1_{[\tau]}$, implying that $\tau$ is not itself a predictable time.

Suppose there were a sequence $(\tilde\tau_n)$ with $[\tau]=\cup_n[\tilde\tau_n]$. Fix $n\ge 1$. Because $1_{[\tau]}\ge 1_{[\tilde\tau_n]}$, the predictable projections of these two processes are likewise ordered. That is, ${1\over 2}1_{[\tau_1]}\ge1_{[\tilde\tau_n]}$. This forces $1_{[\tilde\tau_n]}\equiv 0$. This yields $\cup_n[\tilde\tau_n]=\emptyset$, and a contradiction.