Let $A_t$ be a predictable càdlàg process, and let $a > 0$. Define $$ T = \inf \{t \geq 0 : \Delta A_t > a \} \ .$$ I wish to show that $T$ is a predictable stopping time, so we must show that there exists a sequence of stopping times $T_n \nearrow T$ and $T_n < T$ when $T > 0$. My guess is that the sequence is given by $$T_n = \inf \left\{ t \geq 0 : \Delta A_t > a - \frac{1}{n} \right\} \ .$$ To show that this sequence is increasing, let $\varepsilon > 0$. By the definition of the infimum, we know that there exists $t < T_{n+1} + \varepsilon$, such that $t$ satisfies $$ \Delta A_t > a - \frac{1}{n+1} \ .$$ We have $$ \Delta A_t > a - \frac{1}{n+1} > a - \frac{1}{n} \ ,$$ so $t \geq T_n$, and thus we have the inequality $T_n \leq T_{n+1} + \varepsilon$. Since this holds for all $\varepsilon > 0$, we have $T_n \leq T_{n+1}$. One can also show, in the same way, that $T_n \leq T$ for all $n$. I would like to show that $$ \lim_{n \to \infty} T_n = T,$$ however I am having problems since $A_t$ is only càdlàg. In particular, we have an increasing sequence $T_n$, yet $A_t$ is right continuous, so I am unable to use the right continuity. Any hints?

  • $\begingroup$ you should have a look at George Lowther 's blog he might have results that could help you : almostsure.wordpress.com $\endgroup$
    – TheBridge
    Feb 22, 2017 at 10:48
  • $\begingroup$ I think that $T_n$ does, in general, not converge to $T$. Just consider for instance a Poisson process and $a=1$, then $T=\infty$ but $T_n = \tau_1:= \inf\{t; \Delta A_t = 1\}$ for all $n$. $\endgroup$
    – saz
    Feb 22, 2017 at 16:24

1 Answer 1


$T=\inf\{t\ge 0:\Delta A_t>a\}$ is a predictable stopping time: Since $A$(and $\Delta A$) is a predictable process, $B\stackrel{\text{def}}=\{(t,\omega)\in \mathbb{R}_+\times\Omega:\Delta A_t(\omega)>a\}$ is a predictable set, denote it by $B\in\mathscr{P}$. Since $A$ is càdlàg and $a>0$, there are at most finite numbers of $t$, for which $\Delta A_t>a$, in every finite interval $[0,N]$, this means that $\Delta A_T>a$ on $\{T<\infty\}$ and $$ [\![T]\!](=\{(T(\omega),\omega)\in \mathbb{R}_+\times\Omega\})\subset B.$$ Hence $[\![T]\!]=[\![0,T]\!]\cap B\in \mathscr{P}$, since both $[\![0,T]\!]$ and $B$ are predictable sets. Therefore $T$ is a predictable stoping time.


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