# Predictability of jumping times of increasing cad lag processes

The following is a remark that appears just at the beginning of the proof of Proposition 16.23 of the book Stochastic Processes of Richard Bass at page 121. I have not been able to prove yet that remark.

The statement If $A_{t}$ is an increasing cad-lag process, and we denote $U_{mi}$ the ith time $\Delta A_{t} \in (2^{-m}, 2^{-m+1}]$, then $U_{im}$ are predictable stopping times.

Comment and doubt The author says that this comes from the exercise 16.5 at page 128 that claims:

"If $A$ is a predictable increasing processes, and let $S_{k}$ the $k$th time $A$ jumps more than $\epsilon > 0$, then $S_{k}$ is a predictable stopping time for each $k$"

For me it looks like that kind of hint doesn't work, especially when the author uses exercises where the hypothesis are what the reader wants to prove about the proposition 16.23.

Any hint, suggestion or reference about how to prove this will be welcomed

Definition: A stopping time $T$ is predictable if there exists a sequence of stopping times $T_{n}$ such that for all $w$:

1) $T_{1}(w) \leq T_{2}(w) \leq \cdots$

2) $\lim_{n \to \infty} T_{n}(w) = T(w)$

3) If $T(w) > 0$, then $T_{n}(w) < T(w)$ for each $n$

Definition: $\Delta A_{t} := A_{t} - A_{t-}$

Proposition 16.23: If $A_{t}$ is a cad-lag increasing process such that

1) $\Delta A_{T} = 0$ whenever $T$ is a totally inaccessible stopping time, and

2) $\Delta A_{T} = 0$ is $\mathcal{F}_{T-}$ measurable whenever $T$ is a predictable stopping time.

Then $A$ is predictable.

• Could you please specify what predictable means in this setting? Also, part of the sentence "If A is a ..." seems to be missing. – Sayantan Apr 22 '18 at 5:00
• Sayantan thanks for your feedback – Ivan Apr 22 '18 at 14:08