A cadlag predictable finite variation process is of locally integrable variation?

265 Views Asked by At

I am trying to understand a line in the proof of Theorem 32 from Chapter 3 of Protter's Stochastic Integration and Differential Equations. Here, given a semimartingale $X$, if $X$ has a decomposition $X_t = X_0 + M_t + A_t$ with $M_0=A_0=0, M$ a local martingale, $A$ an FV process (process with paths of finite variation on compacts) and with $A$ predictable, then $X$ is said to be a special semimartingale.

In the proof, it says that since $X$ is special $A$ of its canonical decomposition is of locally integrable variation. However, I cannot see why this is the case. How can we find a reducing sequence $T_n$ of stopping times such that $E \int_0^{T_n} |dA_s| < \infty$ for each $n$?

enter image description here

enter image description here

enter image description here

1

There are 1 best solutions below

0
On

The key is to use that all càdlàg predictable processes are locally bounded.

As $(A_t)_{t\geq0}$ is predictable, the process $(|A|_t)_{t\geq0}$ is predictable as well. Thus $T_n:=\inf\{t\geq0: |A|_t\geq n \}$ is a predictable stopping time and for each $n$ there exists a sequece of stopping times $(T_n^m)_{m\in\Bbb N}$ with $T_n^m<T_n$. Now take the sequence of stopping times $\sigma_k:=\max(T_1^k,T_2^k,...,T_k^k)$. This is the one which makes the process $(|A|_t)_{t\geq0}$ locally bounded. You can read the full proof for this here. And as $(|A|_t)_{t\geq0}$ is locally bounded it is also locally integrable.