Decomposition of a stochastic process into a continuous and pure jump part

We work on a filtered probability space $$(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\in [0,T]},P)$$. Let $$X$$ be an adapted làdlàg stochastic process (i.e. the left and right limits exist). For $$0 denote by $$\Delta_+ X_t := X_{t+}-X_t$$ its right and by $$\Delta X_t := X_t -X_{t-}$$ its left jumps.

In this paper on page 1894 it is claimed that the continuous part of $$X$$ is given by $$\begin{equation} X_t^c:=X_t -\sum_{s However, since $$\sum_{s<\cdot}\Delta_+ X_s$$ is right-continuous, shouldn't we take the left limit of this sum, i.e. $$\begin{equation} X_t^c:=X_t -\left(\sum_{s so that $$X^c$$ is continuous?

• The sum $\sum_{s<\cdot} \Delta_+X_s$ is left continuous. Jan 13 at 19:32
• @JohnDawkins Thank you. The fact that the sum $\sum_{s<\cdot}\Delta_+ X_s$ is left-continuous explains that both formulations in my question are actually correct. Do you have a reference for a proof of the left-continuity? Jan 14 at 14:53
• @hannah better than a reference is trying to prove it ourselves. See my edit of the answer. Jan 14 at 15:07

The simplest example $$\tag{1} X_t=a1_{\{0\}}(t)+1_{(0,+\infty)}(t)$$ of a làdlàg path should show what $$X_t^c$$ should be. Obviously, \begin{align} X_{t-}&=1_{(0,+\infty)}(t), \\[3mm] X_{t+}&=1_{[0,+\infty)}(t),\\[3mm] \Delta_+X_t&=X_{t+}-X_t=(1-a)1_{\{0\}}(t),\\[3mm] \Delta_-X_t&=X_t-X_{t-}=a1_{\{0\}}(t)\,. \end{align} Further, considering when the single jump occurs it is easy to see that \begin{align} \sum_{s As commented by John Dawkins, (2) is left continuous. Now we see that \begin{align} X^c_t&=X_t-(1-a)1_{(0,+\infty)}(t)-a1_{[0,+\infty)}(t)\\ &=\Big(a1_{\{0\}}+1_{(0,+\infty)}-(1-a)1_{(0,+\infty)}-a1_{[0,+\infty)}\Big)(t)\\ &\equiv0\, \end{align} which is continuous.
The proof that (2) is always left continuous goes as follows: Let $$(t_n)_{n\in\mathbb N}$$ be a sequence with $$t_n Clearly,
$$\bigcup_{t_n Therefore, $$\sum_{s because every $$s is in all intervals $$(-\infty,t_n)$$ provided that $$n$$ is large enough.
• Thank you for the proof. One comment: The notation "$\uparrow$" suggests that the sequence on the left is increasing. However, I don't think this is true in general for $\sum_{s<t_n}\Delta_+ X_s$ ($X$ can have negative right jumps). Jan 14 at 15:37