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

Question

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 $0<t<T$ 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<t}\Delta_+ X_s -\sum_{s\leq t}\Delta X_s. \end{equation} 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<t}\Delta_+ X_s\right)_- -\sum_{s\leq t}\Delta X_s \end{equation} so that $X^c$ is continuous?

Answer 1

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<t}\Delta_+X_s&=(1-a)1_{(0,+\infty)}(t)\,,\tag{2}\\ \sum_{s\le t}\Delta_-X_s&=a1_{[0,+\infty)}(t)\,.\tag{3} \end{align} 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 $t_n<t\,,t_n\uparrow t\,.$ Clearly,

$$ \bigcup_{\scriptstyle t_n<t\atop\scriptstyle t_n\uparrow t}(-\infty,t_n)=(-\infty,t)\,. $$ Therefore, $$ \sum_{s<t_n}\Delta_+X_s\quad\to\quad \sum_{s<t}\Delta_+X_s $$ because every $s<t$ is in all intervals $(-\infty,t_n)$ provided that $n$ is large enough.