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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?

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    $\begingroup$ The sum $\sum_{s<\cdot} \Delta_+X_s$ is left continuous. $\endgroup$ Jan 13 at 19:32
  • $\begingroup$ @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? $\endgroup$
    – hannah
    Jan 14 at 14:53
  • $\begingroup$ @hannah better than a reference is trying to prove it ourselves. See my edit of the answer. $\endgroup$
    – Kurt G.
    Jan 14 at 15:07
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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.

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  • $\begingroup$ 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). $\endgroup$
    – hannah
    Jan 14 at 15:37
  • $\begingroup$ @hannah indeed ! Thanks for pointing that out. $\endgroup$
    – Kurt G.
    Jan 14 at 16:26

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