following corollary to Theorem \[theorem:reflection\_and\_absorbing\].

\[cor:reflection\_and\_absorbing\] Let $(X,\mathbb{F})$ be a solution of Eq. (\[eqn:SDE\_mvd\]) with right (resp., left, symmetric) local time and arbitrary initial condition $X_0$. Then the following statements are satisfied:

\(i) If $x \in \mathbb{R}$ is such that $\nu(\{x\}) \geq 1/2$ (resp., $\nu(\{x\}) \leq -1/2$, $|\nu(\{x\})| \geq 1$), then on $\{\tau_x < +\infty\}$ it holds $$\begin{split}
                                                        &X_{\tau_x + t} \geq x, \ t \geq 0, \ \mathbf{P}\text{-a.s}\phantom{\Bigr)} \\
                              \Bigl(\text{resp., } &X_{\tau_x + t} \leq x,  \ t \geq 0, \ \mathbf{P}\text{-a.s},\\
                                                                 &X_{\tau_x + t} \leq x,  \ t \geq 0,\ \mathbf{P}\text{-a.s} \text{ if } \nu(\{x\}) \leq -1
                                                    \text{ and } X_{\tau_x + t} \geq x, \ t \geq 0, \ \mathbf{P}\text{-a.s} \text{ if } \nu(\{x\}) \geq 1 \Bigr).
                 \end{split}$$ (ii) If $x \in \mathbb{R}$ is such that $\nu(\{x\}) > 1/2$ (resp., $\nu(\{x\}) < -1/2$, $|\nu(\{x\})| > 1$), then it holds $$X_{\tau_x + t} = x, \qquad t \geq 0, \ \mathbf{P}\text{-a.s. on } \{\tau_x < +\infty\}.$$ In particular, if $\mathbf{P}(\{\tau_x < +\infty\})>0$, then $b(x) = 0$ must be fulfilled.

We only give the idea of the proof. The details are left to the reader. If $\mathbf{P}(\{\tau_x < +\infty\}) > 0$ is satisfied, then we use the trace of 