**Appendix A. Proof of Proposition 1**

To prove this result, we can adapt the methodology used in the proof of Proposition 1 of (Renaud and Zhou 2007) (for the case *k* = 1 in that paper). Let us define *κp* as the time of Parisian ruin with rate *p* for *X* or, said differently, the time of Parisian ruin when the pay-no-dividend strategy, i.e., the strategy *π* with *<sup>L</sup><sup>π</sup>t* ≡ 0, is implemented. More precisely, define

$$\kappa^p = \inf\left\{ t > 0 \colon t - \lg > \mathbf{e}\_p^{\mathbb{R}\_t} \text{ and } X\_t < 0 \right\},$$

where *gt* = sup {0 ≤ *s* ≤ *t*: *Xs* ≥ <sup>0</sup>}. Let us also define, for *a* ∈ R, the stopping time

$$\mathfrak{r}\_a^+ = \inf\left\{ t > 0 \colon X\_t > a \right\} \dots$$

It is known that (see e.g., Equation (16) in (Lkabous and Renaud 2019)), for *x* ≤ *a*,

$$\mathbb{E}\_{\mathbf{x}}\left[\mathbf{e}^{-q\tau\_{\mathbf{z}}^{+}}\mathbf{1}\_{\{\tau\_{\mathbf{z}}^{+} < \mathbf{x}^{p}\}}\right] = \frac{Z\_{\boldsymbol{q}}(\mathbf{x}, \Phi(p+q))}{Z\_{\boldsymbol{q}}(\mathbf{a}, \Phi(p+q))}.\tag{A1}$$

As in Renaud and Zhou (2007), we can show that

$$\begin{cases} \boldsymbol{v}\_{b}(\boldsymbol{b}) + \frac{1}{n} \right) \mathbb{E}\_{\boldsymbol{b} - 1/n} \left[ \mathbf{e}^{-q\tau\_{b}^{+}} \mathbf{1}\_{\left\{ \tau\_{b}^{+} < \kappa^{p} \right\}} \right] \\ \quad + \boldsymbol{v}\_{b}(\boldsymbol{b}) \le \left( \boldsymbol{v}\_{b}(\boldsymbol{b}) + \frac{1}{n} \right) \mathbb{E}\_{\boldsymbol{b}} \left[ \mathbf{e}^{-q\tau\_{b+1/n}^{+}} \mathbf{1}\_{\left\{ \tau\_{b+1/n}^{+} < \kappa^{p} \right\}} \right] + o(1/n). \end{cases} \tag{A2}$$

The result for *x* = *b* follows by taking a limit and then the result for 0 ≤ *x* ≤ *b* follows by using again the identity in (A1). Finally, if *x* < 0, then using (1) we have

$$v\_b(\mathbf{x}) = \mathbf{e}^{\Phi(p+q)\mathbf{x}} \frac{Z\_{\eta}(0, \Phi(p+q))}{Z'\_{\eta}(b, \Phi(p+q))} = \frac{Z\_{\eta}(\mathbf{x}, \Phi(p+q))}{Z'\_{\eta}(b, \Phi(p+q))}.$$
