**2. More on the Value Function**

Please note that for *π* ∈ <sup>Π</sup>*p* and *x* < 0, using the strong Markov property and the spectral negativity of *X*, we can easily verify that

$$v\_{\pi}(\mathbf{x}) = \mathbb{E}\_{\mathbf{x}}\left[\mathbf{e}^{-q\tau\_{0}^{+}}\mathbf{1}\_{\{\tau\_{0}^{+} < \mathbf{e}\_{p}\}}\right]v\_{\pi}(0) = \mathbf{e}^{\Phi(p+q)\mathbf{x}}v\_{\pi}(0),\tag{1}$$

where *τ*+0 = inf {*t* > 0: *Xt* > 0} and where **<sup>e</sup>***p* is an independent exponentially distributed random variable with mean 1/*p*, thanks to the well-known fluctuation identity (see e.g., (Kyprianou 2014))

$$\mathbb{E}\_{\mathbf{x}}\left[\mathbf{e}^{-r\tau\_b^+}\mathbf{1}\_{\{\tau\_b^+ < \infty\}}\right] = \mathbf{e}^{-\Phi(r)(b-x)}, \; \mathbf{x} \le b,\tag{2}$$

where *τ*+*b*= inf {*t* > 0: *Xt* > *b*}.

Interestingly, we can show that (see the proof of Lemma 1 below), for any *π* ∈ <sup>Π</sup>*p*, we have

$$\upsilon\_{\pi}(\mathbf{x}) = \mathbb{E}\_{\mathbf{x}} \left[ \int\_{0}^{\infty} \mathbf{e}^{-q\mathbf{s} - p\int\_{0}^{\boldsymbol{s}} \mathbf{1}\_{\left(-\infty, 0\right)}(\mathcal{U}\_{r}^{\boldsymbol{\pi}}) d\mathbf{r}} \mathbf{d} \mathcal{L}\_{\mathbf{s}}^{\boldsymbol{\pi}} \right]. \tag{3}$$

Using this last identity, we can argue that using Parisian ruin with rate *p* fills the gap between the model with classical ruin (no delay, *p* → ∞) and the model with no ruin (infinite delays, *p* → 0). Indeed, using (3), we see directly that

$$w\_{\pi}(\mathbf{x}) = \mathbb{E}\_{\mathbf{x}} \left[ \int\_{0}^{\infty} \mathbf{e}^{-qs-p} \int\_{0}^{\imath} \mathbf{1}\_{(-\infty,0)}(\mathcal{U}\_{r}^{\pi}) \mathrm{d}\mathcal{U}\_{s}^{\pi} \right] \underset{p \to 0}{\longrightarrow} \mathbb{E}\_{\mathbf{x}} \left[ \int\_{0}^{\infty} \mathbf{e}^{-qs} \mathrm{d}\mathcal{U}\_{s}^{\pi} \right].$$

On the other hand, as

$$\int\_0^\infty \mathbf{e}^{-qs-p} \, {}^{\operatorname{\mathbf{d}}}\mathbf{1}\_{(-\infty,0)}(\mathrm{d}l\_r^{\pi}) \mathrm{d}\mathbf{r} \, \mathrm{d}L\_s^{\pi} = \int\_0^\infty \mathbf{e}^{-qs} \, \mathbf{1}\_{\{s \le \pi^{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\cdots}{\cdots}{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\cdots}{\cdots}{\cdots}{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\rm{\cdots}}}{\cdots}}}}}}}}}}}}}}}}}}}}}}} \}} $$

where *<sup>σ</sup>π*∞ := inf {*t* > 0: *<sup>U</sup><sup>π</sup>t* < <sup>0</sup>}, we obtain

$$\operatorname{tr}\_{\pi}(\mathbf{x}) = \mathbb{E}\_{\mathbf{x}} \left[ \int\_{0}^{\infty} \mathbf{e}^{-qs - p \int\_{0}^{s} \mathbf{1}\_{(-\infty, 0)}(\mathcal{U}^{\pi}\_{r}) \mathrm{d}\mathbf{r}} \mathrm{d}L\_{\mathbf{s}}^{\pi} \right] \underset{p \to \infty}{\longrightarrow} \mathbb{E}\_{\mathbf{x}} \left[ \int\_{0}^{\sigma^{\mathrm{pr}}\_{\infty}} \mathbf{e}^{-qs} \mathrm{d}L\_{\mathbf{s}}^{\pi} \right].$$

**Remark 2.** *Note also that the expression of the value function given in* (3) *tells us that the current control problem amounts to a control problem with no ruin and in which the dividend payments are penalized by the occupation time of the surplus process. Indeed, from this point of view, the discount factor increases with the time spent below zero by the surplus process.*
