**Lemma 3.**


**Proof.** 1. By Equation (5) together with Equation (26), we note that for *x* ≤ *c*Λ2,

$$\begin{split} \mathbf{z}\_{\delta,\Lambda}^{\prime}(\mathbf{x}) &= \Lambda \left( Z^{(q)}(\mathbf{x}) - \frac{W^{(q)}(\mathbf{x})}{W^{(q)}(c\_{2}^{\Lambda})} Z^{(q)}(c\_{2}^{\Lambda}) \right) + \frac{W^{(q)}(\mathbf{x})}{W^{(q)}(c\_{2}^{\Lambda})} \\ &= \Lambda \mathbb{E}\_{\mathbf{x}} \left[ \mathbf{e}^{-q\tau\_{0}^{-}} \mathbf{1} \left\{ \tau\_{0}^{-} < \tau\_{c\_{2}^{\Lambda}}^{+} \right\} \right] + \mathbb{E}\_{\mathbf{x}} \left[ \mathbf{e}^{-q\tau\_{c\_{2}^{\Lambda}}^{+}} \mathbf{1} \left\{ \tau\_{c\_{2}^{\Lambda}}^{+} < \tau\_{0}^{-} \right\} \right] \\ &\leq \Lambda \left( \mathbb{E}\_{\mathbf{x}} \left[ \mathbf{e}^{-q\tau\_{0}^{-}} \mathbf{1} \left\{ \tau\_{0}^{-} < \tau\_{c\_{2}^{\Lambda}}^{+} \right\} \right] + \mathbb{E}\_{\mathbf{x}} \left[ \mathbf{e}^{-q\tau\_{c\_{2}^{\Lambda}}^{+} \mathbf{1} \left\{ \tau\_{c\_{2}^{\Lambda}}^{+} < \tau\_{0}^{-} \right\} \right] \right) \\ &\leq \Lambda \left( \mathbb{P}\_{x} \left[ \tau\_{0}^{-} < \tau\_{c\_{2}^{\Lambda}}^{+} \right] + \mathbb{P}\_{x} \left[ \tau\_{c\_{2}^{\Lambda}}^{+} < \tau\_{0}^{-} \right] \right) = \Lambda. \end{split}$$

On the other hand, *<sup>v</sup>δ*,<sup>Λ</sup>(*x*) = 1 ≤ Λ for *x* > *c*Λ2 2. Let us consider *c*Λ2 ≥ *x* ≥ *y*. We note that

$$\begin{aligned} v\_{\delta,\Lambda}(\mathbf{x}) - v\_{\delta,\Lambda}(\mathbf{y}) &= \Lambda \left( \overline{Z}^{(q)}(\mathbf{x}) - \overline{Z}^{(q)}(\mathbf{y}) \right) + \left( Z^{(q)}(\mathbf{x}) - Z^{(q)}(\mathbf{y}) \right) \mathbb{J}\_{\Lambda}(c\_2^{\Lambda}) \\ &= \Lambda \left( \overline{Z}^{(q)}(\mathbf{x}) - \overline{Z}^{(q)}(\mathbf{y}) \right) + \left( Z^{(q)}(\mathbf{x}) - Z^{(q)}(\mathbf{y}) \right) \mathbb{G}\_{\Lambda}(c\_1^{\Lambda}, c\_2^{\Lambda}) \end{aligned}$$

.

$$\begin{aligned} &\geq \Lambda \left( \overline{Z}^{(q)}(\mathbf{x}) - \overline{Z}^{(q)}(\mathbf{y}) \right) + (Z^{(q)}(\mathbf{x}) - Z^{(q)}(\mathbf{y})) G\_{\Lambda}(\mathbf{x}, \mathbf{y}) \\ &= \Lambda \left( \overline{Z}^{(q)}(\mathbf{x}) - \overline{Z}^{(q)}(\mathbf{y}) \right) + (Z^{(q)}(\mathbf{x}) - Z^{(q)}(\mathbf{y})) \frac{\mathbf{x} - \mathbf{y} - \delta - \Lambda \left( \overline{Z}^{(q)}(\mathbf{x}) - \overline{Z}^{(q)}(\mathbf{y}) \right)}{Z^{(q)}(\mathbf{x}) - Z^{(q)}(\mathbf{y})} \\ &= \mathbf{x} - \mathbf{y} - \delta. \end{aligned} \tag{27}$$

Now, suppose that *x* ≥ *y* ≥ *c*Λ2, then using Equation (26) we obtain

$$v\_{\delta,\Lambda}(\mathbf{x}) - v\_{\delta,\Lambda}(y) = \mathbf{x} - y \ge \mathbf{x} - y - \delta.$$

Finally, for the case *x* ≥ *c*Λ2 ≥ *y*, by Equation (27), we have

$$v\_{\delta,\Lambda}(\mathbf{x}) - v\_{\delta,\Lambda}(y) = \mathbf{x} - c\_2^{\Lambda} + v\_{\delta,\Lambda}(c\_2^{\Lambda}) - v\_{\delta,\Lambda}(y) \ge \mathbf{x} - c\_2^{\Lambda} + (c\_2^{\Lambda} - y - \delta) = \mathbf{x} - y - \delta. \quad \Box$$

Now, we proceed to the verification theorem that proves the optimality of the (*c*Λ1 , *c*Λ2 )-policy.

**Theorem 1** (Verification Theorem)**.** *Let <sup>V</sup>δ*,Λ*, <sup>v</sup>δ*,<sup>Λ</sup> *be as in Equations* (3) *and* (26)*, respectively. Then, <sup>v</sup>δ*,<sup>Λ</sup>(*x*) = *<sup>V</sup>δ*,<sup>Λ</sup>(*x*) *for all x* ≥ 0*. Hence, the* (*c*Λ1 , *c*Λ2 )*-policy is optimal.*

**Proof.** By the definition of *<sup>V</sup>δ*,Λ, *<sup>v</sup>δ*,<sup>Λ</sup>(*x*) ≤ *<sup>V</sup>δ*,<sup>Λ</sup>(*x*) for all *x* ≥ 0. Let us verify that *<sup>v</sup>δ*,<sup>Λ</sup>(*x*) ≥ *<sup>v</sup>πδ*,<sup>Λ</sup>(*x*) for all admissible *π* ∈ Θ*δ* and for all *x* ≥ 0. Recall that *<sup>v</sup>πδ*,<sup>Λ</sup> is defined in Equation (2). Take *π* = {*L<sup>π</sup>*, *Rπ*} ∈ Θ*δ* fixed and let (*Tn*)*n*∈<sup>N</sup> be the sequence of stopping times where *Tn* := inf{*t* > 0 : *<sup>X</sup><sup>π</sup>t* > *<sup>n</sup>*}. Since *Xπ* = *X* − *Lπ* + *R<sup>π</sup>*, with *X* being a spectrally negative Lévy process, it is a semi-martingale and *<sup>v</sup>δ*,<sup>Λ</sup> is sufficiently smooth on (0, ∞) by Lemma 2, and continuous (respectively, continuously differentiable) at zero for the case of bounded variation (respectively, unbounded variation) by Remark 10, we can use the change of variables/Meyer-Itô's formula (cf. Theorems II.31 and II.32 of Protter (2005)) on the stopped process (e<sup>−</sup>*q*(*<sup>t</sup>*∧*Tn*) *<sup>v</sup>δ*,<sup>Λ</sup>(*X<sup>π</sup>t*∧*Tn* ); *t* ≥ 0) to deduce under P*x* that

$$\begin{split} \mathbf{e}^{-q(t \wedge T\_{\pi})} \, v\_{\delta, \Lambda}(\mathbf{X}\_{t \wedge T\_{\pi}}^{\pi}) - v\_{\delta, \Lambda}(\mathbf{x}) &= \int\_{0}^{t \wedge T\_{\pi}} \mathbf{e}^{-qs}(\mathcal{L} - q) v\_{\delta, \Lambda}(\mathbf{X}\_{s-}^{\pi}) \mathrm{d}s + M\_{t \wedge T\_{\pi}} + l\_{t \wedge T\_{\pi}} \\ &+ \int\_{[0, t \wedge T\_{\pi}]} \mathbf{e}^{-qs} v\_{\delta, \Lambda}'(\mathbf{X}\_{s-}^{\pi}) \mathrm{d}R\_{s}^{\pi, c} .\end{split} \tag{28}$$

where *M* is a local martingale with *M*0 = 0, *R<sup>π</sup>*,*<sup>c</sup>* is the continuous part of *R<sup>π</sup>*, and *J* is a jump process, which is given by

$$\begin{split} \mathcal{J}\_{t} &= \sum\_{0 \leq s \leq t} \mathsf{e}^{-\mathsf{qs}} \left( \upsilon\_{\delta,\Lambda} (X\_{\mathsf{s}-}^{\pi} + \Delta[X + \mathsf{R}^{\pi}]\_{\mathfrak{s}}) - \upsilon\_{\delta,\Lambda} (X\_{\mathsf{s}-}^{\pi} + \Delta X\_{\mathsf{s}}) \right) \mathbf{1}\_{\{\Delta[X + \mathsf{R}^{\pi}]\_{\mathfrak{s}} \neq \mathbf{0}\}} \\ &+ \sum\_{0 \leq s \leq t} \mathsf{e}^{-\mathsf{qs}} \left( \upsilon\_{\delta,\Lambda} (X\_{\mathsf{s}-}^{\pi} + \Delta[X + \mathsf{R}^{\pi}]\_{\mathfrak{s}} - \Delta L\_{\mathfrak{s}}^{\pi}) - \upsilon\_{\delta,\Lambda} (X\_{\mathsf{s}-}^{\pi} + \Delta[X + \mathsf{R}^{\pi}]\_{\mathfrak{s}}) \right) \mathbf{1}\_{\{\Delta L\_{\mathfrak{s}}^{\pi} \neq \mathbf{0}\}}, \text{ for } t \geq 0. \end{split}$$

On the other hand, by Part (1) of Lemma 3, we obtain that

$$\begin{split} &\int\_{[0,t\wedge T\_{n}]} \mathbf{e}^{-qs} \boldsymbol{v}\_{\delta,\Lambda}^{\prime}(\mathbf{X}\_{s-}^{\pi}) \mathrm{d}\mathcal{R}\_{s}^{\pi,c} \\ &+ \sum\_{0 \le s \le t\wedge T\_{n}} \mathbf{e}^{-qs} \left[ \boldsymbol{v}\_{\delta,\Lambda}(\boldsymbol{X}\_{s-}^{\pi} + \Delta[\boldsymbol{X} + \boldsymbol{R}^{\pi}]\_{s}) - \boldsymbol{v}\_{\delta,\Lambda}(\boldsymbol{X}\_{s-}^{\pi} + \Delta\boldsymbol{X}\_{s}) \right] \boldsymbol{1}\_{\{\Delta[\boldsymbol{X} + \boldsymbol{R}^{\pi}]\_{s} \neq 0\}} \\ &\leq \Lambda \int\_{[0,t\wedge T\_{n}]} \mathbf{e}^{-qs} \operatorname{\mathbf{d}\mathcal{R}}\_{s}^{\pi,c} + \Lambda \sum\_{0 \le s \le t\wedge T\_{n}} \mathbf{e}^{-qs} \boldsymbol{\Delta}\mathcal{R}\_{s}^{\pi} = \Lambda \int\_{[0,t\wedge T\_{n}]} \mathbf{e}^{-qs} \operatorname{\mathbf{d}\mathcal{R}}\_{s}^{\pi}. \end{split}$$

Similarly, by Part (2) of Lemma 3,

$$\begin{split} \sum\_{\boldsymbol{0} \leq \boldsymbol{s} \leq \boldsymbol{t} \wedge \boldsymbol{T}\_{\boldsymbol{n}}} \mathbf{e}^{-qs} [\upsilon\_{\delta,\Lambda} (\mathbf{X}\_{\boldsymbol{s}-}^{\pi} + \Delta[\mathbf{X} + \mathbf{R}^{\pi}]\_{\boldsymbol{s}} - \Delta L\_{\boldsymbol{s}}^{\pi}) - \upsilon\_{\delta,\Lambda} (\mathbf{X}\_{\boldsymbol{s}-}^{\pi} + \Delta[\mathbf{X} + \mathbf{R}^{\pi}]\_{\boldsymbol{s}})] \mathbf{1}\_{\{\Delta L\_{\boldsymbol{u}}^{\pi} \neq \boldsymbol{0}\}} \\ \leq - \sum\_{\boldsymbol{0} \leq \boldsymbol{s} \leq \boldsymbol{t} \wedge \boldsymbol{T}\_{\boldsymbol{n}}} \mathbf{e}^{-qs} \Delta L\_{\boldsymbol{s}}^{\pi} + \delta \sum\_{\boldsymbol{0} \leq \boldsymbol{s} \leq \boldsymbol{t} \wedge \boldsymbol{T}\_{\boldsymbol{n}}} \mathbf{e}^{-qs} \mathbf{1}\_{\{\Delta L\_{\boldsymbol{u}}^{\pi} > \boldsymbol{0}\}} \\ = - \int\_{[0,t \wedge T\_{\boldsymbol{n}}]} \mathbf{e}^{-qs} \cdot \mathbf{d} \left( L\_{\boldsymbol{s}}^{\pi} - \delta \sum\_{0 \leq \boldsymbol{u} \leq \boldsymbol{s}} \mathbf{1}\_{\{\Delta L\_{\boldsymbol{u}}^{\pi} > \boldsymbol{0}\}} \right). \end{split}$$

Hence, from Equation (28), we derive that

$$\begin{split} \boldsymbol{\upsilon}\_{\delta,\Lambda}(\mathbf{x}) &\geq -\int\_{0}^{t\wedge T\_{n}} \mathbf{e}^{-q\boldsymbol{s}}(\mathcal{L}-q)\boldsymbol{\upsilon}\_{\delta,\Lambda}(X\_{s-}^{\pi})\mathrm{d}s-\Lambda\int\_{[0,t\wedge T\_{n}]} \mathbf{e}^{-q\boldsymbol{s}}\,\mathrm{d}R\_{\mathbf{s}}^{\pi} \\ &+\int\_{[0,t\wedge T\_{n}]} \mathbf{e}^{-q\boldsymbol{s}}\,\mathrm{d}\left(L\_{\mathbf{s}}^{\pi}-\delta\sum\_{0\leq\boldsymbol{u}\leq\boldsymbol{s}}\mathbf{1}\_{\{\Delta L\_{\mathbf{u}}^{\pi}>0\}}\right)-M\_{\mathbb{t}\wedge T\_{n}}+\mathbf{e}^{-q(\boldsymbol{t}\wedge T\_{n})}\,\boldsymbol{\upsilon}\_{\delta,\Lambda}(X\_{t\wedge T\_{n}}^{\pi}). \end{split}$$

Using Proposition 3 along with Point 3 in the proof of Lemma 6 in Loeffen (2009), and that *<sup>X</sup><sup>π</sup>s*− ≥ 0 a.s. for *s* ≥ 0, we observe that

$$\begin{split} \upsilon\_{\delta,\Lambda}(\mathbf{x}) &\geq \int\_{[0,t\wedge T\_{\mathbf{z}}]} \mathbf{e}^{-qs} \, \mathbf{d} \left( L\_{\mathbf{z}}^{\pi} - \delta \sum\_{0 \leq u \leq s} \mathbf{1}\_{\{\Delta\Gamma\_{u}^{\pi} > 0\}} \right) - \Lambda \int\_{[0,t\wedge T\_{\mathbf{z}}]} \mathrm{d} \mathbf{R}\_{s}^{\pi} - M\_{l\wedge T\_{\mathbf{z}}} + \mathbf{e}^{-q(t\wedge T\_{\mathbf{z}})} \upsilon\_{\delta,\Lambda}(\mathbf{X}\_{l\wedge T\_{\mathbf{z}}}^{\pi}) \\ &\geq \int\_{[0,t\wedge T\_{\mathbf{z}}]} \mathbf{e}^{-qs} \, \mathbf{d} \left( L\_{\mathbf{z}}^{\pi} - \delta \sum\_{0 \leq u \leq s} \mathbf{1}\_{\{\Delta\Gamma\_{u}^{\pi} > 0\}} \right) - \Lambda \int\_{[0,t\wedge T\_{\mathbf{z}}]} \mathrm{d} \mathbf{R}\_{s}^{\pi} - M\_{l\wedge T\_{\mathbf{z}}} \\ &+ \mathbf{e}^{-q(t\wedge T\_{\mathbf{z}})} \left( \frac{\Lambda \mathsf{y}^{\ell}(0+)}{q} + Z^{(q)}(c\_{2}^{\Lambda}) \left( \mathsf{f}(c\_{2}^{\Lambda}) \right) \right), \end{split} \tag{29}$$

where the last inequality follows from Remark 9. In addition, by the compensation formula (see, e.g., Corollary 4.6 of Kyprianou (2014)), (*Mt*∧*Tn* : *t* ≥ 0) is a zero-mean P*x*-martingale. Now, taking expected value in Equation (29) and letting (*t* ∧ *Tn*) ∞ P*x*-a.s., the monotone convergence theorem, applied 

$$\text{isoperately for } \mathbb{E}\_{\mathbf{x}} \left[ \int\_{[0,t \wedge T\_{\mathbf{n}}]} \mathbf{e}^{-q\boldsymbol{\kappa}} \mathbf{d} \left( L\_{\mathbf{s}}^{\pi} - \delta \sum\_{0 \le u \le s} \mathbf{1}\_{\{\Delta L\_{\mathbf{s}}^{\pi} > 0\}} \right) \right] \text{ and } \mathbb{E}\_{\mathbf{x}} \left( \Lambda \int\_{[0,t \wedge T\_{\mathbf{n}}]} \mathbf{e}^{-q\boldsymbol{\kappa}} \mathbf{d} R\_{\mathbf{s}}^{\pi} \right) \text{, gives} $$
 
$$\boldsymbol{v}\_{\delta,\Lambda}(\mathbf{x}) \ge \mathbb{E}\_{\mathbf{x}} \left( \int\_{[0,\infty)} \mathbf{e}^{-q\boldsymbol{\kappa}} \mathbf{d} \left( L\_{\mathbf{s}}^{\pi} - \delta \sum\_{0 \le u \le s} \mathbf{1}\_{\{\Delta L\_{\mathbf{u}}^{\pi} > 0\}} \right) - \Lambda \int\_{[0,\infty)} \mathbf{e}^{-q\boldsymbol{\kappa}} \mathbf{d} R\_{\mathbf{s}}^{\pi} \right) = \boldsymbol{v}\_{\delta,\Lambda}^{\pi}(\mathbf{x}).$$

This completes the proof.

### **5. Optimal Dividends with Capital Injection Constraint**

In this section, we are interested in maximizing the expected NPV of the dividend strategy subject to a constraint in the expected present value of the injected capital. Specifically, we aim to solve

$$V\_{\delta}(\mathbf{x}, \mathcal{K}) := \sup\_{\pi \in \Theta\_{\delta}} \mathbb{E}\_{\mathbf{x}} \left[ \int\_{0}^{\infty} \mathbf{e}^{-qt} \, \mathbf{d} \left( L\_{t}^{\pi} - \delta \sum\_{0 \le s < t} \mathbf{1}\_{\{\Delta L\_{t}^{\pi} > 0\}} \right) \right] \quad \text{s.t.} \,\mathbb{E}\_{\mathbf{x}} \left[ \int\_{0}^{\infty} \mathbf{e}^{-qt} \, \mathbf{d} \mathcal{R}\_{t}^{\pi} \right] \le K,\tag{30}$$

for any *x* ≥ 0 and *K* ≥ 0. Strategies *π* that do not satisfy the capital injection constraint are called infeasible. Recall that the insurance company has to inject capital to ensure the non-negativity of the risk process. Therefore, small values of *K* require very low dividend payments to keep the risk process non-negative, or would even make the problem infeasible. In the latter case, we define the value function as <sup>−</sup>∞.

To solve this problem, we use the solution of the optimal dividend problem with capital injection found in the section above. Thus, for Λ ≥ 0, we define the function

$$v^{\pi}\_{\delta,\Lambda}(\mathfrak{x},\mathsf{K}) := v^{\pi}\_{\delta,\Lambda}(\mathfrak{x}) + \Lambda \mathsf{K}\_{\mathsf{A}}$$

with *<sup>v</sup>πδ*,<sup>Λ</sup> as in Equation (2). It is easy to check that *<sup>V</sup>δ*(*<sup>x</sup>*, *K*) = sup *<sup>π</sup>*∈Θ*δ* inf Λ≥0 *<sup>v</sup>πδ*,<sup>Λ</sup>(*<sup>x</sup>*, *K*) since for infeasible strategies inf Λ≥0 *<sup>v</sup>πδ*,<sup>Λ</sup>(*<sup>x</sup>*, *K*) = <sup>−</sup>∞. By interchanging the sup with the inf we obtain an upper bound for *<sup>V</sup>δ*(*<sup>x</sup>*, *<sup>K</sup>*), the so-called weak duality. Hence, the dual problem of Equation (30) is defined as

$$V^{D}\_{\delta}(\mathbf{x},K) := \inf\_{\Lambda \ge 0} \sup\_{\pi \in \Theta\_{\delta}} v^{\pi}\_{\delta,\Lambda}(\mathbf{x},K) = \inf\_{\Lambda \ge 0} \left\{ \Lambda K + \sup\_{\pi \in \Theta\_{\delta}} v^{\pi}\_{\delta,\Lambda}(\mathbf{x}) \right\} = \inf\_{\Lambda \ge 1} \left\{ \Lambda K + V\_{\delta,\Lambda}(\mathbf{x}) \right\}, \tag{31}$$

with *<sup>V</sup>δ*,<sup>Λ</sup> given in Equation (3). The last equality in Equation (31) is true, since *<sup>V</sup>δ*,<sup>Λ</sup>(*x*) is infinite for any Λ < 1; see Remark 2. The main goal is to prove that *<sup>V</sup>Dδ* (*<sup>x</sup>*, *K*) ≤ *<sup>V</sup>δ*(*<sup>x</sup>*, *<sup>K</sup>*).
