*2.2. Shreve, Lehoczky, and Gaver's Problem*

In Shreve, Lehoczky, and Gaver's problem, capital is injected as soon as it is necessary—that is, to keep the controlled process non-negative, so *τ* = ∞. In this case, the controlled process is never ruined—zero acts as a (lower) reflecting barrier. For this problem, the optimal value function will be denoted by *VSLG k*.

It is well-known that for this problem, double-barrier strategies play an important role. For *b* ≥ 0, a double-barrier strategy with an upper barrier at level *b* is such that 0 ≤ *U*0,*<sup>b</sup> t* = *Xt* − *Dbt* + *C*0*t* ≤ *b* for all *t* ≥ 0. As capital injections are now considered, the process *D<sup>b</sup>* here is different from the one in de Finetti's problem; see (Avram et al. 2007) for details. For such a strategy, the value function is such that *J*(*<sup>x</sup>*, *C*0, *D<sup>b</sup>*) = *V*0,*<sup>b</sup> k* (*x*), where

$$V\_k^{0,b}(\mathbf{x}) := \mathbb{E}\_{\mathbf{x}} \left[ \int\_0^\infty \mathbf{e}^{-qt} \left( \mathbf{d}D\_t^b - k \mathbf{d} \mathbf{C}\_t^0 \right) \right],$$

and the optimal value function is such that *VSLG k* (*x*) = sup*b*≥0 *V*0,*<sup>b</sup> k* (*x*).

It is well-known that for a SNLP,

$$V\_k^{0,b}(\mathbf{x}) = \begin{cases} k\left(\mathbb{Z}\_\emptyset(\mathbf{x}) + \frac{p}{q}\right) + Z\_\emptyset(\mathbf{x})H\_k^{SL.G}(b), & \mathbf{x} \le b, \\ \mathbf{x} - b + V\_k^{0,b}(b), & \mathbf{x} > b, \end{cases} \tag{9}$$

where

$$Z\_{\!\!\!\!-1}(\mathbf{x}) = 1 + q \int\_0^\mathbf{x} \mathcal{W}\_{\!\!\!\!-1}(\mathbf{y}) \, \mathbf{d}y \qquad \text{and} \qquad \mathbb{Z}\_{\!\!\!-1}(\mathbf{x}) = \int\_0^\mathbf{x} Z\_{\!\!\!\!-1}(\mathbf{y}) \, \mathbf{d}y,\tag{10}$$

and where, for *x* ≥ 0, the *barrier function* is defined by

$$H\_k^{SLG}(b) = \frac{1 - kZ\_q(b)}{q\mathcal{W}\_q(b)}, \ b > 0. \tag{11}$$

The next proposition—namely, Proposition 1—contains new results, as well as results taken from Lemma 2 of (Avram et al. 2007). In particular, we provide a new relationship (see (13)) between the value functions of de Finetti's and Shreve, Lehoczky, and Gaver's problems.

The main object in this next proposition is the function *k f* : [0, ∞) → [*k*0, ∞) defined by

$$k\_f(b) := \frac{\mathcal{W}'\_q(b)}{\mathcal{Z}\_{\overline{q}}(b)\mathcal{W}'\_{\overline{q}}(b) - q\mathcal{W}^2\_{\overline{q}}(b)},\tag{12}$$

where

$$k\_0 := k\_f(0\_+) = \frac{\mathcal{W}\_q'(0\_+)}{\mathcal{W}\_q'(0\_+) - q\mathcal{W}\_q^2(0\_+)} = \begin{cases} 1, & \text{if } X \text{ is of unbounded variation,} \\ 1 + \frac{q}{\Pi(0,\infty)}, & \text{if } X \text{ is of bounded variation.} \end{cases}$$

The function *k f* is increasing, thanks to (Avram et al. 2004, Theorem 1)1. Indeed, it is known that

$$\mathbb{E}\_{\mathbf{x}}\left[\mathbf{e}^{-q\tau^b}\right] = Z\_{\mathbb{q}}(\mathbf{x}) - q \frac{\mathcal{W}\_{\mathbb{q}}(b)}{\mathcal{W}\_{\mathbb{q}}'(b)} \mathcal{W}\_{\mathbb{q}}(\mathbf{x})\_{\prime\prime}$$

so *k f*(*b*) = 1 <sup>E</sup>*b*<sup>e</sup><sup>−</sup>*q<sup>τ</sup><sup>b</sup>* . The statement follows from the fact that the map *b* → E*b* e<sup>−</sup>*q<sup>τ</sup><sup>b</sup>* is decreasing.

The monotonicity allows us to re-parametrize the problem in terms of the optimal barrier, *bk* associated to a fixed cost, *k*.

**Proposition 1.** *Assume X is a SNLP. We have the following results:*


$$\begin{split} V\_{k\_f(b)}^{0,b}(\mathbf{x}) = k\_f(b) \left[ \mathbf{Z}\_{\emptyset}(\mathbf{x}) + \frac{p}{q} - Z\_{\emptyset}(\mathbf{x}) V^b(b) \right] \\ &= k\_f(b) \left[ Z\_{\emptyset}^{(1)}(\mathbf{x}) + Z\_{\emptyset}(\mathbf{x}) \left( \frac{p}{q} - V^b(b) \right) \right], \end{split} \tag{13}$$

*where <sup>V</sup><sup>b</sup>*(*x*) *is defined in* (6) *and*

$$Z\_{\eta}^{(1)}(x) := \int\_{0}^{x} \left( Z\_{\eta}(y) - p\mathcal{W}\_{\eta}(y) \right) \, \mathrm{d}y. \tag{14}$$

*(c) For fixed k, the barrier function HSLG k has a unique point of maximum bk* ≥ 0*. It is decreasing, and thus bk* = 0 *if, and only if k* ∈ (1, *k*0]*. Finally, if bk* > 0*, then k* = *k <sup>f</sup>*(*bk*)*.*

**Remark 1.** *Note that*

$$Z\_{\eta}^{(1)}(x) = \frac{\partial Z\_{\eta\mathcal{A}}(x)}{\partial\theta}\Big|\_{\theta=0}$$

<sup>1</sup> Some papers refer to this as the log-convexity of *Zq*(*x*).

*where Zq*,*<sup>θ</sup>* (*x*) = (*ψ*(*θ*) − *q*) ∞0 e<sup>−</sup>*<sup>θ</sup>yWq*(*x* + *y*)d*y. The function Zq*,*<sup>θ</sup> was introduced simultaneously in (Avram et al. 2015; Ivanovs and Palmowski 2012) (but was already present implicitly in (Avram et al. 2004, Theorem 1), where it was presented as an Esscher transform of Zq*(*x*)*). It was first used as a generating function for Gerber–Shiu penalty functions induced by polynomial rewards* 1, *x*, *<sup>x</sup>*2*, which were denoted respectively by Zq*, *Z*(1) *q* , *Z*(2) *q* , ... *, and started also being used intensively in exponential Parisian ruin problems following the work of (Albrecher et al. 2016). See (Avram et al. 2019) for more information.*

**Proof.** (a) The result follows from the fact that *k* → *V*0,*<sup>b</sup> k* (*x*) is decreasing (by definition) and because *VSLG k* (*x*) is obtained by a maximization of *V*0,*<sup>b</sup> k* (*x*) over all barrier levels *b* (chosen independently of *k*). (b)Recalling(9),weneedtoshowthat

> − *HSLG kf*(*b*)(*b*) = *k <sup>f</sup>*(*b*)*V<sup>b</sup>*(*b*). (15)

Indeed, it is easy to check that the equality

$$\frac{kZ\_q(b) - 1}{q\mathcal{W}\_q(b)} = k \frac{\mathcal{W}\_q(b)}{\mathcal{W}\_q'(b)}$$

holds for *k* = *k <sup>f</sup>*(*b*).

(c) It is well-known (see Avram et al. 2007, Lemma 2) that *HSLG k* is an increasing-decreasing function in *b*, with a unique maximum *bk* ≥ 0. For the sake of completeness, let us reproduce this proof. The derivative of the barrier function (11) satisfies

$$q\frac{H'W\_q^2}{W\_q'}(b) = f(b) := k\frac{\Delta\_q^{(ZW)}(b)}{W\_q'(b)} - 1 = k\operatorname{\mathbb{E}}\_b[e^{-q\tau^b}] - 1 = \frac{k}{k\_f(b)} - 1,\tag{16}$$

where

$$
\Delta\_q^{(Z\mathcal{W})}(b) := Z^{(q)}(b)\mathcal{W}\_q'(b) - \left(Z^{(q)}\right)'(b)\mathcal{W}\_q(b). \tag{17}
$$

Therefore, the sign of the derivative of the barrier function (11) coincides with that of *f* . Clearly, the latter function *f* is decreasing in *b* from lim*b*→<sup>0</sup> *f*(*b*) = *kk*0− 1 to −1.

**Remark 2.** *In conclusion, if k* ≤ *k*0*, then the barrier function HSLG k reaches its unique maximum at bk* = 0 *and, if k* > *k*0*, then bk is such that k* = *k <sup>f</sup>*(*bk*)*.*

**Remark 3.** *The previous proposition suggests the following new (and short) proof of the Løkka–Zervos alternative in the Brownian motion case. It is easy to verify that*

$$Z\_q^{(1)}(\mathbf{x}) + Z\_q(\mathbf{x}) \left(\frac{p}{q} - V^b(b)\right) = Z\_q^{(1)}(\mathbf{x}) = \frac{\sigma^2}{2} \mathcal{W}\_\emptyset(b)\_\prime$$

*which yields V*0,*b*<sup>∗</sup> *k f*(*b*∗)(*x*) = *<sup>V</sup>dF*(*x*)*, where b*∗ *denotes the optimal barrier level in de Finetti's problem, as defined in Section 2.1. Then, use the monotonicity of VSLG k*(*x*) *in k.*

Similar computations below will establish the Løkka–Zervos alternative in the Cramér-Lundberg case with exponential jumps.

### **3. The Løkka–Zervos Alternative for a Cramér–Lundberg Model with Exponential Jumps**

Here is our main result. **Theorem 1.** *For a Cramér–Lundberg process with exponentially distributed jumps, the Løkka–Zervos alternative holds with two regimes separated by the threshold k* = *k f*(*b*∗)*—that is, for all x* ≥ 0*,*

$$V\_k^{SL.G}(\mathbf{x}) \ge V^{dF}(\mathbf{x}) \quad \text{if, and only if} \quad k \le k\_f(b^\*).$$

**Proof.** By Proposition 1, we know that for fixed *x* and *b*, the function *k* → *VSLG k* (*x*) = sup*b*≥0 *V*0,*<sup>b</sup> k* (*x*) is non-increasing. One deduces that, for all *x* ≥ 0,

$$V\_k^{SLG}(\mathfrak{x}) \ge V\_{k\_f(b^\*)}^{0,b^\*}(\mathfrak{x}) \quad \text{if and only if} \quad k \le k\_f(b^\*).$$

Therefore, the Løkka–Zervos alternative follows from Lemma 1, given below, where it is proved that

$$V\_{k\_f(b^\*)}^{0,b^\*} (\mathfrak{x}) = V^{dF} (\mathfrak{x}). \tag{18}$$

Recall that we assume *q* > 0.

**Lemma 1.** *For a Cramér–Lundberg process with exponentially distributed jumps, we have:*

$$\begin{aligned} \text{(a)} \qquad & Z\_{\emptyset}(\mathfrak{x}) + \mu Z\_{\emptyset}^{(1)}(\mathfrak{x}) = c \mathcal{W}\_{\emptyset}(\mathfrak{x}), \text{for all } \mathfrak{x} > 0;\\ \text{(b)} \qquad & \mathbf{1} \qquad \stackrel{c}{\longrightarrow} \mathcal{W}\_{\emptyset}'(\mathfrak{b}^\*) \ . \end{aligned}$$

*;*

$$\mathcal{C}(b) \qquad \frac{1}{k\_f(b^\*)} = \frac{cW\_q(b)}{\mu}$$

*(c) V*0,*b*<sup>∗</sup> *k f*(*b*∗)(*x*) = *<sup>V</sup>dF*(*x*)*, for all x* > 0*.*

**Proof.** (a) This can be verified by taking Laplace transforms and using (2). Letting *<sup>F</sup>*(*s*) denote the Laplace transform of the tail distribution function *<sup>F</sup>*(*z*) = 1 − *<sup>F</sup>*(*z*), it amounts to checking

$$\frac{c}{\Psi(s) - q} = \frac{c - \lambda \widehat{\overline{F}}(s)}{\Psi(s) - q} + \mu \frac{c - \lambda \widehat{\overline{F}}(s) - p}{s(\psi(s) - q)} \Longleftrightarrow \lambda \widehat{\overline{F}}(s) = \mu \lambda \frac{\mu^{-1} - \widehat{\overline{F}}(s)}{s}.$$

which holds true for exponential jumps.

> (b) Manipulating the Kolmogorov IDE for *Zq*, we can reduce it to

$$cZ\_q^{\eta}(\mathbf{x}) + (c\mu - \lambda - q)Z\_q'(\mathbf{x}) - q\mu Z\_q(\mathbf{x}) = q\mathcal{W}\_q'(\mathbf{x}) \left( c + (c\mu - \lambda - q) \frac{\mathcal{W}\_q(\mathbf{x})}{\mathcal{W}\_q'(\mathbf{x})} - \mu \frac{Z\_q(\mathbf{x})}{\mathcal{W}\_q'(\mathbf{x})} \right) = 0.1$$

At *x* = *b*<sup>∗</sup>, using the fact that

$$V^{dF}(b^\*) = \frac{p}{q} - \frac{1}{\mu} \tag{19}$$

(see, for example, Equation (5.24) in Gerber et al. 2006) together with simple algebraic manipulations yields

$$\frac{Z\_q(b^\*)}{\mathcal{W}'\_q(b^\*)} = q\left(\frac{p}{q} - \frac{1}{\mu}\right)^2 + \frac{c}{\mu}.$$

This is equivalent to the result.

(c) From (13) and part (a), we ge<sup>t</sup>

$$V\_{k\_f(b^\*)}^{0,b^\*} (\mathbf{x}) = \frac{k\_f(b^\*)}{\mu} \left(\mu Z\_q^{(1)}(\mathbf{x}) + Z\_\emptyset(\mathbf{x})\right) = \frac{ck\_f(b^\*)}{\mu} W\_\emptyset(\mathbf{x}).$$

Then the result follows from (b), as well as the fact that *VdF* = *Vb*<sup>∗</sup> .
