**Remark 4.**

*1. (Kyprianou 2006, pp. 157, 168) κ*ˆ *is also the Laplace exponent of the (possibly killed) bivariate descending ladder subordinator* (*L*ˆ −1, *H*ˆ )*, where L*ˆ *is a local time at the minimum, and the descending ladder heights process H* ˆ = *XL*ˆ −1 *(on* {*L*ˆ −1 < ∞}*;* +∞ *otherwise) is X sampled at its right-continuous inverse L*ˆ <sup>−</sup>1*:*

$$\mathbb{E}[\mathfrak{e}^{-\mathfrak{a}\mathbb{L}\_1^{-1}-\beta\mathbb{H}\_1}\mathbb{1}\_{\{1<\tilde{l},\infty\}}] = \mathfrak{e}^{-\mathbb{A}(\mathfrak{a},\beta)}, \quad \{\mathfrak{a},\beta\} \subset \overline{\mathbb{C}^{\rightarrow}}.$$

*2. As for the strict ascending ladder heights subordinator H*∗ := *XL*∗−<sup>1</sup> *(on L*∗−<sup>1</sup> < ∞*;* +∞ *otherwise), L*∗−<sup>1</sup> *being the right-continuous inverse of L*<sup>∗</sup>*, and L*∗ *denoting the amount of time X has spent at a new maximum, we have, thanks to the skip-free property of X, as follows. Since* P(*Th* < ∞) = *e*<sup>−</sup><sup>Φ</sup>(0)*h, X stays at a newly achieved maximum each time for an* Exp(*λ*(R))*-distributed amount of time, departing it to achieve a new maximum later on with probability e*<sup>−</sup><sup>Φ</sup>(0)*h, and departing it, never to achieve a new maximum thereafter, with probability* 1 − *e*<sup>−</sup><sup>Φ</sup>(0)*h. It follows that the Laplace exponent of H*∗ *is given by:*

$$-\log \mathbb{E}[e^{-\beta H\_1} \mathbb{I}(H\_1 < +\infty)] = (1 - e^{-\beta h})\lambda(\mathbb{R})e^{-\Phi(0)h} + \lambda(\mathbb{R})(1 - e^{-\Phi(0)h}) = \lambda(\mathbb{R})(1 - e^{-(\beta + \Phi(0))h})$$

*(where β* ∈ R+*). In other words, H*∗/*h is a killed Poisson process of intensity <sup>λ</sup>*(R)*e*<sup>−</sup><sup>Φ</sup>(0)*<sup>h</sup> and with killing rate λ*(R)(1 − *<sup>e</sup>*<sup>−</sup><sup>Φ</sup>(0)*<sup>h</sup>*)*.*

Again thanks to the skip-free nature of *X*, we can expand on the contents of Proposition 3, by offering further details of the Wiener-Hopf factorization. Indeed, if we let *Nt* := *Xt*/*h* and *Tk* := *Tkh* (*t* ≥ 0, *k* ∈ N0) then clearly *T* := (*Tk*)*k*≥0 are the arrival times of a renewal process (with a possibly defective inter-arrival time distribution) and *N* := (*Nt*)*t*≥0 is the 'number of arrivals' process. One also has the relation: *G*∗*t* = *TNt* , *t* ≥ 0 (P-a.s.). Thus the random variables entering the Wiener-Hopf factorization are determined in terms of the renewal process (*<sup>T</sup>*, *<sup>N</sup>*).

Moreover, we can proceed to calculate explicitly the Wiener-Hopf factors as well as *κ*ˆ and *<sup>κ</sup>*<sup>∗</sup>. Let *p* > 0. First, since *Xep*/*h* is a geometrically distributed random variable, we have, for any *β* ∈ C<sup>→</sup>:

$$\mathbb{E}[e^{-\beta \overline{X}\_{\mathcal{C}p}}] = \sum\_{k=0}^{\infty} e^{-\beta hk} (1 - e^{-\Phi(p)h}) e^{-\Phi(p)hk} = \frac{1 - e^{-\Phi(p)h}}{1 - e^{-\beta h - \Phi(p)h}}.\tag{4}$$

Note here that <sup>Φ</sup>(*p*) > 0 for all *p* > 0. On the other hand, using conditioning (for any *α* ≥ 0):

$$\begin{split} \mathbb{E}\left[e^{-a\widetilde{\mathbf{C}}\_{\nu\_{p}}^{\omega}}\right] &=& \mathbb{E}\left[\left((\boldsymbol{\mu},t)\mapsto\sum\_{k=0}^{\infty}\mathbbm{1}\_{[0,\infty)}(t\_{k})e^{-at\_{k}}\mathbbm{1}\_{[t\_{k},t\_{k+1})}(\boldsymbol{\mu})\right)\odot(\boldsymbol{e}\_{p},T)\right] \\ &=& \mathbb{E}\left[\left(t\mapsto\sum\_{k=0}^{\infty}\mathbbm{1}\_{[0,\infty)}(t\_{k})e^{-at\_{k}}(e^{-pt\_{k}}-e^{-pt\_{k+1}})\right)\odot T\right],\text{ since }e\_{p}\perp T \\ &=& \mathbb{E}\left[\sum\_{k=0}^{\infty}\mathbbm{1}\_{\{T\_{k}<\infty\}}\left(e^{-(p+a)T\_{k}}-e^{-(p+a)T\_{k}}e^{-p(T\_{k+1}-T\_{k})}\right)\right] \\ &=& \mathbb{E}\left[\sum\_{k=0}^{\infty}e^{-(p+a)T\_{k}}\mathbbm{1}\_{\{T\_{k}<\infty\}}\left(1-e^{-p(T\_{k+1}-T\_{k})}\right)\right]. \end{split}$$

Now, conditionally on *Tk* < <sup>∞</sup>, *Tk*+<sup>1</sup> − *Tk* is independent of *Tk* and has the same distribution as *T*1. Therefore, by (1) and the theorem of Fubini:

$$\mathbb{E}[\varepsilon^{-a\overline{G}^\*\_{\varepsilon\_p}}] = \sum\_{k=0}^{\infty} \varepsilon^{-\Phi(p+a)hk} (1 - e^{-\Phi(p)h}) = \frac{1 - e^{-\Phi(p)h}}{1 - e^{-\Phi(p+a)h}}.\tag{5}$$

We identify from (4) for any *β* ∈ C<sup>→</sup>: *<sup>κ</sup>*<sup>∗</sup>(*p*,<sup>0</sup>) *<sup>κ</sup>*<sup>∗</sup>(*p*,*β*) = 1−*e*<sup>−</sup><sup>Φ</sup>(*p*)*<sup>h</sup>* 1−*e*<sup>−</sup>*βh*−<sup>Φ</sup>(*p*)*<sup>h</sup>* and therefore for any *α* ≥ 0: *<sup>κ</sup>*<sup>∗</sup>(*p*+*α*,<sup>0</sup>) *<sup>κ</sup>*<sup>∗</sup>(*p*+*α*,*β*) = 1−*e*<sup>−</sup><sup>Φ</sup>(*p*+*α*)*<sup>h</sup>* 1−*e*<sup>−</sup>*βh*−<sup>Φ</sup>(*p*+*α*)*<sup>h</sup>* . We identify from (5) for any *α* ≥ 0: *<sup>κ</sup>*<sup>∗</sup>(*p*,<sup>0</sup>) *<sup>κ</sup>*<sup>∗</sup>(*p*+*α*,<sup>0</sup>) = 1−*e*<sup>−</sup>*h*Φ(*p*) 1−*e*<sup>−</sup><sup>Φ</sup>(*p*+*α*)*<sup>h</sup>* . Therefore, multiplying the last two equalities, for *α* ≥ 0 and *β* ∈ C<sup>→</sup>, the equality:

$$\frac{\kappa^\*(p,0)}{\kappa^\*(p+a,\beta)} = \frac{1 - e^{-\Phi(p)h}}{1 - e^{-\beta h - \Phi(p+a)h}}\tag{6}$$

obtains. In particular, for *α* > 0 and *β* ∈ C<sup>→</sup>, we recognize for some constant *k*∗ ∈ (0, ∞): *<sup>κ</sup>*<sup>∗</sup>(*<sup>α</sup>*, *β*) = *k*∗(1 − *<sup>e</sup>*<sup>−</sup>(*β*+<sup>Φ</sup>(*α*))*<sup>h</sup>*). Next, observe that by independence and duality (for *α* ≥ 0 and *θ* ∈ R):

$$\begin{split} &\mathbb{E}[\exp\{-a\widetilde{\mathbf{G}}\_{\varepsilon\_{p}}^{\*} + i\theta\mathbf{\overline{X}}\_{\varepsilon\_{p}}\}]\mathbb{E}[\exp\{-a\underline{\mathbf{G}}\_{\varepsilon\_{p}} + i\theta\underline{X}\_{\varepsilon\_{p}}\}] = \int\_{0}^{\infty} dt p e^{-pt} \mathbb{E}[\exp\{-at + i\theta\mathbf{X}\_{l}\}] = \\ &\int\_{0}^{\infty} dt p e^{-pt - at + \Psi(\theta)t} = \frac{p}{p + a - \Psi(\theta)}. \end{split}$$

Therefore:

$$(p+\mathfrak{a}-\mathfrak{v}(i\theta))\frac{\mathfrak{k}(p,0)}{\mathfrak{k}(p+\mathfrak{a},i\theta)} = p\frac{1-\mathfrak{c}^{i\theta\mathfrak{h}-\Phi(p+\mathfrak{a})\mathfrak{h}}}{1-\mathfrak{c}^{-\Phi(p)\mathfrak{h}}}.$$

Both sides of this equality are continuous in *θ* ∈ C↓ and analytic in *θ* ∈ C↓. They agree on R, hence agree on C↓ by analytic continuation. Therefore (for all *α* ≥ 0, *β* ∈ C→):

$$\hat{\chi}(p+\mathfrak{a}-\mathfrak{v}(\beta))\frac{\hat{\kappa}(p,0)}{\hat{\kappa}(p+\mathfrak{a},\beta)} = p\frac{1-e^{\mathfrak{f}\mathfrak{h}-\Phi(p+\mathfrak{a})\mathfrak{h}}}{1-e^{-\Phi(p)\mathfrak{h}}},\tag{7}$$

i.e., for all *β* ∈ C→ and *α* ≥ 0 for which *p* + *α* = *ψ*(*β*) one has:

$$\mathbb{E}\left[\exp\left\{-\kappa \underline{G}\_{\mathfrak{c}\_p} + \beta \underline{X}\_{\mathfrak{c}\_p}\right\}\right] = \frac{p}{p + \kappa - \psi(\beta)} \frac{1 - e^{(\beta - \Phi(p+a))h}}{1 - e^{-\Phi(p)h}}.$$

Moreover, for the unique *β*0 > 0, for which *ψ*(*β*0) = *p* + *α*, one can take the limit *β* → *β*0 in the above to obtain: <sup>E</sup>[exp{−*αGep* + *β*0*Xep* }] = *ph ψ*(*β*0)(<sup>1</sup> − *e*<sup>−</sup><sup>Φ</sup>(*p*)*<sup>h</sup>*) = *ph*Φ(*p* + *α*) 1 − *e*<sup>−</sup><sup>Φ</sup>(*p*)*<sup>h</sup>* . We also recognize from (7) for *α* > 0 and *β* ∈ C→ with *α* = *ψ*(*β*), and some constant ˆ*k* ∈ (0, ∞): *<sup>κ</sup>*<sup>ˆ</sup>(*<sup>α</sup>*, *β*) = ˆ*k <sup>α</sup>*−*ψ*(*β*) 1−*<sup>e</sup>*(*β*−<sup>Φ</sup>(*α*))*<sup>h</sup>* . With *β*0 = <sup>Φ</sup>(*α*) one can take the limit in the latter as *β* → *β*0 to obtain: *<sup>κ</sup>*<sup>ˆ</sup>(*<sup>α</sup>*, *β*0) = ˆ*<sup>k</sup>ψ*(*β*0)/*<sup>h</sup>* = ˆ *k h*Φ(*α*).
