De Finetti's control problem with Parisian ruin for spectrally negative L\'evy processes

TL;DR

This paper studies an optimal control problem involving spectrally negative Lévy processes with Parisian ruin, establishing barrier strategies as optimal under certain conditions and comparing them to classical ruin scenarios.

Contribution

It introduces a generalized de Finetti control problem with Parisian ruin for spectrally negative Lévy processes and characterizes the optimal barrier strategy.

Findings

Optimal barrier strategy is always below the classical ruin barrier.

Necessary and sufficient conditions for zero-level barrier optimality.

Optimal barrier level is finite and depends on the Lévy measure.

Abstract

We consider de Finetti's stochastic control problem when the (controlled) process is allowed to spend time under the critical level. More precisely, we consider a generalized version of this control problem in a spectrally negative L\'evy model with exponential Parisian ruin. We show that, under mild assumptions on the L\'evy measure, an optimal strategy is formed by a barrier strategy and that this optimal barrier level is always less than the optimal barrier level when classical ruin is implemented. Also, we give necessary and sufficient conditions for the barrier strategy at level zero to be optimal.

Equations

$${\int }_0^{\infty }{\mathrm{e}}^{-\theta x}{W}^{(q)}(x)\mathrm{d}x={\left(\psi (\theta )-q\right)}^{-1},$$

$$\psi (\theta )=\gamma \theta +\frac{1}{2}{\sigma }^2{\theta }^2+{\int }_0^{\infty }\left({\mathrm{e}}^{-\theta z}-1+\theta z{1}_{(0,1]}(z)\right)\nu (\mathrm{d}z),$$

$${\int }_0^{\infty }(1\wedge x^2)\nu (\mathrm{d}x)<\infty .$$

$${\sigma }_p^{\pi }=\inf \left\{t>0:t-g_t^{\pi }>{\mathbf{e}}_p^{g_t^{\pi }}\text{and}{U}_t^{\pi }<0\right\},$$

$$v_{\pi }(x)={\mathbb{E}}_x\left[{\int }_0^{{\sigma }_p^{\pi }}{\mathrm{e}}^{-qt}\mathrm{d}{L}_t^{\pi }\right],x\in \mathbb{R}.$$

$$v_{\pi }(x)={\mathbb{E}}_x\left[{\mathrm{e}}^{-q{\tau }_0^{+}}{1}_{\{{\tau }_0^{+}<{\mathbf{e}}_p\}}\right]v_{\pi }(0)={\mathrm{e}}^{\Phi (p+q)x}{v}_{\pi }(0),$$

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

$$v_{\ast }(x)=\underset{\pi \in {\Pi }_p}{\sup }{v}_{\pi }(x)$$

$$v_{{\pi }_{\ast }}(x)=v_{\ast }(x),$$

$$v_b(x)={\mathbb{E}}_x\left[{\int }_0^{{\sigma }_p^b}{\mathrm{e}}^{-qt}\mathrm{d}{L}_t^b\right],$$

$$Z_q(x,\theta )={\mathrm{e}}^{\theta x}\left(1-(\psi (\theta )-q){\int }_0^x{\mathrm{e}}^{-\theta y}{W}^{(q)}(y)\mathrm{d}y\right).$$

$$Z_q(x,\Phi (p+q))={\mathrm{e}}^{\Phi (p+q)x}\left(1-p{\int }_0^x{\mathrm{e}}^{-\Phi (p+q)y}{W}^{(q)}(y)\mathrm{d}y\right),$$

$$Z_q^{\prime }(x,\Phi (p+q))=\Phi (p+q)Z_q(x,\Phi (p+q))-p{W}^{(q)}(x).$$

$$Z_q(x,\Phi (p+q))=p{\int }_0^{\infty }{\mathrm{e}}^{-\Phi (p+q)y}{W}^{(q)}(x+y)\mathrm{d}y.$$

$$Z_q^{\prime }(x,\Phi (p+q))=p{\int }_0^{\infty }{\mathrm{e}}^{-\Phi (p+q)y}{W}^{(q)\prime }(x+y)\mathrm{d}y,$$

$$v_b(x)=\{\begin{array}{ll}\frac{Z_q(x,\Phi (p+q))}{Z_q^{\prime }(b,\Phi (p+q))}& \text{for x∈(−∞,b],}\\ x-b+v_b(b)& \text{for x∈(b,∞).}\end{array}$$

$${\mathbb{E}}_b\left[{\int }_0^{\infty }{\mathrm{e}}^{-qt}\mathrm{d}{L}_t^b\right]=\frac{1}{\Phi (q)}.$$

$${\mathbb{E}}_x\left[{\int }_0^{\infty }{\mathrm{e}}^{-qt}\mathrm{d}{L}_t^b\right]=\frac{{\mathrm{e}}^{-\Phi (q)(b-x)}}{\Phi (q)},$$

$$v_b(x)=\frac{Z_q(x,\Phi (p+q))}{Z_q^{\prime }(b,\Phi (p+q))}\underset{p\to 0}{⟶}\frac{Z_q(x,\Phi (q))}{Z_q^{\prime }(b,\Phi (q))}=\frac{{\mathrm{e}}^{\Phi (q)x}}{\Phi (q){\mathrm{e}}^{\Phi (q)b}}=\frac{1}{\Phi (q)}{\mathrm{e}}^{-\Phi (q)(b-x)}.$$

$${\mathbb{E}}_b\left[{\int }_0^{{\sigma }_{\infty }^b}{\mathrm{e}}^{-qt}\mathrm{d}{L}_t^b\right]=\frac{W^{(q)}(x)}{W^{(q)\prime }(b)},x\le b,$$

$$v_b(x)=\frac{Z_q(x,\Phi (p+q))}{Z_q^{\prime }(b,\Phi (p+q))}=\frac{\Phi (p+q){\int }_0^{\infty }{\mathrm{e}}^{-\Phi (p+q)y}{W}^{(q)}(x+y)\mathrm{d}y}{\Phi (p+q){\int }_0^{\infty }{\mathrm{e}}^{-\Phi (p+q)y}{W}^{(q)\prime }(b+y)\mathrm{d}y}\underset{p\to \infty }{⟶}\frac{W^{(q)}(x)}{W^{(q)\prime }(b)},$$

$$Z_q(x,\Phi (p+q))={\mathrm{e}}^{\Phi (p+q)x}\underset{p\to \infty }{⟶}0=W^{(q)}(x).$$

$$b_{\infty }^{\ast }=\sup \left\{b\ge 0:W^{(q)\prime }(b)\le W^{(q)\prime }(x),\text{for all x≥0}\right\}.$$

$$b_p^{\ast }=\sup \left\{b\ge 0:Z_q^{\prime }(b,\Phi (p+q))\le Z_q^{\prime }(x,\Phi (p+q)),\text{for all x≥0}\right\}.$$

$$\frac{c\Phi (p+q)}{p}\left(\Phi (p+q)-\frac{p}{c}\right)<\frac{q+\nu (0,\infty )}{c},$$

$$\Phi (p+q)=\frac{1}{{\sigma }^2}\left(\sqrt{c^2+2{\sigma }^2(p+q)}-c\right).$$

$$\frac{c\Phi (p+q)}{p}\mathbb{E}\left[{\int }_0^{{\sigma }_{\infty }^0}{\mathrm{e}}^{-qt}\mathrm{d}{L}_t^0\right]<\mathbb{E}\left[{\int }_0^{{\sigma }_p^0}{\mathrm{e}}^{-qt}\mathrm{d}{L}_t^0\right]=v_0(0).$$

$$\Gamma v(x)=\gamma v^{\prime }(x)+\frac{{\sigma }^2}{2}{v}^{\prime \prime }(x)+{\int }_0^{\infty }\left(v(x-z)-v(x)+v^{\prime }(x)z{1}_{(0,1]}(z)\right)\nu (\mathrm{d}z),$$

$$\left(\Gamma -q-p{1}_{(-\infty ,0)}\right)v_{\hat{\pi }}(x)\le 0$$

$$\begin{gathered} {\mathrm{e}}^{-qt-p{\int }_0^t{1}_{(-\infty ,0)}(U_r^{\pi })\mathrm{d}r}w\left(U_t^{\pi }\right)-w\left(U_0^{\pi }\right) \\ ={\int }_0^t{\mathrm{e}}^{-qs-p{\int }_0^s{1}_{(-\infty ,0)}(U_r^{\pi })\mathrm{d}r}\left[(\Gamma -q)w\left(U_s^{\pi }\right)-p{1}_{(-\infty ,0)}\left(U_s^{\pi }\right)w\left(U_s^{\pi }\right)\right]\mathrm{d}s \\ -{\int }_0^t{\mathrm{e}}^{-qs-p{\int }_0^s{1}_{(-\infty ,0)}(U_r^{\pi })\mathrm{d}r}{w}^{\prime }\left(U_{s-}^{\pi }\right)\mathrm{d}{L}_s^{\pi }+M_t^{\pi } \\ +\sum _{0<s\le t}{\mathrm{e}}^{-qs-p{\int }_0^s{1}_{(-\infty ,0)}(U_r^{\pi })\mathrm{d}r}\left[w\left(U_{s-}^{\pi }-\Delta L_s^{\pi }\right)-w\left(U_{s-}^{\pi }\right)+w^{\prime }\left(U_{s-}^{\pi }\right)\Delta L_s^{\pi }\right], \end{gathered}$$

Full text

1. Introduction and main result

On a filtered probability space $\left(\Omega,\mathcal{F},\mathbb{F},\mathbb{P}\right)$, let $X=\left\{X_{t},t\geq 0\right\}$ be a spectrally negative Lévy process with Laplace exponent $\theta\mapsto\psi(\theta)$ and with $q$-scale functions $\left\{W^{(q)},q\geq 0\right\}$ given by

[TABLE]

for all $\theta>\Phi(q)=\sup\left\{\lambda\geq 0\colon\psi(\lambda)=q\right\}$. Recall that

[TABLE]

where $\gamma\in\mathbb{R}$ and $\sigma\geq 0$, and where $\nu$ is a $\sigma$-finite measure on $(0,\infty)$, called the Lévy measure of $X$, satisfying

[TABLE]

For more details on spectrally negative Lévy processes and scale functions, see e.g. [kuznetsov-e-tal_2012, kyprianou_2014].

In what follows, we will use the following notation: the law of $X$ when starting from $X_{0}=x$ is denoted by $\mathbb{P}_{x}$ and the corresponding expectation by $\mathbb{E}_{x}$. We write $\mathbb{P}$ and $\mathbb{E}$ when $x=0$.

1.1. Problem formulation

Let the spectrally negative Lévy process $X$ be the underlying surplus process. A dividend strategy $\pi$ is represented by a non-decreasing, left-continuous and adapted stochastic process $L^{\pi}=\left\{L^{\pi}_{t},t\geq 0\right\}$, where $L^{\pi}_{t}$ represents the cumulative amount of dividends paid up to time $t$ under this strategy, and such that $L^{\pi}_{0}=0$. For a given strategy $\pi$, the corresponding controlled surplus process $U^{\pi}=\left\{U^{\pi}_{t},t\geq 0\right\}$ is defined by $U^{\pi}_{t}=X_{t}-L^{\pi}_{t}$. The stochastic control problem considered in this paper involves the time of Parisian ruin (with rate $p>0$) for $U^{\pi}$ defined by

[TABLE]

where $g_{t}^{\pi}=\sup\left\{0\leq s\leq t\colon U^{\pi}_{s}\geq 0\right\}$, with $\mathbf{e}_{p}^{g_{t}^{\pi}}$ an independent random variable, following the exponential distribution with mean $1/p$, associated to the corresponding excursion below [math] (see [baurdoux-et-al_2016] for more details). Note that, without loss of generality, we have chosen [math] to be the critical level.

A strategy $\pi$ is said to be admissible if a dividend payment is not larger than the current surplus level, i.e. $L^{\pi}_{t+}-L^{\pi}_{t}\leq U^{\pi}_{t}$, for all $t<\sigma^{\pi}_{p}$, and if no dividends are paid when the controlled surplus is negative, i.e. $t\mapsto L^{\pi}_{t}\,\mathbf{1}_{(-\infty,0)}(U^{\pi}_{t})\equiv 0$. The set of admissible dividend strategies will be denoted by $\Pi_{p}$.

Fix a discounting rate $q\geq 0$. The value function associated to an admissible dividend strategy $\pi\in\Pi_{p}$ is defined by

[TABLE]

In particular, for $\pi\in\Pi_{p}$ and $x<0$, using the strong Markov property and the spectral negativity of $X$, we can easily verify that

[TABLE]

where $\tau_{0}^{+}=\inf\left\{t>0\colon X_{t}>0\right\}$ and where $\mathbf{e}_{p}$ is an independent exponentially distributed random variable with mean $1/p$, thanks to the well-known fluctuation identity (see e.g. [kyprianou_2014])

[TABLE]

where $\tau_{b}^{+}=\inf\left\{t>0\colon X_{t}>b\right\}$.

The goal is to find the optimal value function $v_{\ast}$ defined by

[TABLE]

and an optimal strategy $\pi_{\ast}\in\Pi_{p}$ such that

[TABLE]

for all $x\in\mathbb{R}$. Because of the Parisian nature of the time of ruin considered in this control problem, we have to deal with possibly negative starting capital.

1.2. Main result and organization of the paper

Let us introduce the family of horizontal barrier strategies, also called reflection strategies. For $b\in\mathbb{R}$, the (horizontal) barrier strategy at level $b$ is the strategy denoted by $\pi^{b}$ and with cumulative amount of dividends paid until time $t$ given by $L_{t}^{b}=\left(\sup_{0<s\leq t}X_{s}-b\right)_{+}$, for $t>0$. If $X_{0}=x>b$, then $L^{b}_{0+}=x-b$. Note that, if $b\geq 0$, then $\pi_{b}\in\Pi_{p}$. The corresponding value function is thus given by

[TABLE]

for all $x\in\mathbb{R}$, where $\sigma^{b}_{p}$ is the time of Parisian ruin (with rate $p>0$) for the controlled process $U^{b}_{t}=X_{t}-L_{t}^{b}$.

Before stating the main result of this paper, recall that the tail of the Lévy measure is the function $x\mapsto\nu(x,\infty)$, where $x\in(0,\infty)$, and that a function $f\colon(0,\infty)\to(0,\infty)$ is log-convex if the function $\log(f)$ is convex on $(0,\infty)$.

Theorem 1.1**.**

Fix $q\geq 0$ and $p>0$. If the tail of the Lévy measure is log-convex, then an optimal strategy for the control problem is formed by a barrier strategy.

The original version of de Finetti’s optimal dividends problem, i.e. when the time of ruin is the first passage time below the critical level (intuitively, when $p\to\infty$), has been extensively studied. In a spectrally negative Lévy model, following the work of Avram, Palmowski & Pistorius [avram-et-al_2007], an important breakthrough was made by Loeffen [loeffen_2008]; in the latter, a sufficient condition, on the Lévy measure $\nu$, is given for a barrier strategy to be optimal. This condition was further relaxed by Loeffen & Renaud [loeffen-renaud_2010]; in this other paper, it is shown that if the tail of the Lévy measure is log-convex then a barrier strategy is optimal for de Finetti’s optimal dividends problem with an affine penalty function at ruin (if we set $S=K=0$ in that paper, we recover the classical problem). To the best of our knowledge, this still stands as the mildest condition for the optimality of a barrier strategy in a spectrally negative Lévy model. Finally, note that Czarna & Palmowski [czarna-palmowski_2014] have considered de Finetti’s control problem with deterministic Parisian delays.

The rest of the paper is organized as follows. First, we compute the value function of an arbitrary horizontal barrier strategy and then find the optimal barrier level $b^{\ast}_{p}$ (see the definition in (8)). Second, we derive the appropriate verification lemma for this control problem and prove that the barrier strategy at level $b^{\ast}_{p}$ is optimal.

2. Horizontal barrier strategies

Before computing the value function of an arbitrary barrier strategy at level $b$, we have to define another family of scale functions, also called second $q$-scale functions of $X$.

2.1. Second family of scale functions

For each $x\in\mathbb{R}$ and $q,\theta\geq 0$, define as in [albrecher-et-al_2016]

[TABLE]

Note that, for $x\leq 0$ or for $\theta=\Phi(q)$, we have $Z_{q}(x,\theta)=\mathrm{e}^{\theta x}$. In what follows, $Z_{q}^{\prime}(x,\theta)$ will represent the derivative with respect to the first argument. Consequently, for $x>0$, we have $Z_{q}^{\prime}(x,\theta)=\theta Z_{q}(x,\theta)-(\psi(\theta)-q)W^{(q)}(x)$ and, for $x<0$, we have $Z_{q}^{\prime}(x,\theta)=\theta\mathrm{e}^{\theta x}$. Note that $Z_{q}^{\prime}(0-,\theta)=Z_{q}^{\prime}(0+,\theta)=\theta$ if and only if $W^{(q)}(0)=0$, i.e. if and only if $X$ has paths of unbounded variation.

In this paper, we will encounter the function $Z_{q}$ when $\theta=\Phi(p+q)$, that is the function

[TABLE]

from which we deduce that, for $x>0$,

[TABLE]

Consequently, set $Z_{q}^{\prime}(0,\Phi(p+q))=\Phi(p+q)-pW^{(q)}(0+)$. Since we assume that $p>0$, we have that $\Phi(p+q)>\Phi(q)$ and we can write

[TABLE]

Then, for $x>0$, we have

[TABLE]

which is well defined since $W^{(q)}$ is differentiable almost everywhere (see e.g. Lemma 2.3 in [kuznetsov-e-tal_2012]). Clearly, $x\mapsto Z_{q}(x,\Phi(p+q))$ is a non-decreasing continuous function.

2.2. Value function of a barrier strategy

Here is the value of an arbitrary admissible barrier strategy:

Proposition 2.1**.**

For $q,b\geq 0$, the value function associated to $\pi_{b}$ is given by

[TABLE]

Proof.

See the proof in Appendix A. ∎

Using Parisian ruin with rate $p$ allows to fill in the spectrum of possibilities between classical ruin (no delay, $p\to\infty$) and no ruin at all (infinite delays, $p\to 0$). To illustrate this, let us have a look at the behaviour of $v_{b}$, which depends implicitly on the Parisian rate $p$, when:

1. (1)

$p\to 0$ (infinite delays, no ruin); 2. (2)

$p\to\infty$ (no delay, classical ruin).

First, it is known that (see e.g. Equation (3.15) in [avram-et-al_2007]), if there is no ruin, then

[TABLE]

Using the strong Markov property and the spectral negativity of $X$, for $-\infty<x\leq b$, we thus have

[TABLE]

where we used again the identity in (2).

On the other hand, for $-\infty<x\leq b$, using Equation (7), we get

[TABLE]

Second, recall that the value of a barrier strategy at level $b$ subject to classical ruin is given by

[TABLE]

where $\sigma^{b}_{\infty}:=\inf\left\{t>0\colon U_{t}^{b}<0\right\}$; see e.g. [avram-et-al_2007]. Since $W^{(q)}(x)=0$ if $x<0$, this last identity is valid for all $-\infty<x\leq b$. On the other hand, for $-\infty<x\leq b$, using Equation (7) and the fact that $\lim_{p\to\infty}\Phi(p+q)=\infty$, we get

[TABLE]

where, in the last step, we used the Initial Value Theorem for Laplace transforms (see e.g. [feller_1971]). Note that, when $x<0$, we have directly

[TABLE]

2.3. Optimal barrier level

As defined in [loeffen_2008, loeffen-renaud_2010], the optimal barrier level in de Finetti’s classical control problem is given by

[TABLE]

Similarly, let us define the candidate for the optimal barrier level for the current version of this control problem by

[TABLE]

Proposition 2.2**.**

Fix $q\geq 0$ and $p>0$. Suppose the tail of the Lévy measure is log-convex. Then, we have that $0\leq b_{p}^{\ast}\leq b_{\infty}^{\ast}$. Further, $b_{p}^{\ast}>0$ if and only if one of the following three cases hold:

1. (a)

$\sigma>0$* and $\left(\Phi(p+q)\right)^{2}/p<2/\sigma^{2}$;* 2. (b)

$\sigma=0$* and $\nu(0,\infty)=\infty$;* 3. (c)

$\sigma=0$, $\nu(0,\infty)<\infty$ and

[TABLE]

where $c=\gamma+\int_{0}^{1}x\nu(\mathrm{d}x)$.

Proof.

See the proof in Appendix B. ∎

First of all, note from Proposition 2.2 that the optimal barrier level $b_{p}^{\ast}$, when Parisian ruin with rate $p$ is implemented, is always lower than the optimal barrier level $b_{\infty}^{\ast}$ when classical ruin is used.

Also, from (a) and (c), when $\sigma>0$ or $\nu(0,\infty)<\infty$, we have that if the Parisian rate $p>0$ is small enough (large delays), then the barrier strategy at level zero is optimal. If $X_{t}=ct+\sigma B_{t}$ is a Brownian motion with drift, then

[TABLE]

In this case, using (a), we can verify that if the Brownian coefficient $\sigma$ is large enough, then the barrier strategy at level zero is optimal.

Interestingly, in (c), the condition can be re-written as follows:

[TABLE]

See e.g. Equation (3.14) in [avram-et-al_2007].

3. Verification lemma and proof of the main result

Define the operator $\Gamma$ associated with $X$ by

[TABLE]

where $v$ is a function defined on $\mathbb{R}$ such that $\Gamma v(x)$ is well defined.

Next is the verification lemma of our stochastic control problem. As the controlled process is now allowed to spend time below the critical level, it is different from the classical verification lemma (see [loeffen_2008]).

Lemma 3.1**.**

Let $\Gamma$ be the operator defined in (9). Suppose that $\hat{\pi}\in\Pi_{p}$ is such that $v_{\hat{\pi}}$ is sufficiently smooth and that, for all $x\in\mathbb{R}$,

[TABLE]

and, for all $x>0$, $v^{\prime}_{\hat{\pi}}(x)\geq 1$. In this case, $\hat{\pi}$ is an optimal strategy for the control problem.

Proof.

Set $w:=v_{\hat{\pi}}$ and let $\pi\in\Pi_{p}$ be an arbitrary admissible strategy. As $w$ is sufficiently smooth, applying an appropriate change-of-variable/version of Ito’s formula to the joint process $\left(t,\int_{0}^{t}\mathbf{1}_{(-\infty,0)}(U^{\pi}_{r})\mathrm{d}r,U^{\pi}_{t}\right)$ yields

[TABLE]

where $M^{\pi}=\left\{M^{\pi}_{t},t\geq 0\right\}$ is a (local) martingale.

Let $\mathcal{F}_{\infty}$ denote the sigma-field generated by the whole trajectories of our processes. Consider an independent (of $\mathcal{F}_{\infty}$) Poisson process with intensity measure $p\,\mathrm{d}t$ and jump times $\left\{T^{p}_{i},i\geq 1\right\}$. Therefore, we can write

[TABLE]

and consequently

[TABLE]

where we used the definition of a Riemann-Stieltjes integral and the monotone convergence theorem for conditional expectations.

Now, as for all $x\in\mathbb{R}$,

[TABLE]

and, for all $x>0$, $w^{\prime}(x)\geq 1$, using standard arguments (see e.g. [loeffen_2008]) and our definition of an admissible strategy, e.g. that $L^{\pi}$ is identically zero when $U^{\pi}$ is below zero, we get

[TABLE]

This concludes the proof. ∎

The rest of this section is devoted to proving Theorem 1.1, i.e. proving that an optimal strategy for the control problem is formed by the barrier strategy at level $b^{\ast}:=b_{p}^{\ast}$.

By the definition of $b^{\ast}$ given in (8), for $0\leq x\leq b^{\ast}$, we have

[TABLE]

By the definition of $v_{b^{\ast}}$, for $x>b^{\ast}$, we have $v_{b^{\ast}}^{\prime}(x)=1$. This means $v_{b^{\ast}}^{\prime}(x)\geq 1$, for all $x\geq 0$.

Note that, for any $x\in\mathbb{R}$, we have

[TABLE]

Consequently, for $x<0$, we have

[TABLE]

and, for $x\geq 0$, using (5), we have

[TABLE]

since $\left(\Gamma-q\right)W^{(q)}(x)=0$ for all $x>0$ (see e.g. [biffis-kyprianou_2010]). As a consequence, and since $v_{b^{\ast}}$ is smooth in $x=b^{\ast}$, we have

[TABLE]

All is left to verify now is that $\left(\Gamma-q\right)v_{b^{\ast}}(x)\leq 0$, for all $x>b^{\ast}$. It can be done following the same steps as in the proof of Theorem 2 in [loeffen_2008], thanks to the smoothness of the scale function $Z_{q}(\cdot,\Phi(p+q))$. The details are left to the reader.

Acknowledgements

Funding in support of this work was provided by the Natural Sciences and Engineering Research Council of Canada (NSERC).