*Examples*

The scale functions for the below processes are taken from review article of Hubalek and Kyprianou (2011). Throughout this section, E(*λ*) denotes the exponential distribution with parameter *λ* > 0.

**Example 1** (Compound Poisson Process)**.** *Assume we have a compound Poisson process with negative exponentially distributed jumps,*

$$X\_t = ct - \sum\_{k=0}^{N\_t^\lambda} \mathfrak{J}\_{k\prime} \quad \mathfrak{J}\_k \quad \text{i.i.d. and } \sim \mathcal{E}(\mu), \quad c - \lambda/\mu > 0.$$

*We get*

$$W(x) = \frac{1}{c} \left( 1 + \frac{\lambda}{c\mu - \lambda} (1 - e^{-(\mu - \lambda/c)x}) \right).$$

*Clearly, W* ∈ *C*<sup>1</sup>(0, <sup>∞</sup>)*,*

$$W'(x) = \frac{\lambda}{c^2} e^{-(\mu - \lambda/c)x}.$$

*Therefore, by Theorem 4*

$$\mathcal{M}^\delta \sim \mathcal{E}\left(\frac{\lambda/\mathfrak{c}}{\varepsilon^{\delta(\mu-\lambda/\mathfrak{c})} - \frac{\lambda/\mathfrak{c}}{\mu-\lambda/\mathfrak{c}}}\right), \quad \lim\_{\delta \downarrow 0} \mathbb{E}[\mathcal{M}^\delta] = \lambda/\mathfrak{c} > 0, \quad \lim\_{\delta \uparrow \infty} \mathbb{E}[M^\delta] = \mu - \lambda/\mathfrak{c} < \infty.$$

Unlike the previous example, the following two examples exhibit the same qualitative dependence on the threshold *δ*, as the standard Brownian motion, where *M<sup>δ</sup>* ∼ E(1/*δ*): when *δ* → 0, the average maximum at drawdown of size *δ* tends to 0, and when *δ* → <sup>∞</sup>, this average goes to infinity.

**Example 2** (Brownian motion with drift)**.** *A Brownian motion with drift μ* > 0 *and volatility σ,*

$$X\_t = \mu t + \sigma B\_t$$

*has scale function*

$$\mathcal{W}(\mathbf{x}) \sim e^{-\mu \mathbf{x}/\sigma^2} \sinh(\sqrt{\mu} \mathbf{x}/\sigma^2).$$

*Hence,*

$$\frac{\mathcal{W}'(\mathbf{x})}{\mathcal{W}(\mathbf{x})} = \frac{-\mu/\sigma^2 \sinh(\sqrt{\mu}\mathbf{x}/\sigma^2) + \sqrt{\mu}/\sigma^2 \cosh(\sqrt{\mu}\mathbf{x}/\sigma^2)}{\sinh(\sqrt{\mu}\mathbf{x}/\sigma^2).}$$

*Therefore, by Theorem 4 (see, e.g., Golub et al. (2016)),*

$$\mathcal{M}^{\delta} \sim \mathcal{E}\left(\mu/\sigma^2 \left(\coth(\sqrt{\mu}\delta/\sigma^2) - 1\right)\right).$$

**Example 3** (Caballero and Chaumont (2006))**.** *This is a Lévy process without diffusion component, defined by its Lévy measure*

$$\nu(d\xi) = \frac{\mathfrak{e}^{(\beta - 1)\xi}}{(\mathfrak{e}^{\xi} - 1)^{\beta + 1}}, \quad \xi < 0, 1$$

*where β* ∈ (1, <sup>2</sup>)*, and its Laplace exponent,*

$$\Psi(\theta) = \frac{\Gamma(\theta + \beta)}{\Gamma(\theta)\Gamma(\beta)}, \quad \theta > 0.$$

*The process exhibits Infinite variation jumps, and drifts to* <sup>−</sup>∞*, because* Ψ(0) < 0*. The scale function is*

$$W(x) = (1 - e^{-x})^{\beta - 1}.$$

*Using Theorem 4, we thus get*

$$M^\delta \sim \mathcal{E}\left(\frac{\beta - 1}{\varepsilon^\delta - 1}\right), \quad \mathbb{E}[M^\delta] = \frac{\varepsilon^\delta - 1}{\beta - 1}.$$

The asymptotic behaviour of the logarithmic derivative of the scale function of a spectrally negative Lévy process can be characterized using the asymptotic behaviour of *W* and *W*, cf. (Kuznetsov et al. 2012, Chapter 3). For instance, *W*(0) = *W*(0+) = 0, if and only if the process is of infinite variation. In the case of finite variation, we can write the process as *δt* − *Jt*, where *J* is a subordinator; and then *W*(0) = 1/*δ* > 0. Furthermore, *W*(0+) = <sup>∞</sup>, if a diffusion component is present, or if the Lévy measure is infinite. These general findings are consistent with the three examples.

**Funding:** This research received no external funding.

**Acknowledgments:** I thank John Appleby, Florin Avram, Huayuan Dong, Friedrich Hubalek, Andreas Kyprianou and two anonymous referees for useful comments.

**Conflicts of Interest:** The author declares no conflict of interest.
