*Remarks On Perry et al. (2004)*

For another "sanity check" of Theorem 4, we compute the Laplace transform Equation (1) independently when *μ* = 0 and *x* = 0. In this case, the reflected Brownian motion is equal to |*σ<sup>B</sup>*| in law, where *B* is a standard Brownian motion. But then *τδ* is equal in distribution to

$$\widetilde{\mathfrak{r}}\_{\delta} := \inf \{ \mathfrak{s} > 0 \mid B\_{\mathfrak{s}} \in \{ \pm \frac{\delta}{\sigma} \} \}.$$

Now, it is well known that the Laplace transform of *τδ* is given by

$$\mathbb{E}[\mathbf{e}^{-\bar{\theta}\mathbf{r}\_{\delta}} \mid \mathbf{X}\_{0} = \mathbf{x}] = \frac{1}{\cosh(\frac{\delta}{\mathcal{C}}\sqrt{2\theta})},\tag{2}$$

which indeed coincides with Equation (1) for *μ* → 0 and *x* = 0.

Perry et al. (2004), Formula (5.2), state a different Laplace transforms than our Theorem 4. Letting *μ* → 0 in Perry et al. (2004), Formula (5.2) indeed yields for *σ*<sup>2</sup> = 1 and *x* = 0,

$$\mathbb{E}[e^{-\theta \tau\_{\delta}} \mid \mathcal{X}\_0 = \mathfrak{x}] = \frac{1}{\cosh(\delta \sqrt{\theta})} \gamma$$

which contradicts Equation (2). The proof of Perry et al. (2004), Lemma 5.1, cannot be rectified, however, by merely fixing the (obviously) missing factor of 1/2 for *α*2 in the second line of their proof. Indeed, in the same line, they forget a factor *e*<sup>−</sup>*<sup>κ</sup><sup>W</sup>*(*s*) in the second integrand; thus, by inserting special values of *κ* into the process in line 2, one does not ge<sup>t</sup> rid of the local-time term, as claimed.

### **3. Diffusions with Exponentially Distributed Gains Before Fixed Drawdowns**

Let *X* be a diffusion process on the interval [−*a*, <sup>∞</sup>), satisfying the SDE

$$dX\_l = \mu(X\_l)dt + \sigma(X\_l)dW\_l, \quad X\_0 = 0,\tag{3}$$

where *μ*(*x*) and *σ*(*x*) are locally Lipschitz continuous functions of linear growth on [−*a*, <sup>∞</sup>), and *σ*(*x*) > 0 thereon.

For a threshold 0 < *δ* ≤ *a*, we define *M<sup>δ</sup>* as the maximum of *X*, prior to a drawdown of size *δ*, that is

$$M^\delta = M(\tau^\delta), \quad \text{where} \quad M(t) := \max\_{s \le t} X\_{s}, \quad \text{and} \quad \tau^\delta := \inf \{ t > 0 \mid M\_t - X\_t = \delta \}.$$

We use the abbreviation <sup>Φ</sup>(*x*) := *e*<sup>−</sup><sup>2</sup> *x*0 *γ*(*u*)*du*, where *γ*(*x*) = *<sup>μ</sup>*(*x*)/*σ*<sup>2</sup>(*x*). The following is due to Lehoczky (1977):
