1 Erratum to: Probab Theory Relat Fields (2013) 157:1–45 DOI 10.1007/s00440-012-0448-x

In this Erratum, we correct an error in our paper “Asymptotic behaviour of first passage time distributions for Lévy processes” published in Probab Theory Relat Fields, 157(1–2):1–45, 2013.

Let X be a real valued Lévy process with law \(\mathbb {P},\) characteristic exponent \(\Psi \) and characteristic triplet \((a,\sigma ,\Pi ).\) We assume that X is in the domain of attraction of a stable distribution without centering, that is there exists a deterministic function \(c:(0,\infty )\rightarrow (0,\infty )\) such that

$$\begin{aligned} \frac{X_{t}}{c(t)}\xrightarrow []{\mathcal {D}}Y_{1},\quad \text {as}\ \,\,t\rightarrow \infty , \end{aligned}$$
(1)

with \(Y_{1}\) a strictly stable random variable of parameter \(0<\alpha \le 2,\) and positivity parameter \(\rho =\mathbb {P}(Y_{1}>0).\) In this case we will use the notation \(X\in D(\alpha ,\rho ),\) and put \(\overline{\rho }=1-\rho .\) Hereafter \((Y_{t},t\ge 0)\) will denote an \(\alpha \)-stable Lévy process with positivity parameter \(\rho =\mathbb {P}(Y_{1}>0).\) We write f for the density of \(Y_{1},\) and \(\Psi _{\alpha }\) for its characteristic exponent.

In the original paper we provided sharp estimates for the local behaviour of the distribution of the first passage time of X below 0,  i.e. \(T_{0}=\inf \{t>0:X_{t}<0\},\) under \(\mathbb {P}_{x}(\cdot ),\) for \(x>0,\) both in the event of creeping and non-creeping. The proof of our results has been based in the validity of the following result, Proposition 13 in the original paper.

Proposition 1

Assume that \(X\in D(\alpha ,\rho ).\) Then uniformly in \(0<\Delta <\Delta _{0},\) with \(\Delta _{0}>0,\) and \(x\in \mathbb {R},\)

$$\begin{aligned} c(t)\mathbb {P(}X_{t}\in (x,x+\Delta ])=\Delta \left( f\left( \frac{x}{c(t)}\right) +o(1)\right) \,\,\text { as }\,\,t\rightarrow \infty . \end{aligned}$$
(2)

Consequently, given any \(\Delta _{0}>0\) there are constants \(k_{0}\) and \( t_{0} \) such that

$$\begin{aligned} c(t)\mathbb {P(}X_{t}\in (x,x+\Delta ])\le k_{0}\Delta ,\quad \text { for all }\,\, t\ge t_{0}\,\,\text { and }\,\,\Delta \in (0,\Delta _{0}]. \end{aligned}$$
(3)

We claimed that this result can easily be proved by repeating the argument used for non-lattice random walks in [3], with very minor changes. This fact is indeed true to some extent, but the uniformity in \(\Delta \) is only true in general on intervals [a, b] with \(0<a<b<\infty ;\) and further assumptions are needed to obtain the uniformity on \(0<\Delta <\Delta _{0}\), for \(\Delta _{0}>0.\) Hence for our results in original paper to be valid we require two extra assumptions, namely (H2) and (H3) below.

  1. (H1)

    \(\displaystyle X\in D(\alpha ,\rho ).\)

  2. (H2)

    There exists a \(t_{0}>0\) such that

    $$\begin{aligned} \int _{|\lambda |>1}e^{-t_{0}\mathfrak {R}\Psi (\lambda )}d\lambda <\infty . \end{aligned}$$
  3. (H3)

    X is strongly non-lattice,

    $$\begin{aligned} m=\liminf _{|\lambda |\rightarrow \infty }\mathfrak {R}\Psi (\lambda )>0. \end{aligned}$$

Observe that the assumption (H2) implies that the law of \(X_{t}\) has a density for all \(t>t_{0},\) see e.g. Proposition 2.5 in [2].

Under these assumptions we have the following Lemma which replaces the Proposition 13 in the original paper.

Lemma 2

Assume (H1–3) hold. We have that

$$\begin{aligned} c(t)\mathbb {P}(X_{t}\in (x,x+\Delta ])=\Delta \left( f\left( \frac{x}{c(t)}\right) +o(1)\right) ,\quad \text {as}\,\,\ t\rightarrow \infty , \end{aligned}$$

uniformly in \(x\in \mathbb {R}\), and \(0\le \Delta <\Delta _{0}.\) Consequently (3) holds.

Proof

We would like to estimate

$$\begin{aligned} c(t)\mathbb {P}(X_{t}\in (x,x+\Delta ]) \end{aligned}$$

uniformly in \(x\in \mathbb {R},\) and uniformly in \(|\Delta |<\Delta _{0}\) for \(\Delta _{0}\) fixed. For \(\Delta >0,\) the function

$$\begin{aligned} g_{\Delta ,t}(x)=\frac{1}{\Delta }\mathbb {P}(X_{t}\in (x,x+\Delta ]),\quad x\in \mathbb {R}, \end{aligned}$$

is a probability density function, that of \(X_{t}+U_{\Delta },\) with \(U_{\Delta }\) an independent r.v. with uniform distribution over \((-\Delta ,0).\) Its characteristic function is given by

$$\begin{aligned} \widehat{g}_{\Delta , t}(\lambda ):=\int _{\mathbb {R}}e^{i\lambda x}g_{\Delta ,t}(x)dx=\mathbb {E}\left( e^{i\lambda X_{t}}\right) \frac{(1-e^{-i\lambda \Delta })}{i\lambda \Delta }= e^{-t\Psi (\lambda )}\frac{(1-e^{-i\lambda \Delta })}{i\lambda \Delta }. \end{aligned}$$

The integrability assumption (H2) implies that for \(t>t_{0},\) and \(\Delta >0\)

$$\begin{aligned} \int ^{\infty }_{-\infty }|\widehat{g}_{\Delta , t}(\lambda )|d\lambda <\infty . \end{aligned}$$

By the Fourier inversion theorem we have that for \(\Delta >0, t> t_{0}\)

$$\begin{aligned}&\Delta c(t)g_{\Delta ,t}(x)-\Delta f(x/c(t))\\&\quad =\frac{\Delta }{2\pi }\int ^{\infty }_{-\infty }d\lambda e^{-ix\lambda /c(t)}\left[ e^{-t\Psi (\lambda /c(t))}- e^{-\Psi _{\alpha }(\lambda )}\right] \left( \frac{1-e^{-i\lambda \Delta /c(t)}}{i\lambda \Delta /c(t)}\right) \\&\qquad +\,\frac{\Delta }{2\pi }\int ^{\infty }_{-\infty }d\lambda e^{-ix\lambda /c(t)}e^{-\Psi _{\alpha }(\lambda )}\left( \frac{1- e^{-i\lambda \Delta /c(t)}}{i\lambda \Delta /c(t)}-1\right) . \end{aligned}$$

To estimate this expression we will use among other things the inequalities

$$\begin{aligned} \left| \frac{e^{iu}-1}{iu}\right| \le 1,\quad \left| \frac{e^{iu}-1-iu}{iu}\right| \le \frac{|u|}{2}, \quad u\in \mathbb {R}, \end{aligned}$$

see [2] Lemma 8.6. Using the latter, the second term in the above expression can be estimated by

$$\begin{aligned}&\frac{\Delta }{2\pi }\left| \int ^{\infty }_{-\infty }d\lambda e^{-ix\lambda /c(t)}e^{-\Psi _{\alpha }(\lambda )}\left( \frac{1-e^{-i\lambda \Delta /c(t)}}{i\lambda \Delta /c(t)}-1\right) \right| \\&\quad \le \frac{\Delta }{4\pi }\left( \frac{\Delta _{0}}{c(t)}\right) \int ^{\infty }_{-\infty }d\lambda e^{-\mathfrak {R}\Psi _{\alpha }(\lambda )}|\lambda |. \end{aligned}$$

Since \(\mathfrak {R}\Psi _{\alpha }(\lambda )=|\lambda |^{\alpha }c_{\alpha }\), with \(c_{\alpha }\in (0,\infty )\) a constant, the latter integral is finite, and hence its product with \(\Delta /c(t)\) tends to zero uniformly in x and \(\Delta ,\) as long as \(\Delta \) remains bounded.

Because \(X\in D(\alpha ,\rho )\) we have that

$$\begin{aligned} \lim _{t\rightarrow \infty }|\exp \{-t\Psi (\lambda /c(t))\}-e^{-\Psi _{\alpha }(\lambda )}|=0, \end{aligned}$$

uniformly over closed intervals \([-A,A],\) and also \(\left| \frac{1-e^{-i\lambda \Delta /c(t)}}{i\lambda \Delta /c(t)}\right| \le 1.\) It follows that for any \(A>0\)

$$\begin{aligned} \Delta \left| \int ^{A}_{-A}d\lambda e^{-ix\lambda /c(t)}\left[ e^{-t\Psi (\lambda /c(t))}-e^{-\Psi _{\alpha }(\lambda )}\right] \left( \frac{1-e^{-i\lambda \Delta /c(t)}}{i\lambda \Delta /c(t)}\right) \right| \xrightarrow []{} 0, \quad t\rightarrow \infty , \end{aligned}$$

uniformly in x and \(\Delta <\Delta _{0}\). To finish it will be sufficient to prove that

$$\begin{aligned} \Delta \int _{(-A,A)^{c}}d\lambda e^{-t\mathfrak {R}\Psi (\lambda /c(t))}\rightarrow 0, \end{aligned}$$

uniformly in x, and \(\Delta <\Delta _{0}.\) We proceed as follows. Because the function \(\lambda \mapsto \mathfrak {R}\Psi (\lambda )\) is regularly varying at 0, the Potter bounds, [1] Theorem 1.5.6, ensure that for any \(\alpha >\epsilon >0\) there exists constant K and a \(B_{1}\) such that

$$\begin{aligned} \mathfrak {R}\Psi (\lambda )\ge K\lambda ^{\alpha -\epsilon }, \quad 0\le \lambda <B_{1}. \end{aligned}$$

We apply this inequality to infer

$$\begin{aligned} \Delta \int ^{B_{1}c(t)}_{A}d\lambda e^{-t\mathfrak {R}\Psi (\lambda /c(t))}\le \Delta \int ^{\infty }_{0}d\lambda e^{-K t\left( \frac{\lambda }{c(t)}\right) ^{\alpha -\epsilon }}. \end{aligned}$$

An application of the monotone convergence theorem shows that the latter term tends to 0,  as \(t\rightarrow \infty ,\) because \(c(t)\in RV^{\infty }_{1/\alpha }\) and therefore \(t/(c(t))^{\alpha -\epsilon }\in RV^{\infty }_{\epsilon /\alpha }.\) The convergence is uniform in x,  and in \(\Delta \) on bounded intervals. By symmetry we also get the convergence

$$\begin{aligned} \Delta \int ^{-A}_{-B_{1}c(t)}d\lambda e^{-t\mathfrak {R}\Psi (\lambda /c(t))}\xrightarrow [t\rightarrow \infty ]{}0, \end{aligned}$$

uniformly in x, and in \(\Delta \) on bounded intervals. The assumption of having X strongly non-lattice, implies that, given \(\epsilon >0\) and small enough, there is a \(B_{2}\) such that \(\mathfrak {R}\Psi (\lambda )>m-\epsilon >0\) for all \(|\lambda |>B_{2}.\) By the continuity of \(\mathfrak {R}\Psi (\lambda )\) and the fact that this function does not take the value zero in \(\mathbb {R}{\setminus }\{0\}\), since X is non-lattice, we can assume that \(B_{2}=B_{1},\) maybe at the price of replacing \(m-\epsilon \) by \(0<\widetilde{m}\le m-\epsilon .\) We get that for \(t>t_{0}>0\)

$$\begin{aligned} \Delta \int _{(B_{2}c(t),\infty )}d\lambda e^{-t\mathfrak {R}\Psi (\lambda /c(t))}\le & {} \Delta c(t)\int ^{\infty }_{B_{2}} \exp \{-t\mathfrak {R}\Psi (\lambda )\}{d\lambda }\\\le & {} \Delta c(t)\exp \{-(t-t_{0})\widetilde{m}\} \int ^{\infty }_{B_{2}} \exp \{-t_{0}\mathfrak {R}\Psi (\lambda )\}{d\lambda }. \end{aligned}$$

The rightmost term in the above inequality tends to zero as \(t\rightarrow \infty ,\) because the function \(c(\cdot )\) is regularly varying and hence its growth is at most polynomial. By symmetry we deduce the convergence

$$\begin{aligned} \Delta \int _{(-\infty , -B_{2}c(t))}d\lambda e^{-t\mathfrak {R}\Psi (\lambda /c(t))}\xrightarrow [t\rightarrow \infty ]{}0, \end{aligned}$$

uniformly in x,  and in \(\Delta \) on bounded intervals. The result follows. \(\square \)