Law of the First Passage Triple of a Spectrally Positive Strictly Stable Process

Abstract

For a spectrally positive and strictly stable process with index in (1, 2), a series representation is obtained for the joint distribution of the “first passage triple” that consists of the time of first passage and the undershoot and the overshoot at first passage. The result leads to several corollaries, including (1) the joint law of the first passage triple and the pre-passage running supremum, and (2) at a fixed time point, the joint law of the process’ value, running supremum, and the time of the running supremum. The representation can be decomposed as a sum of strictly positive functions that allow exact sampling of the first passage triple.

Appendix

On the connection between (9) and (10) When $\alpha =2$, $\sin (\pi k/\alpha )$ is 0 if k is even and is $(-1)^j$ is $k = 2j + 1$ for integer $j\ge 0$. Then, the series in (9) can be written as

$$\begin{aligned} \frac{1}{\pi } \sum _{j=0,n=1}^\infty \frac{\Gamma (j+1/2+n)}{(2j+1)!(2n-1)!} (-1)^{j+n+1} [(2j+1)c + (2n-1)(c-x)](c-x)^{2j} c^{2n-2}. \end{aligned}$$

Write $n=l+1$ and $m=j+l$. Then, the series becomes

$$\begin{aligned}&\frac{1}{\pi } \sum _{j,l=0}^\infty \frac{\Gamma (j + l+3/2)}{(2j+1)!(2l+1)!} (-1)^{j+l} [(2j+1)(c-\!x)^{2j} c^{2l+\!1} \!+\! (2l+1) (c-\!x)^{2j+1} c^{2l}]\\&= \frac{1}{\pi } \sum _{m=0}^\infty \frac{\Gamma (m+3/2)}{(2m+1)!} (-1)^m \sum ^m_{j=0}\left[ \frac{(c-x)^{2j} c^{2m-2j+1}}{(2j)!(2m-2j+1)!} + \frac{(c-x)^{2j+1} c^{2m-2j}}{(2j+1)!(2m-2j)!} \right] \\&= \frac{1}{\pi } \sum _{m=0}^\infty \frac{\Gamma (m+3/2)}{(2m+1)!} (-1)^m \sum ^{2m+1}_{s=0} \frac{(2m+1)!}{s!(2m+1-s)!} (c-x)^s c^{2m+1-s}\\&= \frac{1}{\pi } \sum _{m=0}^\infty \frac{\sqrt{\pi }}{2^{2m+1} m!}(-1)^m (2c-x)^{2m+1}\\&= \frac{2c-x}{2\sqrt{\pi }}\exp \left\{ -\frac{(2c-x)^2}{4}\right\} . \end{aligned}$$

Since $(X_t)_{t\ge 0}\sim (W_{2t})_{t\ge 0}$, this is essentially the same result as (10). $\square$

Proof of Eq. (16) We need the following refined version of Lemma 16(a).

Lemma 21

There is a constant $M>0$, such that for all $0<x<1/2$ and all $t>0$,

$$\begin{aligned} h_1(1-x,t) \le M x(t+t^{1-{1/\alpha }}). \end{aligned}$$

Assume the lemma is true for now. Then, given $c>0$, by scaling, for all $0<x<c/2$, $h_c(c-x, t)\le M x(t+t^{1-{1/\alpha }})$ for some $M = M(c)>0$. Then, by Lemma 16(a) and dominated convergence, for each $q>0$,

$$\begin{aligned} \int _0^\infty m(c,t) e^{-q t}\,\mathrm {d}t = \lim _{x\rightarrow 0} \frac{1}{x}\int _0^\infty h_c(c-x,t) e^{-q t}\,\mathrm {d}t. \end{aligned}$$

However, by scaling (6) and Proposition 10, for $0<x<c/2$,

$$\begin{aligned} \frac{1}{x}\int _0^\infty h_c(c-x,t) e^{-q t}\,\mathrm {d}t&= \frac{e^{-q^{1/\alpha }x}-1}{x} \sum _{n=1}^\infty \frac{q^{n-1} c^{\alpha n-1}}{\Gamma (\alpha n)}\\&\quad + \sum _{n=1}^\infty \frac{q^{n-1} [c^{\alpha n - 1} -(c-x)^{\alpha n-1}]}{x\Gamma (\alpha n)}. \end{aligned}$$

As a result,

$$\begin{aligned} \int _0^\infty m(c,t) e^{-q t}\,\mathrm {d}t = \sum _{n=1}^\infty \frac{q^{n-1} c^{\alpha n - 2}}{\Gamma (\alpha n-1)} - q^{{1/\alpha }} \sum _{n=1}^\infty \frac{q^{n-1} c^{\alpha n-1}}{\Gamma (\alpha n)}. \end{aligned}$$

Provided that $\beta > q^{1/\alpha }$, integration term by term of the r.h.s. yields

$$\begin{aligned} \int _0^\infty \left( \int _0^\infty m(c,t) e^{-q t}\,\mathrm {d}t\right) e^{-\beta c}\,\mathrm {d}c = \sum _{n=1}^\infty \frac{q^{n-1}}{\beta ^{\alpha n-1}} - q^{1/\alpha }\sum _{n=1}^\infty \frac{q^{n-1}}{\beta ^{\alpha n}} = \frac{\beta - q^{1/\alpha }}{\beta ^\alpha - q}. \end{aligned}$$

By analytic extension, the equality still holds for $0\le \beta \le q^{1/\alpha }$. Then, by (15), the proof is complete. $\square$

Proof of Lemma 21

By (22) and integral by parts,

$$\begin{aligned} h_1(1-x,t) = g_t(1-x) - g_t(1) + \int ^t_0 \overline{F}_{-x}(t-s) \frac{\partial g_s(1)}{\partial s}\,\mathrm {d}s, \end{aligned}$$

(45)

where $\overline{F}_{-x}(t) = \int ^\infty _t f_{-x}(s)\,\mathrm {d}s = \mathbb {P}\{\tau _{-x} > t\}$. For $0<x<1/2$, $g_t(1-x) - g_t(1) = - g'_t(z) x$ for some $z\in (1-x,1)$. Clearly, $z>1/2$. It is not hard to show that $M_1 := \sup _{y>0} [y^{\alpha + 2} |g'_1(y)|]<\infty$ ([18], p. 88). On the other hand, by $g_t(z) = t^{-{1/\alpha }} g_1(t^{-{1/\alpha }} z)$, $g'_t(z) = t^{-2/\alpha } g'_1(t^{-{1/\alpha }} z)$. Then,

$$\begin{aligned} |g_t(1-x) - g_t(1)| = x t^{-2/\alpha } |g'_1(t^{-{1/\alpha }} z)| \le x t^{-2/\alpha } M_1 (t^{-{1/\alpha }} z)^{-\alpha -2} \le M_1 2^{\alpha + 2} x t. \end{aligned}$$

(46)

Next, by $g_s(1) = s^{-{1/\alpha }} g_1(s^{-{1/\alpha }})$, $|\partial g_s(1)/\partial s| \le ({1/\alpha }) [s^{-{1/\alpha }-1} g_1(s^{-{1/\alpha }}) + s^{-2/\alpha -1} |g'_1(s^{-{1/\alpha }})|]$ is bounded. Then, for some $M_2>0$,

$$\begin{aligned} \left| \int ^t_0 \overline{F}_{-x}(t-s) \frac{\partial g_s(1)}{\partial s}\,\mathrm {d}s \right| \le M_2 \int ^t_0 \overline{F}_{-x}(s)\,\mathrm {d}s = M_2 x^\alpha \int ^{x^{-\alpha } t}_0 \overline{F}_{-1}(s) \,\mathrm {d}s, \end{aligned}$$

where the equality is due to $\overline{F}_{-x}(s) = \overline{F}_{-1}(x^{-\alpha } s)$ and change of variable. Because $\overline{F}_{-1}(s)$ is decreasing with $\overline{F}_{-1}(0)=1$ and is slowly varying at $\infty$ with index $-{1/\alpha }$, there is a constant $M_3>0$ such that $\int ^y_0 \overline{F}_{-1}(s)\,\mathrm {d}s \le M_3 y^{1-{1/\alpha }}$ for all $y>0$. It follows that

$$\begin{aligned} \left| \int ^t_0 \overline{F}_{-x}(t-s) \frac{\partial g_s(1)}{\partial s}\,\mathrm {d}s \right| \le M_2 M_3 x t^{1-{1/\alpha }}. \end{aligned}$$

(47)

Then, the proof is complete by combining (45)–(47). $\square$

Proof of Eq. (20) Denote the r.h.s. of (20) by $v^q(x)$. The task is to show $\widehat{v^q} = \widehat{u^q}$, where, for example, $\widehat{v^q}(x) = {v^q}(-x)$. Since $v^q$ is a version of the q-resolvent density, according to the proof of Proposition I.13 of [2], $(r-q) U^r \widehat{v^q}\uparrow \widehat{u^q}$ as $r\rightarrow \infty$, where $U^r$ is the r-resolvent operator. For $r>q$,

$$\begin{aligned} U^r \widehat{v^q}(x)&= \int _0^\infty e^{-r t} \mathbb {E}^x[\widehat{v^q}(X_t)]\,\mathrm {d}t\\&= \int _0^\infty e^{-r t} \left[ \int \left( \int _0^\infty e^{-q s} g_s(-y)\,\mathrm {d}s\right) g_t(y-x)\,\mathrm {d}y\right] \,\mathrm {d}t\\&= \int _0^\infty \int _0^\infty e^{-r t -q s} g_{s+t}(-x)\,\mathrm {d}s\,\mathrm {d}t\\&= (r-q)^{-1} \int _0^\infty (1-e^{(q-r) s}) e^{-q s}g_s(-x)\,\mathrm {d}s. \end{aligned}$$

Then, by monotone convergence, $(r-q) U^r \widehat{v^q}(x)\rightarrow \widehat{v^q}(x)$, giving $\widehat{v^q}(x) = \widehat{u^q}(x)$. $\square$