Subordinators at First Passage and Renewal Measures

Abstract

In this chapter, we look at subordinators. These are Lévy processes which have paths that are non-decreasing. In addition, we consider killed subordinators, that is, subordinators which are sent to a “cemetery state” at an independent time that is exponentially distributed. Principally, we are interested in first passage over a fixed level and some asymptotic features thereof, as the level tends to infinity. In particular, we will study the (asymptotic) law of the overshoot and undershoot, as well as the phenomenon of crossing a level by hitting it. These three points of interest turn out to be very closely related to renewal measures. The results obtained in this chapter will be of significance later on when we consider first passage over a fixed level of a general Lévy process. As part of the presentation on asymptotic first passage, we will review some basic facts about regular variation. Regular variation will also be of use in later chapters. We conclude with a brief introduction to the theory of special subordinators which, amongst other things, permits the construction of a number of concrete examples of some of the theory discussed earlier in the chapter.

Exercises

5.1

In this exercise, we shall derive the form of the Laplace exponent of a killed subordinator given in (5.1), without appealing to the Lévy–Itô decomposition. To this end, suppose that X={X _t :t≥0} is a [0,∞]-valued stochastic process which has non-decreasing, right-continuous paths with left limits. Here, +∞ serves as a cemetery state. Denote its lifetime by

$$\zeta= \inf\{t>0 : X_t =\infty\}. $$

Suppose that under measure $\mathbb{P}$, X has the property that, for all s,t≥0, on the event {t<ζ}, the increment X _t+s −X _t is independent of {X _u :u≤t} and equal in distribution to X _s .

• (i) By agreeing to write e^−∞=0, show that

  $$\mathbb{E}\bigl(\mathrm{e}^{-\theta X_{t+s}}\bigr) = \mathbb{E}\bigl(\mathrm {e}^{-\theta X_t}\bigr)\mathbb{E}\bigl(\mathrm{e}^{-\theta X_s}\bigr), $$

  where θ,s,t≥0 and $\mathbb{E}$ denotes expectation with respect to $\mathbb{P}$. Hence, deduce that, for θ,t≥0,

  $$\mathbb{E}\bigl(\mathrm{e}^{-\theta X_t}\bigr) = \mathrm {e}^{-\varPhi (\theta) t}, $$

  where $\varPhi(\theta) = -\log\mathbb{E}(\mathrm{e}^{-\theta X_{1}})$.

• (ii) Prove that, for θ≥0,

  $$\frac{\varPhi(\theta)}{\theta} = \lim_{n\uparrow\infty}\int _0^\infty {\mathrm{e}}^{-\theta x} n\mathbb{P}(X_{1/n}\geq x)\mathrm{d}x. $$

• (iii) Hence, deduce that, for θ≥0,

  $$\varPhi(\theta) = \eta+ \delta\theta+ \int_{(0,\infty)} \bigl(1- \mathrm{e}^{-\theta x}\bigr)\varPi(\mathrm{d}x), $$

  where $\varPi(\mathrm{d}x) = -\overline{\varPi}(\mathrm{d}x)$, δ≥0 and η≥0 are uniquely identified.

• (iv) Explain why ∫_(0,∞)(1∧x)Π(dx)<∞.

5.2

Suppose that, under $\mathbb{P}$, X={X _t :t≥0} is a (killed) subordinator with Laplace exponent Φ, just as in part (iii) of Exercise 5.1. Define for q≥0,

$$\frac{\mathrm{d}\mathbb{P}^q}{\mathrm{d}\mathbb{P}} \bigg\vert _{\mathcal{F}_t} := \exp\bigl\{ - q X_t +\varPhi(q)t\bigr\} , \quad t\geq 0, $$

where, as usual, $\{\mathcal{F}_{t}: t\geq0\}$ is the natural filtration generated by X. Show that $(X, \mathbb{P}^{q})$ is a subordinator without killing, the same drift coefficient as $(X, \mathbb{P})$ and Lévy measure given by e^−qx Π(dx), x>0, where Π is the Lévy measure of $(X, \mathbb{P})$.

5.3

Prove Theorem 5.7.

5.4

Suppose that Y is a spectrally positive Lévy process of bounded variation drifting to −∞, with Laplace exponent written in the usual form

$$\log\mathbb{E}\bigl(\mathrm{e}^{-\theta Y_1}\bigr) = \psi(\theta ) = {\delta} \theta- \int_{(0,\infty)}\bigl(1- \mathrm{e}^{-\theta x}\bigr) \nu({\mathrm{d}}x), $$

where necessarily δ>0, ∫_(0,∞)(1∧x)ν(dx)<∞ and ψ′(0+)>0. Define $\sigma^{+}_{x} = \inf\{t>0 : Y_{t} >x\}$ and $\overline{Y}_{t} = \sup_{s\leq t}Y_{s}$.

• (i) Suppose that X={X _t :t≥0} is a compound Poisson subordinator with jump distribution (δ−ψ′(0+))⁻¹ ν(x,∞)dx. By following similar reasoning to the explanation of the Pollaczek–Khintchine formula in Chap. 4, show that

  $$\begin{aligned} &\mathbb{P}\bigl(Y_{\sigma^+_x}- x \in{\mathrm{d}}u, x - \overline{Y}_{\sigma^+_x -} \in{\mathrm{d}}y \,|\, \sigma _x^+<\infty\bigr) \\ &\quad= \mathbb{P}(X_{\tau^+_x} - x \in{\mathrm{d}}u, x- X_{\tau ^+_x -}\in{\mathrm{d}}y). \end{aligned}$$

• (ii) Deduce that if $\int_{0}^{\infty}x\nu(x,\infty){\mathrm{d}}x<\infty$, then, for u,y>0, in the sense of weak convergence,

  $$\begin{aligned} &\lim_{x\uparrow\infty} \mathbb{P} \bigl(Y_{\sigma^+_x}- x \in{\mathrm{d}}u, x -\overline{ Y}_{\sigma ^+_x -} \in {\mathrm{d}}y | \sigma_x^+<\infty\bigr) \\ &\quad= \frac{1}{ \int_0^\infty x\nu(x,\infty){\mathrm{d}}x}\nu(u+y,\infty){\mathrm{d}}u\, {\mathrm{d}}y. \end{aligned}$$

• (iii) Give an interpretation of the result in (ii) in the context of modelling insurance claims.

5.5

Suppose that X is a finite mean subordinator and that its associated potential measure U does not have lattice support. Suppose that Z is a random variable whose distribution is equal to that of the limiting distribution of $X_{\tau^{+}_{x} } - X_{\tau ^{+}_{x} -}$ as x↑∞. Suppose further that (V,W) is a bivariate random variable whose distribution is equal to the limiting distribution of $(X_{\tau^{+}_{x}} - x, x- X_{\tau^{+}_{x} -})$ as x↑∞, and that U is independent of V,W,Z and uniformly distributed on [0,1]. Show that (V,W) is equal in distribution to ((1−U)Z,UZ).

5.6

Let X and Y be two (possibly correlated) subordinators killed independently at the rate η≥0. Denote their bivariate jump measure by Π(⋅,⋅). Define their bivariate renewal function

$$\mathcal{U}({\mathrm{d}}x,{\mathrm{d}}y) = \int_0^\infty{\mathrm{d}}t \cdot\mathbb{P}(X_t \in{\mathrm{d}}x, Y_t \in{\mathrm{d}}y), \quad x,y \geq0, $$

and suppose, as usual, that

$$\tau^+_x = \inf\{t>0 : X_t >x\}, \quad x\geq0. $$

Use a generalised version of the compensation formula to establish the following quadruple law

$$\begin{aligned} &P(\varDelta X_{\tau^+_x} \in{\mathrm{d}}t, X_{\tau^+_x -}\in{\mathrm {d}}s, x - Y_{\tau^+_x -}\in{\mathrm{d}}y, Y_{\tau^+_x} - x \in{\mathrm {d}}u) \\ &\quad=\mathcal{U}({\mathrm{d}}s,x- {\mathrm{d}}y)\boldsymbol{\Pi } ({\mathrm{d}}t, {\mathrm{d}}u + y), \end{aligned}$$

for u>0, y∈[0,x] and s,t≥0. This formula will be of use later on when considering the first passage of a general Lévy process over a fixed level.

5.7

Let X be any subordinator with Laplace exponent Φ, drift coefficient δ≥0 and recall that $\tau^{+}_{x} = \inf\{t>0 : X_{t} > x\}$. Let e _α be an exponentially distributed random variable with rate α, which is independent of X.

• (i) By applying the strong Markov property at time $\tau^{+}_{x}$ in the expectation $\mathbb{E}(\mathrm{e}^{-\beta X_{\mathbf{\mathrm{e}}_{\alpha}}} \mathbf{1}_{(X_{\mathbf{\mathrm{e}}_{\alpha}} >x)})$, show that, for all α,β,x≥0, we have

  $$ \mathbb{E} \bigl( \mathrm{e}^{-\alpha\tau_{x}^{+}-\beta X_{\tau _{x}^{+}}} \bigr) =\bigl(\alpha+\varPhi(\beta) \bigr)\int_{(x,\infty )}\mathrm{e}^{-\beta z}U^{(\alpha)}({ \mathrm{d}}z), $$

  (5.37)

  for all x>0.

• (ii) Show further, with the help of the identity in (i), that, when q>0 and β≥0,

  $$\int_{0}^{\infty}\mathrm{e}^{-qx} \mathbb{E} \bigl( \mathrm{e}^{-\alpha\tau_{x}^{+}-\beta (X_{\tau_{x}^{+}}-x)} \bigr) {\mathrm{d}}x \\ =\frac{1}{q-\beta} \biggl( 1-\frac{\alpha+\varPhi( \beta ) }{\alpha+\varPhi( q ) } \biggr). $$

• (iii) Deduce, with the help of Theorem 5.9, that

  $$\mathbb{E}\bigl(\mathrm{e}^{-\alpha\tau^+_x}\mathbf{1}_{(X_{\tau^+_x} = x)} \bigr)= { \delta}u^{(\alpha)}(x), $$

  where, if δ=0, the term u ^(α)(x) may be taken as equal to zero and, otherwise, the potential measure U ^(α) has a density such that u ^(α) is a continuous and strictly positive version thereof.

• (iv) Show that for this version of the density, u ^(α)(0+)=1/δ, where δ is the drift of X.

5.8

Suppose that X is a stable subordinator with parameter α∈(0,1), thus having Laplace exponent Φ(θ)=cθ ^α, for θ≥0 and some c>0. In this exercise, we will take c=1.

• (i) Show from Exercise 1.4 that the precise expression for the jump measure is

  $$\varPi({\mathrm{d}}x) = \frac{x^{-(1+\alpha)}}{-\varGamma(-\alpha )} {\mathrm{d} }x,\quad x>0. $$

• (ii) By considering the Laplace transform of the potential measure U, show that

  $$U({\mathrm{d}}x) = \frac{x^{\alpha- 1}}{\varGamma(\alpha)}{\mathrm {d}}x, \quad x\geq0. $$

• (iii) Hence deduce that

  $$\begin{aligned} &\mathbb{P}(X_{\tau^+_x} - x\in{\mathrm{d}}u, x - X_{\tau^+_x -}\in{\mathrm{d}}y) \\ &\quad= \frac{\alpha\sin\alpha\pi}{\pi} (x-y)^{\alpha-1} (y+u)^{-(\alpha + 1)}{\mathrm{d}}u\,{\mathrm{d}}y, \end{aligned}$$

  for u>0 and y∈[0,x]. Note further that the distribution of the pair

  $$ \biggl( \frac{x - X_{\tau^+_x -}}{x}, \frac{X_{\tau^+_x} -x}{x} \biggr) $$

  (5.38)

  is independent of x.

• (iv) Show directly that stable subordinators do not creep.

5.9

Suppose that X is any subordinator.

• (i) Use the joint law of the overshoot and undershoot to deduce that, for β,γ≥0 and q>0,

  $$\begin{aligned} &\int_0^\infty\mathrm{d}x\cdot\mathrm{e}^{-q x} \mathbb{E}\bigl(\mathrm{e}^{-\beta X_{\tau^+_x - } - \gamma (X_{\tau ^+_x} - x)}\mathbf{ 1}_{(X_{\tau^+_x }>x)}\bigr) \\ &\quad= \frac{1}{q-\gamma} \biggl(\frac{\varPhi(q)- \varPhi(\gamma)}{\varPhi(q+\beta)} \biggr) - \frac{\delta }{\varPhi (q+\beta)}. \end{aligned}$$

• (ii) Taking account of creeping, use part (i) to deduce that

  $$\int_0^\infty{\mathrm{d}}x\cdot\mathrm{e}^{-q x} \mathbb{E}\bigl(\mathrm{e}^{-\beta(X_{\tau^+_{tx} - }/t) - \gamma (X_{\tau^+_{tx}} - tx)/t }\bigr) = \frac{1}{(q-\gamma)} \frac{\varPhi(q/t)- \varPhi(\gamma/t)}{\varPhi((q+\beta)/t)}, $$

  for β,γ≥0 and q>0.

• (iii) Show that if $\varPhi\in\mathcal{R}_{0}(\alpha)$ (resp. $\varPhi\in\mathcal{R}_{\infty}(\alpha)$) with α equal to 0 or 1, then the limiting distribution of the pair in (5.38) is trivial as x tends to infinity (resp. zero).

• (iv) It is possible to show that, if for a given measurable function f:[0,∞)→(0,∞), there exists a g:(0,λ)→(0,∞) such that

  $$\lim\frac{f(\lambda t)}{f(t)}=g(\lambda), $$

  for all λ>0, as t tends to zero (resp. infinity), then f must be regularly varying. Roughly speaking, the reason for this is that, for λ,μ>0,

  $$g(\lambda\mu) = \lim\frac{f(\mu\lambda t)}{f(\lambda t)}\frac{f(\lambda t)}{f(t)} = g(\mu)g(\lambda) $$

  showing that g is a multiplicative function. With a little measure theory, one can show that g(λ)=λ ^ρ, for some $\rho\in\mathbb{R}$. See Theorem 1.4.1 of Bingham et al. (1987) for the full details.

  Use the above remarks to deduce that, if (5.38) has a limiting distribution as x tends to infinity (resp. zero), then necessarily $\varPhi\in\mathcal{R}_{0}(\alpha)$ (resp. $\varPhi\in\mathcal{R}_{\infty}(\alpha)$) with α∈[0,1]. Hence conclude that (5.38) has a non-trivial limiting distribution if and only if α∈(0,1).

5.10

Suppose that F is a probability distribution function. Write $\overline{F}(x) = 1- F(x)$. Then F belongs to $\mathcal{L}^{(\alpha)}$, where α≥0, if the support of F is non-lattice in [0,∞), $\overline{F}(x)>0$ for all x≥0 and, for all y>0,

$$\lim_{x\uparrow\infty}\frac{\overline{F}(x+y)}{\overline{F(x)}} = \mathrm{e}^{-\alpha y}. $$

Note that the requirement that F is a probability measure can be weakened to a finite measure, as one may always normalise by its total mass to fulfil the conditions given earlier.

We are interested in establishing an asymptotic conditional distribution for the overshoot of a killed subordinator. To this end, we assume that X is a killed subordinator with killing rate η>0, Laplace exponent Φ, jump measure Π, drift δ≥0 and potential measure U which is assumed to belong to class $\mathcal{L}^{(\alpha)}$, for some α≥0 such that Φ(−α)<∞.

• (i) Show that, for x>0,

  $$\mathbb{P}\bigl(\tau^+_x <\infty\bigr) = \eta U(x,\infty), $$

  where $\tau^{+}_{x} = \inf\{t>0 :X_{t} >x\}$.

• (ii) Show that, for all β≥0,

  $$\mathbb{E}\bigl(\mathrm{e}^{-\beta(X_{\tau^+_x} - x) } | \tau^+_x <\infty\bigr) = \frac{\varPhi(\beta)}{\eta U(x,\infty)} \int_{(x,\infty)} \mathrm {e}^{-\beta(y-x)} U({\mathrm{d}}y). $$

• (iii) Applying integration by parts, deduce that

  $$\lim_{x\uparrow\infty}\mathbb{E}\bigl(\mathrm{e}^{-\beta(X_{\tau^+_x} - x) } | \tau^+_x <\infty\bigr) = \frac{\varPhi(\beta)}{\eta} \biggl( \frac{\alpha}{\alpha+ \beta} \biggr). $$

• (iv) Now take the distribution G on [0,∞), defined by its tail

  $$G(x,\infty) = \frac{\mathrm{e}^{-\alpha x}}{\eta} \biggl\{ \varPhi (-\alpha) + \int _{(x,\infty)} \bigl(\mathrm{e}^{\alpha y} - \mathrm{e}^{\alpha x}\bigr)\varPi({\mathrm{d}}y) \biggr\} . $$

  Show that G has an atom at zero and

  $$\int_{(0,\infty)} \mathrm{e}^{-\beta y}G({\mathrm{d}}y) = \frac{\varPhi(\beta)}{\eta} \biggl( \frac{\alpha}{\alpha +\beta} \biggr) - \frac{{\delta}\alpha}{\eta}. $$

• (v) Deduce that, for all u>0,

  $$\lim_{x\uparrow\infty} \mathbb{P}\bigl(X_{\tau^+_x} - x >u | \tau^+_x <\infty\bigr) = {G}(u,\infty) $$

  and

  $$\lim_{x\uparrow\infty} \mathbb{P}\bigl(X_{\tau^+_x}=x | \tau^+_x <\infty\bigr) = \frac{\delta\alpha}{\eta}. $$

5.11

Suppose that f is a completely monotone function.

• (i) If f is another completely monotone function show that αf+βg is completely monotone for all α,β≥0 as well as fg.

• (ii) Suppose that Φ is a Bernstein function. Show that f∘Φ is completely monotone.

5.12

This exercise gives two more examples of complete subordinators for which the analysis in Sect. 5.6 is completely tractable.

• (i) Consider the, apparently trivial, Bernstein function

  $$\varPhi(\theta) = \eta+ \delta\theta, \quad\theta\geq0, $$

  where δ,η>0. This corresponds to the subordinator which is a deterministic linear drift killed at rate η. Show that

  $$U(x) = \frac{1}{\eta} \bigl(1 - \mathrm{e}^{- x\eta/\delta}\bigr), \quad x \geq0, $$

  and hence deduce that δ ^∗=η ^∗=0, Π ^∗(x,∞)=δ ⁻¹e^−xη/δ and U ^∗(x)=δ+ηx for x≥0.

• (ii) Now consider, for ν∈(0,1) and γ>0,

  $$\varPhi(\theta): = \eta+ \lambda\biggl( 1- \biggl(\frac{\gamma }{\gamma +\theta} \biggr)^\nu\biggr),\quad\theta\geq0, $$

  where η,λ>0. Show that Φ is a complete Bernstein with components δ=0, the killing rate is η,

  $$\varPi(\mathrm{d}x) = \frac{\lambda\gamma^\nu}{\varGamma(\nu )}x^{\nu-1}{\mathrm {e}}^{-\gamma x}\mathrm{d} x,\quad x>0, $$

  and

  $$U(x) =\frac{1}{\lambda+\eta} + \frac{\rho\gamma^\nu}{\lambda +\eta} \int_0^x \mathrm{e}^{-\gamma y} y^{\nu-1}{\mathrm{E}}_{\nu,\nu}\bigl( \rho\gamma^\nu y^\nu\bigr)\mathrm{d}y, $$

  where ρ=λ/(λ+η).

  Hence deduce that η ^∗=0, δ ^∗=1/(η+λ),

  $$\varPi^*(x,\infty) =\frac{\rho\gamma^\nu}{\lambda+\eta} {\mathrm {e}}^{-\gamma x} x^{\nu-1}{\mathrm{E}}_{\nu,\nu}\bigl(\rho\gamma^\nu x^\nu\bigr) $$

  and

  $$U^*(x) = \eta x + \frac{\lambda\gamma^\nu}{\varGamma(\nu)}\int_0^x \biggl\{ \int_y^\infty z^{\nu-1} { \mathrm{e}}^{-\gamma z}\mathrm{d}z \biggr\} \mathrm{d}y. $$

5.13

This exercise concerns another transformation for Bernstein functions which produces again a Bernstein function. The origins and more details of this transformation can be found in Urbanik (2005), Gnedin (2010) and Chazal et al. (2012).

• (i) Suppose that Φ is the Laplace exponent of a subordinator with triple (η,δ,Π). Show that, for θ,β≥0,

  $$\phi(\theta) : =\frac{\theta}{\theta+\beta}\varPhi(\theta+ \beta), \quad\theta\geq0, $$

  is the Laplace exponent of a subordinator.

• (ii) Show moreover that the triple of ϕ is equal to (0,δ,ν), where

  $$\nu(\mathrm{d}x) = \beta{\mathrm{e}}^{-\beta x}\varPi(x,\infty )\mathrm{d}x + \mathrm{e}^{-\beta x}\varPi(\mathrm{d}x) + \eta\beta{\mathrm {e}}^{-\beta x}\mathrm{d}x, $$

  for x>0.

5.14

In this exercise, we show that the proof of Theorem 5.9 is relatively straightforward for special subordinators. To this end, suppose that Φ is a special Bernstein function with representation (5.1). Assume that Π(0,∞)=∞. Recall from Theorem 5.19 that its conjugate has a potential density on (0,∞), denoted by u ^∗(x), which satisfies u ^∗(x)=η+Π(x,∞).

• (i) By considering the factorisation θ ⁻¹=Φ(θ)⁻¹×Φ(θ)/θ for θ>0, show that, for all x>0,

  $$1 = \delta u(x) + \int_0^x u(x-y)u^*(y)\mathrm{d}y. $$

• (ii) Deduce with the help of Theorem 5.6 that, for the killed subordinator X with Laplace exponent Φ,

  $$\mathbb{P}(X_{\tau^+_x} = x) = \delta u(x),\quad x>0. $$