Positive Self-similar Markov Processes

Abstract

In this chapter, our objective is to explore in detail the general class of so-called positive self-similar Markov processes. Emphasis will be placed on the bijection between this class and the class of Lévy processes which are killed at an independent and exponentially distributed time. This bijection can be expressed through a straightforward space-time transformation and, thereby it, we are able to explore a number of specific examples of positive self-similar Markov processes, which illuminate a variety of explicit and semi-explicit fluctuation identities for Lévy processes. Our first such family of examples will be positive self-similar Markov processes that are obtained when considering path transformations of stable processes and conditioned stable processes. Here, the underlying associated Lévy processes are known as Lamperti-stable processes. Known properties of stable processes, when transferred through the aforementioned space-time transform, will give us explicit fluctuation identities for Lamperti-stable processes; in particular, we will obtain their Wiener–Hopf factorisation. Another family of examples we will consider is continuous-state branching processes and continuous-state branching processes with immigration, which are also self-similar.

Whilst our exposition of general positive self-similar Markov processes will, in the beginning, insist that their initial value lies in (0,∞), we will also look at the more complicated case that the point of issue is the origin. This discussion leads us to the concept of recurrent extensions of positive self-similar Markov processes. With this theory in hand, we will conclude the chapter by looking at elements of fluctuation theory for positive self-similar Markov processes associated with spectrally negative Lévy processes.

Exercises

13.1

Suppose that Y={Y _t :t≥0} is an α-stable process. Define the occupation time of (0,∞),

$$A_t = \int_0^t \mathbf{1}_{(X_s > 0)} \mathrm{d}s , \quad t\geq0, $$

and let γ(t):=inf{s≥0:A _s >t} be its right-continuous inverse. Show that $Y_{\gamma(t)} \mathbf{1}_{(t<T_{0})}$, t≥0, is a positive self-similar Markov process, where T ₀=inf{t>0:Y _γ(t)=0}.

13.2

Suppose that X={X _t :t≥0} is an α-stable subordinator, so that (necessarily) α∈(0,1). Use the method of looking at the asymptotic overshoot of X at first entry into (y,∞), as y↑∞, to deduce that the Lévy process that appears in the second Lamperti transform is a subordinator with no killing, no drift and jump measure ν satisfying

$$\nu(\mathrm{d}x) = c\frac{\mathrm{e}^x}{(\mathrm{e}^x - 1)^{\alpha +1}}{\mathrm{d}}x, \quad x>0, $$

where c>0 is a constant.

13.3

Consider the case that B={B _t :t≥0} is a standard Brownian motion. Use first-passage problems to deduce the following facts.

• (i) Set $\tau^{-}_{0} = \inf\{t>0: B_{t}<0\}$. The process $\{ B_{t}\mathbf{1}_{(t<\tau^{-}_{0})} : t\geq0\}$ is a positive self-similar Markov process driven through the second Lamperti transform by a constant multiple of a standard Brownian motion with drift −1/2.

• (ii) Fix x>0. Suppose that $\mathbb{P}^{\uparrow}_{x}$ is the law of B conditioned to stay positive. Show that the process $(B, \mathbb{P}_{x}^{\uparrow})$ is a positive self-similar Markov process driven through the second Lamperti transform by a constant multiple of standard Brownian motion with drift 1/2.

In both cases, how might one deduce that the unspecified constant is equal to 1?

13.4

Suppose that $(X,\mathbb {P})$ is a Lévy process which satisfies lim sup _t↑∞ X _t =∞. We exclude the case that X is a compound Poisson process. Fix 0<y≤x. Write, as usual, $\tau^{-}_{y} = \inf\{t>0: X_{t} <y\}$. By considering the event $\{\tau^{-}_{y} <\infty\}$ under $\mathbb {P}^{\uparrow}_{x}$ and the computation in (13.13), show that

$$\mathbb{P}^\uparrow_x\Bigl(\inf_{s\geq0}X_s \geq y\Bigr) = \frac{\widehat {U}(x-y)}{\widehat{U}(x)}, \quad x,y\geq0. $$

Deduce that the law of the global minimum of a standard Brownian motion conditioned to stay positive, with initial value x>0, is uniformly distributed on (0,x).

13.5

In this exercise, our aim is to follow Chaumont (1996) in constructing the law of a stable process conditioned to be absorbed continuously at the origin. To this end, we suppose that Y={Y _t :t≥0} is an α-stable process and that, for all $x\in\mathbb{R}$, $\tau^{-}_{x} = \inf\{t>0: Y_{t}<x\}$. Following our usual notation, we shall also write $\underline{Y}_{t} = \inf_{s\leq t}Y_{s}$.

• (i) Show that, for any ε,x>0,

  $$\mathbb{P}_x(\underline{Y}_{\tau^-_0 - }\leq\varepsilon) \propto \int_0^\varepsilon\frac{(x-u)^{\alpha(1-\rho)-1}}{u^{\alpha (1-\rho)}}{\mathrm{d}}u. $$

• (ii) Deduce that, for x,y>0,

  $$\lim_{\varepsilon\downarrow0} \frac{\mathbb{P}_x(\underline {Y}_{\tau^-_0 - }\leq\varepsilon)}{\mathbb{P}_y(\underline {Y}_{\tau^-_0 - }\leq\varepsilon)}= \biggl(\frac{x}{y} \biggr)^{\alpha(1-\rho)-1}. $$

• (iii) Now suppose that A belongs to the sigma-algebra generated by {Y _s :s≤t}. Show, using the Markov property, that, for all x,t,η>0 and 0<ε<η,

  $$\mathbb{P}_x\bigl(A, \, t<\tau^-_\eta| \underline{Y}_{\tau^-_0 - } \leq \varepsilon\bigr) = \mathbb{E}_x \biggl( \mathbf{1}_{(A, \, t<\tau_{(0,\eta)})}\frac{\mathbb{P}_{Y_{\tau _{(0,\eta)}}}(\underline{Y}_{\tau^-_0 -} \leq\varepsilon)}{\mathbb {P}_x(\underline{Y}_{\tau^-_0 -}\leq\varepsilon)} \biggr), $$

  where τ _(0,η)=inf{t>0:Y _t ∈(0,η)}.

• (iv) Now assume that, for all x,t>0, $\mathbb {E}_{x}(Y_{t}^{\alpha(1-\rho)-1}\mathbf{1}_{(t<\tau^{-}_{0})}) = x^{\alpha (1-\rho)-1}$. Show that

  $$\begin{aligned} \lim_{\varepsilon\downarrow0}\mathbb{P}_x\bigl(A, \, t< \tau^-_\eta| \underline{Y}_{\tau^-_0 - } \leq\varepsilon\bigr) =& \mathbb{E}_x \biggl(\mathbf{1}_{(A, \,t<\tau_{(0,\eta)})} \frac{ X^{\alpha(1-\rho)-1}_{t} }{x^{\alpha(1-\rho)-1}} \biggr). \end{aligned}$$

13.6

This exercise is concerned with the proof of Lemma 13.4. Suppose that ξ is a Lévy process which is killed at rate q≥0.

• (i) Suppose that q>0. Using pathwise arguments, explain why it is trivial that $\mathbb{P}(I_{\infty}<\infty) =1$.

• (ii) Now suppose that q=0 and lim sup _t↑∞ ξ _t <∞. Use Theorem 7.2 to deduce that $\mathbb {P}(I_{\infty}<\infty)=1$.

• (iii) Keeping with the case that q=0, suppose that lim _t↑∞ ξ _t =∞. Use the same hint as in part (ii) to deduce that $\mathbb{P}(I_{\infty}<\infty) =0$.

• (iv) Finally, in the case that q=0 and ξ oscillates, define the sequence of stopping times

  $$T_1 =\inf\{t>0 : \xi_t >2\}\quad \text{and}\quad S_1 = \inf\{t> T_1 : \xi_t < 1\} $$

  and for n≥2

  $$T_n = \inf\{t>S_{n-1}: \xi_t >2\}\quad \text{and}\quad S_n = \inf\{t> T_n : \xi_t <1\}. $$

  Show that

  $$I_\infty\geq{\mathrm{e}}^{\alpha} \sum _{n\geq1} (S_n - T_n) $$

  and hence, by comparing the random variable T ₁−S ₁ to $\tau ^{-}_{0}=\inf\{t>0: X_{t}<0\}$ under $\mathbb{P}_{1}$, show that $\mathbb {P}(I_{\infty}= \infty)=1$.

13.7

In the notation of Sect. 13.4.2, let $\tau^{+,\downarrow}_{x}= \inf\{t> 0: \xi^{\downarrow}_{t} >x\}$ and $\tau^{-,\downarrow}_{x} = \inf\{t> 0: \xi^{\downarrow}_{t}<x\}$ for any $x\in\mathbb{R}$. Fix −∞<v<0<u<∞.

• (i) Show that, for θ≥0,

  $$\begin{aligned} &\mathbf{P}^\downarrow \bigl(\xi^\downarrow_{\tau ^{+,\downarrow}_u}- u \in{\mathrm{d}}\theta; \tau^{+,\downarrow }_u< \tau^{-,\downarrow}_v \bigr) \\ &\quad = \frac{\sin\pi\alpha(1-\rho)}{\pi} \bigl(\mathrm{e}^u-1\bigr)^{\alpha (1-\rho)} \bigl(1-\mathrm{e}^v\bigr)^{\alpha\rho} \\ &\qquad {}\times \bigl(\mathrm{e}^{u+\theta} \bigr)^{\alpha\rho} \bigl(\mathrm{e}^{u+\theta} -\mathrm{e}^u\bigr)^{ -\alpha(1-\rho)} \bigl(\mathrm{e}^{u+\theta} -\mathrm{e}^v\bigr)^{-\alpha\rho} \bigl(\mathrm {e}^{u+\theta} -1\bigr)^{-1}{\mathrm{d}}\theta. \end{aligned}$$

• (ii) Show moreover that, for θ≥0,

  $$\begin{aligned} &\mathbf{P}^\downarrow \bigl(v-\xi^\downarrow_{\tau ^{-,\downarrow}_v} \in{\mathrm{d}}\theta; \tau^{+,\downarrow}_u> \tau^{-,\downarrow}_v \bigr) \\ &\quad =\frac{\sin\pi\alpha(1-\rho)}{\pi} \bigl(\mathrm{e}^u-1\bigr)^{\alpha (1-\rho)} \bigl(1-\mathrm{e}^v\bigr)^{\alpha\rho} \\ &\qquad {}\times \bigl( \mathrm{e}^{v-\theta} \bigr)^{\alpha\rho} \bigl(\mathrm{e}^{v} -\mathrm {e}^{v-\theta}\bigr)^{ -\alpha\rho} \bigl(\mathrm{e}^u -\mathrm{e}^{v-\theta}\bigr)^{-\alpha(1-\rho)} \bigl(1-\mathrm {e}^{v-\theta}\bigr)^{-1}{\mathrm{d}}\theta. \end{aligned}$$

13.8

We use here the notation of Sect. 13.4 and consider the case of scale functions for spectrally negative Lamperti-stable processes; see Patie (2009a) and Chaumont et al. (2009).

• (i) Show that, for z≥0, $\mathbf{P}^{\uparrow}(-\underline{\xi}^{\uparrow}_{\infty}\leq z) = (1-\mathrm {e}^{-z})^{\alpha(1-\rho)}$. Hence deduce that there exists a spectrally negative Lévy process with Laplace exponent

  $$\psi^\uparrow(\theta) = \frac{\varGamma(\theta+ \alpha )}{\varGamma(\theta)}, \quad\theta\geq0, $$

  whose associated 0-scale function, say W ^↑, is given by

  $$W^\uparrow(x) =\frac{1}{\varGamma(\alpha)}\bigl(1-\mathrm {e}^{-x} \bigr)^{\alpha-1}, \quad x\geq0. $$

• (ii) Show that, for z≥0, $\mathbf{P}^{\downarrow}(\overline{\xi}^{\downarrow}_{\infty}\leq z) =(1-\mathrm {e}^{-z})^{\alpha\rho}$. Hence deduce that there exists a spectrally negative Lévy process with Laplace exponent

  $$\psi^\downarrow(\theta) = \frac{\varGamma(\theta-1+\alpha )}{\varGamma(\theta-1)}, \quad\theta\geq0, $$

  whose associated 0-scale function, say W ^↓, is written

  $$W^\downarrow(x) =\frac{1}{\varGamma(\alpha)}\bigl(1-\mathrm {e}^{-x} \bigr)^{\alpha- 1}{\mathrm{e}}^x, \quad x\geq0. $$

13.9

Consider the case of an α-stable process conditioned to stay positive, as discussed in Sect. 13.4.1. As usual, it is denoted by {Y _t :t≥0} with probabilities $\{\mathbb{P}_{x}^{\uparrow}: x>0\}$.

• (i) Let b>x>0. Use the quintuple law applied to the Lévy process ξ ^↑ to deduce that for u∈[0,b−x], v∈[u,b) and y>0,

  $$\begin{aligned} &\mathbb{P}_x^{\uparrow}( b- \overline{Y}_{\tau^+_{b}-}\in{\mathrm{d}}u, \, b-Y_{\tau^+_{b}-}\in{ \mathrm{d}}v, \,Y_{\tau^+_{b}}-b\in {\mathrm{d}}y) \\ &\quad =\frac{\sin(\pi\alpha\rho)}{\pi}\frac{\varGamma(\alpha +1)}{\varGamma(\alpha\rho)\varGamma(\alpha(1-\rho))} \\ &\qquad {}\times \frac{(b-x-u)^{\alpha\rho-1}(v-u)^{\alpha(1-\rho)-1}(b-v)^{\alpha \rho}(y+b)^{\alpha(1-\rho)}}{(b-u)^{\alpha}(y+v)^{\alpha +1}}{\mathrm{d}}u\, \mathrm{d}v\, \mathrm{d}y. \end{aligned}$$

• (ii) Deduce from the previous part of the question that, for u∈[0,b−x], v∈[u,b) and y>0,

  $$\begin{aligned} &\mathbb{P}_x\bigl( b- \overline{Y}_{\tau^+_{b}-}\in{ \mathrm{d}}u, \, b-Y_{\tau^+_{b}-}\in{\mathrm{d}}v, \,Y_{\tau^+_{b}}-b\in { \mathrm{d}}y, \, \tau^+_b<\tau^-_0\bigr) \\ &\quad =\frac{\sin(\pi\alpha\rho)}{\pi}\frac{\varGamma(\alpha +1)}{\varGamma(\alpha\rho)\varGamma(\alpha(1-\rho))} \\ &\qquad {}\times \frac{x^{\alpha\rho}(b-x-u)^{\alpha\rho-1}(v-u)^{\alpha(1-\rho )-1}(b-v)^{\alpha\rho}(y+b)^{\alpha(1-2\rho)}}{(b-u)^{\alpha }(y+v)^{\alpha+1}}{\mathrm{d}}u\, \mathrm{d}v\, \mathrm{d}y. \end{aligned}$$

13.10

An alternative definition of the Bessel process uses the second Lamperti transform. Specifically, we define a Bessel process of dimension d>0, say R={R _t :t≥0}, to be a positive self-similar Markov process with index of self-similarity 2, whose driving Lévy process, ξ={ξ _t :t≥0}, is given by

$$\xi_t := B_t + \biggl(\frac{\mathtt{d}}{2}-1 \biggr)t, \quad t\geq0, $$

where {B _t :t≥0} is a Brownian motion. Note in particular that the resulting process must have continuous paths.

In the case d≥2, we have that lim sup _t≥0 ξ _t =∞. Hence R never hits the origin and has an entrance law at the origin. When d∈(0,2) then the process R visits the origin in an almost surely finite time. Moreover, since

$$\mathbb{E}\bigl(\mathrm{e}^{\lambda\xi_1}\bigr) = \exp \biggl\{ \lambda^2/2 + \biggl(\frac{\mathtt{d}}{2}-1 \biggr)\lambda \biggr\} , \quad \lambda\in\mathbb{R} $$

it follows that Φ(0)=(2−d)<2 and hence there exists a recurrent extension which leaves the origin continuously.

• (i) Verify that the generalised Ciesielski–Taylor identity proved in Theorem 13.11 confirms the original result for Bessel processes.

• (ii) Now suppose that q>0 is a constant. Use the second Lamperti transform or otherwise to show that {(R _t )^q:t≥0} is a positive self-similar Markov process. In particular, show that its index of self-similarity is 2/q.