Lévy Processes and Applications

Abstract

We define and characterise the class of Lévy processes. To illustrate the variety of processes captured within the definition of a Lévy process, we explore briefly the relationship between Lévy processes and infinitely divisible distributions. We also discuss some classical applied probability models, which are built on the strength of well-understood path properties of elementary Lévy processes. We hint at how generalisations of these models may be approached using more sophisticated Lévy processes. At a number of points later on in this text, we handle these generalisations in more detail.

Exercises

1.1

Prove that, in order to check for stationary and independent increments of the process {X _t :t≥0}, it suffices to check that, for all $n\in\mathbb{N}$ and 0≤s ₁≤t ₁≤⋯≤s _n ≤t _n <∞ and $\theta_{1},\ldots, \theta_{n}\in\mathbb{R}$,

$$\mathbb{E} \Biggl[\prod_{j=1}^n{\rm e}^{\mathrm{i}\theta_j (X_{t_j}-X_{s_j}) } \Biggr] = \prod_{j=1}^n \mathbb{E} \bigl[{\rm e}^{\mathrm{i}\theta_j X_{t_j-s_j} } \bigr]. $$

Show, moreover, that the sum of two (or indeed any finite number of) independent Lévy processes is again a Lévy process.

1.2

Suppose that S={S _n :n≥0} is any random walk and Γ _p is an independent random variable with a geometric distribution on {0,1,2,…}, with parameter p.

• (i) Show that Γ _p is infinitely divisible.

• (ii) Show that $S_{\boldsymbol{\Gamma}_{p}}$ is infinitely divisible.

1.3

(Proof of Lemma 1.7)

In this exercise, we derive the Frullani identity.

• (i) Show for any function f, such that f′ exists and is continuous and f(0) and f(∞) are finite, that

  $$\int_0^\infty\frac{ f(ax)-f(bx)}{x}{\mathrm{d}}x = \bigl(f(0)- f(\infty)\bigr)\log\biggl(\frac{b}{a} \biggr), $$

  where b>a>0.

• (ii) By choosing f(x)=e^−x, a=α>0 and b=α−z, where z<0, show that

  $$ \frac{1}{(1- z/\alpha)^\beta} = \mathrm{e}^{-\int_0^\infty (1-\mathrm{e}^{zx}) \frac{\beta}{x}\mathrm{e}^{-\alpha x}{\mathrm {d}}x} $$

  (1.24)

  and hence, by analytic extension, show that the above identity is still valid for all $z\in\mathbb{C}$ such that ℜz≤0.

1.4

Establishing formulae (1.9) and (1.10) from the Lévy measure given in (1.11) is the result of a series of technical manipulations of special integrals. In this exercise, we work through them. In the following text, we will use the gamma function Γ(z), defined by

$$\varGamma(z) = \int_0^\infty t^{z-1} \mathrm{e}^{-t}{\mathrm{d}}t, $$

for z>0. Note the gamma function can also be analytically extended so that it is also defined on $\mathbb{R}\backslash\{ 0,-1,-2,\ldots\}$ (see Lebedev 1972). Whilst the specific definition of the gamma function for negative numbers will not play an important role in this exercise, the following two facts, which can be derived from it, will. For $z\in\mathbb{R}\backslash\{0,-1,-2,\ldots\}$ the gamma function observes the recursion Γ(1+z)=zΓ(z) and $\varGamma(1/2)=\sqrt{\pi}$.

• (i) Suppose that 0<α<1. Prove that for u>0,

  $$ \int_{0}^{\infty}\bigl(\mathrm{e}^{-ur}-1 \bigr)r^{-\alpha-1}{\mathrm{d}}r=\varGamma(-\alpha)u^{\alpha} $$

  and show that the same equality is valid when −u is replaced by any complex number w≠0 with ℜw≤0. Conclude, by considering w=i, that

  $$ \int_{0}^{\infty}\bigl(1- \mathrm{e}^{\mathrm{i}r} \bigr)r^{-\alpha -1}{\mathrm{d}}r= -\varGamma(-\alpha)\mathrm{e}^{-\mathrm{i}\pi \alpha/2} $$

  (1.25)

  and similarly for the complex conjugate of both sides of (1.25). Deduce (1.9) by considering the integral

  $$ \int_{0}^{\infty}\bigl(1 - \mathrm{e}^{\mathrm{i}\xi\theta r} \bigr)r^{-\alpha-1}{\mathrm{d}}r $$

  for ξ=±1 and $\theta\in\mathbb{R}$. Note that you will have to take $a=\eta-\int_{\mathbb{R}}x\mathbf{1}_{(|x|<1)} \varPi ( {\mathrm{d}}x )$, which you should check is finite.

• (ii) Now suppose that α=1. First prove that

  $$\int_{|x|<1}\mathrm{e}^{\mathrm{i}\theta x}\bigl(1-|x|\bigr){\mathrm{d}}x = 2 \biggl( \frac{1- \cos\theta}{\theta^2} \biggr), $$

  for $\theta\in\mathbb{R}$. Hence by, Fourier inversion, show that

  $$\int_0^\infty\frac{1- \cos r}{r^2}{\mathrm{d}}r = \frac{\pi}{2}. $$

  Use this identity to show that for z>0,

  $$\int_0^\infty\bigl(1- \mathrm{e}^{\mathrm{i}rz} + \mathrm{i}zr\mathbf{1}_{(r<1)}\bigr)\frac{1}{r^2}{\mathrm{d}}r = \frac{\pi}{2}z + \mathrm{i}z\log z - \mathrm{i}kz, $$

  for some constant $k\in\mathbb{R}$. By considering the complex conjugate of the above integral, establish the expression in (1.10). Note that you will need a different choice of a to part (i).

• (iii) Now suppose that 1<α<2. Integrate (1.25) by parts to get

  $$ \int_{0}^{\infty }\bigl(\mathrm{e}^{\mathrm{i}r}-1- \mathrm{i}r\bigr)r^{-\alpha -1}{\mathrm{d}}r=\varGamma(-\alpha)\mathrm{e}^{-\mathrm{i}\pi \alpha/2}. $$

  Deduce the identity (1.9) in a similar manner to the proof of (i) and (ii).

1.5

For any $\theta\in\mathbb{R}$, prove that

$$\exp\bigl\{ \mathrm{i}\theta X_t + t \varPsi(\theta)\bigr\} , \quad t \geq0, $$

is a martingale where {X _t :t≥0} is a Lévy process with characteristic exponent Ψ.

1.6

In this exercise, we will work out in detail some features of the inverse Gaussian process discussed earlier on in this chapter. Recall that τ={τ _s :s≥0} is a non-decreasing Lévy process defined by τ _s =inf{t≥0:B _t +bt>s}, s≥0, where B={B _t :t≥0} is a standard Brownian motion and b>0.

• (i) Argue along the lines of Exercise 1.5 to show that, for each λ>0,

  $$\mathrm{e}^{\lambda B_t - \frac{1}{2}\lambda^2 t},\quad t\geq0, $$

  is a martingale. Use Doob’s Optional Sampling Theorem to obtain

  $$\mathbb{E}\bigl(\mathrm{e}^{-(\frac{1}{2}\lambda^2 + b\lambda )\tau_s}\bigr) = \mathrm{e}^{-\lambda s}. $$

  Use analytic extension to deduce further that τ _s has characteristic exponent

  $$\varPsi_s(\theta)= s\bigl(\sqrt{-2\mathrm{i}\theta+ b^2} - b\bigr), $$

  for all $\theta\in\mathbb{R}$.

• (ii) Defining the measure $\varPi({\mathrm{d}}x)=(2\pi x^{3})^{-1/2} \mathrm{e}^{-xb^{2}/2}{\mathrm{d}}x$ on x>0, check, using (1.25) from Exercise 1.4, that

  $$\int_0^\infty\bigl(1-\mathrm{e}^{\mathrm{i}\theta x} \bigr)\varPi({\mathrm{d}}x) = \varPsi(\theta), $$

  for all $\theta\in\mathbb{R}$. Confirm that the triple (a,σ,Π), appearing in the Lévy–Khintchine formula, is thus σ=0, Π as above and $a= - 2sb^{-1}\int_{0}^{b} (2\pi)^{-1/2} \mathrm{e}^{-y^{2}/2}{\mathrm{d}}y$.

• (iii) Taking

  $$\mu_s({\mathrm{d}}x) = \frac{s}{\sqrt{2\pi x^3}} \mathrm{e}^{sb} \mathrm{e}^{-\frac{1}{2}(s^2 x^{-1} + b^2 x)}{\mathrm{d}}x, \quad x>0, $$

  show that

  $$\begin{aligned} \int_0^\infty\mathrm{e}^{-\lambda x} \mu_s ({\mathrm{d}}x) =& \mathrm{e}^{bs - s\sqrt{b^2 + 2\lambda}} \int _0^\infty\frac{s}{\sqrt{2\pi x^3}}\mathrm{e}^{-\frac{1}{2}(\frac {s}{\sqrt{x}} - \sqrt{(b^2 + 2\lambda)x} )^2 } {\mathrm{d}}x \\ =& \mathrm{e}^{bs - s\sqrt{b^2 + 2\lambda}} \int_0^\infty \sqrt{\frac{2\lambda+ b^2}{2\pi u}} \mathrm{e}^{-\frac{1}{2} (\frac{s}{\sqrt{u}} - \sqrt{(b^2 + 2\lambda)u})^2}{\mathrm{d}}u. \end{aligned}$$

  Hence, by adding the last two integrals together deduce that

  $$\int_0^\infty\mathrm{e}^{-\lambda x} \mu_s ({\mathrm{d}}x) = \mathrm{e}^{-s(\sqrt{b^2 + 2\lambda} -b)}, $$

  thereby confirming both that μ _s (dx) is a probability distribution on (0,∞), and that it is the probability distribution of τ _s .

1.7

Show that for a simple Brownian motion B={B _t :t>0} the first passage process τ={τ _s :s>0}, where τ _s =inf{t≥0:B _t ≥s}, is a stable process with parameters α=1/2 and β=1.

1.8

(Proof of Theorem 1.9)

As we shall see in this exercise, the proof of Theorem 1.9 follows from the proof of a more general result given by the conclusion of parts (i)–(iv) below for random walks.

• (i) Suppose that S={S _n :n≥0} is a random walk with S ₀=0 and jump distribution Q on $\mathbb{R}$. By considering the variables $S^{*}_{k} := S_{n} - S_{n-k}$ for k=0,1,…,n and noting that the joint distributions of (S ₀,…,S _n ) and $(S^{*}_{0},\ldots,S^{*}_{n})$ are identical, show that for all y>0 and n≥1,

  $$\begin{aligned} &P(S_n \in{\mathrm{d}}y \text{ and }S_n> S_j \text{ for } j= 0, \ldots, n-1) \\ &\quad= P(S_n \in{\mathrm{d}}y \text{ and }S_j >0 \text{ for } j=1,\ldots,n). \end{aligned}$$

  Hint: it may be helpful to draw a diagram of the path of the first n steps of S and to rotate it by 180^∘.

• (ii) Define

  $$T^-_0 = \inf\{n> 0 : S_n \leq0 \}\quad\text{and}\quad T^+_0 = \inf\{n> 0 : S_n > 0\}. $$

  By summing both sides of the equality

  $$\begin{aligned} & P(S_1>0, \ldots, S_n >0, S_{n+1}\in{\mathrm{d}}x ) \\ &\quad= \int_{(0,\infty)} P(S_1>0, \ldots, S_n>0, S_n \in{\mathrm{d}}y) Q({\mathrm{d}}x-y) \end{aligned}$$

  over n, show that for x≤0,

  $$P(S_{T^-_0}\in{\mathrm{d}}x)= \int_{[0,\infty)} V({\mathrm{d}}y) Q({\mathrm{d}}x - y), $$

  where, for y≥0,

  $$V({\mathrm{d}}y)= \delta_0 ({\mathrm{d}}y) + \sum_{n\geq1}P(H_n \in{\mathrm{d}}y ) $$

  and H={H _n :n≥0} is a random walk with H ₀=0 and step distribution given by $P(S_{T^{+}_{0}}\in{\mathrm{d}}z)$, for z≥0.

• (iii) Embedded in the Cramér–Lundberg model is a random walk S whose increments are equal in distribution to that of c e _λ −ξ ₁, where e _λ is an independent exponential random variable with mean 1/λ. Noting (using obvious notation) that c e _λ has the same distribution as e _β , where β=λ/c, show that the step distribution of this random walk satisfies

  $$Q(z,\infty)= \biggl(\int_0^\infty \mathrm{e}^{-\beta u} F({\mathrm{d}}u) \biggr)\mathrm{e}^{-\beta z}, \quad z\geq 0, $$

  and

  $$Q(-\infty, -z) = E\bigl(\overline{F}(\mathbf{e}_\beta+ z)\bigr),\quad z>0, $$

  where $\overline{F}(x) = 1-F(x)$, for all x≥0, and E is expectation with respect to the random variable e _β .

• (iv) Since upward jumps are exponentially distributed in this random walk, use the lack-of-memory property to reason that

  $$V({\mathrm{d}}y)= \delta_0 ({\mathrm{d}}y) + \beta{\mathrm {d}}y,\quad y\geq0. $$

  Hence deduce from parts (ii) and (iii) that

  $$P( - S_{T^-_0}> z)=E \biggl(\overline{F}(\mathbf{e}_\beta) + \int_x^\infty\beta\overline{F}( \mathbf{e}_\beta+ z){\mathrm{d}}z \biggr) $$

  and so, by writing out the above identity with the density of the exponential distribution, show that the conclusions of Theorem 1.9 hold.

1.9

(Proof of Theorem 1.11)

Suppose that X is a compound Poisson process of the form

$$X_t = t - \sum_{i=1}^{N_t} \xi_i, \quad t\geq0, $$

where the process N={N _t :t≥0} is a Poisson process with rate λ>0 and {ξ _i :i≥1} positive, independent and identically distributed with common distribution F having finite mean μ.

• (i) Show by direct computation that, for all θ≥0, $\mathbb{E}(\mathrm{e}^{\theta X_{t}})= \mathrm{e}^{\psi(\theta)t}$, where

  $$\psi(\theta)=\theta- \lambda\int_{(0,\infty)} \bigl(1- \mathrm{e}^{-\theta x}\bigr)F({\mathrm{d}}x). $$

  Show that ψ is strictly convex, is equal to zero at the origin and tends to infinity at infinity. Further, show that ψ(θ)=0 has one additional root in [0,∞) other than θ=0 if and only if ψ′(0+)<0.

• (ii) Show that $\{\exp\{\theta^{*} X_{t\wedge\tau^{+}_{x}} \}: t\geq0\}$ is a martingale, where $\tau^{+}_{x} = \inf\{t > 0: X_{t} >x\}$, x>0 and θ ^∗ is the largest root described in the previous part of the question. Show further that

  $$\mathbb{P}(\overline{X}_\infty> x) = \mathrm{e}^{-\theta^* x}, $$

  for all x>0.

• (iii) Show that for all t≥0,

  $$\int_0^t \mathbf{1}_{(W_s = 0 )}{\mathrm{d}}s = ( \overline{X}_t- w)\vee0, $$

  where $W_{t} = (w\vee\overline{X}_{t}) - X_{t}$.

• (iv) Deduce that $I: = \int_{0}^{\infty}\mathbf{1}_{(W_{s} = 0 )}{\mathrm{d}}s=\infty$ if λμ≤1.

• (v) Assume that λμ>1. Show that

  $$\mathbb{P}\bigl(I\in{\mathrm{d}}x; \tau^+_w =\infty| W_0 =w \bigr) = \bigl(1- \mathrm{e}^{-\theta^* w}\bigr)\delta_0 ({\mathrm{d}}x), \quad x\geq0. $$

  Next use the lack-of-memory property to deduce that

  $$\mathbb{P}\bigl(I\in{\mathrm{d}}x; \tau^+_w <\infty| W_0 =w \bigr) = \theta^* \mathrm{e}^{-\theta^*(w+x)}{\mathrm{d}}x. $$

1.10

Here, we solve a considerably simpler optimal stopping problem than (1.20). Suppose, as in the aforementioned problem, that X is a linear Brownian motion with scaling parameter σ>0 and drift $\gamma\in\mathbb{R}$. Fix K>0 and let

$$ v(x) = \sup_{a\in\mathbb{R}}\mathbb{E}_x\bigl( \mathrm{e}^{-q \tau^-_a}\bigl(K - \mathrm{e}^{X_{\tau^-_a}} \bigr)^+\bigr), $$

(1.26)

where

$$\tau^-_a = \inf\{t>0 : X_t < a\}. $$

• (i) Following similar arguments to those in Exercises 1.5 and 1.9, show that {exp{θX _t −ψ(θ)t}:t≥0} is a martingale, where ψ(θ)=σ ² θ ²/2+γθ.

• (ii) By considering the martingale in part (i) at the stopping time $t\wedge\tau^{+}_{x}$ and then letting t↑∞, deduce that

  $$\mathbb{E}\bigl(\mathrm{e}^{-q\tau^{+}_x}\bigr) = \mathrm{e}^{ - x(\sqrt{\gamma^2 + 2\sigma^2 q} - \gamma)/\sigma^2} $$

  and hence deduce that for a≥0,

  $$\mathbb{E}\bigl(\mathrm{e}^{-q\tau^{-}_{-a}}\bigr) = \mathrm{e}^{ - a(\sqrt{\gamma^2 + 2\sigma^2 q} +\gamma)/\sigma^2}. $$

• (iii) Let $v(x,a) = \mathbb{E}_{x}(\mathrm{e}^{-q\tau^{-}_{-a}} (K - \exp\{X_{\tau^{-}_{-a}}\}))$. For each fixed x differentiate v(x,a) in the variable a and show that the solution to (1.26) is the same as the solution given in Theorem 1.13.

1.11

In this exercise, we characterise the Laplace exponent of the continuous-time Markov branching process, Y, described in Sect. 1.3.4.

• (i) Show that for ϕ>0 and t≥0 there exists some function u _t (ϕ)>0 satisfying

  $$E_y\bigl(\mathrm{e}^{-\phi Y_t}\bigr) = \mathrm{e}^{- y u_t (\phi)}, $$

  where y∈{0,1,2,…}.

• (ii) Show that for s,t≥0,

  $$u_{t+s}(\phi) = u_s\bigl(u_t(\phi)\bigr). $$

• (iii) Appealing to the infinitesimal behaviour of the Markov chain Y, show that

  $$\frac{\partial u_t (\phi)}{\partial t} = \psi\bigl(u_t (\phi)\bigr) $$

  and u ₀(ϕ)=ϕ, where

  $$\psi(q) = \lambda\int_{[-1,\infty)} \bigl(1 - \mathrm{e}^{-qx} \bigr) F({\mathrm{d}}x) $$

  and F is given in (1.22).