Part C L´evy Processes and Finance · L´evy processes and indicate some of their applications. By doing so, we will review some results from BS3a Applied Probability and B10 Martingales

Part C

Levy Processes and Finance

Matthias Winkel1

University of Oxford

HT 2007

1Departmental lecturer (Institute of Actuaries and Aon Lecturer in Statistics) at the Department ofStatistics, University of Oxford

MS3 Levy Processes and FinanceMatthias Winkel – 16 lectures HT 2007

Prerequisites

Part A Probability is a prerequisite. BS3a/OBS3a Applied Probability or B10 Martin-gales and Financial Mathematics would be useful, but are by no means essential; somematerial from these courses will be reviewed without proof.

Aims

Levy processes form a central class of stochastic processes, contain both Brownian motionand the Poisson process, and are prototypes of Markov processes and semimartingales.Like Brownian motion, they are used in a multitude of applications ranging from biologyand physics to insurance and finance. Like the Poisson process, they allow to modelabrupt moves by jumps, which is an important feature for many applications. In the lastten years Levy processes have seen a hugely increased attention as is reflected on theacademic side by a number of excellent graduate texts and on the industrial side realisingthat they provide versatile stochastic models of financial markets. This continues tostimulate further research in both theoretical and applied directions. This course willgive a solid introduction to some of the theory of Levy processes as needed for financialand other applications.

Synopsis

Review of (compound) Poisson processes, Brownian motion (informal), Markov property.Connection with random walks, [Donsker’s theorem], Poisson limit theorem. SpatialPoisson processes, construction of Levy processes.

Special cases of increasing Levy processes (subordinators) and processes with onlypositive jumps. Subordination. Examples and applications. Financial models drivenby Levy processes. Stochastic volatility. Level passage problems. Applications: optionpricing, insurance ruin, dams.

Simulation: via increments, via simulation of jumps, via subordination. Applications:option pricing, branching processes.

Reading

• J.F.C. Kingman: Poisson processes. Oxford University Press (1993), Ch.1-5, 8

• A.E. Kyprianou: Introductory lectures on fluctuations of Levy processes with Ap-

plications. Springer (2006), Ch. 1-3, 8-9

• W. Schoutens: Levy processes in finance: pricing financial derivatives. Wiley (2003)

Further reading

• J. Bertoin: Levy processes. Cambridge University Press (1996), Sect. 0.1-0.6, I.1,III.1-2, VII.1

• K. Sato: Levy processes and infinite divisibility.Cambridge University Press (1999),Ch. 1-2, 4, 6, 9

Lecture 1

Introduction

Reading: Kyprianou Chapter 1

Further reading: Sato Chapter 1, Schoutens Sections 5.1 and 5.3

In this lecture we give the general definition of a Levy process, study some examples ofLevy processes and indicate some of their applications. By doing so, we will review someresults from BS3a Applied Probability and B10 Martingales and Financial Mathematics.

1.1 Definition of Levy processes

Stochastic processes are collections of random variables Xt, t ≥ 0 (meaning t ∈ [0,∞)as opposed to n ≥ 0 by which means n ∈ N = 0, 1, 2, . . .). For us, all Xt, t ≥ 0, takevalues in a common state space, which we will choose specifically as R (or [0,∞) or R

d

for some d ≥ 2). We can think of Xt as the position of a particle at time t, changing ast varies. It is natural to suppose that the particle moves continuously in the sense thatt 7→ Xt is continuous (with probability 1), or that it has jumps for some t ≥ 0:

∆Xt = Xt+ − Xt− = limε↓0

Xt+ε − limε↓0

Xt−ε.

We will usually suppose that these limits exist for all t ≥ 0 and that in fact Xt+ = Xt,i.e. that t 7→ Xt is right-continuous with left limits Xt− for all t ≥ 0 almost surely. Thepath t 7→ Xt can then be viewed as a random right-continuous function.

Definition 1 (Levy process) A real-valued (or Rd-valued) stochastic process X =

(Xt)t≥0 is called a Levy process if

(i) the random variables Xt0 , Xt1 −Xt0 , . . . , Xtn −Xtn−1 are independent for all n ≥ 1and 0 ≤ t0 < t1 < . . . < tn(independent increments),

(ii) Xt+s − Xt has the same distribution as Xs for all s, t ≥ 0 (stationary increments),

(iii) the paths t 7→ Xt are right-continuous with left limits (with probability 1).

It is implicit in (ii) that P(X0 = 0) = 1 (choose s = 0).

1

2 Lecture Notes – MS3b Levy Processes and Finance – Oxford HT 2008

Figure 1.1: Variance Gamma process and a Levy process with no positive jumps

Here the independence of n random variables is understood in the following sense:

Definition 2 (Independence) Let Y (j) be an Rdj -valued random variable for j =

1, . . . , n. The random variables Y (1), . . . , Y (n) are called independent if, for all (Borelmeasurable) C(j)

⊂ Rdj

P(Y (1)∈ C(1), . . . , Y (n)

∈ C(n)) = P(Y (1)∈ C(1)) . . . P(Y (n)

∈ C(n)). (1)

An infinite collection (Y (j))j∈J is called independent if Y (j1), . . . , Y (jn) are independent for

every finite subcollection. Infinite-dimensional random variables (Y(1)

i )i∈I1, . . . , (Y(n)

i )i∈In

are called independent if (Y(1)

i )i∈F1, . . . , (Y(n)

i )i∈Fnare independent for all finite Fj ⊂ Ij .

It is sufficient to check (1) for rectangles of the form C(j) = (a(j)1

, b(j)1

] × . . . × (a(j)dj

, b(j)dj

].

1.2 First main example: Poisson process

Poisson processes are Levy processes. We recall the definition as follows. An N(⊂ R)-valued stochastic process X = (Xt)t≥0 is called a Poisson process with rate λ ∈ (0,∞) ifX satisfies (i)-(iii) and

(iv)PoiP(Xt = k) = (λt)k

k!e−λt, k ≥ 0, t ≥ 0 (Poisson distribution).

The Poisson process is a continuous-time Markov chain. We will see that all Levy pro-cesses have a Markov property. Also recall that Poisson processes have jumps of size 1(spaced by independent exponential random variables Zn = Tn+1−Tn, n ≥ 0, with param-eter λ, i.e. with density λe−λs, s ≥ 0). In particular, t ≥ 0 : ∆Xt 6= 0 = Tn, n ≥ 1and ∆XTn

= 1 almost surely (short a.s., i.e. with probability 1). We can define moregeneral Levy processes by putting

Ct =Xt∑

k=1

Yk, t ≥ 0,

for a Poisson process (Xt)t≥0 and independent identically distributed Yk, k ≥ 1. Suchprocesses are called compound Poisson processes. The term “compound” stems from therepresentation Ct = S Xt = SXt

for the random walk Sn = Y1 + . . . + Yn. You maythink of Xt as the number of claims up to time t and of Yk as the size of the kth claim.Recall (from BS3a) that its moment generating function, if it exists, is given by

E(expγCt) = exp−λt(E(eγY1− 1).

This will be an important building block of a general Levy process.

Lecture 1: Introduction 3

Figure 1.2: Poisson process and Brownian motion

1.3 Second main example: Brownian motion

Brownian motion is a Levy process. We recall (from B10b) the definition as follows. AnR-valued stochastic process X = (Xt)t≥0 is called Brownian motion if X satisfies (i)-(ii)and

(iii)BM the paths t 7→ Xt are continuous almost surely,

(iv)BMP(Xt ≤ x) =

∫ x

−∞1√2πt

exp

−y2

2t

dy, x ∈ R, t > 0. (Normal distribution).

The paths of Brownian motion are continuous, but turn out to be nowhere differentiable(we will not prove this). They exhibit erratic movements at all scales. This makesBrownian motion an appealing model for stock prices. Brownian motion has the scaling

property (√

cXt/c)t≥0 ∼ X where “∼” means “has the same distribution as”.Brownian motion will be the other important building block of a general Levy process.The canonical space for Brownian paths is the space C([0,∞), R) of continuous real-

valued functions f : [0,∞) → R which can be equipped with the topology of locally

uniform convergence, induced by the metric

d(f, g) =∑

k≥1

2−k mindk(f, g), 1, where dk(f, g) = supx∈[0,k]

|f(x) − g(x)|.

This metric topology is complete (Cauchy sequences converge) and separable (has acountable dense subset), two attributes important for the existence and properties oflimits. The bigger space D([0,∞), R) of right-continuous real-valued functions with leftlimits can also be equipped with the topology of locally uniform convergence. The spaceis still complete, but not separable. There is a weaker metric topology, called Skorohod’stopology, that is complete and separable. In the present course we will not developthis and only occasionally use the familiar uniform convergence for (right-continuous)functions f, fn : [0, k] → R, n ≥ 1:

supx∈[0,k]

|fn(x) − f(x)| → 0, as n → ∞,

which for stochastic processes X, X(n), n ≥ 1, with time range t ∈ [0, T ] takes the form

supt∈[0,T ]

|X(n)

t − Xt| → 0, as n → ∞,

and will be as a convergence in probability or as almost sure convergence (from BS3a orB10a) or as L2-convergence, where Zn → Z in the L2-sense means E(|Zn − Z|

2) → 0.


1.4 Markov property

The Markov property is a consequence of the independent increments property (and thestationary increments property):

Proposition 3 (Markov property) Let X be a Levy process and t ≥ 0 a fixed time,

then the pre-t process (Xr)r≤t is independent of the post-t process (Xt+s−Xt)s≥0, and the

post-t process has the same distribution as X.

Proof: By Definition 2, we need to check the independence of (Xr1, . . . , Xrn) and (Xt+s1−

Xt, . . . , Xt+sm− Xt). By property (i) of the Levy process, we have that increments over

disjoint time intervals are independent, in particular the increments

Xr1, Xr2 − Xr1 , . . . , Xrn− Xrn−1 , Xt+s1 − Xt, Xt+s2 − Xt+s1 , . . . , Xt+sm

− Xt+sm−1 .

Since functions (here linear transformations from increments to marginals) of independentrandom variables are independent, the proof of independence is complete. Identicaldistribution follows first on the level of single increments from (ii), then by (i) and lineartransformation also for finite-dimensional marginal distributions. 2

1.5 Some applications

Example 4 (Insurance ruin) A compound Poisson process (Zt)t≥0 with positive jumpsizes Ak, k ≥ 1, can be interpreted as a claim process recording the total claim amountincurred before time t. If there is linear premium income at rate r > 0, then also thegain process rt− Zt, t ≥ 0, is a Levy process. For an initial reserve of u > 0, the reserveprocess u + rt − Zt is a shifted Levy process starting from a non-zero initial value u.

Example 5 (Financial stock prices) Brownian motion (Bt)t≥0 or linear Brownian mo-tion σBt +µt, t ≥ 0, was the first model of stock prices, introduced by Bachelier in 1900.Black, Scholes and Merton studied geometric Brownian motion exp(σBt + µt) in 1973,which is not itself a Levy process but can be studied with similar methods. The EconomicsNobel Prize 1997 was awarded for their work. Several deficiencies of the Black-Scholesmodel have been identified, e.g. the Gaussian density decreases too quickly, no variationof the volatility σ over time, no macroscopic jumps in the price processes. These deficien-cies can be addressed by models based on Levy processes. The Variance gamma modelis a time-changed Brownian motion BTs

by an independent increasing jump process, aso-called Gamma Levy process with Ts ∼ Gamma(αs, β). The process BTs

is then also aLevy process itself.

Example 6 (Population models) Branching processes are generalisations of birth-and-death processes (see BS3a) where each individual in a population dies after an ex-ponentially distributed lifetime with parameter µ, but gives birth not to single children,but to twins, triplets, quadruplet etc. To simplify, it is assumed that children are onlyborn at the end of a lifetime. The numbers of children are independent and identicallydistributed according to an offspring distribution q on 0, 2, 3, . . .. The population sizeprocess (Zt)t≥0 can jump downwards by 1 or upwards by an integer. It is not a Levyprocess but is closely related to Levy processes and can be studied with similar meth-ods. There are also analogues of processes in [0,∞), so-called continuous-state branchingprocesses that are useful large-population approximations.

Lecture 2

Levy processes and random walks

Reading: Kingman Section 1.1, Grimmett and Stirzaker Section 3.5(4)

Further reading: Sato Section 7, Durrett Sections 2.8 and 7.6, Kallenberg Chapter 15

Levy processes are the continuous-time analogues of random walks. In this lecture weexamine this analogy and indicate connections via scaling limits and other limiting results.We begin with a first look at infinite divisibility.

2.1 Increments of random walks and Levy processes

Recall that a random walk is a stochastic process in discrete time

S0 = 0, Sn =

n∑

j=1

Aj, n ≥ 1,

for a family (Aj)j≥1 of independent and identically distributed real-valued (or Rd-valued)

random variables. Clearly, random walks have stationary and independent increments.Specifically, the Aj , j ≥ 1, themselves are the increments over single time units. We referto Sn+m − Sn as an increment over m time units, m ≥ 1.

While every distribution may be chosen for Aj , increments over m time units are sumsof m independent and identically distributed random variables, and not every distributionhas this property. This is not a deep observation, but it becomes important when movingto Levy processes. In fact, the increment distribution of Levy processes is restricted: anyincrement Xt+s − Xt, or Xs for simplicity, can be decomposed, for every m ≥ 1,

Xs =

m∑

j=1

(Xjs/m − X(j−1)s/m)

into a sum of m independent and identically distributed random variables.

Definition 7 (Infinite divisibility) A random variable Y is said to have an infinitely

divisible distribution if for every m ≥ 1, we can write

Y ∼ Y(m)

1+ . . . + Y (m)

m

for some independent and identically distributed random variables Y(m)

1, . . . , Y

(m)

m .

We stress that the distribution of Y(m)

j may vary as m varies, but not as j varies.

5


The argument just before the definition shows that increments of Levy processes areinfinitely divisible. Many known distributions are infinitely divisible, some are not.

Example 8 The Normal, Poisson, Gamma and geometric distributions are infinitelydivisible. This often follows from the closure under convolutions of the type

Y1 ∼ Normal(µ, σ2), Y2 ∼ Normal(ν, τ 2) ⇒ Y1 + Y2 ∼ Normal(µ + ν, σ2 + τ 2)

for independent Y1 and Y2 since this implies by induction that for independent

Y(m)

1, . . . , Y (m)

m ∼ Normal(µ/m, σ2/m) ⇒ Y(m)

1+ . . . + Y (m)

m ∼ Normal(µ, σ2).

The analogous arguments (and calculations, if necessary) for the other distributions areleft as an exercise. The geometric(p) distribution here is P(X = n) = pn(1 − p), n ≥ 0.

Example 9 The Bernoulli(p) distribution, for p ∈ (0, 1), is not infinitely divisible. As-sume that you can represent a Bernoulli(p) random variable X as Y1+Y2 for independentidentically distributed Y1 and Y2. Then

P(Y1 > 1/2) > 0 ⇒ 0 = P(X > 1) ≥ P(Y1 > 1/2, Y2 > 1/2) > 0

is a contradiction, so we must have P(Y1 > 1/2) = 0, but then

P(Y1 > 1/2) = 0 ⇒ p = P(X = 1) = P(Y1 = 1/2)P(Y2 = 1/2) ⇒ P(Y1 = 1/2) =√

p.

Similarly,

P(Y1 < 0) > 0 ⇒ 0 = P(X < 0) ≥ P(Y1 < 0, Y2 < 0) > 0

is a contradiction, so we must have P(Y1 < 0) = 0 and then

1 − p = P(X = 0) = P(Y1 = 0, Y2 = 0) ⇒ P(Y1 = 0) =√

1 − p > 0.

This is impossible for several reasons. Clearly,√

p +√

1 − p > 1, but also

0 = P(X = 1/2) ≥ P(Y1 = 0)P(Y2 = 1/2) > 0.

2.2 Central Limit Theorem and Donsker’s theorem

Theorem 10 (Central Limit Theorem) Let (Sn)n≥0 be a random walk with E(S2

1) =

E(A2

1) < ∞. Then, as n → ∞,

Sn − E(Sn)√

Var(Sn)=

Sn − nE(A1)√

nVar(A1)→ Normal(0, 1) in distribution.

This result as a result for one time n → ∞ can be extended to a convergence of pro-cesses, a convergence of the discrete-time process (Sn)n≥0 to a (continuous-time) Brownianmotion, by scaling of both space and time. The processes

S[nt] − [nt]E(A1)√

nVar(A1), t ≥ 0,

where [nt] ∈ Z with [nt] ≤ nt < [nt]+1 denotes the integer part of nt, are scaled versionsof the random walk (Sn)n≥0, now performing n steps per time unit (holding time 1/n),centred and each only a multiple 1/

√

nVar(A1) of the original size. If E(A1) = 0, youmay think that you look at (Sn)n≥0 from further and further away, but note that spaceand time are scaled differently, in fact so as to yield a non-trivial limit.

Lecture 2: Levy processes and random walks 7

Figure 2.1: Random walk converging to Brownian motion

Theorem 11 (Donsker) Let (Sn)n≥0 be a random walk with E(S2

1) = E(A2

1) < ∞.

Then, as n → ∞,

S[nt] − [nt]E(A1)√

nVar(A1)→ Bt locally uniformly in t ≥ 0,

“in distribution”, for a Brownian motion (Bt)t≥0.

Proof: [only for A1 ∼ Normal(0, 1)] This proof is a coupling proof. We are not going towork directly with the original random walk (Sn)n≥0, but start from Brownian motion(Bt)t≥0 and define a family of embedded random walks

S(n)

k := Bk/n, k ≥ 0, n ≥ 1.

Then note using in particular E(A1) = 0 and Var(A1) = 1 that

S(n)

1∼ Normal(0, 1/n) ∼

S1 − E(A1)√

nVar(A1),

and indeed

(

S(n)

[nt]

)

t≥0

∼

(

S[nt] − [nt]E(A1)√

nVar(A1)

)

t≥0

.

To show convergence in distribution for the processes on the right-hand side, it suffices toestablish convergence in distribution for the processes on the left-hand side, as n → ∞.

To show locally uniform convergence we take an arbitrary T ≥ 0 and show uniformconvergence on [0, T ]. Since (Bt)0≤t≤T is uniformly continuous (being continuous on acompact interval), we get a.s.

sup0≤t≤T

∣

∣S[nt] − Bt

∣

∣

≤ sup0≤s≤t≤T :|s−t|≤1/n

|Bs − Bt| → 0

as n → ∞. This establishes a.s. convergence, which “implies” convergence in distributionfor the embedded random walks and for the original scaled random walk. This completesthe proof for A1 ∼ Normal(0, 1). 2

Note that the almost sure convergence only holds for the embedded random walks(S

(n)

k )k≥0, n ≥ 1. Since the identity in distribution with the rescaled original randomwalk only holds for fixed n ≥ 1, not jointly, we cannot deduce almost sure convergence inthe statement of the theorem. Indeed, it can be shown that almost sure convergence willfail. The proof for general increment distribution is much harder and will not be given inthis course. If time permits, we will give a similar coupling proof for another importantspecial case where P(A1 = 1) = P(A1 = −1) = 1/2, the simple symmetric random walk.


2.3 Poisson limit theorem

The Central Limit Theorem for Bernoulli random variables A1, . . . , An says that for largen, the number of 1s in the sequence is well-approximated by a Normal random variable.In practice, the approximation is good if p is not too small. If p is small, the Bernoullirandom variables count rare events, and a different limit theorem is relevant:

Theorem 12 (Poisson limit theorem) Let Wn be binomially distributed with param-

eters n and pn = λ/n (or if npn → λ, as n → ∞). Then we have

Wn → Poi(λ), in distribution, as n → ∞.

Proof: Just calculate that, as n → ∞,(

n

k

)

pkn(1 − pn)n−k =

n(n − 1) . . . (n − k + 1)

k!

(npn)k

nk

(

1 −npn

n

)n

(1 − pn)k→

λk

k!e−λ.

2

Theorem 13 Suppose that S(n)

k = A(n)

1+ . . . + A

(n)

k , k ≥ 0, is the sum of independent

Bernoulli(pn) random variables for all n ≥ 1, and that npn → λ ∈ (0,∞). Then

S(n)

[nt] → Nt “in the Skorohod sense” as functions of t ≥ 0,

“in distribution” as n → ∞, for a Poisson process (Nt)t≥0 with rate λ.

The proof of so-called finite-dimensional convergence for vectors (S(n)

[nt1], . . . , S

(n)

[ntm]) is

not very hard but not included here. One can also show that the jump times (T(n)

m )m≥1

of (S(n)

[nt])t≥0 converge to the jump times of a Poisson process. E.g.

P(T(n)

1> t) = (1 − pn)[nt] =

(

1 −

[nt]pn

[nt]

)[nt]

→ exp−λt,

since [nt]/n → t (since (nt − 1)/n → t and nt/n = t) and so [nt]pn → tλ. The generalstatement is hard to make precise and prove, certainly beyond the scope of this course.

2.4 Generalisations

Infinitely divisible distributions and Levy processes are precisely the classes of limits thatarise for random walks as in Theorems 10 and 12 (respectively 11 and 13) with differentstep distributions. Stable Levy processes are ones with a scaling property (c1/αXt/c)t≥0 ∼

X for some α ∈ R. These exist, in fact, for α ∈ (0, 2]. Theorem 10 (and 11) for suitabledistributions of A1 (depending on α and where E(A2

1) = ∞ in particular) then yield

convergence in distribution

Sn − nE(A1)

n1/α→ stable(α) for α ≥ 1, or

Sn

n1/α→ stable(α) for α ≤ 1.

Example 14 (Brownian ladder times) For a Brownian motion B and a level r > 0,the distribution of Tr = inft ≥ 0 : Bt > r is 1/2-stable, see later in the course.

Example 15 (Cauchy process) The Cauchy distribution with density a/(π(x2 + a2)),x ∈ R, for some parameter c ∈ R is 1-stable, see later in the course.

Lecture 3

Spatial Poisson processes

Reading: Kingman 1.1 and 2.1, Grimmett and Stirzaker 6.13, Kyprianou Section 2.2

Further reading: Sato Section 19

We will soon construct the most general nonnegative Levy process (and then generalreal-valued ones). Even though we will not prove that they are the most general, wehave already seen that only infinitely divisible distributions are admissible as incrementdistributions, so we know that there are restrictions; the part missing in our discussionwill be to show that a given distribution is infinitely divisible only if there exists a Levyprocess X of the type that we will construct such that X1 has the given distribution.Today we prepare the construction by looking at spatial Poisson processes, objects ofinterest in their own right.

3.1 Motivation from the study of Levy processes

Brownian motion (Bt)t≥0 has continuous sample paths. It turns out that (σBt + µt)t≥0

for σ ≥ 0 and µ ∈ R is the only continuous Levy process. To describe the full class ofLevy processes (Xt)t≥0, it is vital to study the process (∆Xt)t≥0 of jumps.

Take e.g. the Variance Gamma process. In Assignment 1.2.(b), we introduce thisprocess as Xt = Gt − Ht, t ≥ 0, for two independent Gamma Levy processes G and H .But how do Gamma Levy processes evolve? We could simulate discretisations (and willdo!) and get some feeling for them, but we also want to understand them mathematically.Do they really exist? We have not shown this. Are they compound Poisson processes?Let us look at their moment generating function (cf. Assignment 2.4.):

E(expγGt) =

(

β

β − γ

)αt

= exp

αt

∫ ∞

0

(eγx− 1)

1

xe−βxdx

.

This is almost of the form of a compound Poisson process of rate λ with non-negativejump sizes Yj, j ≥ 1, that have a probability density function h(x) = hY1(x), x > 0:

E(expγCt) = expλt

∫ ∞

0

(eγx− 1)h(x)dx

To match the two expressions, however, we would have to put

λh(x) = λ0h(0)(x) =

α

xe−βx, x > 0,

9


and h(0) cannot be a probability density function, because αxe−βx is not integrable at

x ↓ 0. What we can do is e.g. truncate at ε > 0 and specify

λεh(ε)(x) =

α

xe−βx, x > ε, h(ε)(x) = 0, x ≤ ε.

In order for h(ε) to be a probability density, we just put λε =∫∞

εαxe−βxdx, and notice

that λε → ∞ as ε ↓ 0. But λε is the rate of the Poisson process driving the compoundPoisson process, so jumps are more and more frequent as ε ↓ 0. On the other hand, theaverage jump size, the mean of the distribution with density h(ε) tends to zero, so mostof these jumps are very small. In fact, we will see that

Gt =∑

s≤t

∆Gs,

as an absolutely convergent series of infinitely (but clearly countably) many positive jumpsizes, where (∆Gs)s≥0 is a Poisson point process with intensity g(x) = α

xe−βx, x > 0, the

collection of random variables

N((a, b] × (c, d]) = #t ∈ (a, b] : ∆Gt ∈ (c, d], 0 ≤ a < b, 0 < c < d

a Poisson counting measure (evaluated on rectangles) with intensity function λ(t, x) =g(x), x > 0, t ≥ 0; the random countable set (t, ∆Gt) : t ≥ 0 and ∆Ct 6= 0 a spatialPoisson process with intensity λ(t, x). Let us now formally introduce these notions.

3.2 Poisson counting measures

The essence of one-dimensional Poisson processes (Nt)t≥0 is the set of arrival (“event”)times Π = T1, T2, T3, . . ., which is a random countable set. The increment N((s, t]) :=Nt − Ns counts the number of points in Π ∩ (s, t]. We can generalise this concept tocounting measures of random countable subsets on other spaces, say R

d. Saying directlywhat exactly (the distribution of) random countable sets is, is quite difficult in general.Random counting measures are a way to describe the random countable sets implicitly.

Definition 16 (Spatial Poisson process) A random countable subset Π ⊂ Rd is called

a spatial Poisson process with (constant) intensity λ if the random variables N(A) =#Π∩A, A ⊂ R

d (Borel measurable, always, for the whole course, but we stop saying this

all the time now), satisfy

(a) for all n ≥ 1 and disjoint A1, . . . , An ⊂ Rd, the random variables N(A1), . . . , N(An)

are independent,

hom(b) N(A) ∼ Poi(λ|A|), where |A| denotes the volume (Lebesgue measure) of A.

Here, we use the convention that X ∼ Poi(0) means P(X = 0) = 1 and X ∼ Poi(∞)means P(X = ∞) = 1. This is consistent with E(X) = λ for X ∼ Poi(λ), λ ∈ (0,∞).This convention captures that Π does not have points in a given set of zero volume a.s.,and it has infinitely many points in given sets of infinite volume a.s.

In fact, the definition fully specifies the joint distributions of the random set functionN on subsets of R

d, since for any non-disjoint B1, . . . , Bm ⊂ Rd we can consider all

Lecture 3: Spatial Poisson processes 11

intersections of the form Ak = B∗1∩ . . . ∩ B∗

m, where each B∗j is either B∗

j = Bj or B∗j =

Bcj = R

d\ Bj . They form n = 2m disjoint sets A1, . . . , An to which (a) of the definition

applies. (N(B1), . . . , N(Bm)) is a just a linear transformation of (N(A1), . . . , N(An)).Grimmett and Stirzaker collect a long list of applications including modelling stars in

a galaxy, galaxies in the universe, weeds in the lawn, the incidence of thunderstorms andtornadoes. Sometimes the process in Definition 16 is not a perfect description of such asystem, but useful as a first step. A second step is the following generalisation:

Definition 16 (Spatial Poisson process, continued) A random countable subset Π ⊂

D ⊂ Rd is called a spatial Poisson process with (locally integrable) intensity function

λ : D → [0,∞), if N(A) = #Π ∩ A, A ⊂ D, satisfy

(a) for all n ≥ 1 and disjoint A1, . . . , An ⊂ D, the random variables N(A1), . . . , N(An)are independent,

inhom(b) N(A) ∼ Poi(∫

Aλ(x)dx

)

.

Definition 17 (Poisson counting measure) A set function A 7→ N(A) that satisfies(a) and inhom(b) is referred to as a Poisson counting measure with intensity function λ(x).

It is sufficient to check (a) and (b) for rectangles Aj = (a(j)1

, b(j)1

] × . . . × (a(j)d , b

(j)d ].

The set function Λ(A) =∫

Aλ(x)dx is called the intensity measure of Π. Definitions

16 and 17 can be extended to measures that are not integrals of intensity functions.Only if Λ(x) > 0, we would require P(N(x) ≥ 2) > 0 and this is incompatible withN(x) = #Π ∩ x for a random countable set Π, so we prohibit such “atoms” of Λ.

Example 18 (Compound Poisson process) Let (Ct)t≥0 be a compound Poisson pro-cess with independent jump sizes Yj, j ≥ 1 with common probability density h(x), x > 0,at the times of a Poisson process (Xt)t≥0 with rate λ > 0. Let us show that

N((a, b] × (c, d]) = #t ∈ (a, b] : ∆Ct ∈ (c, d]

defines a Poisson counting measure. First note N((a, b]× (0,∞)) = Xb −Xa. Now recall

Thinning property of Poisson processes: If each point of a Poisson pro-cess (Xt)t≥0 of rate λ is of type 1 with probability p and of type 2 with prob-ability 1 − p, independently of one another, then the processes X(1) and X(2)

counting points of type 1 and 2, respectively, are independent Poisson processeswith rates pλ and (1 − p)λ, respectively.

Consider the thinning mechanism, where the jth jump is of type 1 if Yj ∈ (c, d]. Then,the process counting jumps in (c, d] is a Poisson process with rate λP(Y1 ∈ (c, d]), and so

N((a, b] × (c, d]) = X(1)

b − X(1)

a ∼ Poi((b − a)λP(Y1 ∈ (c, d])).

We identify the intensity measure Λ((a, b] × (c, d]) = (b − a)λP(Y1 ∈ (c, d]).For the independence of counts in disjoint rectangles A1, . . . , An, we cut them into

smaller rectangles Bi = (ai, bi]×(ci, di], 1 ≤ i ≤ m such that for any two Bi and Bj either(ci, di] = (cj , dj] or (ci, di] ∩ (cj, dj] = ∅. Denote by k the number of different intervals(ci, di], w.l.o.g. (ci, di] for 1 ≤ i ≤ k. Now a straightforward generalisation of the thinningproperty to k types splits (Xt)t≥0 into k independent Poisson processes X(i) with ratesλP(Y1 ∈ (ci, di]), 1 ≤ i ≤ k. Now N(B1), . . . , N(Bm) are independent as increments ofindependent Poisson processes or of the same Poisson process over disjoint time intervals.


3.3 Poisson point processes

In Example 18, the intensity measure is of the product form Λ((a, b] × (c, d]) = (b −a)ν((c, d]) for a measure ν on D0 = (0,∞). Take D = [0,∞)×D0 in Definition 16. Thismeans, that the spatial Poisson process is homogeneous in the first component, the timecomponent, like the Poisson process.

Proposition 19 If Λ((a, b]×A0) = (b− a)∫

A0g(x)dx for a locally integrable function g

on D0 (or = (b − a)ν(A0) for a locally finite measure ν on D0), then no two points of Πshare the same first coordinate.

Proof: If ν is finite, this is clear, since then Xt = N([0, t] × D0), t ≥ 0, is a Poissonprocess with rate ν(D0). Let us restrict attention to D0 = R

∗ = R \ 0 for simplicity– this is the most relevant case for us. The local integrability condition means that wecan find intervals (In)n≥1 such that

⋃

n≥1In = D0 and ν(In) < ∞, n ≥ 1. Then the

independence of N((tj−1, tj ]×In), j = 1, . . . , m, n ≥ 1, implies that X(n)

t = N([0, t]×In),t ≥ 0, are independent Poisson processes with rates ν(In), n ≥ 1. Therefore any two of

the jump times (T(n)

j , j ≥ 1, n ≥ 1) are jointly continuously distributed and take differentvalues almost surely:

P(T(n)

j = T(m)

i ) =

∫ ∞

0

∫ x

x

fT

(n)j

(x)fT

(m)i

(y)dydx = 0 for all n 6= m.

[Alternatively, show that T(n)

j − T(m)

i has a continuous distribution and hence does nottake a fixed value 0 almost surely].

Finally, there are only countably many pairs of jump times, so almost surely no twojump times coincide. 2

Let Π be a spatial Poisson process with intensity measure Λ((a, b] × (c, d]) = (b −

a)∫ d

cg(x)dx for a locally integrable function g on D0 (or = (b − a)ν((c, d]) for a locally

finite measure ν on D0), then the process (∆t)t≥0 given by

∆t = 0 if Π ∩ t × D = ∅, ∆t = x if Π ∩ t = (t, x)

is a Poisson point process in D0 ∪ 0 with intensity function g on D0 in the sense of thefollowing definition.

Definition 20 (Poisson point process) Let g be locally integrable on D0 ⊂ Rd−1

\0(or ν locally finite). A process (∆t)t≥0 in D0 ∪ 0 such that

N((a, b] × A0) = #t ∈ (a, b] : ∆t ∈ A0, 0 ≤ a < b, A0 ⊂ D0 (measurable),

is a Poisson counting measure with intensity Λ((a, b] × A0) = (b − a)∫

A0g(x)dx (or

Λ((a, b] × A0) = (b − a)ν(A0)), is called a Poisson point process with intensity g (or

intensity measure ν).

Note that for every Poisson point process, the set Π = (t, ∆t) : t ≥ 0, ∆t 6= 0is a spatial Poisson process. Poisson random measure and Poisson point process arerepresentations of this spatial Poisson process. Poisson point processes as we have definedthem always have a time coordinate and are homogeneous in time, but not in their spatialcoordinates.

In the next lecture we will see how one can do computations with Poisson pointprocesses, notably relating to

∑

∆t.

Lecture 4

Spatial Poisson processes II

Reading: Kingman Sections 2.2, 2.5, 3.1; Further reading: Williams Chapters 9 and 10

In this lecture, we construct spatial Poisson processes and study sums∑

s≤t f(∆s) overPoisson point processes (∆t)t≥0. We will identify

∑

s≤t ∆s as Levy process next lecture.

4.1 Series and increasing limits of random variables

Recall that for two independent Poisson random variables X ∼ Poi(λ) and Y ∼ Poi(µ)we have X + Y ∼ Poi(λ + µ). Much more is true. A simple induction shows that

Xj ∼ Poi(µj), 1 ≤ j ≤ m, independent ⇒ X1 + . . . + Xm ∼ Poi(µ1 + . . . + µm).

What about countably infinite families with µ =∑

m≥1µm < ∞? Here is a general

result, a bit stronger than the convergence theorem for moment generating functions.

Lemma 21 Let (Zm)m≥1 be an increasing sequence of [0,∞)-valued random variables.

Then Z = limm→∞ Zm exists a.s. as a [0,∞]-valued random variable. In particular,

E(eγZm) → E(eγZ) = M(γ) for all γ 6= 0.

We have

P(Z < ∞) = 1 ⇐⇒ limγ↑0

M(γ) = 1

and P(Z = ∞) = 1 ⇐⇒ M(γ) = 0 for all (one) γ < 0.

Proof: Limits of increasing sequences exist in [0,∞]. Hence, if a random sequence(Zm)m≥1 is increasing a.s., its limit Z exists in [0,∞] a.s. Therefore, we also haveeγZm

→ eγZ∈ [0,∞] with the conventions e−∞ = 0 and e∞ = ∞. Then (by mono-

tone convergence) E(eγZm) → E(eγZ).If γ < 0, then eγZ = 0 ⇐⇒ Z = ∞, but E(eγZ) is a mean (weighted average) of

nonnegative numbers (write out the definition in the discrete case), so P(Z = ∞) = 1 ifand only if E(eγZ) = 0. As γ ↑ 0, we get e−γZ

↑ 1 if Z < ∞ and e−γZ = 0 → 0 if Z = ∞,so (by monotone convergence)

E(eγZ) ↑ E(1Z<∞) = P(Z < ∞)

and the result follows. 2

13


Example 22 For independent Xj ∼ Poi(µj) and Zm = X1 + . . . + Xm, the randomvariable Z = limm→∞ Zm exists in [0,∞] a.s. Now

E(eγZm) = E((eγ)Zm) = e(eγ−1)(µ1+...+µm)→ e−(1−eγ

)µ

shows that the limit is Poi(µ) if µ =∑

m→∞ µm < ∞. We do not need the lemma forthis, since we can even directly identify the limiting moment generating function.

If µ = ∞, the limit of the moment generating function vanishes, and by the lemma, weobtain P(Z = ∞) = 1. So we still get S ∼ Poi(µ) within the extended range 0 ≤ µ ≤ ∞.

4.2 Construction of spatial Poisson processes

The examples of compound Poisson processes are the key to constructing spatial Poissonprocesses with finite intensity measure. Infinite intensity measures can be decomposed.

Theorem 23 (Construction) Let Λ be an intensity measure on D ⊂ Rd and suppose

that there is a partition (In)n≥1 of D into regions with Λ(In) < ∞. Consider independently

Nn ∼ Poi(Λ(In)), Y(n)

1, Y

(n)

2, . . . ∼

Λ(In ∩ ·)

Λ(In), i.e. P(Y

(n)

j ∈ A) =Λ(In ∩ A)

Λ(In)

and define Πn = Y(n)

j : 1 ≤ j ≤ Nn. Then Π =⋃

n≥1Πn is a spatial Poisson process

with intensity measure Λ.

Proof: First fix n and show that Πn is a spatial Poisson process on In

Thinning property of Poisson variables: Consider a sequence of inde-pendent Bernoulli(p) random variables (Bj)j≥1 and independent X ∼ Poi(λ).Then the following two random variables are independent:

X1 =

X∑

j=1

Bj ∼ Poi(pλ) and X2 =

X∑

j=1

(1 − Bj) ∼ Poi((1 − p)λ).

To prove this, calculate the joint probability generating function

E(rX1sX2) =

∞∑

n=0

P(X = n)E(rB1+...+Bnsn−B1−...−Bn)

=

∞∑

n=0

λn

n!e−λ

n∑

k=0

(

n

k

)

pk(1 − p)n−krksn−k

=∞∑

n=0

λn

n!e−λ(pr + (1 − p)s)n = e−λp(1−r)e−λ(1−p)(1−s),

so the probability generating function factorises giving independence and werecognise the Poisson distributions as claimed.

For A ⊂ In, consider X = Nn and the thinning mechanism, where Bj = 1Y (n)j ∈A ∼

Bernoulli(P(Y(n)

j ∈ A)), then we get property (b):

Nn(A) = X1 is Poisson distributed with parameter P(Y(n)

j ∈ A)Λ(In) = Λ(A).

Lecture 4: Spatial Poisson processes II 15

For property (a), disjoint sets A1, . . . , Am ⊂ In, we apply the analogous thinning

property for m + 1 types Y(n)

j ∈ Ai i = 0, . . . , m, where A0 = In \ (A1 ∪ . . . ∪ Am) todeduce the independence of Nn(A1), . . . , Nn(Am). Thus, Πn is a spatial Poisson process.

Now for N(A) =∑

n≥1Nn(A∩ In), we add up infinitely many Poisson variables and,

by Example 22, obtain a Poi(µ) variable, where µ =∑

n≥1Λ(A∩In) = Λ(A), i.e. property

(b). Property (a) also holds, since Nn(Aj ∩ In), n ≥ 1, j = 1, . . . , m, are all independent,and N(A1), . . . , N(Am) are independent as functions of independent random variables.

2

4.3 Sums over Poisson point processes

Recall that a Poisson point process (∆t)t≥0 with intensity function g : D0 → [0,∞) –focus on D0 = (0,∞) first but this can then be generalised – is a process such that

N((a, b] × (c, d]) = #a < t ≤ b : ∆t ∈ (c, d] ∼ Poi

(

(b − a)

∫ d

c

g(x)dx

)

,

0 ≤ a < b, (c, d] ⊂ D0, defines a Poisson counting measure on D = [0,∞) × D0. Thismeans that

Π = (t, ∆t) : t ≥ 0 and ∆t 6= 0

is a spatial Poisson process. Thinking of ∆s as a jump size at time s, let us studyXt =

∑

0≤s≤t ∆s, the process performing all these jumps. Note that this is the situationfor compound Poisson processes X; in Example 18, g : (0,∞) → [0,∞) is integrable.

Theorem 24 (Exponential formula) Let (∆t)t≥0 be a Poisson point process with lo-cally integrable intensity function g : (0,∞) → [0,∞). Then for all γ ∈ R

E

(

exp

γ∑

0≤s≤t

∆s

)

= exp

t

∫ ∞

0

(eγx− 1)g(x)dx

.

Proof: Local integrability of g on (0,∞) means in particular that g is integrable onIn = (2n, 2n+1], n ∈ Z. The properties of the associated Poisson counting measure Nimmediately imply that the random counting measures Nn counting all points in In,n ∈ Z, defined by

Nn((a, b] × (c, d]) = a < t ≤ b : ∆t ∈ (c, d] ∩ In, 0 ≤ a < b, (c, d] ⊂ (0,∞),

are independent. Furthermore, Nn is the Poisson counting measure of jumps of a com-pound Poisson process with (b−a)

∫ d

cg(x)dx = (b−a)λnP(Y

(n)

1∈ (c, d]) for 0 ≤ a < b and

(c, d] ⊂ In (cf. Example 18), so λn =∫

Ing(x)dx and (if λn > 0) jump density hn = λ−1

n gon In, zero elsewhere. Therefore, we obtain

E

(

exp

γ∑

0≤s≤t

∆(n)

s

)

= exp

t

∫

In

(eγx− 1)g(x)dx

, where ∆(n)

s =

∆s if ∆s ∈ In

0 otherwise


Now we have

Zm =m∑

n=−m

∑

0≤s≤t

∆(n)

s ↑

∑

0≤s≤t

∆s as m → ∞,

and (cf. Lemma 21 about finite or infinite limits), the associated moment generatingfunctions (products of individual moment generating functions) converge as required:

m∏

n=−m

exp

t

∫

2n+1

2n

(eγx− 1)g(x)dx

→ exp

t

∫ ∞

0

(eγx− 1)g(x)dx

.

2

4.4 Martingales (from B10a)

A discrete-time stochastic process (Mn)n≥0 in R is called a martingale if for all n ≥ 0

E(Mn+1|M0, . . . , Mn) = Mn, i.e. if E(Mn+1|M0 = x0, . . . , Mn = xn) = xn for all xj .

This is the principle of a fair game. What can I expect from the future if my current stateis Mn = xn? No gain and no loss, on average, whatever the past. The following importantrules for conditional expectations are crucial to establish the martingale property

• If X and Y are independent, then E(X|Y ) = E(X).

• If X = f(Y ), then E(X|Y ) = E(f(Y )|Y ) = f(Y ) for functions f : R → R for whichthe conditional expectations exist.

• Conditional expectation is linear E(αX1 + X2|Y ) = αE(X1|Y ) + E(X2|Y ).

• More generally: E(g(Y )X|Y ) = g(Y )E(X|Y ) for functions g : R → R for which theconditional expectations exist.

These are all not hard to prove for discrete random variables. The full statements (con-tinuous analogues) are harder. Martingales in continuous time can also be defined, but(formally) the conditioning needs to be placed on a more abstract footing. Denote by Fs

the “information available up to time s ≥ 0”, for us just the process (Mr)r≤s up to times – this is often written Fs = σ(Mr, r ≤ s). Then the four bullet point rules still hold forY = (Mr)r≤s or for Y replaced by Fs.

We call (Mt)t≥0 a martingale if for all s ≤ t

E(Mt|Fs) = Ms.

Example 25 Let (Ns)s≥0 be a Poisson process with rate λ. Then Ms = Ns − λs is amartingale: by the first three bullet points and by the Markov property (Proposition 3)

E(Nt − λt|Fs) = E(Ns + (Nt − Ns) − λt|Fs) = Ns + (t − s)λ − λt = Ns − λs.

Also Es = expγNs − λs(eγ− 1) is a martingale since by the first and last bullet points

above, and by the Markov property

E(Et|Fs) = E(expγNs + γ(Nt − Ns) − λt(eγ− 1)|Fs)

= expγNs − λt(eγ− 1)E(expγ(Nt − Ns))

= expγNs − λt(eγ− 1) exp−λ(t − s)(eγ

− 1) = Es.

We will review relevant martingale theory when this becomes relevant.

Lecture 5

The characteristics of subordinators

Reading: Kingman Section 8.4

We have done the leg-work. We can now harvest the fruit of our efforts and proceed toa number of important consequences. Our programme for the next couple of lectures is:

• We construct Levy processes from their jumps, first the most general increasingLevy process. As linear combinations of independent Levy processes are Levyprocesses (Assignment A.1.2.(a)), we can then construct Levy processes such asVariance Gamma processes of the form Zt = Xt − Yt for two increasing X and Y .

• We have seen martingales associated with Nt and expNt for a Poisson process N .Similar martingales exist for all Levy processes (cf. Assignment A.2.3.). Martin-gales are important for finance applications, since they are the basis of arbitrage-freemodels (more precisely, we need equivalent martingale measures, but we will as-sume here a “risk-free” measure directly to avoid technicalities).

• Our rather restrictive first range of examples of Levy processes was obtained fromknown infinitely divisible distributions. We can now model using the intensity func-tion of the Poisson point process of jumps to get a wider range of examples.

• We can simulate these Levy processes, either by approximating random walks basedon the increment distribution, or by constructing the associated Poisson point pro-cess of jumps, as we have seen, from a collection of independent random variables.

5.1 Subordinators and the Levy-Khintchine formula

We will call (weakly) increasing Levy processes “subordinators”. Recall “ν(dx)=g(x)dx”.

Theorem 26 (Construction) Let a ≥ 0, and let (∆t)t≥0 be a Poisson point process

with intensity measure ν on (0,∞) such that∫

(0,∞)

(1 ∧ x)ν(dx) < ∞,

then the process Xt = at +∑

s≤t ∆s is a subordinator with moment generating function

E(expγXt) = exptΨ(γ), where

Ψ(γ) = aγ +

∫

(0,∞)

(eγx− 1)ν(dx).

17


Proof: Clearly (at)t≥0 is a deterministic subordinator and we may assume a = 0 inthe sequel. Now the Exponential formula gives the moment generating function of Xt =∑

s≤t ∆s. We can now use Lemma 21 to check whether Xt < ∞ for t > 0:

P(Xt < ∞) = 1 ⇐⇒ E (exp γXt) = exp

t

∫ ∞

0

(eγx− 1)ν(dx)

→ 1 as γ ↑ 0.

This happens, by monotone convergence, if and only if for some (equivalently all) γ < 0∫ ∞

0

(1 − eγx)ν(dx) < ∞ ⇐⇒

∫ ∞

0

(1 ∧ x)ν(dx) < ∞.

It remains to check that (Xt)t≥0 is a Levy process. Fix 0 ≤ t0 < t1 < . . . < tn. Since(∆s)s≥0 is a Poisson point process, the processes (∆s)tj−1≤s<tj , j = 1, . . . , n, are inde-pendent (consider the restrictions to disjoint domains [tj−1, tj) × (0,∞) of the Poissoncounting measure

N((a, b] × (c, d]) = a ≤ t < b : ∆t ∈ (c, d], 0 ≤ a < b, 0 < c < d),

and so are the sums∑

tj−1≤s<tj∆s as functions of independent random variables. Fix

s < t. Then the process (∆s+r)r≥0 has the same distribution as (∆s)s≥0. In particular,∑

0≤r≤t ∆s+t ∼∑

0≤r≤t ∆r. The process t 7→∑

s≤t ∆s is right-continuous with left limits,since it is a random increasing function where for each jump time T , we have

limt↑T

∑

s≤t

∆s = limt↑T

∑

s<T

∆s1s≤t =∑

s<T

∆s and limt↓T

∑

s≤t

∆s = limt↓T

∑

s≤T+1

∆s1s≤t =∑

s≤T

∆s,

by monotone convergence, because each of the terms ∆s1s≤t in the sums converges. 2

Note also that, due to the Exponential formula, P(Xt < ∞) > 0 already impliesP(Xt < ∞) = 1. We shall now state but not prove the Levy-Khintchine formula fornonnegative random variables.

Theorem 27 (Levy-Khintchine) A nonnegative random variable Y has an infinitely

divisible distribution if and only if there is a pair (a, ν) such that for all γ ≤ 0

E(expγY ) = exp

aγ +

∫

(0,∞)

(eγx− 1)ν(dx)

, (1)

where a ≥ 0 and ν is such that∫

(0,∞)(1 ∧ x)ν(dx) < ∞.

Corollary 28 Given a nonnegative random variable Y with infinitely divisible distribu-

tion, there exists a subordinator (Xt)t≥0 with X1 ∼ Y .

Proof: Let Y have an infinitely divisible distribution. By the Levy-Khintchine theorem,its moment generating function is of the form (1) for parameters (a, ν). Theorem 26constructs a subordinator (Xt)t≥0 with X1 ∼ Y . 2

This means that the class of subordinators can be parameterised by two parameters,the nonnegative “drift parameter” a ≥ 0, and the “Levy measure” ν, or its density, the“Levy density” g : (0,∞) → [0,∞). The parameters (a, ν) are referred to as the “Levy-Khintchine characteristics” of the subordinator (or of the infinitely divisible distribution).Using the Uniqueness theorem for moment generating functions, it can be shown that aand ν are unique, i.e. that no two sets of characteristics refer to the same distribution.

Lecture 5: The characteristics of subordinators 19

5.2 Examples

Example 29 (Gamma process) The Gamma process, where Xt ∼ Gamma(αt, β), isan increasing Levy process. In Assignment A.2.4. we showed that

E (exp γXt) =

(

β

β − γ

)αt

= exp

t

∫ ∞

0

(eγx− 1)αx−1e−βxdx

, γ < β.

We read off the characteristics a = 0 and g(x) = αx−1e−βx, x > 0.

Example 30 (Poisson process) The Poisson process, where Xt ∼ Poi(λt), has

E (exp γXt) = exp tλ(eγ− 1)

This corresponds to characteristics a = 0 and ν = λδ1, where δ1 is the discrete unit pointmass in (jump size) 1.

Example 31 (Increasing compound Poisson process) The compound Poisson pro-cess Ct = Y1 + . . .+YXt

, for a Poisson process X and independent identically distributednonnegative Y1, Y2, . . . with probability density function h(x), x > 0, satisfies

E (exp γCt) = exp

t

∫ ∞

0

(eγx− 1)λh(x)dx

,

and we read off characteristics a = 0 and g(x) = λh(x), x > 0. We can add a drift and

consider ˜Ct = at + Ct for some a > 0 to get a compound Poisson process with drift.

Example 32 (Stable subordinator) The stable subordinator is best defined in termsof its Levy-Khintchine characteristics a = 0 and g(x) = x−α−1. This gives for γ ≤ 0

E (exp γXt) = exp

t

∫ ∞

0

(eγx− 1)x−α−1dx

= exp

tΓ(1 − α)

α(−γ)α

.

Note that E(expγc1/αXt/c) = E(expγXt), so that (c1/αXt/c)t≥0 ∼ X. More generally,we can also consider e.g. tempered stable processes with g(x) = x−α−1 exp−ρx, ρ > 0.

Figure 5.1: Examples: Poisson process, Gamma process, stable subordinator

5.3 Aside: nonnegative Levy processes

It may seem obvious that a nonnegative Levy process, i.e. one where Xt ≥ 0 a.s. for all t ≥0, is automatically increasing, since every increment Xs+t −Xs has the same distributionXt and is hence also nonnegative. Let us be careful, however, and remember that thereis a difference between something never happening at a fixed time and something neverhappening at any time. We have e.g. for a (one-dimensional) Poisson process (Nt)t≥0

P(∆Nt 6= 0) =∑

n≥1

P(Tn = t) = 0 for all t ≥ 0, but P(∃t : ∆Nt 6= 0) = 1.


Here we can argue that if f(t) < f(s) for some s < t and a right-continuous function,then there are also two rational numbers s0 < t0 for which f(t0) < f(s0), so

P(∃s, t ∈ (0,∞), s < t : Xt − Xs < 0) > 0 ⇒ P(∃s0, t0 ∈ (0,∞) ∩ Q : Xt0 − Xs0 < 0) > 0

However, the latter can be bounded above (by subadditivity P(⋃

n An) ≤∑

n P(An))

P(∃s0, t0 ∈ (0,∞) ∩ Q : Xt0 − Xs0 < 0) ≤∑

s0,t0∈(0,∞)∩Q

P(Xt0−s0 < 0) = 0.

Another instance of such delicate argument is the following: if Xt ≥ 0 a.s. for onet > 0 and a subordinator X, then Xt ≥ 0 a.s. for all t ≥ 0. It is true, but to say ifP(Xs < 0) > 0 for some s < t then P(Xt < 0) > 0 may not be all that obvious. Itis, however, easily justified for s = t/m, since then P(Xt < 0) ≥ P(Xtj/m − Xt(j−1)/m <0 for all j = 1, . . . , m) > 0. We have to apply a similar argument to get P(Xtq < 0) = 0for all rational q > 0. Then we use again right-continuity to see that a function that isnonnegative at all rationals cannot take a negative value at an irrational either, so weget

P(∃s ∈ [0,∞) : Xs < 0) = P(∃s ∈ [0,∞) ∩ Q : Xs < 0) ≤∑

s∈[0,∞)∩Q

P(Xs < 0) = 0.

5.4 Applications

Subordinators have found a huge range of applications, but are not directly models fora lot of real world phenomena. We can now construct more general Levy processes ofthe form Zt = Xt − Yt for two subordinators X and Y . Let us here indicate somesubordinators as they are used/arise in connection with other Levy processes.

Example 33 (Subordination) For a Levy process X and an independent subordinatorT , the process Ys = XTs

, s ≥ 0, is also a Levy process (we study this later in the course).The rough argument is that (XTs+u −XTs

)u≥0 is independent of (Xr)r≤Tsand distributed

as X, by the Markov property. Hence XTs+r−XTs

is independent of XTsand distributed

as XTr. A rigorous argument can be based on calculations of joint moment generating

functions. Hence, subordinators are a useful tool to construct Levy processes, e.g. fromBrownian motion X. Many models of financial markets are of this type. The operationYs = XTs

is called subordination – this is where subordinators got their name from.

Example 34 (Level passage) Let Zt = at − Xt where a = E(X1). It can be shownthat τs = inft ≥ 0 : Zt > s < ∞ a.s. for all s ≥ 0 (from the analogous random walkresult). It turns out (cf. later in the course) that (τs)s≥0 is a subordinator.

Example 35 (Level set) Look at the zero set Z = t ≥ 0 : Bt = 0 for Brownianmotion (or indeed any other centred Levy process) B. Z is unbounded since B crosseszero at arbitrarily large times so as to pass beyond all s and −s. Recall that (tB1/t)t≥0

is also a Brownian motion. Therefore, Z also has an accumulation point at t = 0, i.e.crosses zero infinitely often at arbitrarily small times. In fact, it can be shown thatZ is the closed range Xr, r ≥ 0cl of a subordinator (Xr)r≥0. The Brownian scalingproperty (

√

cBt/c)t≥0 ∼ B shows that Xr/c, r ≥ 0cl∼ Z, and so X must have a scaling

property. In fact, X is a stable subordinator of index 1/2. Similar results, with differentsubordinators, hold not just for all Levy processes but even for most Markov processes.

Lecture 6

Levy processes with no negative

jumps

Reading: Kyprianou 2.1, 2.4, 2.6, Schoutens 2.2; Further reading: Williams 10-11

Subordinators X are processes with no negative jumps. We get processes that can de-crease by adding a negative drift at for a < 0. Also, Brownian motion B has no negativejumps. A guess might be that Xt + at + σBt is the most general Levy process withno negative jumps, but this is false. It turns out that even a non-summable amount ofpositive jumps can be incorporated, but we will have to look at this carefully.

6.1 Bounded and unbounded variation

The (total) variation of a right-continuous function f : [0, t] → R with left limits is

||f ||TV := sup

n∑

j=1

|f(tj) − f(tj−1)| : 0 = t0 < t1 < . . . < tn = t, n ∈ N

.

Clearly, for an increasing function with f(0) = 0 this is just f(t) and for a differencef = g−h of two increasing functions with g(0) = h(0) = 0 this is at most g(t)+h(t) < ∞,so all differences of increasing functions are of bounded variation. There are, however,functions of infinite variation, e.g. Brownian paths: they have finite quadradic variation

2n∑

j=1

|Btj2−n − Bt(j−1)2−n |2→ t in the L2 sense

since

E

(

2n∑

j=1

|Btj2−n − Bt(j−1)2−n |2

)

= 2nE(B2

t2−n) = t

and

E

(

2n∑

j=1

|Btj2−n − Bt(j−1)2−n |2− t

)2

= Var

(

2n∑

j=1

|Btj2−n − Bt(j−1)2−n |2

)

≤ 2n(2−nt)2Var(B2

1) → 0,

21


but then assuming finite total variation with positive probability, the uniform continuityof the Brownian path implies

2n∑

j=1

|Btj2−n − Bt(j−1)2−n |2≤

(

supj=1,...,2n

|Btj2−n − Bt(j−1)2−n |

) 2n∑

j=1

|Btj2−n − Bt(j−1)2−n | → 0

with positive probability, but this is incompatible with convergence to t, so the assump-tion of finite total variation must have been wrong.

Here is how jumps influence total variation:

Proposition 36 Let f be a right-continuous function with left limits and jumps (∆fs)0≤s≥t.

Then

||f ||TV ≥

∑

0≤s≤t

|∆fs|

Proof: Enumerate the jumps in decreasing order of size by (Tn, ∆fTn)n≥0. Fix N ∈ N

and δ > 0. Choose ε > 0 so small that⋃

[Tn − ε, Tn] is a disjoint union and such that|f(Tn − ε)− f(Tn−)| < δ/N . Then for Tn − ε, Tn : n = 1, . . . , N = t1, . . . , t2N+1 suchthat 0 = t0 < t1 < . . . < t2N+1 < t2N+2 = t, we have

2N+2∑

j=1

|f(tj) − f(tj−1)| ≥N∑

n=1

∆f(Tn) − δ.

Since N and δ were arbitrary, this completes the proof, whether the right-hand side isfinite or infinite. 2

6.2 Martingales (from B10a)

Three martingale theorems are of central importance. We will require in this lecturejust the maximal inequality, but we formulate all three here for easier reference. Theyall come in several different forms. We present the L2-versions as they are most easilyformulated and will suffice for us.

A stopping time is a random time T such that for every s ≥ 0 the information Fs

allows to decide whether T ≤ s. More formally, if the event T ≤ s can be expressedin terms of (Mr, r ≤ s) (is measurable with respect to Fs). The prime example of astopping time is the first entrance time TA = inft ≥ 0 : Mt ∈ A. Note that

T ≤ s = Mr 6∈ A for all r ≤ s

(and at least for closed sets A we can drop the irrational r ≤ s and see measurability,then approximate open sets.)

Theorem 37 (Optional stopping) Let (Mt)t≥0 be a martingale and T a stopping time.

If supt≥0E(M2

t ) < ∞, then E(MT ) = E(M0).

Theorem 38 (Convergence) Let (Mt)t≥0 be a martingale such that supt≥0E(M2

t ) <∞, then Mt → M∞ almost surely.

Theorem 39 (Maximal inequality) Let (Mt)t≥0 be a martingale. Then E(supM2

s :0 ≤ s ≤ t) ≤ 4E(M2

t ).

Lecture 6: Levy processes with no negative jumps 23

6.3 Compensation

Let g : (0,∞) → [0,∞) be the intensity function of a Poisson point process (∆t)t≥0. Ifg is not integrable at infinity, then #0 ≤ s ≤ t : ∆s > 1 ∼ Poi(

∫∞1

g(x)dx) = Poi(∞),and it is impossible for a right-continuous function with left limits to have accumulationpoints in the set of such jumps (lower and upper points of a sequence of jumps will thenhave different limit points). If however g is not integrable at zero, we have to investigatethis further.

Proposition 40 Let (∆t)t≥0 be a Poisson point process with intensity measure ν on

(0,∞).

(i) If∫∞0

xν(dx) < ∞, then

E

(

∑

s≤t

∆s

)

= t

∫ ∞

0

xν(dx).

(ii) If∫∞0

x2ν(dx) < ∞, then

Var

(

∑

s≤t

∆s

)

= t

∫ ∞

0

x2ν(dx).

Proof: These are the two leading terms in the expansion with respect to γ of the Expo-nential formula: the first moment can always be obtained from the moment generatingfunction by taking ∂

∂γ|γ=0, here

∂

∂γexp

t

∫ ∞

0

(eγx− 1)ν(dx)

∣

∣

∣

∣

γ=0

= t

∫ ∞

0

xeγxν(dx)

∣

∣

∣

∣

γ=0

= t

∫ ∞

0

ν(dx),

and the second moment follows from the second derivative in the same way. 2

Consider compound Poisson processes, with a drift that turns them into martingales

Zεt =

∑

s≤t

∆s1ε<∆s≤1 − t

∫

1

ε

xν(dx) (1)

We have deliberately excluded jumps in (1,∞). These are easier to handle separately.What integrability condition on ν do we need for Zε

t to converge as ε ↓ 0?

Lemma 41 Let (∆t)t≥0 be a Poisson point process with intensity measure ν on (0, 1).

With Zε defined in (1), Zεt converges in L2 if

∫

1

0x2ν(dx) < ∞.

Proof: We only do this for ν(dx) = g(x)dx. Note that for 0 < δ < ε < 1, by Proposition40(ii) applied to gδ,ε(x) = g(x)1δ≤x<ε,

E(|Zεt − Zδ

t |2) = t

∫ ε

δ

x2g(x)dx

so that (Zεt )0<ε<1 is a Cauchy family as ε ↓ 0, for the L2-distance d(X, Y ) =

√

E((X − Y )2).By completeness of L2-space, there is a limiting random variable Zt as required. 2

We can slightly tune this argument to establish a general existence theorem:

Theorem 42 (Existence) There exists a Levy process whose jumps form a Poisson

point process with intensity measure ν on (0,∞) if and only if∫

(0,∞)(1 ∧ x2)ν(dx) < ∞.


Proof: The “only if” statement is a consequence of a Levy-Khintchine type characteri-sation of infinitely divisible distributions on R, cf. Theorem 44, which we will not prove.Let us prove the “if” part in the case where ν(dx) = g(x)dx.

By Proposition 40(i), E(Zεt −Zδ

t ) = 0. By Assignment A.2.3.(c), the procecss Zεt −Zδ

t

is a martingale, and the maximal inequality (Theorem 39) shows that

E

(

sup0≤s≤t

∣

∣Zεt − Zδ

t

∣

∣

)

≤ 4E(|Zεt − Zδ

t |2) = 4t

∫ ε

δ

x2g(x)dx

so that (Zεs , 0 ≤ s ≤ t)0<ε<1 is a Cauchy family as ε ↓ 0, for the uniform L2-distance

d[0,t](X, Y ) =√

E(sup0≤s≤t |Xs − Ys|

2). By completeness of L2-space, there is a limiting

process (Z(1)

s )0≤s≤t, which as the uniform limit (in L2) of (Zεs )0≤s≤t is right-continuous

with left limits. Also consider the independent compound Poisson process

Z(2)

t =∑

s≤t

∆s1∆s>1 and set Z = Z(1) + Z(2).

It is not hard to show that Z is a Levy process that incorporates all jumps (∆s)0≤s≤t. 2

Example 43 Let us look at a Levy density g(x) = |x|−5/2, x ∈ [−3, 0). Then the

compensating drifts∫

3

εxg(x)dx take values 0.845, 2.496, 5.170 and 18.845 for ε = 1,

ε = 0.3, ε = 0.1 and ε = 0.01. In the simulation, you see that the slope increases (toinfinity, actually as ε ↓ 0), but the picture begins to stabilise and converge to a limit.

Figure 6.1: Approximation of a Levy process with no positive jumps – compensating drift

Lecture 7

General Levy processes and

simulation

Reading: Schoutens Sections 8.1, 8.2, 8.4

For processes with no negative jumps, we compensated jumps by a linear drift and incor-porated more and more smaller jumps while letting the slope of the linear drift tend tonegative infinity. We will now construct the most general real-valued Levy process as thedifference of two such processes (and a Brownian motion). For explicit marginal distribu-tions, we can simulate Levy processes by approximating random walks. In practice, weoften only have explicit characteristics (drift coefficient, Brownian coefficient and Levymeasure). We will also simulate Levy processes based on the characteristics.

7.1 Construction of Levy processes

The analogue of Theorem 27 for real-valued random variables is as follows.

Theorem 44 (Levy-Khintchine) A real-valued random variable X has an infinitely

divisible distribution if there are parameters a ∈ R, σ2≥ 0 and a measure ν on R \ 0

with∫∞−∞(1 ∧ x2)ν(dx) <∞ such that E(eiλX) = e−ψ(λ), where

ψ(λ) = −iaλ +1

2σ2λ2

−

∫ ∞

−∞(eiλx − 1 − iλx1|x|≤1)ν(dx), λ ∈ R.

Levy processes are parameterised by their Levy-Khintchine characteristics (a, σ2, ν),where we call a the drift coefficient, σ2 the Brownian coefficient and ν the Levy measure

or jump measure. ν(dx) will often be of the form g(x)dx, and we then refer to g as theLevy density or jump density.

Theorem 45 (Existence) Let (a, σ2, ν) be Levy-Khintchine characteristics, (Bt)t≥0 a

standard Brownian motion and (∆t)t≥0 an independent Poisson point process of jumps

with intensity measure ν. Then there is a Levy process

Zt = at+ σBt +Mt + Ct, where Ct =∑

s≤t∆s1|∆s|>1,

25


is a compound Poisson process (of big jumps) and

Mt = limε↓0

(

∑

s≤t∆s1ε<|∆s|≤1 − t

∫

x∈R:ε<|x|≤1xν(dx)

)

is a martingale (of small jumps – compensated by a linear drift).

Proof: The construction of Mt = Pt −Nt can be made from two independent processesPt and Nt with no negative jumps as in Theorem 42. Nt will be built from a Poisson pointprocess with intensity measure ν((c, d]) = ν([−d,−c)), 0 < c < d ≤ 1 (or g(y) = g(−y),0 < y < 1).

We check that the characteristic function of Zt = at + σBt + Pt − Nt + Ct is ofLevy-Khintchine type with parameters (a, σ, ν). We have five independent components.Evaluate at t = 1 to get

E(eγa) = eγa

E(eγσB1) = exp1

2γ2σ2

E(eγP1) = exp

∫

1

0

(eγx − 1 − γx)ν(dx)

E(e−γN1) = exp

∫

1

0

(e−γy − 1 + γy)ν(dy)

= exp

∫

0

−1

(eγx − 1 − γx)ν(dx)

E(eiλC1) = exp

∫

|x|>1

(eiλx − 1)ν(dx)

.

The last formula is checked in analogy with the moment generating function computationof Assignment A.1.3 (in general, the moment generating function will not be well-definedfor this component). For the others, now “replace” γ by iλ. A formal justification canbe obtained by analytic continuation, since the moment generating functions of thesecomponents are entire functions of γ as a complex parameter. Now the characteristicfunction of Z1 is the product of characteristic functions of the independent components,and this yields the formula required. 2

We stress in particular, that every Levy process is the difference of two processes withonly positive jumps. In general, these processes are not subordinators, but of the formin Theorem 42 plus a Brownian motion component. They can then both take positiveand negative values.

Example 46 (Variance Gamma process) We introduced the Variance Gamma pro-cess as difference X = G − H of two independent Gamma subordinators G and H .We can generalise the setting of Exercise A.1.2.(b) and allow G1 ∼ Gamma(α+, β+) andH1 ∼ Gamma(α−, β−). The moment generating function of the Variance Gamma processis

E(eγXt) = E(eγGt)E(e−γHt) =

(

β+

β+ − γ

)α+t( β−β− + γ

)α−t

= exp

t

∫ ∞

0

(eγx − 1)α+x−1e−β+xdx

exp

t

∫ ∞

0

(e−γy − 1)α−y−1e−β−ydy

= exp

t

∫ ∞

0

(eγx − 1)α+|x|−1e−β+|x|dx+ t

∫

0

−∞(eγx − 1)α−|x|

−1e−β−|x|dx

.

Lecture 7: General Levy processes and simulation 27

and this is in Levy-Khintchine form with ν(dx) = g(x)dx with

g(x) =

α+|x|−1e−β+|x| x > 0

α−|x|−1e−β−|x| x < 0

The process (∆Xt)t≥0 is a Poisson point process with intensity function g.

Example 47 (CGMY process) Theorem 45 encourages to specify Levy processes bytheir characteristics. As a natural generalisation of the Variance Gamma process, Carr,Geman, Madan and Yor (CGMY) suggested the following for financial price processes

g(x) =

C+ exp−G|x||x|−Y−1 x > 0C− exp−M |x||x|−Y−1 x < 0

for parameters C± > 0, G > 0, M > 0, Y ∈ [0, 2). While the Levy density is a nicefunction, the probability density function of an associated Levy process Xt is not availablein closed form, in general. The CGMY model contains the Gamma model for Y = 0.When this model is fitted to financial data, there is usually significant evidence againstY = 0, so the CGMY model is more appropriate than the Variance Gamma model.

We can construct Levy processes from their Levy density and will also simulate fromLevy densities. Note that this way of modelling is easier than searching directly forinfinitely divisible probability density functions.

7.2 Simulation via embedded random walks

“Simulation” usually refers to the realisation of a random variable using a computer.Most mathematical and statistical packages provide functions, procedures or commandsfor the generation of sequences of pseudo-random numbers that, while not random, showfeatures of independent and identically distributed random variables that are adequatefor most purposes. We will not go into the details of the generation of such sequences,but assume that we a sequence (Uk)k≥1 of independent Unif(0, 1) random variables.

If the increment distribution is explicitly known, we simulate via time discretisation.

Method 1 (Time discretisation) Let (Xt)t≥0 be a Levy process so that Xt has prob-ability density function ft. Fix a time lag δ > 0. Denote Ft(x) =

∫ x

−∞ ft(y)dy and

F−1

t (u) = infx ∈ R : Ft(x) > u. Then the process

X(1,δ)t = S[t/δ], where Sn =

n∑

k=1

Yk and Yk = F−1

δ (Uk),

is called the time discretisation of X with time lag δ.

One usually requires numerical approximation for F−1

t , even if ft is available in closedform. That the approximations converge, is shown in the following proposition.

Proposition 48 As δ ↓ 0, we have X(1,δ)t → Xt in distribution.

Proof: We can employ a coupling proof: t is a.s. not a jump time of X, so we haveX[t/δ]δ → Xt a.s., and so convergence in distribution for X

(1,δ)t ∼ X[t/δ]δ. 2


0 2 4 6 8 10

0.00.2

0.40.6

0.81.0

Gamma process with shape parameter 0.1 and scale parameter 0.1time

0 2 4 6 8 10

01

23

45

67

Gamma process with shape parameter 1 and scale parameter 1time

0 2 4 6 8 10

02

46

8


0 2 4 6 8 10

02

46

810


Figure 7.1: Simulation of Gamma processes from random walks with Gamma increments

Example 49 (Gamma processes) For Gamma processes, Ft is an incomplete Gammafunction, which has no closed-form expression, and F−1

t is also not explicit, but numericalevaluations have been implemented in many statistical packages. There are also Gammagenerators based on more uniform random variables. We display a range of parameterchoices. Since for a Gamma(1, 1) process X, the process (β−1Xαt)t≥0 is Gamma(α, β):

E(expγβ−1Xαt) =

(

1

1 − γβ−1

)αt

=

(

β

β − γ

)αt

,

we chose α = β (keeping mean 1 and comparable spatial scale) but a range of parametersα ∈ 0.1, 1, 10, 100 on a fixed time interval [0, 10]. We “see” convergence to a lineardrift as α → ∞ (for fixed t this is due to the laws of large numbers).

0 2 4 6 8 10

02

46

810

Approximation of Gamma(1,1) process, delta=1time33

0 2 4 6 8 10

02

46

810

Approximation of Gamma(1,1) process, delta=0.1time22

0 2 4 6 8 10

02

46

810

Approximation of Gamma(1,1) process, delta=0.01time1

Figure 7.2: Random walk approximation to a Levy process, as in Proposition 48

Lecture 7: General Levy processes and simulation 29

Example 50 (Variance Gamma processes) We represent the Variance Gamma pro-cess as the difference of two independent Gamma processes and focus on the sym-metric case, so achieve mean 0 and fix variance 1 by putting β = α2/2; we considerα ∈ 1, 10, 100, 1000. We “see” convergence to Brownian motion as α → ∞ (for fixed tdue to the Central Limit Theorem).

0 2 4 6 8 10

−2.0

−1.5

−1.0

−0.5

0.0

Variance Gamma process with shape parameter 0.5 and scale parameter 1time

0 2 4 6 8 100

12

34

5

Variance Gamma process with shape parameter 50 and scale parameter 10time

0 2 4 6 8 10

0.00.5

1.01.5

2.0

Variance Gamma process with shape parameter 5000 and scale parameter 100time

0 2 4 6 8 10

−0.5

0.00.5

1.01.5

2.0

Variance Gamma process with shape parameter 5e+05 and scale parameter 1000time

Figure 7.3: Simulation of Variance Gamma processes as differences of random walks

7.3 R code – not examinable

The following code is posted on the course website as gammavgamma.R.

psum <- function(vector)

b=vector;

b[1]=vector[1];

for (j in 2:length(vector)) b[j]=b[j-1]+vector[j]; b

gammarw <- function(a,p)

unif=runif(10*p,0,1)

pos=qgamma(unif,a/p,a);

space=psum(pos);

time=(1/p)*1:(10*p);

plot(time,space,

pch=".",

sub=paste("Gamma process with shape parameter",a,"and scale parameter",a))


vgammarw <- function(a,p)

unifpos=runif(10*p,0,1)

unifneg=runif(10*p,0,1)

pos=qgamma(unifpos,a*a/(2*p),a);

neg=qgamma(unifneg,a*a/(2*p),a);

space=psum(pos-neg);

time=(1/p)*1:(10*p);

plot(time,space,

pch=".",

sub=paste("Variance Gamma process with shape parameter",a*a/2,

"and scale parameter",a))

Now you can try various values of parameters a > 0 and steps per time unit p = 1/δin gammarw(a,p), e.g.

gammarw(10,100)

vgammarw(10,1000)

Lecture 8

Simulation II

Reading: Ross 11.3, Schoutens Sections 8.1, 8.2, 8.4

In practice, the increment distribution is often not known, but the Levy characteristicsare, so we have to simulate Poisson point processes of jumps, by “throwing away the smalljumps” and then analyse (and correct) the error committed.

8.1 Simulation via truncated Poisson point processes

Example 51 (Compound Poisson process) Let (Xt)t≥0 be a compound Poisson pro-cess with Levy density g(x) = λh(x), where h is a probability density function. DenoteH(x) =

∫ x

−∞ h(y)dy and H−1(u) = infx ∈ R : H(x) > u. Let Yk = H−1(U2k) and

Zk = −λ−1 ln(U2k−1), k ≥ 1. Then the process

X(2)

t = SNt, where Sn =

n∑

k=1

Yk, Tn =

n∑

k=1

Zk, Nt = #n ≥ 1 : Tn ≤ t,

has the same distribution as X.

Method 2 (Throwing away the small jumps) Let (Xt)t≥0 be a Levy process withcharacteristics (a, 0, g), where g is not the multiple of a probability density function. Fixa jump size threshold ε > 0 so that λε =

∫

x∈R:|x|>ε g(x)dx > 0, and write

g(x) = λεhε(x), |x| > ε, hε(x) = 0, |x| ≤ ε,

for a probability density function hε. Denote H(x) =∫ x

−∞ h(y)dy and H−1(u) = infx ∈

R : H(x) > u. Let Yk = H−1(U2k) and Zk = −λ−1 ln(U2k−1), k ≥ 1. Then the process

X(2,ε)t = SNt

− bεt, where Sn =

n∑

k=1

Yk, Tn =

n∑

k=1

Zk, Nt = #n ≥ 1 : Tn ≤ t,

and bε = a−∫

x∈R:ε<|x|≤1 xg(x)dx, is called the process with small jumps thrown away.

For characteristics (a, σ2, g) we can now simulate Lt = σBt +Xt by σB(1,δ)t +X

(2,ε)t .

The following proposition says that such approximations converge as ε ↓ 0 (and δ ↓ 0).This is illustrated in Figure 6.3.

31


Proposition 52 As ε ↓ 0, we have X(2,ε)t → Xt in distribution.

Proof: For a process with no negative jumps and characteristics (0, 0, g), this is aconsequence of the stronger Lemma 41, which gives a coupling for which convergenceholds in the L2 sense. For a general Levy process with characteristics (a, 0, g) thatargument can be adapted, or we write Xt = at+ Pt −Nt and deduce the result:

E(expiλX(2,ε)t ) = eiatE(expiλP

(2,ε)t )E(exp−iλN

(2,ε)t )

→ eiatE(expiλPt)E(exp−iλNt) = E(eiλXt).

2

Example 53 (Symmetric stable processes) Symmetric stable processes X are Levyprocesses with characteristics (0, 0, g), where g(x) = c|x|−α−1, x ∈ R \ 0 for someα ∈ (0, 2) (cf. Assignment 3.2.). We decompose X = P − N for two independentprocesses with no negative jumps and simulate P and N . By doing this, we have

λε =

∫ ∞

ε

g(x)dx =c

αε−α, Hε(x) = 1 − (ε/x)α and H−1

ε (u) = ε(1 − u)−1/α.

For the simulation we choose ε = 0.01. We compare α ∈ 0.5, 1, 1.5, 1.8. All processesare centred with infinite variance. Big jumps dominate the plots for small α. Recall thatE(eiλXt) = e−b|λ|

α

→ e−bλ2

as α ↑ 2, and we get, in fact, convergence to Brownian motion,the stable process of index 2.

0 2 4 6 8 10

−600

−400

−200

0

Stable process with index 0.5 and cplus= 1 and cminus= 1time

0 2 4 6 8 10

−40

−20

020

Stable process with index 1 and cplus= 1 and cminus= 1time

0 2 4 6 8 10

−20

24

68

1012


0 2 4 6 8 10

−20

24

68


Figure 8.1: Simulation of symmetric stable processes from their jumps

Lecture 8: Simulation II 33

0 2 4 6 8 10

010

000

2000

030

000


0 2 4 6 8 10

020

4060

80


0 2 4 6 8 10

−10

−50

510

1520


0 2 4 6 8 10

02

46

810


Figure 8.2: Simulation of stable processes with no negative jumps

Example 54 (Stable processes with no negative jumps) For stable processes withno negative jumps, we have g(x) = c+x

−α−1, x > 0. The subordinator case α ∈ (0, 1)was discussed in Assignment A.3.1. – Xt =

∑

s≤t ∆s. The case α ∈ [1, 2), where com-

pensation is required, is such that E(X1) = 0, i.e. a = −

∫∞1xg(x)dx. We choose ǫ = 0.1

for α ∈ 0.5, 0.8 and ε = 0.01 for α ∈ 1.5, 1.8.

Strictly speaking, we take as triplet (a, σ2, g) in Theorem 45 g as given, but for

α ∈ (0, 1) we take a =∫

1

0xg(x)dx so that

E(eiλXt) = exp

t

∫ ∞

0

(eiλx − 1)g(x)dx

= exp

−t

(

−iλa−

∫ ∞

0

(eiλx − 1 − iλx1|x|≤1)g(x)dx

)

,

since compensation of small jumps is not needed and we obtain a subordinator if we donot compensate, and for α ∈ (1, 2), we take a = −

∫∞1xg(x)dx so that

E(eiλXt) = exp

t

∫ ∞

0

(eiλx − 1 − iλx)g(x)dx

= exp

−t

(

−iλa−

∫ ∞

0

(eiλx − 1 − iλx1|x|≤1)g(x)dx

)

,

since we can compensate all jump and achieve E(Xt) = 0. Only with these choices weobtain the Levy processes with no negative jumps that satisfy the scaling property.

This discussion shows that the representation in Theorem 45 is artificial and repre-sentations with different compensating drifts are often more natural.


8.2 Generating specific distributions

In this course, we will not go into the computational details of simulations. However,we do point out some principles here that lead to improved simulations, and we discusssome of the resulting modifications to the methods presented.

Often, it is not efficient to compute the inverse cumulative distribution function. Fora number of standard distributions, other methods have been developed. We will herelook at standard Normal generators. The Gamma distribution is discussed in AssignmentA.4.2.

Example 55 (Box-Muller generator) Consider the following procedure

1. Generate two independent random numbers U ∼ Unif(0, 1) and V ∼ Unif(0, 1).

2. Set X =√

−2 ln(U) cos(2πV ) and Y =√

−2 ln(U) sin(2πV ).

3. Return the pair (X, Y ).

The claim is the X and Y are independent standard Normal random variables. Theproof is an exercise on the transformation formula. First, the transformation is clearlybijective from (0, 1)2 to R

2. The inverse transformation can be worked out fromX2+Y 2 =−2 ln(U) and Y/X = tan(2πV ) as

(U, V ) = T−1(X, Y ) = (e−(X2+Y 2

)/2, (2π)−1 arctan(Y/X))

(with an appropriate choice of the branch of arctan, which is not relevant here). TheJacobian of the inverse transformation is

J =

(

−xe−(x2+y2)/2

−ye−(x2+y2)/2

−1

2πyx2

1

1+y2/x21

2π1

x1

1+y2/x2

)

⇒ | det(J)| =1

2πe−(x2

+y2)/2

and so, as required,

fX,Y (x, y) = fU,V (T−1(x, y))| det(J)| =1

2πe−(x2

+y2)/2.

For a more efficient generation of standard Normal random variables, it turns outuseful to first generate uniform random variables on on the disk of radius 1:

Example 56 (Uniform distribution on the disk) For U1 ∼ Unif(0, 1) and U2 ∼

Unif(0, 1) independent, we have that (V1, V2) = (2U1 − 1, 2U2 − 1) is uniformly dis-tributed on the square (−1, 1)2 centered at (0, 0), which contains the disk D = (x, y) ∈R

2 : x2 +y2 < 1, and we have, in particular P((V1, V2) ∈ D) = π/4. Now, for all A ⊂ R2,

we have

P((V1, V2) ∈ A|(V1, V2) ∈ D) =area(A ∩D)

π,

so the conditional distribution of (V1, V2) given (V1, V2) ∈ D is uniform on D. By thefollowing lemma, this conditioning can be turned into an algorithm by repeated trials:

1. Generate two independent random numbers U1 ∼ Unif(0, 1) and U2 ∼ Unif(0, 1).

Lecture 8: Simulation II 35

2. Set (V1, V2) = (2U1 − 1, 2U2 − 1).

3. If (V1, V2) ∈ D, go to 4., else go to 1.

4. Return the numbers (V1, V2).

The pair of numbers returned will be uniformly distributed on the disk D.

Lemma 57 (Conditioning by repeated trials) Let X,X1, X2, . . . be independent and

identically distributed d-dimensional random vectors. Also let A ⊂ Rd such that p =

P(X ∈ A) > 0. Denote N = infn ≥ 1 : Xi ∈ A. Then N ∼ geom(p) is independent

of XN , and XN has as its (unconditional) distribution the conditional distribution of Xgiven X ∈ A, i.e.

P(XN ∈ B) = P(X ∈ B|X ∈ A) for all B ⊂ Rd.

Proof: We calculate the joint distribution

P(N = n,Xn ∈ B) = P(X1 6∈ A, . . . , Xn−1 6∈ A,Xn ∈ A ∩B)

= (1 − p)n−1P(Xn ∈ A ∩B) = (1 − p)n−1pP(Xn ∈ B|Xn ∈ A).

2

We now get the following modification of the Normal generator:

Example 58 (Polar method) The following is a more efficient method to generate twoindependent standard Normal random variables:

1. Generate two independent random numbers U1 ∼ Unif(0, 1) and U2 ∼ Unif(0, 1).

2. Set (V1, V2) = (2U1 − 1, 2U2 − 1) and S = V 2

1+ V 2

2.

3. If S ≤ 1, go to 4., else go to 1.

4. Set P =√

−2(ln(S))/S

5. Return the pair (X, Y ) = (PV1, PV2).

The gain in efficiency mainly stems from the fact that no sine and cosine need to be com-puted. The method works because in polar coordinates (V1, V2) = (R cos(Θ), R sin(Θ),we have independent S = R2

∼ Unif(0, 1) and Θ ∼ Unif(0, 2π) (as is easily checked), sowe can choose U = S and 2πV = Θ in the Box-Muller generator.


8.3 R code – not examinable

The following code is posted on the course website as stable.R.

stableonesided <- function(a,c,eps,p)

f=c*eps^(-a)/a;

n=rpois(1,10*f);

t=runif(n,0,10);

y=(eps^(-a)-a*f*runif(n,0,1)/c)^(-1/a);

ytemp=1:n;res=(1:(10*p))/100;

for (k in 1:(10*p))for (j in 1:n)

if(t[j]<=k/p)ytemp[j]<-y[j] else ytemp[j]<-0; res[k]<-sum(ytemp);

res

stable <- function(a,cp,cn,eps,p)

pos=stableonesided(a,cp,eps,p);

neg=stableonesided(a,cn,eps,p);

space=pos-neg;time=(1/p)*1:(10*p);

plot(time,space,

pch=".",

sub=paste("Stable process with index",a,"and cplus=",cp,"and cminus=",cn))

stableonesidedcomp <- function(a,c,eps,p)

f=(c*eps^(-a))/a;

n=rpois(1,10*f);

t=runif(n,0,10);

y=(eps^(-a)-a*f*runif(n,0,1)/c)^(-1/a);

ytemp=1:n;

res=(1:(10*p))/100; for (k in 1:(10*p))if (n!=0)for (j in 1:n)

if(t[j]<=k/p)ytemp[j]<-y[j] else ytemp[j]<-0;

if (n!=0)res[k]<-sum(ytemp)-(c*k/(p*(a-1)))*(eps^(1-a))

else res[k]<--c*k/(p*(a-1))*(eps^(1-a));

res

Lecture 9

Simulation III

Reading: Ross 11.3, Schoutens Sections 8.1, 8.2, 8.4; Further reading: Kyprianou

Section 3.3

9.1 Applications of the rejection method

Lemma 57 can be used in a variety of ways. A widely applicable simulation methodis the rejection method. Suppose you have an explicit probability density function f ,but the inverse distribution function is not explicit. If h ≥ cf , for some c < 1 is aprobability density function whose inverse distribution function is explicit (e.g. uniformor exponential) or from which we can simulate by other means, then the procedure

1. Generate a random variable X with density h.

2. Generate an independent uniform variable U .

3. If Uh(X) ≤ cf(X), go to 4., else go to 1.

4. Return X.

Proposition 59 The procedure returns a random variable with density f .

Proof: Denote p = P(Uh(X) ≤ cf(X)). By Lemma 57 (applied to the vector (X,U)),the procedure returns a random variable with distribution

P(X ≤ x|Uh(X) ≤ cf(X)) =P(X ≤ x, Uh(X) ≤ cf(X))

p

=1

p

∫ x

−∞h(z)P(U ≤ cf(z)/h(z))dz =

c

p

∫ x

−∞f(z)dz,

and letting x→ ∞ shows c = p. 2

Example 60 (Gamma distribution) Note that the Gamma density for α > 1 satisfies

1

Γ(α)βαxα−1e−βx ≤

βα−1

Γ(α)βe−βx

so we can apply the procedure with f(x) = βe−βx and c = Γ(α)/βα−1.

37


It is important that c is not too small, since otherwise lots of iterations are needed untilthe random variable is returned. The number of iterations is geometrically distributed(first success in a sequence of independent Bernoulli trials) with success parameter c, soon average 1/c trials are required.

For the simulation via Poisson point processes (with truncation at ε, say), we can useproperties of Poisson point processes to simulate separately jumps of sizes in intervalsIn, n = 1, . . . , n0 and can choose intervals In so that the intensity function g is almostconstant.

Example 61 (Distribution on In) Suppose that we simulate a Poisson point process

on a bounded spatial interval In = (a, b], with some intensity function g : In → [0,∞).Then we can take uniform

h(x) = 1/(b− a) and c =

∫ b

ag(x)dx

(b− a) maxg(x) : a < x ≤ b.

and simulate Exp(∫ b

ag(x)dx)-spaced times Tn and spatial coordinates ∆Tn

by the rejection

method with h and c as given.

9.2 “Errors increase in sums of approximated terms.”

Methods 1 and 2 are based on sums of many, mostly small, independent identicallydistributed random variables. As δ ↓ 0 or ε ↓ 0, these are more and more smaller andsmaller random variables. If each is affected by a small error, then adding up these errorsmakes the approximations worse whereas the precision should increase.

For Method 1, this can often be prevented by suitable conditioning, e.g. on theterminal value:

Example 62 (Poisson process) A Poisson process with intensity λ on the time inter-val [0, 1] can be generated as follows:

1. Generate a Poisson random variable N with parameter λ.

2. Given N = n, generate n independent Unif(0, 1) random variables U1, . . . , Un.

3. Return Xt = #1 ≤ i ≤ N : Ui ≤ t.

Clearly, this process is a Poisson process, since Xs (and indeed (Xt1 , . . . , Xtn−tn−1)) isobtained from X1 ∼ Poi(λ) by thinning as in Section 4.2.

This is not of much practical use since we would usually simulate a Poisson randomvariable by evaluating a unit rate Poisson process (simulated from standard exponentialinterarrival times) at λ. In the case of Brownian motion (and the Gamma process, seeAssignment A.4.3.), however, such conditioning is very useful and can then be iterated,e.g. in a dyadic scheme:

Example 63 (Brownian motion) Consider the following method to generate Brown-ian motion on the time interval [0, 1].

Lecture 9: Simulation III 39

1. Set X0 = 0 and generate X1 ∼ Normal(0, 1) hence specifying Xk2−n for n = 0,k = 0, . . . , 2n.

2. For k = 1, . . . , 2n, conditionally given X(k−1)2−n = x and Xk2−n = z, generate

X(2k−1)2−n−1 ∼ Normal

(

x+ z

2, 2−n−2

)

3. If the required precision has been reached, stop, else increase n by 1 and go backto 2.

This process is Brownian motion, since the following lemma shows that Brownian motionhas these conditional distributions. Specifically, the n = 0, k = 1 case of 2. is obtaineddirectly for s = 1/2, t = 1. For n ≥ 1, k = 1, . . . , 2n, note that X(2k−1)2−n−1 −X(k−1)2−n

is independent of X(k−1)2−n and so, we are really saying that for Brownian motion

X(2k−1)2−n−1 −X(k−1)2−n ∼ Normal

(

z − x

2, 2−n−2

)

,

conditionally given Zk2−n − Z(k−1)2−n = z − x, which is equivalent to the specification in2.

A further advantage of this method is that δ = 2−n can be decreased without havingto start afresh. Previous less precise simulations can be refined.

Lemma 64 Let (Xt)t≥0 be Brownian motion and 0 < s < t. Then, the conditional

distribution of Xs given Xt = z is Normal(zs/t, s(t− s)/t).

Proof: Note that Xs ∼ Normal(0, s) and Xt −Xs ∼ Normal(0, t− s) are independent.By the transformation formula (Xs, Xt) has joint density

fXs,Xt(x, z) =

1

2π√

s(t− s)exp

−

x2

2s−

(z − x)2

2(t− s)

,

and so the conditional density is

fXs|Xt=z(x) =fXs,Xt

(x, z)

fXt(z)

=1

√

2π√

s(t− s)/texp

−

x2

2s−

(z − x)2

2(t− s)+z2

2t

=1

√

2π√

s(t− s)/texp

−

(x− zs/t)2

2s(t− s)/t

.

2

For Method 2, we can achieve similar improvements by simulating (∆t)t≥0 in stages.Choose a strictly decreasing sequence ∞ = a0 > a1 > a2 > . . . > 0 of jump size thresholdswith an ↓ 0 as n→ ∞

∆(k)t = ∆t1ak≤∆t<ak−1, ∆

(−k)t = ∆t1−ak≥∆t>−ak−1, k ≥ 1, t ≥ 0.

Simulate the Poisson counting processes N (k) associated with ∆(k) as in Example 62 andotherwise construct

Z(k)t =

∑

s≤t∆(k)s − t

∫ ak−1

ak

x10<x<1g(x)dx

as in Method 2 and include so many k = ±1,±2, . . . as precision requires.


9.3 Approximation of small jumps by Brownian mo-

tion

Theorem 65 (Asmussen-Rosinski) Let (Xt)t≥0 be a Levy process with characteristics

(a, 0, g). Denote

σ2(ε) =

∫ ε

−εx2g(x)dx

If σ(ε)/ε→ ∞ as ε ↓ 0, then

Xt −X(2,ε)t

σ(ε)→ Bt in distribution as ε ↓ 0

for an independent Brownian motion (Bt)t≥0

If σ(ε)/ε→ ∞, it is well-justified to adjust Method 2 to set

X(2+,ε)t = X

(2,ε)t + σ(ε)Bt

for an independent Brownian motion. In other words, we may approximate the smalljumps by an independent Brownian motion.

Example 66 (CGMY process) The CGMY process is a popular process in Mathe-matical Finance. It is defined via its characteristics (0, 0, g), where

g(x) = C exp−G|x||x|−Y−1, x < 0, g(x) = C exp−M |x||x|−Y−1, x > 0.

for some C ≥ 0, G > 0, M > 0 and Y < 2. Let (Xt)t≥0 be a CGMY process. Wecalculate

σ2(ε) =

∫ ε

−εx2g(x)dx ≤ C

∫ ε

−ε|x|1−Y dx =

2C

2 − Yε2−Y

and for every given δ > 0 and all ε > 0 small enough, the same quantity with C replacedby C − δ is a lower bound, so that

σ(ε)

ε∼

√

2C

2 − Yε−Y/2 → ∞ ⇐⇒ Y > 0

Hence an approximation of the small jumps of size (−ε, ε) thrown away by a Brownianmotion σ(ε)Bt is appropriate if and only if Y > 0. In fact, for Y < 0, the process hasfinite jump intensity, so all jumps can be simulated. Therefore, only the case Y = 0 isproblematic. This is the Variance Gamma process (and its asymmetric companions).

Whether or not we can approximate small jumps by a Brownian motion, we haveto decide what value of ε to choose. By the independence properties of Poisson pointprocesses, the remainder term that Xt − X2,ε

t is a (zero mean, for ε < 1) Levy processwith intensity function g on [−ε, ε] and variance

σ2(ε) = Var(Xt −X(2,ε)t )

(let δ ↓ 0 in the proof of Lemma 41). We can choose ε so that the accuracy of X(2,ε)t

is within an agreed deviation h, i.e. e.g. 2σ(ε) = h. In the setting of Theorem 65, this

means that a deviation of Xt from X(2,ε)t by more than h would happen with probability

about 0.05.

Lecture 9: Simulation III 41

9.4 Appendix: Consolidation on Poisson point pro-

cesses

This section and and the next should not be necessary this year, because the relevant

material has been included in earlier lectures. They may still be useful as a reminder of

key concepts.

We can consider Poisson point processes (∆t)t≥0 in very general spaces, e.g. (topo-logical) spaces (E,O) where we have a collection/notion of open sets O ∈ O (and anassociated Borel σ-algebra B = σ(O), the smallest σ-algebra that contains all open sets,for which we also require that x ∈ B for all x ∈ E and G = (x, x) : x ∈ E ∈ B ⊗ B).We just require that (∆t)t≥0 is a family of ∆t ∈ E ∪ 0 such that there is an intensity(Borel) measure ν on E with ν(x) = 0 for all x ∈ E,

(a) for disjoint A1 = (a1, b1] × O1, . . . , An = (an, bn] × On, Oi ∈ O, the counts

N(Ai) = N((a1, b1] × Oi) = #t ∈ (ai, bi] : ∆t ∈ Oi, i = 1, . . . , n

are independent random variables and

inhom(b) N(Ai) ∼ Poi((bi − ai)ν(Oi)).

For us, E = R \ 0 is the space of jump sizes, and ν(Oi) = ν((ci, di)) =∫ di

cig(x)dx

for an intensity function g : R \ 0 → [0,∞). Property (a) for all open sets is thenequivalent to property (a) for all measurable sets or all half-open intervals or all closedintervals etc. (all that matters is that the collection generates the Borel σ-algebra). It isan immediate consequence of the definition (and this discussion) that for (measurable)disjoint B1, B2, . . . ⊂ R \ 0, the “restricted” processes

∆(i)t = ∆t1∆t∈Bi, t ≥ 0,

are also Poisson point processes with the restriction of g to Bi as intensity function,and they are independent. We used this fact crucially and repeatedly in two forms.Firstly, for B1 = (0,∞) and B2 = (−∞, 0) (and B3 = B4 = . . . = ∅), we considerPoisson point processes of positive points (jump sizes) and of negative points (jumpsizes). We constructed from them independent Levy processes. Secondly, for a sequence∞ = a0 > a1 > a2 > . . ., we considered Bi = [ai, ai−1), i ≥ 1, so as to simulate separatelyindependent Levy processes (in fact compound Poisson processes with linear drift) withjump sizes only in Bi.

9.5 Appendix: Consolidation on the compensation

of jumps

The general Levy process requires compensation of small jumps in its approximation byprocesses with no jumps in (−ε, ε), as ε ↓ 0. This is reflected in its characteristic functionof the form

E(eiλXt) = e−tψ(λ), ψ(λ) = −ia1λ+1

2σ2λ2

−

∫ ∞

−∞(eiλx−1−iλx1|x|≤1)ν(dx), λ ∈ R, (1)


where usually ν(dx) = g(x)dx. This is a parametrisation by (a1, σ2, ν) or (a1, σ

2, g),where we require the (weak) integrability condition

∫ ∞−∞(1 ∧ x2)ν(dx) <∞.

The first class of Levy processes that we constructed systematically were subordina-tors, where no compensation was necessary. We parametrised them by parameters a2 ≥ 0and g : (0,∞) → [0,∞) (or ν measure on (0,∞)) so that the moment generating functionis of the form

E(eγXt) = etΨ(γ), Ψ(γ) = a2γ +

∫ ∞

0

(eγx − 1)ν(dx), γ ≤ 0. (2)

We required the stronger integrability condition∫ ∞0

(1 ∧ x)ν(dx) < ∞. Similarly, fordifferences of subordinators, we have a characteristic function


2σ2λ2

−

∫ ∞

−∞(eiλx − 1)ν(dx), (3)

under the stronger integrability condition∫ ∞−∞(1 ∧ |x|)ν(dx) < ∞. Compensation in (1)

is only done for small jumps. This is, because, in general the indicator 1|x|≤1 cannot beomitted. However, if

∫ ∞−∞(x ∧ x2)ν(dx) <∞, then we can also represent


2σ2λ2

−

∫ ∞

−∞(eiλx − 1 − iλx)ν(dx). (4)

Equations (1), (3) and (4) are compatible whenever any two integrability conditions arefulfilled, since the linear (in λ) terms under the integral can be added to a1 to give

a2 = a1 +

∫

|x|≤1xν(dx) and a3 = a1 −

∫

|x|>1xν(dx).

Note that then (by differentiation at λ = 0), we get a3 = E(X1). If a3 = 0, then (Xt)t≥0

is a martingale. On the other hand, for processes with finite jump intensity, i.e. underthe even stronger integrability condition

∫ ∞−∞ g(x)dx < ∞, we get a1 as the slope of the

paths of X between the jumps. Both a1 and a3 are therefore natural parameterisations,but not available, in general. a2 is available in general, but does not have such a naturalinterpretation.

We use characteristic functions for similar reasons: in general, moment generat-ing functions do not exist. If they do, i.e. under a strong integrability condition∫ ∞1eγxν(dx) <∞ for some γ > 0 or

∫ −1

−∞ eγxν(dx) <∞ for some γ < 0, we get

E(eγXt) = etΨ(γ), Ψ(γ) = a1γ +1

2σ2γ2 +

∫ ∞

−∞(eγx − 1 − γx1|x|≤1)g(x)dx. (5)

Moment generating functions are always defined on an interval I possibly including endpoints γ− ∈ [−∞, 0] and/or γ+ ∈ [0,∞], we always have 0 ∈ I, but maybe γ− = γ+ = 0.If 1 ∈ I and a1 is such that Ψ(1) = 0, then (eXt)t≥0 is a martingale.

Lecture 10

Levy markets and incompleteness

Reading: Schoutens Chapters 3 and 6

10.1 Arbitrage-free pricing (from B10b)

By Donsker’s Theorem, Brownian motion is the scaling limit of most random walks andin particular of the simple symmetric random walk Rn = X1 + . . .+Xn where X1, X2, . . .are independent with P(Xi = 1) = P(Xi = −1) = 1/2.

Corollary 67 For simple symmetric random walk (Rn)n≥0, we have eR[nt]/√n→ eBt ,

geometric Brownian motion, in distribution as n→ ∞.

Proof: First note that E(X1) = 0 and Var(X1) = 1. Now for all x > 0

P(eR[nt]/√n≤ x) = P(R[nt]/

√

n ≤ ln(x)) → P(Bt ≤ ln(x)) = P(eBt≤ x),

by the Central Limit Theorem. 2

This was convergence of for fixed t. Stronger convergence, locally uniformly in t can alsobe shown. Note that (Rn)n≥0 is a martingale, and so is (Bt)t≥0. However,

E(eRn) =

(

1

2e−1 +

1

2e

)n

→ ∞

Proposition 68 For non-symmetric simple random walk (Rn)n≥0 with P(Xi = 1) = p,the process (eRn)n≥0 is a martingale if and only if p = 1/(1 + e).

Proof: By the fourth and first rules for conditional expectations, we have

E(eRn+1|eR0 , . . . , eRn) = E(eRneXn+1

|eR0 , . . . , eRn) = eRnE(eXn+1)

and so, (eRn)n≥0 is a martingale if and only if

1 = E(eXn+1) = pe+ (1 − p)e−1⇐⇒ p(e− 1/e) = 1 − 1/e ⇐⇒ p = 1/(e+ 1). 2

The argument works just assuming that Rn = X1 + . . .+Xn, n ≥ 0, satisfies P(|Xn+1| =1|R0 = r0, . . . , Rn = rn) = 1. Among all joint distributions, the non-symmetric exponen-tiated random walk with p = 1/(1 + e) is the only martingale.

The concept of arbitrage-free pricing in binary models is leaving aside any randomness.We will approach the Black-Scholes model from discrete models.

43


Suppose we have a risky asset with random price process S per unit and a risk-free

asset with deterministic value function A per unit. Consider portfolios (U, V ) of U unitsof the risky asset and V units of the risk-free asset. We allow that (Ut, Vt) depends onthe performance of (Ss)0≤s<t but not on (Ss)s≥t. We denote the value of the portfolio attime t by Wt = UtSt+VtAt. The composition of the portfolio may change with time, butwe consider only self-financing ones, for which any risky asset bought is paid for fromthe risk-free asset holdings and vice versa. We say that arbitrage opportunities exist ifthere is a self-financing portfolio process (U, V ) and a time t so that P(W0 = 0) = 1,P(Wt ≥ 0) = 1 and P(Wt > 0) > 0. We will be interested in models where no arbitrageopportunities exist.

Example 69 (One-period model) There are two scenarios “up” and “down” (to whichwe may later assign probabilities p ∈ (0, 1) and 1 − p). The model consists of (S0, S1)only, where S0 changes to S1(up) or S1(down) < S1(up) after one time unit. The risk-freeasset will evolve from A0 to A1. At time 0, we have W0 = U0S0 + V0A0. At time 1, thevalue will change to either

W1(up) = U0S1(up) + V0A1 or W1(down) = U0S1(down) + V0A1. (1)

It is easily seen that arbitrage opportunities occur if and only if A1/A0 ≥ S1(up)/S0 orS1(down)/S0 ≥ A1/A0, i.e. if one asset is uniformly preferable to the other.

A derivative security (or contingent claim) with maturity t is a contract that providesthe owner with a payoff Wt dependent on the performance of (Ss)0≤s≤t. If there is aself-financing portfolio process (U, V ) with value Wt at time t, then such a portfolioprocess is called a hedging portfolio process replicating the contingent claim. The valueW0 = U0S0 + V0A0 of the hedging portfolio at time 0 is called the arbitrage-free price ofthe derivative security. It is easily seen that there would be an arbitrage opportunity, ifthe derivative security was available to buy and sell at any other price (as an additionalasset in the model). In general, not all contingent claims can be hedged.

Example 69 (One-period model, continued) Consider any contingent claim, i.e. apayoff of W1(up) or W1(down) according to whether scenario “up” or “down” take place.Equations (1) can now be used to set up a hedging portfolio (U0, V0) and calculate theunique arbitrage-free price. Note that the arbitrage-free price is independent of proba-bilities p ∈ (0, 1) and 1− p that we may assign to the two scenarios as part of our modelspecification. Because of the linearity of (1), there is a unique q ∈ (0, 1) such that for all

contingent claims W1 : up, down → R

W0 = qW1(up) + (1 − q)W1(down).

If we refer to q and 1−q as probabilities of “up” and “down”, then W0 is the expectationof W1 under this distribution. If A0 = A1 = 1, S0 = 1, S1(up) = e, S1(down) = e−1, thenwe identify (for W1 = S1 and hence (U0, V0) = (1, 0)) that q = 1/(1 + e).

The property that every contingent claim can be hedged by a self-financing portfolioprocess is called completeness of the market model.

Lecture 10: Levy markets and incompleteness 45

Example 70 (n-period model) Each of n periods has two scenarios, “up” and “down”,as is the case for the model Sk = eRk for a simple random walk (Rk)0≤k≤n. Then thereare 2n different combinations of “up” (Xk = 1) and “down” (Xk = −1). A contingentclaim at time n is now any function Wn assigning a payoff to each of these combina-tions. By a backward recursion using the one-period model as induction step, we canwork out hedging portfolios (Uk, Vk) and the value Wk of the derivative security at timesk = n − 1, n − 2, . . . , 0, where in each case, (Uk, Vk) and Wk will depend on previous“up”s and “down”s X1, . . . , Xk, so this is a specification of 2k values each. W0 will bethe unique arbitrage-free price for the derivative security. The induction also shows that,for A0 = A1 = . . . = An = 1, it can be worked out as

W0 = E(Wn(X1, . . . , Xn)), where Xk independent with P(Xk = 1) = 1/(1 + e),

and that (Wk)0≤k≤n is a martingale with Wk = E(Wn|X1, . . . , Xk), e.g. (Sk)0≤k≤n. IfAk = (1 + i)k = eδk, we get an arbitrage-free model if and only if −1 < δ < 1, and then

W0 = e−δnE(Wn), where Xk independent with P(Xk = 1) = (e1+δ − 1)/(e2 − 1),

where now (e−δkWk)0≤k≤n is a martingale. In particular, the n-period model is complete.

Example 71 (Black-Scholes model) Let St = S0 expσBt + (µ−1

2σ2)t for a Brow-

nian motion (Bt)t≥0 and two parameters µ ∈ R and σ2 > 0. Also put At = eδt. Itcan be shown that also in this model, every contingent claim can be hedged, i.e. theBlack-Scholes model is complete. Moreover, the pricing of contingent claims Wt can becarried out using the risk-neutral process

Rt = S0 expσBt + (δ − 1

2σ2))t, t ≥ 0,

where the drift parameter is δ, not µ. The discounted process Mt = e−δtRt is a martingaleand has analogous uniqueness properties to the martingale for the n-period model, butthey are much more complicated to formulate here.

The arbitrage-free price of Wt = G((Ss)0≤s≤t) is now (for all µ ∈ R)

W0 = e−δtE(G((Rs)0≤s≤t)).

Examples are G((Ss)0≤s≤t) = (St−K)+ for the European call option and G((Ss)0≤s≤t) =(K − St)

+ for the European put option. We will also consider path-dependent optionssuch as Up-and-out barrier options with payoff (St−K)+1St<H, where St = sup

0≤s≤t Ssand H is the barrier. The option can only be exercised if the stock price does not exceedthe barrier H at any time before maturity.

10.2 Introduction to Levy markets

The Black-Scholes model is widely used for option pricing in the finance industry, largelybecause many options can be priced explicitly and there are computationally efficientmethods also for more complicated derivatives, that can be carried out frequently and forhigh numbers of options. However, its model fit is poor and any price that is obtainedfrom the Black-Scholes model must be adjusted to be realistic. There are several modelsbased on Levy processes that offer better model fit, but the Black-Scholes methods foroption pricing do not transfer one-to-one. Levy processes give a very wide modellingfreedom. For practical applications it is useful to work with parametric subfamilies.Several such families have been suggested.


Example 72 (CGMY process) Carr, Geman, Madan and Yor proposed a model withfour parameters (the first letters of their names. It is defined via its intensity

g(x) = C exp−G|x||x|−Y−1, x < 0, g(x) = C exp−M |x||x|−Y−1, x > 0.

Let (Xt)t≥0 be a CGMY process. If M > 1, then a risk-neutral price process can bemodelled as Rt = R0 expXt − t(φ(1) − δ). Then the discounted process (e−δtRt)t≥0

is a martingale, and it can be shown that arbitrage-free prices for contingent claimsWt = G((Rs)0≤s≤t) can be calculated as

W0 = e−δtE(G((Rs)0≤s≤t)).

It can also be shown, however, that this is not the only way to obtain arbitrage-freeprices, and other prices do not necessarily lead to arbitrage opportunities. Also, notevery contingent claim can be hedged, the model is not complete.

10.3 Incomplete discrete financial markets

Essentially, arbitrage-free discrete models are complete only if the number of possiblescenarios ω0, . . . , ωm (for one period) is the same as the number of assets S

(1)

1, . . . , S

(m)

1:

Ω = ω0, . . . , ωm → R in the model, since this leads to a system of linear equations torelate a contingent claim W1 : ω0, . . . , ωm → R to a portfolio (U (1), . . . , U (m), V )

V0A1 +m

∑

i=1

U(i)0S

(i)1

(ωj) = W1(ωj), j = 0, . . . , m

that can usually be uniquely solved for (U(1)

0, . . . , U

(m)

0, V0), and we can read off

W0 = V0A0 +

m∑

i=1

U(i)0S

(i)0.

If the number of possible scenarios is higher, then the system does not have a solution,in general (and hedging portfolios will not exist, in general). If the number of possiblescenarios is lower, there will usually be infinitely many solutions.

If the system has no solution in general, the model is incomplete, but this does notmean that there is no price. It means that there is not a unique price. We can, ingeneral, get some lower and upper bounds for the price imposed by no-arbitrage. Oneway of approaching this is to add a derivative security to the market as a further assetwith an initial price that keeps the no arbitrage property for the extended model. Onecan, in fact, add more and more assets until the model is complete. Then there existunique probabilities qj = P(ωj), 0 ≤ j ≤ m, that make all discounted assets (A0/A1)S

(j),1 ≤ j ≤ m (including the ones added to complete the market) martingales.

Example 73 (Ternary model) Suppose there are three scenarios, but only two assets.The model with S0 = 1 = A0 < A1 = 2, and 1 = S1(ω0) < 2 = S1(ω1) < 3 = S1(ω2)is easily seen to be arbitrage-free since S1(ω0) < A1 < S1(ω2). The contingent claim0 = W1(ω0) = W1(ω1) < W1(ω2) = 1 can be hedged if and only if

2V0 + U0 = 0, 2V0 + 2U0 = 0, 2V0 + 3U0 = 1,

Lecture 10: Levy markets and incompleteness 47

but the first two equations already imply U0 = V0 = 0 and then the third equation isfalse. Therefore the contingent claim W1 cannot be hedged. The model is not complete.

For the model (W,S,A) to be arbitrage-free we clearly require W0 > 0 since otherwisewe could make arbitrage with a portfolio (1, 0, 0), just “buying” the security. Its cost attime zero is nonpositive and its value at time one is nonnegative and positive for scenarioω2. Now note that (S,A) is arbitrage-free, so any arbitrage portfolio must be of the form(−1/W0, U0, V0) with zero value −1 + U0 + V0 = 0 at time 0 and values at time 1

U0 + 2V0 = 2 − U0, 2U0 + 2V0 = 2 > 0, −1/W0 + 3U0 + 2V0 = −1/W0 + 2 + U0,

so that we need 2 ≥ U0 ≥ 1/W0 − 2, and this is possible if and only if W0 ≥ 1/4.Therefore, the range of arbitrage-free prices is W0 ∈ (0, 1/4).

We can get these prices as expectations under martingale probabilities:

1 =1

2Eq(S1) =

1

2q0 + q1 +

3

2q2, W0 =

1

2Eq(W1) =

1

2q2, q0 + q1 + q2 = 1.

This is a linear system for (q0, q1, q2) that we solve to get

q0 = q2 = 2W0, q1 = 1 − 4W0

and this specifies a probability distributon on all three scenarios iff W0 ∈ (0, 1/4). Sincewe can express every contingent claim as a linear combination of A1, S1,W1, we can nowprice every contingent claim X1 under the martingale probabities as X0 = 1

2Eq(X1).


Lecture 11

Levy markets and time-changes

11.1 Incompleteness and martingale probabilities in

Levy markets

By a Levy market we will understand a model (S,A) of a risky asset St = expXt fora Levy process X = (Xt)t≥0 and a deterministic risk-free bank account process, usuallyAt = eδt, t ≥ 0. We exclude deterministic Xt = µt in the sequel.

Theorem 74 (No arbitrage) A Levy market allows arbitrage if and only if either Xt−

δt is a subordinator or δt−Xt is a subordinator.

Proof: We only prove that these cases lead to arbitrage opportunities. If Xt − δt is asubordinator, then the portfolio (1,−1) is an arbitrage portfolio. 2

The other direction of proof is difficult, since we would need technical definitions ofadmissible portfolio processes and related quantities.

No arbitrage is closely related (almost equivalent) to the existence of martingale prob-abilities. Formally, an equivalent martingale measure Q is a probability measure whichhas the same sets of zero probability as P, i.e. under which the same things are possi-ble/impossible as under P, and under which (e−δtSt)t≥0 is a martingale. For simplicity,we will not bother about this passage to a so-called risk-neutral world that is differentfrom the physical world. Instead, we will consider models where (e−δtSt)t≥0 is alreadya martingale. Prices of the form W0 = e−δtEQ(Wt) are then arbitrage-free prices. Therange of arbitrage-free prices is

e−δtEQ(Wt) : Q martingale measure equivalent to P

.

The proof of incompleteness is also difficult, but the result is not hard to state:

Theorem 75 (Completeness) A Levy market is complete if and only if (Xt)t≥0 is

either a multiple of Brownian motion with drift, Xt = µt + σBt or a multiple of the

Poisson process with drift, Xt = at+ bNt (with (a− δ)b < 0 to get no arbitrage).

Completeness is closely related (almost equivalent) to the uniqueness of martingaleprobabilities. In an incomplete market, there are infinitely many choices for these mar-tingale probabilities. This raises the question of how to make the right choice. Whilewe can determine an arbitrage-free system of prices for all contingent claims, we cannothedge the contingent claim, and this presents a risk to someone selling e.g. options.

49


11.2 Option pricing by simulation

If we are given a risk-neutral price process (martingale) (St)t≥0, we can price contingentclaims G((Ss)0≤s≤t) as expectations

P = e−δtE(G((Ss)0≤s≤t)).

Often, such expectations are difficult to work out theoretically or numerically, particularlyfor path-dependent options such as barrier options. Monte-Carlo simulation always works,by the strong law of large numbers:

1

n

n∑

k=1

G((S(k)s )0≤s≤t) → E(G((Ss)0≤s≤t) almost surely,

as n → ∞, where (S(k)s )0≤s≤t are independent copies of (Ss)0≤s≤t. By simulating these

copies, we can approximate the expectation on the right to get the price of the option.

Figure 11.1: Option pricing by simulation

11.3 Time changes

Levy markets are one way to address shortcomings of the Black-Scholes model. Particu-larly quantities such as one-day return distributions can be fitted well. Other possibilitiesinclude modifications to the Black-Scholes model, where the speed of the market is mod-elled separately. The rationale behind this is to capture days with increased activity (andhence larger price movements) by notions of operational versus real time. In operationaltime, the price process follows a Brownian motion, but in real time, a busy day corre-sponds to several days in operational time, while a quiet day corresponds to a fraction ofa day in operational time.

The passage from operational to real time is naturally modelled by a time-changey 7→ τy, which we will eventually model by a stochastic process built from a Poisson pointprocess. The price process is then (Bτy)y≥0. This stochastic process cannot be observeddirectly in practice, but approximations of quadratic variation permit to estimate thetime change.

The most elementary time change is for τy = f(y), a deterministic continuous strictlyincreasing function f : [0,∞) → [0,∞) with f(0) = 0 and f(∞) = ∞. In this case, the

Lecture 11: Levy markets and time-changes 51

time-changed process Zy = Xf(y), y ≥ 0, visits the same states as X in the same order asX, performing the same jumps as X, but travelling at a different speed. Specifically, iff(y) << y, then, by time y, the process X will have gone toXy, but Z only to Zy = Xf(y).We say that Z has evolved more slowly than X, and faster if instead f(y) >> y. If fis differentiable, we can more appropriately make local speed statements according towhether f ′(y) < 1 or f ′(y) > 1. Note, however, that “speed” really is “relative speed”when comparing X and Z, since X is not “travelling at unit speed” in a sense of rateof spatial displacement; jumps and particularly unbounded variation make such notionsuseless. We easily calculate

E(eiλZy) = E(eiλXf(y)) = e−f(t)ψ(λ), if E(eiλXt) = e−tψ(λ).

and see that Z is a stochastic process with independent increments and right-continuouspaths with left limits, but will only have stationary increments if f(y) = cy for all y ≥ 0and some c ∈ (0,∞).

Example 76 (Foreign exchange rates) Suppose that the EUR/USD-exchange ratetoday is S0 and you wish to model the exchange rate (St)t≥0 over the next couple of days.As a first model you might think of

St = S0 expσBt − tσ2/2,

where B is a standard Brownian motion σ is a volatility parameter that measures themagnitude of variation. This magnitude is related to the amount of activity on theexchange markets and will be much higher for the EUR/USD-exchange rate than e.g.for the EUR/DKK-exchange rate (DansKe Kroner, Danish crowns are not traded sofrequently in such high volumes. Also, DKK is closely aligned with EUR due to strongeconomic ties between Denmark and the Euro countries).

However, in practice, trading activity is not constant during the day. When stockmarkets in the relevant countries are open, activity is much higher than when they areall closed and a periodic function f ′ : [0,∞) → [0,∞) can explain a good deal of thisvariability and provide a better model

St = S0 expσBf(y) − f(y)σ2/2 = S0 exp ˜B ef(y) −˜f(y)/2,

where ˜Bs = σBs/σ2 , s ≥ 0, is also a standard Brownian motion and ˜f(y) = f(y)σ2 makes

the parameter σ redundant – the flexibility for ˜f retains all modelling freedom.

If we weaken the requirement of strict monotonicity to weak monotonicity and f(y) =c, y ∈ [l, r), is constant on an interval, then Zy = Xc, y ∈ [l, r), during this interval.For a financial market model this can be interpreted as time intervals with no marketactivity, when the price will not change.

If we weaken the continuity of f to allow (upward) jumps, then Zy = Xf(y), y ≥ 0,does not evaluate X everywhere. Specifically, if ∆f(y) > 0 is the only jump of f , thenZ will visit the same points as X in the same order until Xf(y−)− and then skip over(Xf(y−)+s)0≤s<∆f(y) to directly jump to Xf(y). In general, this is the behaviour at everyjump of f .


Figure 11.2: piecewise constant volatility

11.4 Quadratic variation of time-changed Brownian

motion

In Section 6.1 we studied quadratic variation of Brownian motion in order to show thatBrownian motion has infinite total variation (and is therefore not the difference of twoincreasing processes). Let us here look at quadratic variation of time-changed Brownianmotion Zy = Bf(y) for an increasing function f : [0,∞) → [0,∞):

[Z]t = p− limn→∞

[Z](n)

t , where [Z](n)

t =

[2nt]∑

j=1

(Zj2−n − Z(j−1)2−n)2

and p − lim denotes a limit of random variables in probability. One may expect that[B]t = t implies that [Z]y = f(y), and this is true under suitable assumptions.

Proposition 77 Let B be Brownian motion and f : [0,∞) → [0,∞) continuous and

increasing with f(0) = 0. Then [Z]y = f(y) for all y ≥ 0.

Proof: The proof (for Z) is the same as for Brownian motion (B) itself, see Section 6.1and Assignment 6. 2

Quadratic variation is accumulated locally. Under the continuity assumption of Brow-nian motion and its time change, it is the wiggly local behaviour of Brownian motionthat generates quadratic variation. In Section 6.1 we showed that under the no-jumpsassumption, positive quadratic variation implies infinite total variation. Hence, still un-der the no-jumps assumption, finite total variation implies zero quadratic variation. Butwhat is the impact of jumps on quadratic variation? It can be shown as in Proposition3 that

[f ]y ≥∑

s≤y|∆fs|

2.

Example 78 Consider a piecewise linear function f : [0,∞) → [0,∞) with slope 0.1and jumps ∆f2k−1 = 1.8, k ≥ 1. Then f(2k) = 2k, but by Proposition 77

[f ]2k ≥∑

s≤2k

|∆fs|2 = k(1.8)2 = 3.24k

Lecture 11: Levy markets and time-changes 53

and, in fact, this is an equality, since

[f ](n)

2k = k(1.8 + 2−n0.1)2 + (2n+1− 1)k(2−n0.1)2

→ k(1.8)2.

Now define Zy = Bf(y) and note that

[Z](n)

2k =k∑

i=1

(

2n−1∑

j=1

(B(2i−2)+j2−n0.1 − B(2i−2)+(j−1)2−n0.1)2

+(B2i−0.1 − B(2i−2)+0.1−2−n)2

+2

n∑

j=1

(B2i−(j−1)2−n0.1 −B2i−j2−n0.1)2

)

→ 2k0.1 +k∑

i=1

(B2i−0.1 − B(2i−2)+0.1)2,

as n→ ∞, which is actually 0.1(2k)+∑

s≤2k |∆Zs|2. Note that this is a random quantity.

In general, quadratic variation consists of a continuous part due to Brownian fluctu-ations and the sum of squared jump sizes.


Lecture 12

Subordination and stochastic

volatility

Subordination is the operation Xτy , y ≥ 0, for a Levy (or more general Markov) process(Xt)t≥0 and a subordinator (τy)y≥0. One distinguishes subordination in the sense ofBochner, where X and τ are independent and subordination in the wide sense where τyis a stopping time for all y ≥ 0. These are both special cases of the more general conceptof time change, where (τy)y≥0 does not have to be a subordinator.

12.1 Bochner’s subordination

Theorem 79 (Bochner) Let (Xt)t≥0 be a Levy process and (τy)y≥0 an independent sub-

ordinator. Then the process Zy = Xτy , y ≥ 0, is a Levy process, and we have

E(eiλZy) = e−yΦ(ψ(λ)), where E(eiλXt) = e−tψ(λ) and E(e−qτy) = e−yΦ(q).

Proof: First calculate by conditioning on τy (assuming that τy is continuous with prob-ability density function fτy)

E(expiλZy) = E(expγXτy) =

∫ ∞

0

fτy(t)E(expiλXt)dt

=

∫ ∞

0

fτy(t) exp−tψ(λ)dt = e−yΦ(−ψ(λ)).

Now, for r, s ≥ 0,

E(expiλZy + iµ(Zy+x − Zy))

=

∫ ∞

0

∫ ∞

0

fτy ,τy+x−τy(t, u)E(expiλXt + iµ(Xt+u −Xt))dtdu

=

∫ ∞

0

∫ ∞

0

fτy(t)fτx(u)e−tψ(λ)e−uψ(µ)dtdu = e−yΦ(ψ(λ))e−xΦ(ψ(µ)),

so that we deduce that Zy and Zy+x − Zy are independent, and that Zy+x − Zy ∼ Zx.For the right-continuity of paths, note that

limε↓0

Zy+ε = limε↓0

Xτy+ε= Xτy = Zy,

55


since τy + δ := τy+ε ↓ τy and therefore Xτy+δ → Xτy . For left limits, the same argumentapplies. 2

Note that ∆Zy = Zy − Zy− = Xτy − Xτy− 6= can be non-zero if either ∆τy 6= 0, or∆Xτy 6= 0, so Z inherits jumps from τ and from X. We have, with probability 1 for ally ≥ 0 that

∆Zy = Xτy −Xτy−− =

(∆X)τy if (∆X)τy 6= 0,Xτy −Xτy− if ∆τy 6= 0.

Note that we claim that Xτy− = Xτy−−, i.e. (∆X)τy− 6= 0 if ∆τy 6= 0, for all y ≥ 0with probability 1. This is due to the fact that the countable set of times τy−, τy : y ≥

0 and ∆τy 6= 0 is a.s. disjoint from t ≥ 0 : ∆Xt 6= 0.

Note also that Xτy = Xτy− is possible with positive probability, certainly in the caseof a compound Poisson process X.

Heuristically, if Xt has density ft and τ Levy density gτ , then Z will have Levy density

g(z) :=

∫ ∞

0

ft(z)gτ (t)dt, z ∈ R, (1)

since every jump of τ of size ∆τy = t leads to a jump Xτy − Xτy− ∼ Xt, and the totalintensity of jumps of size z receives contributions from τ -jumps of all sizes t ∈ (0,∞).We can make this precise as follows:

Proposition 80 Let X be a Levy process with probability density function ft of Xt,

t ≥ 0, τ a subordinator with Levy-Khintchine characteristics (0, gτ), then Zy = Xτy has

Levy-Khintchine characteristics (0, 0, g), where g is given by (1).

Proof: Consider a Poisson point process (∆y)y≥0 with intensity function g, then, by theExponential Formula

E

(

exp

iλ∑

s≤y∆s1|∆s|>ε

)

= exp

y

∫ ∞

−∞(eiλz − 1)g(z)1|z|>εdz

= exp

y

∫ ∞

0

∫ ∞

−∞(eiλz − 1)ft(z)1|z|>εdzgτ (t)dt

→ exp

−y

∫ ∞

0

(1 − e−tψ(λ))gτ(t)dt

= exp −yΦ(ψ(λ)) ,

as ε ↓ 0, and this is the same distribution as we established for Zy. 2

Note that we had to prove convergence in distribution as ε ↓ 0, since we have notstudied integrability conditions for g. This is done on Assignment sheet 6.

Corollary 81 If X is Brownian motion and τ has characteristics (b, gτ ), then Zy = Xτy

has characteristics (0, b, g).

Lecture 12: Subordination and stochastic volatility 57

Proof: Denote Φ0(q) = Φ(q) − bq. Then the calculation in the proof of the propositiondoes not yield a characteristic exponent Φ(ψ(λ)), but Φ0(ψ(λ)). Note that Φ(ψ(λ)) =Φ(1

2λ2) = 1

2bλ2 + Φ0(ψ(λ)), so we consider

bBt +∑

s≤t∆s

for an independent Brownian motion B, which has characteristic exponent as required.2

Example 82 If we define the Variance Gamma process by subordination of Brownianmotion B by a Gamma(α, θ) subordinator τ with Levy density gτ (y) = αy−1e−θy, thenwe obtain a Levy density

g(z) =

∫ ∞

0

ft(z)gτ (t)dt =

∫ ∞

0

1√

2πte−z

2/(2t)αt−1e−θtdt

and we can calculate this integral to get

g(z) = α|z|−1e−√

2θ|z|

as Levy density. The Variance Gamma process as subordinated Brownian motion hasan interesting interpretation when modelling financial price processes. In fact, the stockprice is then considered to evolve according to the Black-Scholes model, but not in realtime, but in operational time τy, y ≥ 0. Time evolves as a Gamma process, with infinitelymany jumps in any bounded interval.

Note that all Levy processes that we can construct as subordinated Brownian motionsare symmetric. However, not all symmetric Levy processes are subordinated Brownianmotions,

12.2 Ornstein-Uhlenbeck processes

Example 83 (Gamma-OU process) Let (Nt)t≥0 be a Poisson process with intensityaλ and jump times (Tk)k≥1, (Xn)n≥1 a sequence of independent Gamma(1, b) randomvariables, Y0 ∼ Gamma(a, b), consider the stochastic process

Yt = Y0e−λt +

Nt∑

k=1

Xne−λ(t−Tk)

We use this model for the speed of the market and think of an initial speed of Y0 whichslows down exponentially, but at times of a Poisson process, events occur that make thespeed jump up at times Tk, k ≥ 0. Each of these also slow down exponentially. In fact,there is a strong equilibrium in that

E(e−qYt) = E(e−qe−λtY0)E

(

exp

−qNt∑

k=1

Xke−λ(t−Tk)

)

=

(

b

b+ qe−λt

)a ∞∑

n=0

(λat)n

n!e−λat

(∫ t

0

1

t

b

b+ qe−λsds

)n

=

(

b

b+ qe−λt

)a(b+ qe−λt

b+ q

)a

=

(

b

b+ q

)a

,


so Yt has the same distribution as Y0. In fact, (Yt)t≥0 is a stationary Markov process.The process Y is called a Gamma-OU process, since it has the Gamma distribution asits stationary distribution. The time change process

τy =

∫ y

0

Ysds

associated with speed Y is called integrated Ornstein-Uhlenbeck process. Note thatthe stationarity of Y implies that τ has stationary increments, but note that τ doesnot have independent increments. The associated stochastic volatility model is now thetime-changed Brownian motion (Bτy)y≥0.

In general, we can define Ornstein-Uhlenbeck processes associated with any subordi-nator Z or rather its Poisson point process (∆Zt)t≥0 of jumps as

Yt = Y0e−λt +

∑

s≤t∆Zse

−λ(t−s),

for any initial distribution for Y0, but a stationary distribution can also be found.We can always associate a stochastic volatility model (Bτy)y≥0 using the integrated

volatility τy =∫

yYsds as time change. Note that, by the discussion of the last section,

we can actually infer the time change from sums of squared increments for a small timelag 2−n, even though the actual time change is not observed. In practice, the so-calledmarket microstructure (piecewise constant prices) destroys model fit for small times, sowe need to choose a moderately small 2−n. In practice, 5 minutes is a good choice.

12.3 Simulation by subordination

Note that we can simulate subordinators using simulation Method 1 (Time discretisa-tion) or Method 2 (Throwing away the small jumps). The latter consisted essentially insimulating the Poisson point process of jumps of the subordinator. Clearly, we can applythis method also to simulate an Ornstein-Uhlenbeck process.

Method 3 (Subordination) Let (τy)y≥0 be an increasing process that we can simulate,and let (Xt)t≥0 be a Levy process with cumulative distribution function Ft of Xt. Fix atime lag δ > 0. Then the process

Z(3,δ)y = S[y/δ], where Sn =

n∑

k=1

Ak and Ak = F−1

τkδ−τ(k−1)δ(Uk)

is the time discretisation of the subordinated process Zy = Xτy .

Example 84 We can use Method 3 to simulate the Variance Gamma process, since wecan simulate the Gamma process τ and we can simulate the Ak. Actually, we can usethe Box-Muller method to generate standard Normal random variables Nk and then use

Ãk ∼√

τkδ − τ(k−1)δNk, k ≥ 1,

instead of Ak, k ≥ 1.

Lecture 13

Level passage problems

Reading: Kyprianou Sections 3.1 and 3.3

13.1 The strong Markov property

Recall that a stopping time is a random time T ∈ [0,∞] such that for every s ≥ 0the information Fs up to time s allows to decide whether T ≤ s. More formally, if theevent T ≤ s can be expressed in terms of (Xr, r ≤ s) (is measurable with respect toFs = σ(Xr, r ≤ s)). The prime example of a stopping time is the first entrance timeTI = inft ≥ 0 : Xt ∈ I into a set I ⊂ R. Note that

T ≤ s = there is r ≤ s such that Xr ∈ I

(for open sets I we can drop the irrational r ≤ s to show measurability.).

We also denote the information FT up to time T . More formally,

FT = A ∈ F : A ∩ T ≤ s ∈ Fs for all s ≥ 0,

i.e. FT contains those events that, if T ≤ s, can be expressed in terms of (Xr, r ≤ s), forall s ≥ 0.

Recall the simple Markov property which we can now state as follows. For a Levy pro-cess (Xt)t≥0 and a fixed time t, the post-t process (Xt+s−Xt)s≥0 has the same distributionas X and is independent of the pre-t information Ft.

Theorem 85 (Strong Markov property) Let (Xt)t≥0 be a Levy process and T a stop-

ping time. Then given T < ∞, the post-T process (XT+s − XT )s≥0 has the same distri-

bution as X and is independent of the pre-T information FT .

Proof: Let 0 < s1 < . . . < sm, C1, . . . , Cm ⊂ R closed, A ∈ FT . Then we need to showthat

P(A, T < ∞, XT+s1 − XT ≤ c1, . . . , XT+sm− XT ≤ cm)

= P(A, T < ∞)P(Xs1 ≤ c1, . . . , Xsm≤ cm).

59


First define stopping times Tn = 2−n([2nT ] + 1), n ≥ 1, that only take countably manyvalues. These are the next dyadic rationals after time T . Note that Tn ↓ T as n → ∞.Now note that A ∩ Tn = k2−n

∈ Fk2−n and the simple Markov property yields

P(A, Tn < ∞, XTn+s1 − XTn≤ c1, . . . , XTn+sm

− XTn≤ cm)

=∞∑

k=0

P(A, Tn = k2−n, Xk2−n+s1− Xk2−n ≤ c1, . . . , Xk2−n+sm

− Xk2−n ≤ cm)

=

∞∑

k=0

P(A, Tn = k2−n)P(Xs1 ≤ c1, . . . , Xsm≤ cm)

= P(A, Tn < ∞)P(Xs1 ≤ c1, . . . , Xsm≤ cm).

Now the right-continuity of sample paths ensures XTn+sj→ XT+sj

as n → ∞ and weconclude

P(A, T < ∞, XT+s1 − XT ≤ c1, . . . , XT+sm− XT ≤ cm)

= limn→∞

P(A, Tn < ∞, XTn+s1 − XTn≤ c1, . . . , XTn+sm

− XTn≤ cm)

= limn→∞

P(A, Tn < ∞)P(Xs1 ≤ c1, . . . , Xsm≤ cm)

= P(A, T < ∞)P(Xs1 ≤ c1, . . . , Xsm≤ cm),

for all (c1, . . . , cm) such that P(XT+sj− XT = cj) = 0, j = 1, . . . , m. Finally note that

(XT+s − XT )s≥0 clearly has right-continuous paths with left limits. 2

13.2 The supremum process

Let X = (Xt)t≥0 be a Levy process. We denote its supremum process by

X t = sup0≤s≤t

Xs, t ≥ 0.

We are interested in the joint distribution of (Xt, X t), e.g. for the payoff of a barrieror lookback option. Moment generating functions are easier to calculate and can benumerically inverted. We can also take such a transform over the time variable, e.g.

q 7→

∫ ∞

0

e−qtE(eγXt)dt =

1

q − Ψ(γ)uniquely identifies E(eγXt),

and the distribution of Xt. But q∫∞0

e−qtE(eγXt)dt = E(eγXτ ) for τ ∼ Exp(q).

Proposition 86 (Independence) Let X be a Levy process, τ ∼ Exp(q) an independent

random time. Then Xτ is independent of Xτ − Xτ .

Proof: We only prove the case where G1 = inft > 0 : Xt > 0 satisfies P(G1 > 0) = 1.In this case we can define successive record times Gn = inft > Gn−1 : Xt > XGn−1, n ≥

2, and also set G0 = 0. Note that, by the strong Markov property at the stopping timesGn we have that XGn

> XGn−1 (otherwise the post-Gn−1 process ˜Xt = XGn−1+t − XGn−1

would have the property ˜G1 = 0, but the strong Markov property yields P( ˜G1 > 0) =

Lecture 13: Level passage problems 61

P(G1 > 0) = 1). So X can only reach new records by upward jumps, Xτ ∈ XGn, n ≥ 0

and more specifically, we will have Xτ = Gn if and only if Gn ≤ τ < Gn+1 so that

E(eβXτ+γ(Xτ−Xτ )) =

∫ ∞

0

qe−qtE(eβXt+γ(Xt−Xt))dt

= qE

( ∞∑

n=0

∫ Gn+1

Gn

e−qteβXt+γ(Xt−Xt)dt

)

= q∞∑

n=0

E

(

eβXGn e−qGn

∫ Gn+1−Gn

0

e−qse−γ(XGn+s−XGn )ds

)

= q

∞∑

n=0

E(

e−qGn+βXGn

)

E

(

∫ eG1

0

e−qs−γ eXsds

)

where we applied the strong Markov property at Gn to split the expectation in the last row– note that

∫ Gn+1−Gn

0e−qs−γ(XGn+s−XGn )ds is a function of the post-Gn process, whereas

e−qGn+βXGn is a function of the pre-Gn process, and the expectation of the product ofindependent random variables is the product of their expectations.

This completes the proof, since the last row is a product of a function of β and afunction of γ, which is enough to conclude. More explicitly, we can put β = 0, γ = 0 andβ = γ = 0, respectively, to see that indeed the required identity holds:

E(eβXτ+γ(Xτ−Xτ )) = E(eβXτ )E(eγ(Xτ−Xτ )).

2

13.3 Levy processes with no positive jumps

Consider stopping times Tx = inft ≥ 0 : Xt ∈ (x,∞), so-called first passage times. ForLevy processes with no positive jumps, we must have XTx

= x, provided that Tx < ∞.This observation allows to calculate the moment generating function of Tx. To preparethis result, recall that the distribution of Xt has moment generating function


2σ2γ2 +

∫

0

−∞(eγx

− 1 − γx1|x|≤1)g(x)dx.

Let us exclude the case where −X is a subordinator, i.e. where σ2 = 0 and a1 −∫

0

−1xg(x)dx ≤ 0, since in that case Tx = ∞. Then note that

Ψ′′(γ) = σ2 +

∫

0

−∞x2eγxg(x)dx > 0,

so that Ψ is convex and hence has at most two zeros, one of which is Ψ(0) = 0. There isa second zero γ0 > 0 if and only if Ψ′(0) = E(X1) < 0, since we excluded the case where−X is a subordinator, and P(Xt > 0) > 0 implies that Ψ(∞) = ∞.

Theorem 87 (Level passage) Let (Xt)t≥0 be a Levy process with no positive jumps

and Tx the first passage time across level x. Then

E(e−qTx1Tx<∞) = e−xΦ(q),

where Φ(q) is the largest γ for which Ψ(γ) = q.


Proof: We only prove this for the case where P(Tx < ∞) = 1, i.e. E(X1) ≥ 0 and γ0 = 0.By Exercise A.2.3.(a) the processes Mt = eγXt−tΨ(γ) are martingales. We will apply theOptional stopping theorem to Tx. Note that E(M2

t ) = et(Ψ(2γ)−2Ψ(γ)) is not such thatsupt≥0

E(M2

t ) < ∞. However, if we put

M(u)

t = Mt if t ≤ u and M(u)

t = Mu if t ≥ u,

then (Mut )t≥0 is a martingale which satisfies supt≥0

E((M(u)

t )2) < ∞. Also, Tx ∧ u is astopping time, so that for γ ≥ γ0 = 0 (so that Ψ(γ) ≥ 0)

1 = E(M(u)

Tx∧u) = E(MTx∧u) → E(MTx) = E(eγx−Ψ(γ)Tx), as u → ∞,

by dominated convergence (MTx∧u ≤ expγx − Ψ(γ)Tx ≤ expγx). We now concludethat

E(e−Ψ(γ)Tx) = e−γx

which for q = Ψ(γ) and Φ(q) the unique γ ≥ γ0 = 0 with Ψ(γ) = q. 2

Corollary 88 Let X be a Levy process with no positive jumps and τ ∼ Exp(q) indepen-

dent. Then Xτ ∼ Exp(Φ(q)).

Proof: P(Xτ > x) = P(Tx ≤ τ) =∫∞0

P(τ ≥ t)fTx(t)dt = E(e−qTx) = e−Φ(q)x. 2

If we combine this with the Independence Theorem of the previous section we obtain.

Corollary 89 Let X be a Levy process with no positive jumps and τ ∼ Exp(q) indepen-

dent. Then

E(e−β(Xτ−Xτ )) =q(Φ(q) − β)

Φ(q)(q − Ψ(β))

Proof: Note that we have from the Independence Theorem that

E(eβXτ )E(e−β(Xτ−Xτ )) = E(eβXτ ) =

∫ ∞

0

qe−qtE(eβXt)dt =

q

q − Ψ(β)

and from the preceding corollary

E(eβXτ ) =Φ(q)

Φ(q) − βand so E(e−β(Xτ−Xτ )) =

q

q − Ψ(β)

Φ(q) − β

Φ(q).

2

13.4 Application: insurance ruin

Proposition 86 splits the Levy process at its supremum into two increments. If you turnthe picture of a Levy process by 180, this split occurs at the infimum, and it can beshown (Exercise A.7.1) that Xτ ∼ Xτ −Xτ . Therefore, Corollary 89 gives E(eβXτ ), alsofor q ↓ 0 if E(X1) > 0, since then

E(eβX∞) = lim

q↓0

q(Φ(q) − β)

Φ(q)(q − Ψ(β))=

βE(X1)

Ψ(β)

since Φ′(0) = 1/Ψ′(0) = 1/E(X1) and note that for an insurance reserve process Rt =u + Xt, the probability of ruin is r(u) = P(X∞ < −u), the distribution funcion of X∞which is uniquely identified by E(eβX

∞).

Lecture 14

Ladder times and storage models

Reading: Kyprianou Sections 1.3.2 and 3.3

14.1 Case 1: No positive jumps

In Theorem 87 we derived the moment generating function of Tx = inft ≥ 0 : Xt > xfor any Levy process with no positive jumps. We also indicated the complication thatTx = ∞ is a possibility, in general. Let us study this is in more detail in our standardsetting


2σ2γ2 +

∫

0

−∞(eγx

− 1 − γx1|x|≤1)g(x)dx.

The important quantity is

E(X1) =∂

∂γE(eγXt)

∣

∣

∣

∣

γ=0

= Ψ′(0) = a1 −

∫

0

−1

xg(x)dx.

The formula that we derived was

E(e−qTx1Tx<∞) = e−xΦ(q)

where for q > 0, Φ(q) > 0 is unique with Ψ(Φ(q)) = q. Letting q ↓ 0, we see that

P(Tx < ∞) = limq↓0

E(e−qTx1Tx<∞) = e−xΦ(0+).

Here the convexity of Ψ that we derived last time implies that Φ(0+) = 0 if and only ifE(X1) = φ′(0) ≥ 0. Therefore, P(Tx < ∞) = 1 if and only if E(X1) ≥ 0.

Part of this could also be deduced by the strong (or weak) law of large numbers.Applied to increments Yk = Xkδ − X(k−1)δ it implies that

Xnδ

nδ=

1

δ

1

n

n∑

k=1

Yk →

1

δE(Y1) =

1

δE(Xδ) = E(X1),

almost surely (or in probability) as n → ∞. We can slightly improve this result to aconvergence as t → ∞ as follows

E(eγXt/t) = etφ(γ/t)→ eγφ′

(0) = eγE(X1)⇒

Xt

t→ E(X1),

63


in probability. We used here that Zt → a in distribution implies Zt → a in probability,which holds only because a is a constant, not a random variable. Note that indeed forall ε > 0, as t → ∞,

P(|Zt − a| > ε) ≤ P(Zt ≤ a − ε) + 1 − P(Zt ≤ a + ε) → 0 + 1 − 1 = 0.

From this, we easily deduce that Xt → ±∞ (in probability) if E(Xt) 6= 0, but the caseE(Xt) = 0 is not so clear from this method. In fact, it can be shown that all convergenceshold in the almost sure sense, here.

By an application of the Strong Markov property we can show the following.

Proposition 90 The process (Tx)x≥0 is a subordinator.

Proof: Let us here just prove that Tx+y − Tx is independent of Tx and has the samedistribution as Ty. The remainder is left as an exercise.

Note first that XTx= x, since there are no positive jumps. The Strong Markov

property at Tx can therefore be stated as ˜X = (XTx+s − x)s≥0 is independent of FTxand

has the same distribution as X. Now note that

Tx + ˜Ty = Tx + infs ≥ 0 : ˜Xs > y = Tx + infs ≥ 0 : XTx+s > x + y

= inft ≥ 0 : Xt > x + y = Tx+y

so that Tx+y − Tx = ˜Ty, and we obtain

P(Tx ≤ s, Tx+y − Tx ≤ t) = P(Tx ≤ s, ˜Ty ≤ t) = P(Tx ≤ s)P(Ty ≤ t),

since Tx ≤ s ∈ FTx. Formally, Tx ≤ s ∩ Tx ≤ r = Tx ≤ s ∧ r ∈ Fr for all r ≥ 0

since Tx is a stopping time. 2

We can understand what the jumps of (Tx)x≥0 are: in fact, X can be split into itssupremum process X t = sup

0≤s≤t Xs and the bits of path below the supremum. Roughly,the times

Tx, x ≥ 0 = t ≥ 0 : Xt = Xt

are the times when the supremum increases. Tx − Tx− > 0 if the supremum processremains constant at height x for an amount of time Tx − Tx−. The process (Tx)x≥0 iscalled “ladder time process”. The process (XTx

)x≥0 is called ladder height process. Inthis case, XTx

= x is not very illuminating. Note that (Tx, XTx)x≥0 is a bivariate Levy

process.

Example 91 (Storage models) Consider a Levy process of bounded variation, repre-sented as At − Bt for two subordinators A and B. We interpret At as the amount ofwork arriving in [0, t] and Bt as the amount of work that can potentially exit from thesystem. Let us focus on the case where A is a compound Poisson process and Bt = tfor a continuously working processor. The quantity of interest is the amount Wt of workwaiting to be carried out and requiring storage, where W0 = w ≥ 0 is an initial amountof work stored.

Lecture 14: Ladder times and storage models 65

Note that Wt 6= w + At − Bt, in general, since w + At − Bt can become negative,whereas Wt ≥ 0. In fact, we can describe as follows: if the storage is empty, then nowork exits from the system. Can we see from At − Bt when the storage will be empty?We can express the first time it becomes empty and the first time it is refilled thereafteras

L1 = inft ≥ 0 : w + At − Bt = 0 and R1 = inft ≥ L1 : ∆At > 0.

On [L1, R1], Xt = Bt−At increases linearly at unit rate from w to w+(R1−L1), whereasW remains constant equal to zero. In fact,

Wt = w − Xt +

∫ t

L1∧t

1ds = w − Xt +

∫ t

0

1Xs=Xs≥wds = (w ∨ X t) − Xt, 0 ≤ t ≤ R1.

An induction now shows that the system is idle if and only if Xt = X t ≥ w, so thatWt = Xt + (w ∨ Xt) for all t ≥ 0.

In this context, (X t−w)+ is the amount of time the system was idle before time t, andTx = inft ≥ 0 : Xt > x is the time by which the system has accumulated time x − win the idle state, x ≥ w, and we see that (x − w)/Tx ∼ x/Tx → 1/E(T1) = 1/Φ′(0) =φ′(0) = E(X1) = 1 − E(A1) in probability, if E(A1) ≤ 1.

Example 92 (Dams) Suppose that the storage model refers more particularly to a damthat releases a steady stream of water at a constant intensity a2. Water is added accordingto a subordinator (At)t≥0. The dam will have a maximal capacity of, say, M > 0. Givenan initial water level of w ≥ 0, the water level at time t is, as before

Wt = (w ∨ X t) − Xt, where Xt = a2t − At.

The time F = inft ≥ 0 : Wt > M, the first time when the dam overflows, is a quantityof interest. We do not pursue this any further theoretically, but note that this can besimulated, since we can simulate X and hence W .

14.2 Case 2: Union of intervals as ladder time set

Proposition 93 If Xt = a2t − Ct for a compound Poisson process (Ct)t≥0 and a drift

a2 ≥ 0 ∨ E(C1), then the ladder time set is a collection of intervals. More precisely,

t ≥ 0 : Xt = Xt is the range σy, y ≥ 0 of a compound Poisson subordinator with pos-

itive drift coefficient.

Proof: Define L0 = 0 and then for n ≥ 0 stopping times

Rn = inft ≥ Ln : ∆Ct > 0, Ln+1 = inft ≥ Rn : Xt = X t.

The strong Markov property at these stopping times show that (Rn−Ln)n≥0 is a sequenceof Exp(λ) random variables where λ =

∫∞0

g(x)dx is the intensity of positive jumps, and(Ln − Rn−1)n≥1 is a sequence of independent identically distributed random variables.Now define Tn = R0 − L0 + . . . + Rn−1 − Ln−1 and (σy)y≥0 to be the compount Poissonprocess with unit drift, jump times Tn, n ≥ 1, and jump heights Ln − Rn−1, n ≥ 1. 2


The ladder height process (Xσy)y≥0 will then also have a positive drift coefficient and

share some jump times with σ (whenever X jumps from below X t− above Xt−, but haveextra jump times when X jumps from X t− upwards, some jump times of (σy)y≥0 are notjump times of (Xσy

)y≥0 – if X reaches X t− again without a jump.)

Example 94 (Storage models) In the context of the previous examples, consider moregeneral subordinators Bt with unit drift. Interpret jumps of B as unfinished work poten-

tially exiting to be carried out elsewhere. We should be explicit and make a conventionthat if the current storage amount Wt is not sufficient for a jump of B, then all remainingwork exits. With this convention, the amount Wt waiting to be carried out is still

Wt = w − Xt +

∫ t

0

1Xs=Xs≥wds t ≥ 0,

but note that the latter integral cannot be expressed in terms of X t so easily, but σy

is still the time by which the system has accumulated time y − w in the idle state, fory ≥ w, so It = infy ≥ 0 : σy > t is the amount of idle time before t.

14.3 Case 3: Discrete ladder time set

If Xt = a2t−Ct for a compound Poisson process (or indeed bounded variation pure jumpprocess) (Ct)t≥0 and a drift a2 < 0, then the ladder time set is discrete. We can still think

of t ≥ 0 : Xt = Xt as the range σy, y ≥ 0 of a compound Poisson subordinator withzero drift coefficient. More naturally, we would define successive ladder times G0 = 0and Gn+1 = inft > Gn : Xt = X t. By the strong Markov property, Gn+1 − Gn, n ≥ 0,is a sequence of independent and identically distributed random variables, and for anyintensity λ > 0, we can specify (σy)y≥0 to be a compound Poisson process with rate λ > 0and jump sizes Gn+1 − Gn, n ≥ 0.

Note that (σy)y≥0 is not unique since we have to choose λ. In fact, once a choice hasbeen made and q > 0, we have σy : y ≥ 0 = σqy, y ≥ 0, not just here, but also inCases 1 and 2. In Cases 1 and 2, however, we identified a natural choice (of q) in eachcase.

14.4 Case 4: non-discrete ladder time set and posi-

tive jumps

The general case is much harder. It turns out that we can still express

t ≥ 0 : Xt = X t = σy : y ≥ 0

for a subordinator (σy)y≥0, but, as in Case 3, there is no natural way to choose this process.It can be shown that the bivariate process (σy, Xσy

)y≥0 is a bivariate subordinator in thisgeneral setting, called the ladder process. There are descriptions of its distribution andrelations between these processes of increasing ladder events and analogous processes ofdecreasing ladder events.

Lecture 15

Branching processes

Reading: Kyprianou Section 1.3.4

15.1 Galton-Watson processes

Let ξ = (ξk)k≥0 be (the probability mass function of) an offspring distribution. Considera population model where each individual gives birth to independent and identicallydistributed numbers of children, starting from Z0 = 1 individual, the common ances-tor. Then the (n + 1)st generation Zn+1 consists of the sum of numbers of childrenNn,1, . . . , Nn,Zn

of the nth generation:

Zn+1 =Zn∑

i=1

Nn,i, where Nn,i ∼ ξ independent, i ≥ 1, n ≥ 0.

Proposition 95 Let ξ be an offspring distribution, g(s) =∑

k≥0ξks

k its generating func-

tion, then

E(sZ1) = g(s), E(sZ2) = g(g(s)), . . . , E(sZn) = g(n)(s),

where g(0)(s) = s, g(n+1)(s) = g(n)(g(s)), n ≥ 0.

Proof: The result is clearly true for n = 0 and n = 1. Now note that

E(sXn+1) = E

(

sPZn

i=1 Nn,i

)

=∞∑

j=0

P(Zn = j)E(

sPj

i=1 Nn,i

)

=

∞∑

j=0

P(Zn = j)(g(s))j = E((g(s))Zn) = g(n)(g(s)).

2

Proposition 96 (Zn)n≥0 is a Markov chain whose transition probabilities are given by

pij = P(N1 + . . . + Ni = j), where N1, . . . , Ni ∼ ξ independent.

In particular, if (Z(1)

n )n≥0 and (Z(2)

n )n≥0 are two independent Markov chains with tran-

sition probabilities (pij)i,j≥0 starting from population sizes k and l, respectively, then

Z(1)

n + Z(2)

n , n ≥ 0, is also a Markov chain with transition probabilities (pij)i,j≥0 starting

from k + l.

67


Proof: Just note that the independence of (Nn,i)i≥1 and (Nk,i)0≤k≤n−1,i≥1 implies that

P(Zn+1 =j|Z0 = i0, . . . , Zn−1 = in−1, Zn = in) = P(Nn,1 + . . . , Nn,in =j|Z0 = i0, . . . , Zn = in)

= P(Nn,1 + . . . , Nn,in = j) = pinj ,

as required. For the second assertion note that

b(i1,i2),j := P(Z(1)

n+1+ Z

(2)

n+1= j|Z(1)

n = i1, Z(2)

n = i2)

= P(N(1)

n,1 + . . . + N(1)

n,i1+ N

(2)

n,1 + . . . + N(2)

n,i2= j) = pi1+i2,j

only depends on i1 + i2 (not i1 or i2 separately) and is of the form required to concludethat

P(Z(1)

n+1+ Z

(2)

n+1= j|Z(1)

n + Z(2)

n = i) =

∑ii1=0

P(Z(1)

n = i1, Z(2)

n = i − i1)b(i1,i−i1),j

P(Z(1)

n + Z(2)

n = i)= pij .

2

The second part of the proposition is called the branching property and expressesthe property that the families of individuals in the same generation evolve completelyindependently of one another.

15.2 Continuous-time Galton-Watson processes

We can also model lifetimes of individuals by independent exponentially distributed ran-dom variables with parameter λ > 0. We assume that births happen at the end of alifetime. This breaks the generations. Since continuously distributed random variablesare almost surely distinct, we will observe one death at a time, each leading to a jumpof size k − 1 with probability ξk, k ≥ 0. It is customary to only consider offspring distri-butions with ξ1 = 0, so that there is indeed a jump at every death time. Note that atany given time, if j individuals are present in the population, the next death occurs at atime

H = minL1, . . . , Lj ∼ Exp(jλ), where L1, . . . , Lj ∼ Exp(λ).

From these observations, one can construct (and simulate!) the associated populationsize process (Yt)t≥0 by induction on the jump times.

Proposition 97 (Yt)t≥0 is a Markov process. If Y (1) and Y (2) are independent Markov

processes with these transition probabilities starting from k and l, then Y (1) +Y (2) is also

a Markov process with the same transition probabilities starting from k + l.

Proof: Based on BS3a Applied Probability, the proof is not difficult. We skip it here.2

(Yt)t≥0 is called a continuous-time Galton-Watson process. In fact, these are the onlyMarkov processes with the branching property (i.e. satisfying the second statement ofthe proposition for all k ≥ 1, l ≥ 1).

Lecture 15: Branching processes 69

Example 98 (Simple birth-and-death processes) If individuals have lifetimes withparameter µ and give birth at rate β to single offspring repeatedly during their lifetime,then we recover the case

λ = µ + β and ξ0 =µ

µ + β, ξ2 =

β

µ + β.

In fact, we have to reinterpret this model by saying each transition is a death, givingbirth to either two or no offspring. These parameters arise since, if only one individualis present, the time to the next transition is the minimum of the exponential birth timeand the exponential death time.

The fact that all jump sizes are independent and identically distributed is reminiscentof compound Poisson processes, but for high population sizes j we have high parameters tothe exponential times between two jumps – the process Y moves faster than a compoundPoisson process at rate λ. Note however that for H ∼ Exp(jλ) we have jH ∼ Exp(λ).Let us use this observation to specify a time-change to slow down Y .

Proposition 99 Let (Yt)t≥0 be a continuous-time Galton-Watson process with offspring

distribution ξ and lifetime distribution Exp(λ). Then for the piecewise linear functions

Jt =

∫ t

0

Yudu, t ≥ 0, ϕs = inft ≥ 0 : Jt > s, 0 ≤ s < J∞,

the process

Xs = Yϕs, 0 ≤ s < J∞,

is a compound Poisson process with jump distribution (ξk+1)k≥−1 and rate λ, run until

the first hitting time of 0.

Proof: Given Y0 = i, the first jump time T1 = inft ≥ 0 : Yt 6= i ∼ Exp(iλ), so

JT1 = iT1 and ϕs = s/i, 0 ≤ s ≤ iT1,

so we identify the first jump of Xs = Ys/i, 0 ≤ s ≤ iT1 at time iT1 ∼ Exp(λ).

Now the strong Markov property (or the lack of memory property of all other life-times) implies that given k offspring are produced at time T1, the process (YT1+t)t≥0 isa continuous-time Galton-Watson process starting from j = i + k − 1, independent of(Yr)0≤r≤T1 . We repeat the above argument to see that T2 − T1 ∼ Exp(jλ), and for j ≥ 1,

JT2 = iT1 + j(T2 − T1) and ϕiT1+s = T1/i + s/j, 0 ≤ s ≤ j(T2 − T1),

and the second jump of XiT1+s = YT1+s/j, 0 ≤ s ≤ j(T2 − T1), happens at time iT1 +j(T2 − T1), where j(T2 − T1) ∼ Exp(λ) is independent of iT1. An induction as long asYTn

> 0 shows that X is a compound Poisson process run until the first hitting time of0. 2


Corollary 100 Let (Xs)s≥0 be a compound Poisson process starting from l ≥ 1 with jump

distribution (ξk+1)k≥−1 and jump rate λ > 0. Then for the piecewise linear functions

ϕs =

∫ s

0

1

Xvdv, 0 ≤ s < T0, and Jt = infs ≥ 0 : ϕs > t, t ≥ 0,

the process

Yt = XJt, t ≥ 0,

is a continuous-time Galton-Watson process with offspring distribution ξ and lifetime

distribution Exp(λ).

15.3 Continuous-state branching processes

Population-size processes with state space N are natural, but for large populations, it isoften convenient to use continuous approximations and use a state space [0,∞). In viewof Corollary 100 it is convenient to define as follows.

Definition 101 (Continuous-state branching process) Let (Xs)s≥0 be a Levy pro-cess with no negative jumps starting from x > 0, with E(exp−γXs) = expsφ(γ).Then for the functions

ϕs =

∫ s

0

1

Xv

dv, 0 ≤ s < T0, and Jt = infs ≥ 0 : ϕs > t, t ≥ 0,

the process

Yt = XJt, t ≥ 0,

is called a continuous-state branching process with branching mechanism φ.

We interpret upward jumps as birth events and continuous downward movement as(infinitesimal) deaths. The behaviour is accelerated at high population sizes, so fluc-tuations will be larger. The behaviour is slowed down at small population sizes, sofluctuations will be smaller.

Example 102 (Pure death process) For Xs = x − cs we obtain

ϕs =

∫ s

0

1

x − cvdv = −

1

clog(1 − cs/x), and Jt =

x

c(1 − e−ct),

and so Yt = xe−ct.

Example 103 (Feller diffusion) For φ(γ) = γ2 we obtain Feller’s diffusion. There arelots of parallels with Brownian motion. There is a Donsker-type result which says thatrescaled Galton-Watson processes converge to Feller’s diffusion. It is the most popularmodel in applications. A lot of quantities can be calculated explicitly.

Proposition 104 Y is a Markov process. Let Y (1) and Y (2) be two independent continuous-

state branching processes with branching mechanism φ starting from x > 0 and y > 0.Then Y (1) + Y (2) is a continuous-state branching process with branching mechanism φstarting from x + y.

Lecture 16

The two-sided exit problem

Reading: Kyprianou Chapter 8, Bertoin Aarhus Notes, Durrett Sections 7.5-7.6

16.1 The two-sided exit problem for Levy processes

with no negative jumps

Let X be a Levy process with no negative jumps. As we have studied processes with nopositive jumps (such as −X) before, it will be convenient to use compatible notation andwrite

E(e−γXt) = etφ(γ), φ(γ) = a−Xγ +1

2σ2γ2 +

∫

0

−∞(eγx

− 1 − γx1|x|≤1)g−X(x)dx

= −aXγ +1

2σ2γ2

−

∫ ∞

0

(1 − e−γx− γx1|x|≤1)gX(x)dx,

where a−X = −aX and gX(x) = g−X(−x), x > 0. Then we deduce from Section 11.4that, if E(X1) < 0,

E(e−βX∞) = −

βE(X1)

φ(β), β ≥ 0. (1)

The two-sided exit problem is concerned with exit from an interval [−a, b], notably thetime

T = T[−a,b]c = inft ≥ 0 : Xt ∈ [−a, b]c

and the probability to exit at the bottom P(XT = −a). Note that an exit from [−a, b]at the bottom happens necessarily at −a, since there are no negative jumps, whereas anexit at the top may be due to a positive jump across the threshold b leading to XT > b.

Proposition 105 For any Levy process X with no negative jumps, all a > 0, b > 0 and

T = T[−a,b]c, we have

P(XT = −a) =W (b)

W (a + b), where W is such that

∫ ∞

0

e−βxW (x)dx =1

φ(β).

71


Proof: We only prove the case E(X1) < 0. By (1), we can identify (the right-continuousfunction) W , since

−

βE(X1)

φ(β)= E(e−βX

∞) =

∫ ∞

0

e−βxfX∞

(x)dx

=

∫ ∞

0

βe−βxP(X∞) ≤ x)dx,

by partial integration, and so by the uniqueness of moment generating functions, we haveW (x) = cP(X∞ ≤ x), where c = −E(X1) > 0.

Now define τa = inft ≥ 0 : Xt < −a and apply the strong Markov property at τa to

get a post-τa process ˜X = (Xτa+s + a)s≥0 independent of (Xr)r≤τa, in particular of Xτa

,so that

cW (b) = P(X∞ ≤ b) = P(Xτa≤ b, ˜X∞ ≤ a + b)

= P(Xτa≤ b)P(X∞ ≤ a + b) = P(Xτa

≤ b)cW (a + b),

and the result follows. 2

Example 106 (Stable processes) Let X be a stable process of index α ∈ (1, 2] withno negative jumps, then we have

∫ ∞

0

e−λxW (x)dx = λ−α⇒ Γ(α)W (x) = xα−1.

We deduce that

P(XT = −a) =

(

b

a + b

)α−1

.

16.2 The two-sided exit problem for Brownian mo-

tion

In the Brownian case, we can push the analysis further without too much effort.

Proposition 107 For Brownian motion B, all a > 0, b > 0 and T = T[−a,b]c, we have

E(e−qT|BT = −a) =

Vq(b)

Vq(a + b)and E(e−qT

|BT = b) =Vq(a)

Vq(a + b),

where

Vq(x) =sinh(

√

2qx)

x.

Lecture 16: The two-sided exit problem 73

Proof: We condition on BT and use the strong Markov property of B at T to obtain

e−a√

2q = E(e−qT−a)

= P(BT = −a)E(e−qT−a

|BT = −a) + P(BT = b)E(e−qT−a

|BT = b)

=b

a + bE(e−qT

|BT = −a) +a

a + bE(e−q(T+ eT

−a−b)|BT = b)

=b

a + bE(e−qT

|BT = −a) +a

a + bE(e−qT

|BT = b)e−(a+b)√

2q

and, by symmetry,

e−b√

2q =a

a + bE(e−qT

|BT = b) +b

a + bE(e−qT

|BT = −a)e−(a+b)√

2q.

These can be written as

b + a

ab= a−1

E(e−qT|BT = −a)ea

√2q + b−1

E(e−qT|BT = b)e−b

√2q

b + a

ab= a−1

E(e−qT|BT = −a)e−a

√2q + b−1

E(e−qT|BT = b)eb

√2q

and suitable linear combinations give, as required,

2 sinh(a√

2q)b + a

ab= 2 sinh((a + b)

√

2q)b−1E(e−qT

|BT = b)

2 sinh(b√

2q)b + a

ab= 2 sinh((a + b)

√

2q)a−1E(e−qT

|BT = −a).

2

Corollary 108 For Brownian motion B, all a > 0 and T = T[−a,a]c, we have

E(e−qT ) =1

cosh(a√

2q)

Proof: Just calculate from the previous proposition

E(e−qT ) =Vq(a)

Vq(2a)= 2

e√

2qa− e−

√2qa

(e√

2qa)2− (e−

√2qa)2

=1

cosh(a√

2q).

2

16.3 Appendix: Donsker’s Theorem revisited

We can now embed simple symmetric random walk (SSRW) into Brownian motion B byputting

T0 = 0, Tk+1 = inft ≥ Tk : |Bt − BTk| = 1, Sk = BTk

, k ≥ 0,

and for step sizes 1/√

n modify T(n)

k+1= inft ≥ T

(n)

k : |Bt − BT

(n)k

| = 1/√

n.


Theorem 109 (Donsker for SSRW) For a simple symmetric random walk (Sn)n≥0

and Brownian motion B, we have

S[nt]√

n→ Bt, locally uniformly in t ≥ 0, in distribution as n → ∞.

Proof: We use a coupling argument. We are not going to work directly with the originalrandom walk (Sn)n≥0, but start from Brownian motion (Bt)t≥0 and define a family ofembedded random walks

S(n)

k := BT

(n)k

, k ≥ 0, n ≥ 1, ⇒

(

S(n)

[nt]

)

t≥0

∼

(

S[nt]√

n

)

t≥0

.

To show convergence in distribution for the processes on the right-hand side, it suffices toestablish convergence in distribution for the processes on the left-hand side, as n → ∞.

To show locally uniform convergence we take an arbitrary T ≥ 0 and show uniformconvergence on [0, T ]. Since (Bt)0≤t≤T is uniformly continuous (being continuous on acompact interval), we get in probability

sup0≤t≤T

∣

∣

∣S

(n)

[nt] − Bt

∣

∣

∣= sup

0≤t≤T

∣

∣

∣

∣

BT

(n)[nt]

− Bt

∣

∣

∣

∣

≤ sup0≤s≤t≤T :|s−t|≤sup0≤r≤T |T (n)

[nr]−r|

|Bs − Bt| → 0

as n → ∞, if we can show (as we do in the lemma below) that sup0≤t≤T |T

(n)

[nt] − t| → 0.This establishes convergence in probability, which “implies” convergence in distribu-

tion for the embedded random walks and for the original scaled random walk. 2

Lemma 110 In the setting of the proof of the theorem, sup0≤t≤T |T

(n)

[nt] − t| → 0 in prob-

ability.

Proof: First for fixed t, we have

E(e−qT

(n)[nt]) =

(

E(e−qT(n)1 ))

[nt]

=1

(cosh(√

2q/n))[nt]→ e−qt,

since cosh(√

2q/n) ∼ 1 + q/n + O(1/n). Therefore, in probability T(n)

[nt] → t. For unifor-

mity, let ε > 0. Let δ > 0. We find n0 ≥ 0 such that for all n ≥ n0 and all tk = kε/2,1 ≤ k ≤ 2T/ε we have

P(|T(n)

[ntk]− tk| > ε/2) < δε/2T,

then

P

(

sup0≤t≤T

|T(n)

[nt] − t| > ε

)

≤ P

(

sup1≤k≤2T/ε

|T(n)

[ntk]− tk| >

ε

2

)

≤

2T/ε∑

k=1

P

(

|T(n)

[ntk]− tk| >

ε

2

)

< δ.

2

We can now describe the recipe for the full proof of Donsker’s Theorem. In fact, wecan embed every standardized random walk ((Sk − kE(S1))/

√

nVarS1)k≥0 in Brownian

motion X, by first exits from independent random intervals [−A(n)

k , B(n)

k ] so that XT

(n)k

∼

(Sk − kE(S1))/√

nVar(S1), and the embedding time change (T(n)

[nt])t≥0 can still be shownto converge uniformly to the identity.

Appendix A

Assignments

Assignment sheets are issued on Tuesdays of weeks 1-7. They are made available on thewebsite of the course at

http://www.stats.ox.ac.uk/∼winkel/ms3b.html.

Classes take place in weeks 2 to 8 at times and locations to be determined probablyon Wednesdays 2.30pm, Thursdays 3.05pm and/or Fridays 8.55am. The class allocationcan be accessed from the course website. Only undergraduates and MSc students inMathematical and Computational Finance can sign up for classes. All others should talkto me after one of the first two lectures.

Scripts are to be handed in probably by Tuesdays 11am or Wednesdays 4pm in theDepartment of Statistics, but one of the classes will probably be in OCIAM.

Exercises on the problem sheets vary in style and difficulty. If you find an exercisedifficult, please do not deduce that you cannot solve the following exercises, but aim atgiving each exercise a serious try. Model solutions will be provided on the course

website.

There are lecture notes available. Please print these so that we can make bestuse of the lecture time. I gradually replace last year’s notes by an updated version.The beginning has not changed much, but some typos and unclear passages have beenimproved.

Below are some comments on the recommended Reading and Further Reading liter-ature.

Kyprianou: Introductory lectures on fluctuations of Levy processes with ap-

plications. Springer 2006

This is the treatment that is closest to the course. It is based on a Masters courseand has been written for Masters students. The book assumes a background in measure-theoretic probability, but is written in a friendly way suitable for a wide range of differentbackgrounds. The text contains some worked examples and exercises.

I

II Assignments – MS3b Levy Processes and Finance – Oxford HT 2008

Kingman: Poisson processes. OUP 1993

This is a gentle introduction to (general, higher-dimensional) Poisson processes and con-tains a thorough discussion of the tools leading up to and including the study increasingLevy processes (Section 8.4). Measure-theoretic arguments are isolated in a few proofs,and the reader can take the measure theory for granted. For consistency of terminologywith other Oxford courses and most of the literature, in our course, we will reserve theterm “Poisson process” to the one-dimensional process. What Kingman calls “Poissonprocess” is the associated counting measure, which we call “Poisson counting measure”.

Schoutens: Levy processes in finance. Wiley 2003

This is not a textbook, but a monograph advertising Levy process for finance applications.All models for financial stock prices that we study in our course, are discussed in detail,both their properties and how they can be calibrated to fit financial market data. Thereare also sections on simulation and option pricing.

Sato: Levy processes and infinitely divisible distributions. CUP 1999

This is a graduate textbook on Levy processes. The focus is on distributional propertiesand analytic methods.

Bertoin: Levy processes. CUP 1996

This is a research monograph on Levy processes. The approach is sample path based.

Grimmett and Stirzaker: Probability and Random Processes. OUP 2001

This is the standard probability reference book used in Oxford, overarching 1st, 2nd and3rd year probability courses and more. It contains a section on spatial Poisson processes,a treatment of martingales.

Williams: Probability with Martingales. CUP 1991

This is the standard reference on measure theory and martingales used in Oxford.

Ross: Applied Probability Models. Academic Press 1989

We only use the simulation chapter.

Durrett: Probability – Theory and Examples. Duxbury 2004

This is a good graduate textbook on Probability. We essentially refer to Durrett only fora proof of Donsker’s theorem, but there is also a section on infinite divisibility.

Kallenberg: Foundations of Modern Probability. Springer 1997

This is a standard reference on Probability. The presentation is extremely concise andeconomical and the amount of material is encyclopaedic. We only refer to it for rigorousstatements and proofs of convergence theorems of random walks to Levy processes.

Assignment 1 – MS3b Levy Processes and Finance – Oxford HT 2008 III

A.1 Infinite divisibility and limits of random walks

If in doubt, hand in scripts by Tuesday 22 January 2007, 11am, Department of Statistics

1. (a) Show that Y ∼ Gamma(α, β) with density

g(x) =βαxα−1

Γ(α)e−βx, x ≥ 0,

is an infinitely divisible distribution and that the independent and identicallydistributed “divisors” Yn,j in Yn,1+ . . .+Yn,n ∼ Y are also Gamma distributed.

(b) Show that the G ∼ geom(p) distribution with probability mass function

P(G = n) = pn(1 − p), n ≥ 0,

is infinitely divisible and that “divisors” are not geometrically distributed.Hint: Study sums of geometric variables and guess the “divisor” distribution.

(c) Show that the uniform distribution on [0, 1] is not infinitely divisible.

2. (a) Let X and Y be independent Levy processes and a, b ∈ R. Show that aX +bYis also a Levy process.

(b) Let C and D be two independent Gamma Levy processes with C1 ∼ D1 ∼

Gamma(α,√

2µ). Determine the moment generating function of Cs − Ds,s ≥ 0.

We will see later that the process C − D has in fact the same distribution as

Zs = BTs, s ≥ 0, for a Brownian motion B and a Gamma Levy process T . It

is called Variance Gamma process, because Var(Bt) = t implies Var(BTs|Ts) =

Ts ∼ Gamma(αs, µ). It is a popular model for financial stock prices.

3. A large number N of policy holders in a given time period make claims indepen-dently of one another with small probability pN . Denote by SN the total number ofpolicy holders who make a claim in the time period. Assume that claim amountsA1, A2, . . . are independent and identically distributed.

(a) State the Poisson limit theorem and use probability generating functions toprove it.

(b) Explain why SN is approximately Poisson distributed and give its parameter.

(c) Calculate the moment generating function of the total amount TN of claims.

(d) Show that the distribution of TN is well-approximated by a compound Poissondistribution, by precisely formulating and proving a limit theorem of the form

TN =

SN∑

n=1

An → T∞ =S∞

∑

n=1

An in distribution, as N → ∞.

4. (a) Let A1, A2, . . . be independent and identically distributed random variableswith µ = E(A1) and σ2 = Var(A1) ∈ (0,∞). Define

Yn,k =Ak − µ

σ√

nand Vn =

n∑

k=1

Yn,k.

IV Assignments – MS3b Levy Processes and Finance – Oxford HT 2008

(i) Formulate the Central Limit Theorem for A1, A2, . . . in terms of Vn.

(ii) Let x > 0. Apply Tchebychev’s inequality to the random variable B1 =|A1 − µ|1|A1−µ|≥σx

√n to show that there is a sequence γn(x) → 0 as

n → ∞ with

P(|A1 − µ| > σx√

n) ≤γn(x)

n.

(iii) Define Mn = max|Yn,1|, . . . , |Yn,n|. Show that P(Mn ≤ x) → 1 asn → ∞, for all x > 0. Deduce that Mn → 0 in probability.

(b) Consider an urn initially containing r red and s black balls, r, s ≥ 1. Oneball is drawn with replacement (stage 1). After this, a black ball is added tothe urn and two balls are drawn, each with replacement (stage 2). After this,another black ball is added and three balls drawn with replacement (stage 3).Continue so that n balls are drawn at stage n followed by the addition of asingle black ball. Let Yn,k = 1 resp. 0 if the kth ball of stage n is red resp.black, 1 ≤ k ≤ n and Wn = Yn,1 + . . . + Yn,n.

(i) Show that Wn → Poi(r).

(ii) Show that P(Yn,k = 0) → 1 as n → ∞.

(iii) Define Mn = max|Yn,1|, . . . , |Yn,n|. Show that P(Mn = 0) → e−r asn → ∞. Deduce that Mn 6→ 0 in probability.

(c) (i) Formulate Donsker’s theorem and the process version of the Poisson limittheorem in the settings of (a) and (b). Hint: Consider only t ∈ [0, 1] and

evaluate the discrete processes V and W at [nt], t ∈ [0, 1].

(ii) Show that in both cases Mn converges in distribution to the size of thebiggest jump of the limit process during the time interval [0, 1].

Hint for 3.(c)-(d): If (Xn)n≥0 or (Xt)t≥0 is a stochastic process and N or T an indepen-

dent random time, then for real-valued functions g for which the expectations exist, we

have

E(g(XN)) =∞

∑

n=0

E(g(Xn))P(N = n) and E(g(XT )) =

∫ ∞

0

E(g(Xt))fT (t)dt.

To prove the first formula for integer-valued (Xn)n≥0 just note that

E(g(XN)) =

∞∑

k=−∞g(k)P(XN = k)

=∞

∑

k=−∞

∞∑

n=0

g(k)P(XN = k, N = n)

=∞

∑

k=−∞

∞∑

n=0

g(k)P(Xn = k)P(N = n)

=

∞∑

n=0

P(N = n)E(g(Xn)).

Assignment 2 – MS3b Levy Processes and Finance – Oxford HT 2008 V

A.2 Poisson counting measures

1. (a) Let (Xt)t≥0 be a Poisson process with rate λ ∈ (0,∞) and arrival timesT1, T2, . . .. Show that N((c, d]) = #n ∈ N : c < Tn ≤ d is a Poissoncounting measure on [0,∞) with constant intensity λ.

(b) Let N be a Poisson counting measure on [0,∞) with time-varying intensityλ(t), t ≥ 0, continuous. Denote Xt = N([0, t]) and Tj = inft ≥ 0 : Xt = j,j ≥ 1.

(i) Show that (Xt)t≥0 has independent increments.

(ii) Show that (Xt)t≥0 has stationary increments if and only if the intensityfunction λ(t) is constant.

(iii) Show that (Xt)t≥0 has right-continuous paths with left limits.

(iv) Calculate the distribution of Xt −Xs.

(v) Calculate the survival function P(T1 > s), s ≥ 0, of T1.

(vi) Show that T2−T1 is independent of T1 if and only if the intensity functionλ is constant. Calculate the joint density of (T1, T2 − T1).

2. Let Π be a spatial Poisson process with constant intensity λ on the ball (x, y, z) ∈R

3 : x2 + y2 + z2≤ 1. Let P be the process given by the (x, y) coordinate of

the points (think of the points as being projected onto the (x, y) plane passingthrough the centre of the ball). Show that P is a spatial Poisson process and findits intensity function. Hint: For a rectangle A in the (x, y) plane, what points of

the ball is P counting?

3. Let (Xt)t≥0 be a Levy process with E(X2

1) <∞. Denote µ = E(X1), σ

2 = Var(X1)and e−ψ(λ) = E(eiλX1). If E(eγX1) < ∞, denote eΨ(γ) = E(eγX1). Show that thefollowing processes are martingales.

(a) expγXt − tΨ(γ), if E(eγX1) < ∞. Hint: First show that E(expγXt) =etΨ(γ) for all t = 1/m, then for all t ∈ Q ∩ [0,∞), and finally, using right-

continuity, for all t ∈ [0,∞).

(b) expiλXt + tψ(λ).

(c) Xt − tµ.

(d) (Xt − tµ)2− tσ2.

4. (a) Show that for β > 0 and γ < β∫ ∞

0

(eγx − 1)1

xe−βxdx = − log

(

1 −

γ

β

)

e.g. by suitable (and well-justified) differentiation under the integral sign.

(b) Let (∆t)t≥0 be a Poisson point process with intensity function αx−1e−βx. Usethe exponential formula for Poisson point processes to show that

Ct =∑

s≤t∆s has a Gamma distribution, density

βαt

Γ(αt)xαt−1e−βx, x ≥ 0.

(c) Show that (Ct)t≥0 as defined in (b) is a Levy process.

VI Assignments – MS3b Levy Processes and Finance – Oxford HT 2008

5. Let X and Y be two independent increasing compound Poisson processes. Denotethe respective jump rates by λX and λY , assume that the jump size distributionsare continuous with densities hX and hY . Denote D = X − Y .

(a) Show that X and Y have no jump times in common.

(b) Show that D has jump times according to a Poisson process with rate λX+λY .

(c) Calculate the distribution of the first jump size of D.

(d) Show that (∆Xt)t≥0 is a Poisson point process and specify its intensity func-tion.

(e) Show that (∆Dt)t≥0 is a Poisson point process and specify its intensity func-tion.

(f) Deduce from (d) and (e) that D is also a compound Poisson process.

(g) Show that every real-valued compound Poisson process C can be writtenuniquely as the difference of two independent increasing compound Poissonprocesses.

Note that the theory of Poisson point processes applied in (d)-(f) is neater than

the conditioning in (c) that could also be developed and iterated to establish (f).Intensity functions just add, jump size distributions are mixtures/weighted averages.

Remark: Interchanging limits and expectation/integration/summation is not always per-

mitted, and while we do not develop in this course the reasons why we may interchange,

we add “by monotone convergence”, whenever we have increasing or decreasing limits

of finite quantities. General measure-theoretic statements have been established in B10a,

special cases for Lebesgue integrals have been established in Part A Integration, which is

also not a prerequisite for this course. As in BS3a, it is enough for our purposes to for-

mulate special cases, whose statements do not require any of the formal technical setup.

For convergence as n→ ∞ these are:

• Zn ↑ Z and E(|Zn|) <∞ for all n ∈ N implies

E(Zn) ↑ E(Z) ∈ R ∪ ∞.

• fn ↑ f and∫

R|fn(x)|dx <∞ for all n ∈ N implies

∫

R

fn(x)dx ↑

∫

R

f(x)dx ∈ R ∪ ∞.

• a(n)

m ↑ am for all m ∈ N and∑∞

m=0|a

(n)

m | <∞ for all n ∈ N implies

∞∑

m=0

a(n)

m ↑

∞∑

m=0

am ∈ R ∪ ∞.

• ∆(n)

s ↑ ∆s and∑

s≤t |∆(n)

s | <∞ for all n ∈ N implies∑

s≤t∆(n)

s ↑

∑

s≤t∆s ∈ R ∪ ∞.

The last statement is useful to show right-continuity and the existence of left limits of

sums of Poisson point processes, such as in 4.(c)

Assignment 3 – MS3b Levy Processes and Finance – Oxford HT 2008 VII

A.3 Construction of Levy processes

1. Let (∆t)t≥0 be a Poisson point process with intensity function g(x) = xκe−x for aparameter κ ∈ R.

(a) Let κ ∈ (−1,∞). Show that

Ct =∑

s≤t

∆s

is a compound Poisson process. Specify its jump rate λ and jump density h.

(b) Let κ ≤ −1. Show that ∆(n)

t = ∆t1∆t>1/n, t ≥ 0, is a Poisson point process.Specify its intensity function. Show that

C(n)

t =∑

s≤t

∆(n)

s

is a compound Poisson process.

(c) For C(n)

t as defined in (b), show that C(n)

t converges to a limit → Ct < ∞ asn→ ∞ if and only if κ > −2. Specify the moment generating function of Ct.

(d) Show that for κ > −2, we have

sups≤t

|C(n)

s − Cs| = |C(n)

t − Ct|.

Deduce that C(n)

t → Ct a.s. (or in probability) locally uniformly.

(e) Show that C(n)

t −E(C(n)

t ) converges for κ > −3. Show that the limit is a Levyprocess and that it has unbounded variation.

2. Let (Xt)t≥0 be a stable subordinator in the sense that (c1/αXt/c)t≥0 ∼ X for allc > 0 (scaling relation) for some α ∈ R.

(a) Show that for all µ ≥ 0 and t ≥ 0, we have E(e−µXt) ∈ (0, 1]. DenoteΦt(µ) = − ln(E(e−µXt)) and Φ = Φ1.

(b) Show that Φt(µ) = tΦ(µ) for all t ≥ 0. Deduce from the scaling relation thatΦ(µ) = Φ(1)µα for all µ ≥ 0.

(c) Show that ∂∂µ

E(e−µY ) ≤ 0 and ∂2

∂µ2 E(e−µY ) ≥ 0 for any nonnegative random

variable Y and for all µ > 0 with equality if and only if P(Y = 0) = 1. Deducethat α ∈ (0, 1] or P(X ≡ 0) = 1.

(d) By letting µ ↓ 0 in (c), show that E(Xt) = ∞ for all t > 0 and α ∈ (0, 1).

(e) For α ∈ (0, 1), calculate g : (0,∞) → (0,∞) such that

Φ(µ) =

∫ ∞

0

(1 − e−µx)g(x)dx.

(f) For every α ∈ (0, 1] and Φ(1) = b > 0, show that there exists a stable subor-dinator (Xt)t≥0.

VIII Assignments – MS3b Levy Processes and Finance – Oxford HT 2008

3. (a) Let (Xt)t≥0 and (Yt)t≥0 be independent stable subordinators with commonindex α and intensities bX = ΦX(1) and bY = ΦY (1). Show that Z = X − Yis also a stable process with index α in the sense that (c1/αXt/c)t≥0 ∼ X forall c > 0.

(b) Let H be a real-valued random variable with symmetric distribution, i.e. H ∼

−H . Show that E(eiλH) ∈ R for all λ ∈ R. Hint: eix = cos(x) + i sin(x).

(c) In the setting of (a), for the special case bX = bY , show that E(expiλZt) =exp−bX |λ|α, λ ∈ R. Hint: Show that Zt is symmetric and that all symmetric

stable processes have a characteristic function of this form for some b > 0. You

may assume without proof that all characteristic functions are continuous in

λ ∈ R and that those of infinitely divisible distributions have no zeros.

(d) Fix ˜b > 0. Show that for α ∈ (0, 2), the function

˜ψ(λ) =

∫ ∞

−∞(cos(λx) − 1)˜b|x|−α−1dx

has the property ˜ψ(λc1/α) = c˜ψ(λ) for all c > 0 and λ ∈ R. Deduce that

˜ψ(λ) = b|λ|α

for some b ∈ R.

(e) Using (d) or otherwise, show that a symmetric stable process R of index α ∈

(0, 2] has bounded variation if and only if α ∈ (0, 1), and deduce from theprevious parts of the exercise that it can then be written as a difference of twostable subordinators. Show that E(Rt) exists if and only if α ∈ (1, 2], and thatthen Var(Rt) <∞ if and only if α = 2.

Warning: Densities of stable processes are only known in closed form for some special

cases α ∈ 1/2, 1, 2. It is known, however, that they all have smooth probability density

functions.

You may use results from the lectures and previous assignment sheets without proof if

you state them clearly, except that the Levy density g of stable processes is to be derived

here, and its form should not be assumed.

Question A.3.1 is the most relevant on this sheet for MSc MCF students – if pushed

for time, please focus on this one.

Assignment 4 – MS3b Levy Processes and Finance – Oxford HT 2008 IX

A.4 Simulation

1. Let U ∼ Unif(0, 1) and F : R → [0, 1] right-continuous (weakly) increasing withF (−∞) = 0 and F (∞) = 1. Define F−1(u) = infx ∈ R : F (x) > u ∈ [−∞,∞]for u ∈ [0, 1]. Show that F−1(U) is a random variable with cumulative distributionfunction F .

2. (a) Let (Xt)t≥0 be a Gamma process with Xt ∼ Gamma(t, 1) for all t > 0. Con-sider A = Xa and B = Xa+b − Xa for some a > 0 and b > 0. Show thatR = A/(A + B) and S = A + B are independent and that R ∼ Beta(a, b),where Gamma and Beta densities are recalled as follows:

fS(s) =sa+b−1e−s

Γ(a + b), s ∈ (0,∞), fR(r) =

Γ(a + b)

Γ(a)Γ(b)ra−1(1 − r)b−1, r ∈ (0, 1).

Deduce, vice versa, the distribution of (SR, S(1 − R)) for independent S ∼

Gamma(1, c) and R ∼ Beta(cp, c(1 − p)), for some c > 0 and p ∈ (0, 1).

(b) Let U ∼ Unif(0, 1) and a > 0. Show that X = U1/a∼ Beta(a, 1).

(c) Let U ∼ Unif(0, 1) and V ∼ Unif(0, 1) be independent and a ∈ (0, 1). Calcu-late for Y = U1/a and Z = V 1/(1−a)

P

(

Y

Y + Z≤ t, Y + Z ≤ 1

)

and deduce that the conditional distribution of W = Y/(Y +Z) given Y +Z ≤ 1is Beta(a, 1− a). Hint: Write both inequalities as constraints on Z to find the

bounds when writing the probability as a double integral.

(d) In the setting of (c), show that the conditional distribution of TW givenY + Z ≤ 1, for an independent T ∼ Exp(1) = Gamma(1, 1) random variable,is Gamma(a, 1).

(e) Consider the following procedure due to Johnk. Let a ∈ (0, 1).

1. Generate two independent random numbers U ∼ Unif(0, 1) and V ∼

Unif(0, 1).

2. Set Y = U1/a and Z = V 1/(1−a).

3. If Y + Z ≤ 1 go to 4., else go to 1.

4. Generate an independent C ∼ Unif(0, 1) and set T == ln(C).

5. Return the number TY/(Y + Z).

What is this procedure doing? Explain its relevance for simulations.

3. (a) In the light of the previous exercise, explain how you can generate a Beta(a, b)random variable from a sequence of Unif(0, 1) random variables, for any a > 0and b > 0. Hint: Consider a ∈ (0, 1) first and use the additivity of Gamma

variables to generate Gamma(a, 1) variables, from which the Beta variable can

be constructed.

X Assignments – MS3b Levy Processes and Finance – Oxford HT 2008

(b) Consider the following method to generate a Gamma process on the timeinterval [0, 1]. Set X0 = 0 and generate X1 ∼ Gamma(1, 1). For n ≥ 0, ifyou have generated Xk2−n, k = 0, . . . , 2n, generate Bk,n ∼ Beta(an, bn) and setX(2k−1)2−n−1 = X(k−1)2−n + Bk,n(X(k+1)2−n − Xk2−n), 1 ≤ k ≤ 2n. For whatchoices of an > 0 and bn > 0 does this procedure yield Gamma distributionsfor all Xk2−n , and for what choice do you get stationary increments? Hint:

an = bn = 2−n−1 works, but are there other choices?

(c) What are the advantages of this method when compared with the plain versionof the time discretisation method (Method 1)?

4. Consider a variant of the Variance Gamma process of the form Vt = at + Gt − Ht

where a ∈ R, G1 ∼ Gamma(α+, β+) and H1 ∼ Gamma(α−, β−)

(a) For what values of a, α±, β± is V a martingale?

(b) Write out the steps needed to simulate Vt

• by Method 1 (using a random walk with increment distribution ∼ Vδ)

• by Method 1 (applied to G and H separately)

• by the refinement of Method 1 given in A.4.3

• by Method 2 (simulating the Poisson point process of jumps truncated atε)

(c) Carry out 9 simulations for a range of parameters α+ ∈ 1, 10, 100 and α− ∈

10, 100, 1000, β± = α2

±/2 and a such that V is a martingale. This part of

this question is optional.

Warning: The incomplete Gamma function Γt(a) =∫ t

0xa−1e−xdx cannot be simplified

into closed form (nor expressed in terms of the Gamma function), except for some special

values of a such as a ∈ N. There are, however, numerical procedures to evaluate Γt(a),which we will not address in this course.

If you have not used R, but would like to, you will find the “First steps with R” at

http://www.stats.ox.ac.uk/∼myers/stats materials/R intro/WA5 R.pdf

useful. Following are brief explanations of the commands used in the sample file

http://www.stats.ox.ac.uk/∼winkel/gammavgamma.R

This is a script file, which has to be run in the command window “R Console” to make

the new commands available, e.g. select “Run all” in the drop-down menu “Edit”.

• runif(n,a,b) generates an n-vector of uniform variables on [a, b].• qgamma(u,a,b) evaluates F−1(u) for the Gamma(a, b) inverse distribution function

F−1 at u. If u is a vector, qgamma is applied to each component.

• 1:n generates the vector (1, 2, . . . , n). Multiplication of vectors v by scalars a can

be written as a*v, similarly for addition and subtraction of vectors.

• plot(x,y,pch=".",sub=paste("text")) produces a scatter plot of pairs (xi, yi)for vectors x and y, with . marking the points, and text in the caption.

• psum <- function(vector)... defines a new command psum that takes a vector

vector as an argument. When this line is executed, the command is just made

available. To execute the command, type psum(v) for a vector v to get the partial

sums of v displayed, or s=psum(v) to create a new vector s containing the partial

sums of v.

Assignment 5 – MS3b Levy Processes and Finance – Oxford HT 2008 XI

A.5 Financial models

1. Consider a one-period model with three assets, a risk-free asset that increases fromA0 = 1 to A1 = eδ and two risky assets B and C that can each move up to or downfrom B0 = C0 = 1, so that there are four scenarios ω1 = (up, up), ω2 = (up, down),ω3 = (down, up) and ω4 = (down, down). Suppose that Bup

1= B1(ω1) = B1(ω2) >

B1(ω3) = B1(ω4) = Bdown

1and Cup

1= C1(ω1) = C1(ω3) > C1(ω2) = C1(ω4) =

Cdown

1. Assume w.l.o.g. that Bup

1< Cup

1.

(a) For a portfolio (T0, U0, V0) of T0 units of A, U0 units of B and V0 units of C,specify the value W0 and W1(ωi) of the portfolio at times 0 and 1, i = 1, 2, 3, 4.

(b) Show that this model is arbitrage-free if and only if B1(ω1) > A1 > B1(ω4) >C1(ω4).

Consider the arbitrage-free case now.

(c) Give an example of a contingent claim W1(ωi) that cannot be hedged.

(d) Show that contingent claims of the form W1(ω1) = W1(ω2), W1(ω3) = W1(ω4)can be hedged and priced as e−δ

E(W1), where you should specify

qB = P(B1 = Bup

1) and 1 − qB = P(B1 = Bdown

1).

In particular, e−δtBt, t = 0, 1, is then a martingale. Is qB unique?

(e) State the result analogous to (d) for contingent claims relating to C only ratherthan B only.

(f) Show that there are infinitely many possibilities to choose

p1 = P(ω1) = P(B1 = Bup

1, C1 = Cup

1), p2 = P(ω2), p3 = P(ω3) and p4 = P(ω4)

so that e−δtBt and e−δtCt, t = 0, 1 are martingales.

(g) Consider the contingent claim W1(ω1) = 1, W1(ω2) = W1(ω3) = W1(ω4) =0. Using the range of possibilities for p = (p1, p2, p3, p4), give the range ofproposed (arbitrage-free) prices W0 = e−δ

Ep(W1).

2. Consider Xt = Nt − µt for a Poisson process N of rate λ ∈ (0,∞) and a driftcoefficient µ ∈ (0,∞). Let

S(ε)n =

n∑

i=1

X(ε)i ,

where (X(ε)i )i≥1 is a sequence of independent random variables with common prob-

ability mass function given by

P(X(ε)i = 1 − µε) = 1 − e−λε =: pε and P(X

(ε)i = −µε) = 1 − pε.

(a) Show that S(ε)[t/ε] → Xt in distribution as ε ↓ 0. Hint: You can either prove this

directly or consider T(ε)n = S

(ε)n + nµε first.

XII Assignments – MS3b Levy Processes and Finance – Oxford HT 2008

(b) Show that the market model (eδεn, eS(ε)n )n≥0 is arbitrage-free if and only if

−µ < δ < 1/ε − µ and then also complete.

Consider the arbitrage-free case now.

(c) Show that the martingale probabilities are

qε = P(˜S(ε)1

= 1 − µε) =eε(δ+µ)

− 1

e − 1.

(d) Show that under the martingale probabilities

˜S(ε)t/ε →

˜Xt = Ñt − µt,

where ( Ñt)t≥0 is a Poisson process with rate (δ + µ)/(e − 1).

(e) Show that in the notation of part (d), the discounted process e−δtRt associated

with Rt = eeNt−µt is a martingale. By conditioning on Nt and Ñt, respectively,

explain briefly why the distribution of eeNt−µt can be seen as providing martin-

gale probabilities for eNt−µt.

(f) Using the following subsets of the set of right-continuous paths with left limits

Dµ = f ∈ D([0, 1], (0,∞)) : ∆ log f(t) = 1 or (log f)′(t) = −µ for all t ≤ 1,

show that (eeNt−µt)0≤t≤1 is the only exponential Levy process that has the

same set of possible paths as (eNt−µt)0≤t≤1 and whose discounted process is amartingale. Remark: In fact, it is the only process that has the same set of

possible paths as (eNt−µt)0≤t≤1 and whose discounted process is a martingale,

so the market model is complete.

Assignment 6 – MS3b Levy Processes and Finance – Oxford HT 2008 XIII

A.6 Time change and subordination

1. Consider Brownian motion (Bt)t≥0 and a continuous increasing function f : [0,∞) →[0,∞) with f(0) = 0. Set Zy = Bf(y), y ≥ 0.

(a) Show that Z has quadratic variation

[Z]y := p-limn→∞

[2ny]

∑

j=1

(Zj2−n − Z(j−1)2−n)2 = f(y),

where p-lim denotes a limit in probability of random variables.

(b) Assume that f is piecewise differentiable on [0,∞) with piecewise constantderivative σ2(s) := f ′(s), say taking values σ2

j on intervals [yj−1, yj) for some0 = y0 < y1 < . . . < yn < . . .. Let (Wy)y≥0 be a Brownian motion. Show thatthe process

˜Zy =

∫ s

0

σ(r)dWr :=

j∑

i=1

σi(Wyi− Wyi−1

) + σj+1(Wy − Wyj),

yj ≤ y < yj+1, has the same distribution as Z.

This result holds in fact for a very wide class of stochastic processes σ. This

is why both time-change models and models where the Brownian motion coef-

ficient varies are called stochastic volatility processes.

(c) Give an example of a Levy process and a function f as in (b) for which Xf(y)

does not have the same distribution as∫ y

0

√

f ′(s)dXs.

In fact, Brownian motion is the only process for which this holds for all such

functions f .

2. Let (Xt)t≥0 be a Levy process with probability density function ft and (τy)y≥0 asubordinator with characteristics (0, gτ) (sum of jumps, no compensation!). Define

g(z) =

∫ ∞

0

ft(z)gτ (t)dt, z ∈ R \ 0.

(a) In the case Var(X1) < ∞ and Var(τ1) < ∞, show that g satisfies the require-ments of a Levy density of a Levy process.

(b) In the case where either τ or X is compound Poisson, show that g also satisfiesthe requirements of a Levy density of a Levy process. More specifically, if X isa compound Poisson process with intensity λ, then we have P(Xt = 0) ≥ e−λt;

assume that, in fact, P(Xt = 0) = e−λt and that P(Xt ∈ (a, b)) =∫ b

aft(x)dx

for (a, b) 6∋ 0.

(c) If X is a stable process of index α > 1, show that g is the Levy density of abounded variation Levy process if

∫ ∞

0

x1/αgτ(x)dx < ∞.

XIV Assignments – MS3b Levy Processes and Finance – Oxford HT 2008

3. Show that all Levy processes X that can be obtained by subordination of Brownianmotion with an independent subordinator are symmetric in the sense X ∼ −X,but that not all symmetric Levy processes can be obtained in this way.

4. (a) Let C and D be two independent Gamma Levy processes with parameters αand

√

2λ for C1 ∼ D1. Let T be a Gamma process with parameters α and λ,and let B be an independent Brownian motion. Show that Xs − Ys ∼ BTs

.This result was mentioned in Question A.1.2. as an explanation for the name

Variance Gamma process.

(b) Let B be Brownian motion, S an independent stable subordinator with indexα ∈ (0, 1). Show that Rt = BSt

, t ≥ 0, is a stable process with index 2α.

(c) Write down procedures to simulate the processes in (a) and (b) using Method3 (Subordination).

5. Let X be a Levy process with E(X1) = µ and Var(X1) = σ2. Let S be an in-dependent subordinator with E(S1) = m and Var(S1) = q2. Denote Zt = XSt

,t ≥ 0.

(a) Show that E(Zt) = mµt, t ≥ 0.

(b) Show that Var(Zt) = (σ2m + µ2q2)t, t ≥ 0. Hint: Consider E(Z2

t ) first.

(c) Check this formula for the Variance Gamma process, using A.6.4.(a). For whatvalues of α and λ is E(Z1) = 0 and Var(Z1) = 1? Show that E(Z4

t ) = 3E(S2

t )and deduce the range of E(Z4

1) for these values of α and λ. Standardized fourth

moments (curtosis) give an indication of heavy tails. They reflect why Levy

processes such as the Variance Gamma process can better fit financial price

processes.

Assignment 7 – MS3b Levy Processes and Finance – Oxford HT 2008 XV

A.7 Level passage events

1. Let (Xs)s≥0 be a Levy process and X t = inf0≤s≤t Xs, t ≥ 0.

(a) For a fixed time t > 0, show that the process ( ˜Xs)0≤s≤t given by

˜X(t)s = Xt− − Xt−s−, 0 ≤ s ≤ t,

is a Levy process with the same distribution as (Xs)0≤s≤t.

(b) Show that this implies that for an independent random time τ with probability

density function fτ (x), x ∈ (0,∞), we have ( ˜X(τ)

s )0≤s≤τ ∼ (Xs)0≤s≤τ in thesense that for all 0 = s0 < s1 < . . . < sn < sn+1 = ∞ and 0 ≤ m ≤ n we have

P(Xs1 ∈ A1, . . . , Xsm∈ Am, τ ∈ [sm, sm+1) ∩ B)

= P( ˜X(τ)

s1∈ A1, . . . , ˜X(τ)

sm∈ Am, τ ∈ [sm, sm+1) ∩ B)

for all intervals A1, . . . , An ⊂ R and B ⊂ [0,∞).

(c) Using results and/or arguments from the lectures show that Xτ is independentof (Xτ − Xτ ) for an independent τ ∼ Exp(q).

(d) Suppose now that X has no positive jumps. Calculate the distribution of Xτ .

2. Let (Xt)t≥0 be an α-stable Levy process with no positive jumps for some α ∈ (1, 2],i.e. such that E(eγXt) = etcγα

. For α = 2 this is Brownian motion, for α ∈ (1, 2),we have a3 = 0 and g(x) = c|x|−α−1. For x ≥ 0 denote Tx = inft ≥ 0 : Xt > x.

(a) Using the strong Markov property of (Xt)t≥0 at Tx, show that (Tx)x≥0 is astable subordinator with index 1/α.

(b) Let Y have probability density function

fb(z) =b

√

2πz3e−b2/(2z), z > 0.

Calculate the distribution of aY and deduce that (fb)b≥0 is the family of den-sities of stable distributions on (0,∞) of index 1/2.

(c) Deduce that there is a constant c > 0 such that

∫ ∞

0

eγxfb(x)dx = e−cb√

γ.

In fact, c =√

2.

3. (a) Let A1, A2, . . . be identically distributed and Sn = A1 + . . .+An the associatedrandom walk. Let (Nm)m≥0 be an independent random walk. Denote themoment generating function of A1 by M(γ) = E(expγA1) and assume thatit is finite for γ ∈ (−ε, ε). Denote the probability generating function of N1

by G(s) = E(sN1). Show that Rm = SNm, m ≥ 0, is also a random walk (with

independent and identically distributed increments).

XVI Assignments – MS3b Levy Processes and Finance – Oxford HT 2008

(b) Let (Xt)t≥0 be a Levy process and (Ts)s≥0 an independent increasing Levyprocess. Show that Ys = XTs

, s ≥ 0, is also a Levy process.

(c) (i) Let (Bt)t≥0 be Brownian motion. For s ≥ 0 define Ts = inft ≥ 0 :Bt + bt > s, where b ≥ 0 is fixed. Using the strong Markov property atTs, show that (Ts)s≥0 is an increasing Levy process.

(ii) Show that expγBt−1

2γ2t is a martingale for all γ ∈ R. Use the Optional

Stopping Theorem to show that

E(expρTs) = exps(b −√

b2− 2ρ).

This distribution is called the inverse Gaussian distribution (note that BTs= s

means that s 7→ Ts is the right inverse of t 7→ Bt.) For an independent

Brownian motion (Xt)t≥0, the process Zs = XTs, s ≥ 0, obtained as in (b) has

the so-called Normal Inverse Gaussian (NIG) distribution. This is another

popular process to model financial price processes.

The last question is optional:

4. Let (Xt)t≥0 be standard Brownian motion with moment generating function E(eγXt) =etγ2/2. Denote X t = sup

0≤s≤t Xs.

(a) For y > 0 denote Ty = inft ≥ 0 : Xt = y. Use the strong Markov propertyto deduce that the process

X∗t =

Xt t ≤ Ty

2y − Xt t ≥ Ty

is a Brownian motion.

(b) Show that

P(Xt ≤ x, X t > y) = P(X∗t > 2y − x), y ∈ (0,∞), x ∈ (−∞, y).

and deduce that

fXt,Xt(x, y) =

2(2y − x)√

2πt3exp

−

(2y − x)2

2t

, y ∈ (0,∞), x ∈ (−∞, y).

(c) Show that

fTx(z) =

x√

2πz3e−x2/(2z), z > 0.

Feedback on the various topics and how you perceived them given your background (no

BS3a, no B10a, MScMCF etc.) will be most gratefully received: [email protected].

Appendix B

Solutions

B.1 Infinite divisibility and limits of random walks

1. (a) Recall that for independent A1 ∼ Gamma(α1, β) and A2 ∼ Gamma(α2, β) wehave A1 + A2 ∼ Gamma(α1 + α2, β). A quick proof can be given using mo-ment generating functions. The Gamma distribution has moment generatingfunction

E(expγA) =

∫ ∞

0

eγx βαxα−1

Γ(α)e−βxdx =

βα

(β − γ)α, γ < β.

We see that

E(expγ(A1 + A2)) = E(expγA1)E(expγA2) =βα1+α2

(β − γ)α1+α2

and recognise the moment generating function of the Gamma(α1 + α2, β)distribution. By the Uniqueness Theorem for moment generating functions,A1 + A2 ∼ Gamma(α1 + α2, β).

If we now choose Yn,1, . . . , Yn,n ∼ Gamma(α/n, β) independent, we obtain, byinduction in n, that Yn,1 + . . . + Yn,n ∼ Gamma(α, β). Since this holds forall n ≥ 1, a random variable Y ∼ Gamma(α, β) has an infinitely divisibledistribution.

(b) First calculate for B1, B2 ∼ geom(p) independent that

P(B1 + B2 = n) =

n∑

k=0

P(B1 = k, B2 = n − k) =

n∑

k=0

pk(1 − p)pn−k(1 − p)

= (n + 1)pn(1 − p)2,

and, e.g. by induction, for Am = B1 + . . . + Bm = Am−1 + Bm a negativebinomial distribution. Alternatively, consider independent Bernoulli trials un-til the mth success, then Am = n means there have been n failures and msuccesses, the m− 1 first successes chosen from the first n + m− 1 trials, andwe get

P(Am = n) =

(

n + m − 1

m − 1

)

pn(1 − p)m =(n + m − 1)!

(m − 1)!n!pn(1 − p)m

=Γ(n + m)

Γ(m)n!pn(1 − p)m.

XVII

XVIII Solutions – MS3b Levy Processes and Finance – Oxford HT 2008

This formula makes sense for m ∈ (0,∞), and we refer to this probability massfunction as NB(m, p). Then we calculate the probability generating functionfor A ∼ NB(m, p)

E(sA) =∑

n≥0

Γ(n + m)

Γ(m)n!(sp)n(1 − p)m =

(1 − p)m

(1 − sp)m, s ∈ [0, 1],

and if B ∼ NB(r, p) is independent, we obtain

E(sA+B) =(1 − p)m+r

(1 − sp)m+r,

the probability generating function of the NB(m + r, p) distribution, so weconclude by the Uniqueness Theorem for probability generating functions thatA + B ∼ NB(m + r, p).

If we now choose Yn,1, . . . , Yn,n ∼ NB(1/n, p) independent, we obtain, by in-duction in n, that Yn,1 + . . . + Yn,n ∼ NB(1, p) = geom(p). Since this holdsfor all n ≥ 1, a random variable Y ∼ geom(p) has an infinitely divisibledistribution.

(c) Assume that a random variable U ∼ Unif(0, 1) can be written as U = Y1 + Y2

for some independent and identically distributed Y1 and Y2. Then for x ∈ [0, 1],

1 − x = P(U ≥ x) ≥ P(Y1 ≥ x/2, Y2 ≥ x/2) ⇒ P(Y1 ≥ x/2) ≤√

1 − x

and

x = P(U ≤ x) ≥ P(Y1 ≤ x/2)2⇒ P(Y1 ≤ x/2) ≤

√

x.

For x = 1 and x = 0, respectively, we deduce P(Y1 ≥ 1/2) = 0 = P(Y1 ≤ 0).Now for x ∈ (0, 1/2)

x = P(U ≤ x) ≤ P(Y1 ≤ x, Y2 ≤ x) ⇐⇒ P(Y1 ≤ x) ≥√

x

and the inequality on the left is an equality if and only if the inequality on theright is an equality. Similarly,

x = P(U ≥ 1 − x) ≤ P(Y1 ≥ 1/2 − x)2⇐⇒ P(Y1 ≥ 1/2 − x) ≥

√

x

For x = 1/4, we get P(Y1 ≤ 1/4) ≥ 1/2 and P(Y1 ≥ 1/4) ≥ 1/2. If bothinequalities were equalities, we would deduce from the left-hand equalities thatP(Y1 ∈ (1/8, 3/8)) = 0 and this is incompatible with P(U ∈ (1/4, 3/8)) > 0,so the assumption that U = Y1 + Y2 must have been wrong.

2. (a) (i) Independence of increments. By the independence of increments of X andY and by the independence of X and Y we have for all 0 ≤ t0 < t1 <. . . < tn that the following random variables are all independent:

Xt0 , Xt1 − Xt0 , . . . , Xtn − Xtn−1 and Yt0 , Yt1 − Yt0 , . . . , Ytn − Ytn−1

Since functions of independent random variables are independent, we canadd take linear combinations and deduce independence of

aXt0 + bYt0 , a(Xt1−Xt0) + b(Yt1−Yt0), . . . , a(Xtn−Xtn−1) + b(Ytn−Ytn−1).

Solutions 1 – MS3b Levy Processes and Finance – Oxford HT 2008 XIX

(ii) Stationarity of increments. We have that Xt+s − Xt and Yt+s − Yt areindependent, and also that Xs and Ys are independent. By the stationarityof increments we have that Xt+s − Xt ∼ Xs and Yt+s − Yt ∼ Ys and sothe joint distributions of (Xt+s − Xt, Yt+s − Yt) is the same as the jointdistribution of (Xs, Ys). If we apply the same linear function to the randomvectors, these will also have the same distribution, i.e.

a(Xt+s − Xt) + b(Yt+s − Yt) ∼ aXs + bYs.

(iii) Right-continuity and left limits of paths. Linear combinations of suchfunctions still have these properties.

(b) We calculated the moment generating function of the Gamma(α, β) distribu-tion in Exercise 1 as

E(expγA) =

∫ ∞

0

eγx βαxα−1

Γ(α)e−βxdx =

βα

(β − γ)α, γ < β.

If C1 ∼ D1 ∼ Gamma(α,√

2µ), then Cs ∼ Ds ∼ Gamma(αs,√

2µ). Hence

E(eγ(Cs−Ds) = E(eγCs)E(e−γDs) =

√

2µαs

(√

2µ − γ)αs

√

2µαs

(√

2µ + γ)αs=

(

µ

µ −1

2γ2

)αs

for all −√

2µ < γ <√

2µ.

3. (a) Let Wn ∼ Binomial(n, pn) with npn → λ, then Wn → Poi(λ) in distributionas n → ∞. To prove this, check

E(sWn) =n∑

k=0

sk

(

n

k

)

pkn(1 − pn)n−k =

(

1 −

npn(1 − s)

n

)n

→ e−λ(1−s),

and this is the probability generating function of Poi(λ). By the UniquenessTheorem and by the Continuity Theorem for probability generating functions,Wn converges in distribution to a Poi(λ) distribution.

(b) Since pN is small, the Poisson limit theorem is appropriate, and since N islarge, it will give a reasonably good approximation. As parameter of thePoisson distribution, NpN is appropriate, since NpN → λ in the limit theoremfor a Poi(λ) limit.

(c) Denote by B1, . . . , BN the Bernoulli random variables so that Bj = 1 if policyholder j makes a claim. Then SN = B1 + . . . + BN ∼ Binomial(N, pN). Wecalculate the moment generating function

E(expγTN) = E

(

exp

γ

SN∑

j=1

Aj

)

=N∑

k=0

E

(

exp

γk∑

j=1

Aj

)

(

N

k

)

pkN(1 − pN)N−k

=N∑

k=0

(

E(

eγA1))k(

N

k

)

pkN(1 − pN )N−k

= (1 − pN + pNE(eγA1))N ,

by the binomial theorem, for all γ ∈ R for which E(eγA1) < ∞.

XX Solutions – MS3b Levy Processes and Finance – Oxford HT 2008

(d) First, we can continue the argument in (b) to suggest that the actual randomvariable

SN∑

n=1

An

has a distribution that is close to the distribution where SN is replaced by aPoisson random variable.

Then we formulate the limit statement. Suppose that SN ∼ Binomial(N, pN)for all N ∈ N, that S∞ ∼ Poi(λ) and that NpN → λ ∈ (0,∞) as N →

∞. Suppose that A1, A2, . . . are nonnegative, independent and identicallydistributed, independent of SN , N ∈ N ∪ ∞. Then

SN∑

n=1

An →

S∞

∑

n=1

An in distribution, as N → ∞.

To prove this, consider the moment generating functions

E(expγTN) =

(

1 −

NpN (1 − E(eγA1))

N

)N

→ exp

−λ(1 − E(eγA1))

,

and this is the moment generating function of the compound Poisson distri-bution, which we calculate as follows

E

(

exp

γ

S∞

∑

j=1

Aj

)

=

∞∑

k=0

E

(

exp

γ

k∑

j=1

Aj

)

λk

k!e−λ

=∞∑

k=0

(E (exp γA1))k λk

k!e−λ

= e−λ exp

λE(eγA1)

= exp

−λ(1 − E(eγA1))

.

4. (a) (i) Note that

∑nk=1

Ak − nE(A1)√

nVar(A1)=

n∑

k=1

Ak − µ

σ√

n=

n∑

k=1

Yn,k = Vn.

Thus, the Central Limit Theorem in terms of Vn states Vn → Normal(0, 1)in distribution as n → ∞.

(ii) Markov’s inequality P(|X| > y) ≤ E(X2)/y2 yields

P(|A1 − µ| > σx√

n) = P(|A1 − µ|1|A1−µ|≥σx√

n > σx√

n)

≤

E(|A1 − µ|21|A1−µ|≥σx√

n)

σ2x2n.

Now note that, as n → ∞,

E(

|A1 − µ|21|A1−µ|<σx√

n)

→ E(|A1 − µ|2) = σ2,

Solutions 1 – MS3b Levy Processes and Finance – Oxford HT 2008 XXI

(by monotone convergence) and so

γn(x) :=1

σ2x2E(|A1 − µ|21|A1−µ|≥σx

√n)

=1

σ2x2

(

σ2− E(|A1 − µ|21|A1−µ|<σx

√n))

→ 0.

(iii) For all x > 0, calculate using (ii)

P(Mn ≤ x) = P(|Yn,1| ≤ x, . . . , |Yn,n| ≤ x) = (P(|Yn,1| ≤ x))n

≥

(

1 −γn(x)

n

)n

→ e0 = 1.

This implies that P(|Mn| > ε) = 1 − P(|Mn| ≤ ε) → 0 for all ε > 0, soMn → 0 in probability.

(b) (i) At stage n there are r red balls and s + n − 1 black balls in the urn. So

Yn,k ∼ Bernoulli

(

r

r + s + n − 1

)

⇒ Wn ∼ Binomial (n, pn) ,

where pn = r/(r + s + n − 1). Note that npn → r, so that the Poissonlimit theorem yields Wn → Poi(r).

(ii) Clearly P(Yn,k = 0) = 1 − pn = 1 − r/(r + s + n − 1) → 1, as n → ∞.

(iii) Now, as n → ∞,

P(Mn = 0) = P(Yn,1 = 0, . . . , Yn,n = 0) = (1 − pn)n =(

1 −

npn

n

)n

→ e−r,

If Mn → 0, then P(|Mn| > ε) = 1− P(Mn = 0) → 0 for all 0 < ε < 1, andthis is incompatible with the limit above. So, Mn 6→ 0 in probability.

(c) (i) Define S(n)

k = Yn,1 + . . . + Yn,k, k ≥ 0, n ≥ 1.

Donsker’s theorem says in the setting of (a), where Vn = S(n)

n , that S(n)

[nt] →

Bt locally uniformly in distribution for a Brownian motion (Bt)t≥0.The process version of the Poisson limit theorem says in the setting of (b),

where Wn = S(n)

n , that S(n)

[nt] → Nt in the Skorohod sense in distribution

for a Poisson process (Nt)0≤t≤1 with rate r.

(ii) Clearly, the size of the biggest jump of Brownian motion is 0, and we haveMn → 0 in probability, hence also in distribution.The number of jumps of (Nt)0≤t≤1 is Poisson distributed with parameterr. The size J of the biggest jump of (Nt)0≤t≤1 is 1 if there is a jump, withprobability P(J = 1) = 1−e−r, and P(J = 0) = e−r is the probability thatthere is no jump. This is the limit distribution that we wish to establish.We have shown that

P(Mn = 0) → e−r = P(J = 0)

and this implies P(Mn = 1) = 1 − P(Mn = 0) → 1 − e−r = P(J = 1), asrequired.

XXII Solutions – MS3b Levy Processes and Finance – Oxford HT 2008

B.2 Poisson counting measures

1. (a) The distribution of Π is specified in terms of the associated counting measure

N((a, b]) = #Π ∩ (a, b] = j ≥ 1 : a < Tj ≤ b = Xb −Xa, 0 ≤ a < b.

Clearly, N satisfies property hom(b) of a Poisson counting measure: N((a, b]) =Xb − Xa ∼ Poi(λ(b − a)) by the stationarity (ii) and Poisson (iv) propertiesfor increments of X, and we identify the constant intensity function λ(t) = λ,t ≥ 0.

N also satisfies (a), since for disjoint intervals (aj, bj ], j = 1, . . . , n, we haveN((aj , bj]) = Xbj − Xaj

increments of X over disjoint time intervals. Byproperty (i) of the Poisson process, these are independent, as required.

(b) (i) Let 0 ≤ t0 < t1 < . . . < tn. Then Xtj − Xtj−1= N((tj−1, tj]). Since the

sets Aj = (tj−1, tj], j = 1, . . . , n, are disjoint, property (a) of the Poissoncounting measure yields the independence of the increments.

(ii) Fix r ≥ 0. For an increment Xs+r−Xs = N((s, s+r]), property inhom(b) ofthe Poisson counting measure yields a Poisson distribution with parameterpr(s) =

∫ s+r

sλ(x)dx. The differentiable function s 7→ pr(s) is constant if

and only if 0 = p′r(s) = λ(s+ r) − λ(s) for all s ≥ 0.Now (Xt)t≥0 has stationary increments if and only if s 7→ pr(s) is constantfor all r ≥ 0 if and only if λ(s) = λ(r+ s) for all r ≥ 0, s ≥ 0. This is thecase if and only if x 7→ λ(x) is constant.

(iii) Clearly t 7→ Xt is an increasing function, so all left and right limits ex-ists. Denote by Π the associated spatial Poisson process, then Π = t ≥0 : ∆Xt > 0 = t ≥ 0 : ∆Xt = 1. The set Π cannot have accu-mulation points since λ is locally integrable, so Π = Tj , j ≥ 1 andXt = N([0, t]) = j for t ∈ [Tj, Tj+1) is right-continuous at jump times,continuous elsewhere.

(iv) Xt −Xs = N((s, t]) = Poi(∫ t

sλ(x)dx), by property inhom(b) of the Poisson

counting measure.

(v) P(T1 > s) = P(N([0, s]) = 0) = exp−∫ s

0λ(x)dx for all s ≥ 0.

(vi) The density of T1 is obtained by differentiating the survival function:

fT1(s) = λ(s) exp

−

∫ s

0

λ(x)dx

.

To calculate the joint distribution of (T1, T2 − T1), first calculate the jointdistribution of (T1, T2), from

P(T1 > s, T2 > t) = P(N([0, s]) = 0, N((s, t]) ≤ 1)

= exp

−

∫ s

0

λ(x)dx

(

1 +

∫ t

s

λ(x)dx

)

exp

−

∫ t

s

λ(x)dx

and differentiation, first with respect to s then with respect to t

fT1,T2(s, t) = λ(s)λ(t) exp

−

∫ t

0

λ(x)dx

Solutions 2 – MS3b Levy Processes and Finance – Oxford HT 2008 XXIII

and the transformation formula for (T1, T2) 7→ (T1, T2 − T1) gives

fT1,T2−T1(s, r) = λ(s)λ(s+ r) exp

−

∫ s+r

0

λ(x)dx

and then

fT2−T1|T1=s(r) = λ(s+ r) exp

−

∫ s+r

s

λ(x)dx

⇒ P(T2 − T1 > r|T1 = s) = exp

−

∫ s+r

s

λ(x)dx

is independent of s for all r ≥ 0 if and only x 7→ λ(x) is constant, by theargument given in (ii).

2. For simplicity think of Π as a spatial Poisson process in R3 with intensity function

λ(x, y, z) = λ if x2 + y2 + z2≤ 1, λ(x, y, z) = 0 otherwise. We check the properties

(a) and inhom(b) of a spatial Poisson process. Denote by N and NP the associatedcounting measures of Π and P . Then note that

NP ((a, b] × (c, d]) = N((a, b] × (c, d] × R) ∼ Poi

(∫ b

a

∫ d

c

∫

R

λ(x, y, z)dzdydx

)

and we see that the intensity function of P will have to be

λP (x, y) =

∫

R

λ(x, y, z)dz =

∫

√

1−x2−y2

−√

1−x2−y2λdz = 2λ

√

1 − x2− y2

for (x, y) ∈ R2 such that x2 + y2

≤ 1, and λP (x, y) = 0 otherwise.

Property (a) also holds since for disjoint (aj , bj ] × (cj , dj], j = 1, . . . , n, the sets(aj, bj ]× (cj , dj]×R are also disjoint, and so independence of NP ((aj, bj ]× (cj , dj]),j = 1, . . . , n, follows from the corresponding property N .

3. (a) First note that eΨ(γ) = E(eγX1) implies that E(eγX1/m) = eΨ(γ)/m since sta-tionarity and independence of increments implies E(eγX1/m)m = eΨ(γ), thenE(eγXq ) = eqΨ(γ), and then the right-continuity of sample paths implies thatXq → Xt almost surely and hence also in distribution, as q ↓ t. Therefore,characteristic functions converge and E(eγXq ) = eqΨ(γ)

→ etΨ(γ).

Now we use the independence and stationarity of increments to see

E(expγXt|Fs) = expγXsE(expγ(Xt −Xs))

= expγXs exp(t− s)Ψ(γ).

(b) The argument in (a) applies, with γ = iλ and ψ instead of Ψ as appropriate.Recall that moment generating functions do not exist for all random variables,but characteristic functions always exist (because x 7→ eiλx is bounded).

XXIV Solutions – MS3b Levy Processes and Finance – Oxford HT 2008

(c) The following argument can more easily be carried out for moment generatingfunctions, but applies more generally if done for characteristic functions.

Differentiate E(expiλXt) = e−tψ(λ) with respect to λ at λ = 0 to getiE(Xt) = −tψ′(0) (see Grimmett-Stirzaker 5.7 for a statement and referenceto the proof). The claim follows since µ = E(X1) must now be the slope ofthis linear function.

Now, we use the independence and stationarity of increments to see

E(Xt − tµ|Fs) = E(Xs + (Xt−Xs) − tµ|Fs) = Xs + (t− s)µ− tµ = Xs − sµ.

(d) Differentiate E(expiλXt) = e−tψ(λ) twice with respect to λ at λ = 0 toget −E(X2

t ) = −t(ψ′′(0) − t(ψ′(0))2), so Var(Xt) = tψ′′(0), where now σ2 =Var(X1) = ψ′′(0).

Now we use the independence and stationarity of increments to see

E((Xt − tµ)2|Fs) = E((Xs − sµ)2 + 2(Xs − sµ)(Xt −Xs − (t− s)µ)

+(Xt −Xs − (t− s)µ)2|Fs)

= (Xs − sµ)2 + 2(Xs − sµ)E(Xt −Xs − (t− s)µ)

+Var(Xt −Xs)

= (Xs − sµ)2 + (t− s)σ2.

4. (a) Fix β > 0. Note that the formula reduces to 0 = 0 for γ = 0. It is thereforesufficient to show that the γ-derivatives of both sides coincide. To differentiatethe left hand side, note that

∂

∂γ(eγx − 1)

1

xe−βx = eγxe−βx ≤ e−βx,

for γ ≤ 0, where x 7→ e−βx is integrable on [0,∞). Therefore, we may in-terchange γ-differentiation and x-integration and have to show that for allγ < 0

∫ ∞

0

eγxe−βxdx =1

1 − γ/β

1

β,

which clearly is true.

The argument works for γ ≤ γ0 if we choose e−(β−γ0) as integrable upperbound. Clearly, for every fixed γ < β, any γ0 ∈ (γ, β) will do.

(b) We apply the exponential formula for Poisson point processes and (a) to obtain

E

(

exp

γ∑

s≤t∆s

)

= exp

t

∫ ∞

0

(eγx − 1)αx−1e−βxdx

=

(

β

β − γ

)αt

.

We recognise the last expression as the moment generating function of theGamma distribution with the required density. By the Uniqueness Theoremfor moment generating functions,

∑

s≤t ∆s has this Gamma distribution.

Solutions 2 – MS3b Levy Processes and Finance – Oxford HT 2008 XXV

(c) Fix 0 ≤ t0 < t1 < . . . < tn. Since (∆s)s≥0 is a Poisson point process, theprocesses (∆s)tj−1<s≤tj , j = 1, . . . , n, are independent (consider the restrictionsto disjoint domains (tj−1, tj ] × (0,∞) of the Poisson counting measure

N((a, b] × (c, d]) = a < t ≤ b : ∆t ∈ (c, d], 0 ≤ a < b, 0 < c < d),

and so are the sums∑

tj−1<s≤tj ∆s as functions of independent random vari-

ables. Fix s < t. Then the process (∆s+r)r≥0 has the same distribution as(∆s)s≥0. In particular,

∑

0≤r≤t ∆s+t ∼∑

0≤r≤t ∆r. The process t 7→∑

s≤t ∆s

is right-continuous with left limits, since it is a random increasing functionwhere for each jump time T , we have (by monotone convergence)

limt↑T

∑

s≤t∆s =

∑

s<T

∆s and limt↓T

∑

s≤t∆s =

∑

s≤T∆s.

5. (a) Denote by Tn ∼ Gamma(n, λX) and T ′m ∼ Gamma(m,λY ) the jump times of

X and Y . These are independent continuously distributed random variablesand so P(Tn = T ′

m) = 0. Therefore, (by subadditivity)

P(Tn, n ≥ 1 ∩ T ′m, m ≥ 1 6= ∅) ≤

∑

m≥1

∑

n≥1

P(Tn = T ′m) = 0.

(b) Denote the Poisson arrival processes of jumps by RX and RY . Then RX +RY

satisfies the four properties of the Poisson process, since (i) RXtj−RX

tj−1+RY

tj−

RYtj−1

, j = 1, . . . , n, are independent as sums of independent random variables,(ii)/(iv) their distributions are Poi(λX(tj − tj−1) + λY (tj − tj−1) as sum oftwo independent Poisson variables only depending on tj − tj−1, (iii) paths areright-continuous with left limits as sums of two such paths.

(c) We condition on whether T1 < T ′1

or T ′1< T1 and get for the first jump size

JD1

of D

P(JD1

∈ A) = P(T1 < T ′1)P(JX

1∈ A|T1 < T ′

1) + P(T1 > T ′

1)P(−JY

1∈ A)

=λX

λX + λYP(JX

1∈ A) +

λYλX + λY

P(−JY1∈ A).

This is a mixture of the jump size distributions of X and Y . We deduce thatthe density is

hD(x) =λX

λX + λYhX(x) +

λYλX + λY

hY (−x) =

λX

λX+λYhX(x) x > 0

λY

λX+λYhY (−x) x < 0

(d) This is bookwork, see Lecture 3, Example 18. The intensity function isλXhX(x), x > 0.

(e) By the previous part, we have two Poisson point process ∆X and ∆Y in (0,∞).It is easy to see that ∆−Y = −∆Y is a Poisson point process in (−∞, 0) withintensity function λY hY (−x), x < 0. It is easy to see that the associatedPoisson counting measures on [0,∞) × (0,∞) and [0,∞) × (−∞, 0) togetherform a Poisson counting measure on [0,∞) × R \ 0 via

N(A× B) = NX(A× (B ∩ (0,∞))) +NY (A× (B ∩ (−∞, 0))).

The intensity function is λXhX(x), x > 0 and λY hY (−x), x < 0.

XXVI Solutions – MS3b Levy Processes and Finance – Oxford HT 2008

(f) Since (∆Dt)t≥0 is a Poisson point process and Dt =∑

s≤t ∆Ds, D is a com-pound Poisson process.

(g) For every real-valued compound Poisson process C we can define the pro-cesses X and Y of positive and negative jumps. Since the associated processes(∆Xt)t≥0 and (∆Yt)t≥0 inherit the properties of Poisson point processes (viatheir Poisson counting measures), this provides the required decompositioninto two independent increasing compound Poisson processes. It is uniquebecause any other decomposition must have more jumps, which must happenat the same time and cancel each other, but by (a), this is incompatible withindependence.

Solutions 3 – MS3b Levy Processes and Finance – Oxford HT 2008 XXVII

B.3 Construction of Levy processes

1. (a) If κ ∈ (−1,∞), then

∫ ∞

0

g(x)dx =

∫ ∞

0

xκe−xdx = Γ(κ+ 1) <∞.

The Poisson point process is hence of the form of Example 18 and so (Ct)t≥0

is a compound Poisson process with intensity Γ(κ + 1) and Gamma(κ + 1, 1)jump distribution with density

h(x) =1

Γ(κ+ 1)xκe−x, x > 0.

(b) The counting measures associated to (∆t)t≥0 and (∆(n)

t )t≥0 are

N((a, b] × (c, d]) = #t ∈ (a, b] : ∆t ∈ (c, d]

∼ Poi

(

(b− a)

∫ d

c

g(x)dx

)

, 0 ≤ a < b, 0 < c < d,

Nn((a, b] × (c, d]) = N((a, b] × ((c, d] ∩ (1/n,∞))

∼ Poi

(

(b− a)

∫ d

c

g(x)1x>1/ndx

)

, 0 ≤ a < b, 0 < c < d.

Nn inherits the properties of a Poisson counting measure from N . We readoff the intensity function gn(x) = g(x), x > 1/n, gn(x) = 0, x ≤ 1/n. The

argument of (a) shows that C(n)

t is a compound Poisson process.

(c) C(n)

t increases as n → ∞. We can study the limit of moment generatingfunctions, whether or not the limit is finite. We get, as n→ ∞,

E(eγC(n)t ) = exp

∫ ∞

1/n

(eγx− 1)g(x)dx

↓ exp

∫ ∞

0

(eγx− 1)g(x)dx

and because for γ < 0

∫ ∞

0

(eγx− 1)g(x)dx <∞ ⇐⇒

∫ ∞

0

(1 ∧ x)g(x)dx <∞,

and by Lemma 21, we need to investigate the right hand condition. We checkthat

∫ ∞

1

g(x)dx <∞, and

∫

1

0

xg(x)dx <∞ ⇐⇒ κ+ 1 > −1,

as required.

(d) We can write

Cs − C(n)

s =∑

r≤s

∆r1∆r≤1/n ≤∑

r≤t

∆r1∆r≤1/n = Ct − C(n)

t ,

XXVIII Solutions – MS3b Levy Processes and Finance – Oxford HT 2008

and putting a supremum over s ≤ t on the left hand side, we get the requiredestimate (as an equality because we can take s = t on the left. Now we showed

in (c) that C(n)

t → Ct a.s., and so we deduce here that

sups≤t

|C(n)

s − Cs| → 0 as n→ ∞,

i.e. that the convergence is locally uniform.

(e) By Proposition 40(ii), we have for m ≤ n

E(|C(n)

t − E(C(n)

t ) − (C(m)

t − E(C(m)

t ))|2) = Var(C(n)

t − C(m)

t )

=

∫

1/m

1/n

x2g(x)dx,

and this decreases to zero as n ≥ m → ∞ if and only if∫

1

0x2g(x)dx < ∞,

i.e. κ > −3. In this case, (C(n)

t − E(C(n)

t ))n≥1 is a Cauchy sequence thatconverges by completeness of R (and the associated L2 space of R-valuedrandom variables).

The limiting process includes all jumps (∆s)s≤t (intuitively, a more formalargument uses uniform convergence that preserves the jumps, see Theorem42), and by (c), these are not summable for κ ∈ (−3,−2]. By Proposition 36,the limiting process has unbounded variation.

2. (a) Just note that for subordinators 0 ≤ Xt < ∞ a.s., and this implies that1 ≥ e−µXt > 0 a.s. and then also 1 ≥ E(e−µXt) > 0 as required. Therefore, Φt

is well-defined.

(b) This follows as in A.2.3, first for rational t ≥ 0 and then, by right-continuityof paths and since a.s. convergence implies convergence in distribution, henceof moment generating functions. The scaling relation for fixed t translates to

Φt/c(c1/αµ) = − ln(E(exp−µc1/αXt/c)) = − ln(E(e−µXt)) = Φt(µ).

and therefore, for t = 1, c = µ−α, as required

µαΦ(1) = Φ1/c(1) = Φ(µ).

(c) Clearly µ 7→ e−µXt is a.s. decreasing and so is hence µ 7→ E(e−µXt), strictlydecreasing if Xt > 0 with positive probability. Now, Φ(µ) = Φ(1)µα is clearlydifferentiable for µ > 0, and so

∂

∂µE(e−µXt) =

∂

∂µe−tΦ(1)µα

= −tΦ(1)αµα−1e−tΦ(1)µα

and this is negative only for α > 0 (or α = 0 but then µ 7→ E(e−µXt) isconstant). To show that also α ≤ 1 note that µ 7→ e−µXt is also a.s. convex,and hence so is µ 7→ E(e−µXt). Now, Φ(µ) is also twice differentiable so that

∂2

∂µ2E(e−µXt) = tΦ(1)αµα−2e−tΦ(1)µα

(tΦ(1)αµα− (α− 1)),

and this is nonnegative for all µ > 0 if and only if α ≤ 1.

Solutions 3 – MS3b Levy Processes and Finance – Oxford HT 2008 XXIX

(d) Note that, (by monotone convergence), as µ ↓ 0,

tΦ(1)αµα−1e−tΦ(1)µα

= E(Xte−µXt) ↑ E(Xt),

where the left-hand side increases to ∞ for α ∈ (0, 1).

(e) Note that Φ(0) = 0 implies that the equation holds for µ = 0 no matter whatg is. Now differentiate both sides with respect to µ to get

Φ(1)αµα−1 =

∫ ∞

0

e−µxxg(x)dx.

Remember that the density of the Gamma(1 − α, µ) distribution is f(x) =(Γ(1−α))−1µ1−αx−αe−µx. Therefore, we can (and have to, by the UniquenessTheorem for moment generating functions) take

g(x) =Φ(1)α

Γ(1 − α)x−α−1, x > 0.

(f) For α ∈ (0, 1), the Construction Theorem for subordinators (Theorem 8) showsthat we can construct the stable subordinator from a Poisson point processwith intensity function g as specified in (e). Note that g satisfies the integra-bility condition

∫ ∞

0

(1 ∧ x)g(x)dx <∞

since x−α−1 is integrable at x = ∞ and x−α is integrable at x = 0.

For α = 1 note that Φt(µ) = Φ(1)tµ. The associated subordinator is thedeterministic drift Xt = Φ(1)t.

3. (a) Just note that

c1/αZt/c = c1/αXt/c − c1/αYt/c ∼ Xt − Yt = Zt

for fixed t, and that, as processes in t ≥ 0, both the left-hand side and theright-hand side are Levy processes. Therefore, the distributions as processescoincide.

(b) H ∼ −H implies

E(cos(λH))+iE(sin(λH)) = E(eiλH) = E(e−iλH) = E(cos(λH))−iE(sin(λH))

and so the imaginary part E(sin(λH)) must vanish for all λ ∈ R.

(c) Clearly Zt = Xt − Yt ∼ Yt − Xt = −Zt, so Zt has a symmetric distribution.By (b), its characteristic function ϕt(λ) = E(eiλZt) is real-valued. By the hint,we may assume that ϕt is continuous, and since Zt is infinitely divisible, thatϕt(λ) 6= 0, so it must stay positive everywhere (note that ϕ(0) = 1). Define

ψt(λ) = − ln(ϕt(λ)), ψ(λ) = ψ1(λ), λ ∈ R.

XXX Solutions – MS3b Levy Processes and Finance – Oxford HT 2008

By A.2.3.(b), we have ψt(λ) = tψ(λ). The scaling relation implies

ψt/c(c1/αλ) = − ln(E(expiλc1/αZt/c)) = − ln(E(eiλZt)) = ψt(λ),

and as in 2.(b), this implies ψ(λ) = ψ(1)λα for all λ ≥ 0. For λ < 0 note that

ψ(λ) = − ln(E(eiλZt)) = − ln(E(e−iλZt)) = ψ(−λ),

so we have ψ(λ) = ψ(1)|λ|α.

(d) Before we start, note that the integral defining ˜ψ(λ) converges for α ∈ (0, 2)since the integrand behaves like x1−α at x = 0 and like x−α−1 at |x| = ∞. Wethen check, by change of variables y = c1/αx (hence x−1dx = y−1dy, that

˜ψ(λc1/α) =

∫ ∞

−∞(cos(λc1/αx) − 1)˜b|x|−α−1dx

=

∫ ∞

−∞(cos(λy) − 1)˜bc|y|−α−1dy = c ˜ψ(λ).

The argument of (c) shows that this implies ˜λ = b|λ|α for some b ≥ 0 – theargument did not depend on α ∈ (0, 1).

(e) Let R be a symmetric stable process R of index α. In (d) we expressed thecharacteristic exponent in terms of the Levy density g(x) = |x|−α−1. Therefore,(Rt)t≥0 can be constructed from a Poisson point process of jumps with thisdensity. We see from the criterion

∫ ∞−∞(1 ∧ |x|)g(x)dx < ∞ that jumps are

absolutely summable if and only if α ∈ (0, 1). In that case, we expressed R asthe difference of two subordinators in (a), so R has indeed bounded variation.If α ∈ [1, 2), jumps are not absolutely summable, so variation is unbounded, byProposition 36. If α = 2, then R is a multiple of Brownian motion, and it wasshown in the lectures, that Brownian motion has also unbounded variation.We differentiate ψ(λ) in λ = 0 to see that E(Xt) = tψ′(0) < ∞ if and only ifα ∈ (1, 2], and that Var(Xt) = tψ′′(0) <∞ if and only if α = 2.

Solutions 4 – MS3b Levy Processes and Finance – Oxford HT 2008 XXXI

B.4 Simulation

1. Note that the definition F−1(u) = infx ∈ R : F (x) > u implies

F (t) > u ⇒ F−1(u) ≤ t ⇒ ∀ε>0F (t + ε > u) ⇒ F (t) ≥ u

Therefore for all t ∈ R

F (t) = P(F (t) > U) ≤ P(F−1(U) ≤ t) ≤ P(F (t) ≥ U) = F (t).

2. (a) Independent A ∼ Gamma(a, 1) and B ∼ Gamma(b, 1) have joint density

fA,B(x, y) =xa−1e−x

Γ(a)

yb−1e−y

Γ(b)

The transformation (R, S) = T (A, B) = (A/(A + B), A + B) is bijective T :(0,∞)2

→ (0, 1) × (0,∞) with inverse transformation (A, B) = T−1(R, S) =(SR, S(1 − R)) that has the Jacobian

J(r, s) =

(

s r−s 1 − r

)

⇒ | det(J(r, s))| = s

and so the transformation formula yields

fR,S(r, s) = | det(J(r, s))|fA,B(T−1(r, s)) = s(sr)a−1e−sr

Γ(a)

(s(1 − r))b−1e−s(1−r)

Γ(b)

=Γ(a + b)

Γ(a)Γ(b)ra−1(1 − r)b−1

sa+b−1e−s

Γ(a + b),

as required.

Vice versa, for c = a + b and p = a/(a + b), we recognise T−1(R, S) = (A, B),which has joint distribution

fA,B(x, y) =xa−1e−x

Γ(a)

yb−1e−y

Γ(b)=

xcp−1e−x

Γ(cp)

yc(1−p)−1e−y

Γ(c(1 − p)).

and so any random variable ( ˜R, ˜S) with joint distribution as (R, S) will be

such that T−1( ˜R, ˜S) ∼ T−1(R, S) = (A, B).

(b) P(X ≤ x) = P(U1/a≤ x) = P(U ≤ xa) = xa, x ∈ (0, 1) and so fX(x) = axa−1,

x ∈ (0, 1). We recognise X ∼ Beta(a, 1).

(c) We calculate

P

(

Y

Y + Z≤ t, Y + Z ≤ 1

)

= P(Y (1 − t)

t≤ Z ≤ 1 − Y )

=

∫ t

0

∫

1−y

y(1−t)/t

aya−1(1 − a)z−adzdy

=

∫ t

0

aya−1(

(1 − y)1−a− y1−a(1 − t)1−ata−1

)

dy.

XXXII Solutions – MS3b Levy Processes and Finance – Oxford HT 2008

We differentiate with respect to t to get

fW |Y +Z≤1(t) =at((1 − a)(1 − t)−ata−1

− (a − 1)(1 − t)1−ata−2)

P(X + Y ≤ 1)

=a(1 − a)ta−1(1 − t)−a((1 − t) + t)

P(X + Y ≤ 1)

and we recognise the density of Beta(a, 1−a), up to the normalisation constant,but we have calculated a conditional density which integrates to 1, so thenormalisation constant must be the one of Beta(a, 1 − a).

(d) Given Y + Z ≤ 1, W is Beta(a, 1 − a)-distributed. Since T is independent of(Y, Z, W ), its conditional distribution given Y +Z ≤ 1 is still Gamma(1, 1) andit is conditionally independent of W given Y + Z ≤ 1. Therefore P(TW ≤

h|Y + Z ≤ 1) = P(SR ≤ h), and we can apply (a) for c = 1 and p = ato deduce that, SR ∼ Gamma(a, 1), i.e. the conditional distribution of WTgiven Y + Z ≤ 1 is Gamma(a, 1).

(e) This procedure generates a Gamma(a, 1) random variable. Specifically, theconditioning on Y + Z ≤ 1 is realised by repeated trials until Y + Z ≤ 1, seeLemma 57. The procedure is easily implemented and gives a more efficient wayof simulating Gamma random variables from uniform random variables thaninverting the distribution function of the Gamma distribution numerically.

3. (a) From 2.(a) we take that we obain A/(A+B) ∼ Beta(a, b) for independent A ∼

Gamma(a, 1) and B ∼ Gamma(b, 1). Johnk’s procedure works for a ∈ (0, 1).To generate Gamma variables for higher parameters, we can write a = [a]+afor integer part and fractional part and then represent

A =

[a]∑

k=1

Ek + A0

where (Ek)1≤k≤[a] is a sequence of independent Exp(1) random variables andA0 ∼ Gamma(a, 1). To summarise, the following procedure generates aBeta(a, b) random variable:

1.-5. Run Johnk’s Gamma generator for parameter a. Set A0 = TY/(Y +Z).

6.-10. Independently of 1.-5., run Johnk’s Gamma generator for parameter b.Set B0 = TY/(Y + Z).

11. Generate independent U1, . . . , U[a]+[b] ∼ Unif(0, 1) and set

A = A0 − ln

[a]∏

k=1

Uk

and B = B0 − ln

[a]+[b]∏

k=[a]+1

Uk

.

12. Return the number A/(A + B).

(b) The procedure generates a stochastic process successively at refining latticesof dyadic times. The key step (for n = 0 and then inductively for n ≥ 1) isto take a 2−n-increment Yk,n = Xk2−n − X(k−1)2−n ∼ Gamma(2−n, 1) and a

Solutions 4 – MS3b Levy Processes and Finance – Oxford HT 2008 XXXIII

Bk,n ∼ Beta(an, bn) random variable to split Yk,n into two increments Bk,nYk,n

and (1 − Bk,n)Yk,n. For 2.(a) to apply, we need an + bn = 2−n, i.e. an = 2−npand bn = 2−n(1− p). In order to get identically Gamma(2−n−1, 1) increments,further p = 1/2. This now yields stationary independent increments.

This is the only way to achieve Gamma distributions for all Xk2−n (at leastin the framework of 2.(a), but in fact, in general) for Beta parameters notdepending on k. If we were to use Bk,n ∼ Beta(ak,n, bk,n) for parametersthat may depend on k, then we get more general processes with independentGamma increments that are not stationary, as soon as we have the consistencycondition that ak,n + bk,n is the parameter of Yk,n.

(c) Johnk’s Gamma generator is more efficient than the inverse distribution func-tion computation. The method is less liable to accumulating errors since time1 is most accurate and errors only accumulate along the dyadic expansions,i.e. with local rather than global impact. Furthermore, we get an iterativeprocedure for which we do not have to fix the time lag δ in advance, but cancontinue to fill in extra points until a satisfactory result is obtained.

4. (a) Since Gt ∼ Gamma(α+t, β+) and Ht ∼ Gamma(α−t, β−), we have

E(at + Gt − Ht) = at +α+t

β+

−

α−t

β−= 0 ⇐⇒ a =

α−β−

−

α+

β+

.

(b) • Denote Fδ(x) = P(Vδ ≤ x).

1. Set S0 = 0 and n = 1.

2. Generate Un ∼ Unif(0, 1).

3. Set Sn = Sn−1 + F−1

δ (Un). If enough steps have been performed, goto 4., otherwise increase n by 1 and go to 2.

4. Return (Sn)n≥0 as simulation of (Vδn)n≥0.

• Denote F (x; α, β) = P(G ≤ x) for G ∼ Gamma(α, β).

1. Set S0 = 0 and n = 1.

2. Generate two independent random numbers U2n−1 ∼ Unif(0, 1) andU2n ∼ Unif(0, 1).

3. Set Sn = Sn−1 + aδ + F−1(U2n−1; α+δ, β+) − F−1(U2n; α−δ, β−). Ifenough steps have been performed, go to 4., otherwise increase n by1 and go to 2.

4. Return (Sn)n≥0 as simulation of (Vδn)n≥0.

• Fix t = 1, iterate for further time units if needed. Denote G(x; a, b) =P(B ≤ x) for B ∼ Beta(a, b).

1. Set V0 = 0 and n = 0.

2. Generate 2 independent random numbers U1 ∼ Unif(0, 1) and U2 ∼

Unif(0, 1).

3. Set P1 = F−1(U1; α+, β+), N1 = F−1(U2; α−, β−) and V1 = a+P1−N1.

4. Generate 2n independent random numbers U2n+1+k ∼ Unif(0, 1), k =1, . . . , 2n.

5. Set Bn,k = G−1(U2n+1+k; 2−n−1α+, 2−n−1α+), k = 1, . . . , 2n−1 and

Cn,k = G−1(U2n+1+2n−1+k; 2−n−1α−, 2−n−1α−), k = 1, . . . , 2n−1

XXXIV Solutions – MS3b Levy Processes and Finance – Oxford HT 2008

6. Set P(2k−1)2−n = Bn,kP(2k−2)2−n + (1 − Bn,k)P(2k)2−n , N(2k−1)2−n =Cn,kN(2k−2)2−n + (1 − Cn,k)N(2k)2−n and V(2k−1)2−n = (2k − 1)2−na +P(2k−1)2−n − N(2k−1)2−n for k = 1, . . . , 2n−1. If the resolution is fineenough, go to 7., otherwise increase n by 1 and go to 4.

7. Return (Vk2−n)k=1,...,2n.

Instead of F−1 and G−1, one can use Johnk’s Gamma generator of A.3.2.and the associated Beta generator of A.3.3.

• Denote H(x; β) =∫ x

εy−1e−βydy/

∫ ∞ε

y−1e−βydy. Also denote

λ = α+

∫ ∞

ε

y−1e−β+ydy + α−

∫ ∞

ε

y−1e−β−

ydy

and p = λ−1α+

∫ ∞

ε

y−1e−β+ydy.

1. Set V0 = 0, T0 = 0 and n = 1.

2. Generate three independent random numbers U3n−2 ∼ Unif(0, 1) andU3n−1 ∼ Unif(0, 1) and U3n ∼ Unif(0, 1).

3. Set Zn = − ln(U3n)/λ.

4. If U3n−1 > p, let Jn = −H−1(U3n; β−), otherwise let Jn = H−1(U3n; β+).

5. Set Tn = Tn−1 + Zn and VTn= VTn−1 + aZn + Jn. If Tn is big enough,

go to 6., otherwise increase n by 1 and go to 2.

6. Return (VTn)n≥0.

(c) Below are 9 simulations for α+ ∈ 1, 10, 100 (rows) and α− ∈ 10, 100, 1000(columns). Note the big positive jumps for α+ = 1, the cases α+ = α− witha = 0 and convergence to Brownian motion from top left to bottom right. Thecode is a similar to the symmetric case and is available on the course website.

0 2 4 6 8 10

−0.5

0.00.5

1.01.5

2.02.5

Variance Gamma process with shape parameters 0.5 and 50 and scale parameters 1 and 10time

spac

e

0 2 4 6 8 10

−3−2

−10

1

Variance Gamma process with shape parameters 0.5 and 5000 and scale parameters 1 and 100time

spac

e

0 2 4 6 8 10

−1.5

−1.0

−0.5

0.00.5

1.0

Variance Gamma process with shape parameters 0.5 and 5e+05 and scale parameters 1 and 1000time

spac

e

0 2 4 6 8 10

−0.2

−0.1

0.00.1

Variance Gamma process with shape parameters 50 and 50 and scale parameters 10 and 10time

spac

e

0 2 4 6 8 10

−3−2

−10

12


spac

e

0 2 4 6 8 10

−10

12

3

Variance Gamma process with shape parameters 50 and 5e+05 and scale parameters 10 and 1000time

spac

e

0 2 4 6 8 10

−5−4

−3−2

−10


spac

e

0 2 4 6 8 10

−1.0

−0.5

0.00.5

1.01.5

2.0


spac

e

0 2 4 6 8 10

−4−3

−2−1

01

Variance Gamma process with shape parameters 5000 and 5e+05 and scale parameters 100 and 1000time

spac

e

Solutions 5 – MS3b Levy Processes and Finance – Oxford HT 2008 XXXV

B.5 Financial models

1. (a) W0 = T0 +U0 +V0, W1(ω1) = T0eδ +U0B

up

1+V0C

up

1, W1(ω2) = T0e

δ +U0Bup

1+

V0Cdown

1, W1(ω3) = T0e

δ + U0Bdown

1+ V0C

up

1and W1(ω4) = T0e

δ + U0Bdown

1+

V0Cdown

1.

(b) By general reasoning, there is arbitrage if one asset is uniformly better thananother asset. In particular:

• If B1(ω1) ≤ A1, then (1,−1, 0) is an arbitrage portfolio, since W0 = 0 andW1 ≥ 0 with W1(ω3) = W1(ω4) > 0.

• If A1 ≤ B1(ω4), then (−1, 1, 0) is an arbitrage portfolio, since W0 = 0 andW1 ≥ 0 with W1(ω1) = W1(ω2) > 0.

• Similarly (1, 0,−1) or (−1, 0, 1) are arbitrage portfolios if C1(ω1) ≤ A1 orA1 ≤ C1(ω4).

These can also be deduced from the standard two-asset binary model (A, B) or(A, C). Now let Bup

1> A1 > Bdown

1and Cup

1> A1 > Cdown

1. Since the model

(A, B) has no arbitrage, there is no arbitrage portfolio of the form (T0, U0, 0).Assume that (T0, U0, 1) is an arbitrage portfolio. Then 0 = W0 = T0 + U0 + 1,W1(ω1) > W1(ω2) ≥ 0 and W1(ω3) > W1(ω4) ≥ 0.

• If U0 ≥ 0, then we have 0 ≤ W1(ω4) = T1A1 + U0Bdown

1+ Cdown

1<

(T1 + U0 + 1)A1 = 0, which is a contradiction.

• If U0 ≤ 0, then we have 0 ≤ W1(ω2) = T1A1 + U0Bup

1+ Cdown

1< (T1 +

U0 + 1)A1 = 0, which is a contradiction.

Similarly, now assume that (T0, U0,−1) is an arbitrage portfolio, then 0 =W0 = T0 + U0 − 1, W1(ω2) > W1(ω1) ≥ 0 and W1(ω4) > W1(ω3) ≥ 0.

• If U0 ≥ 0, then we have 0 ≤ W1(ω3) = T1A1 + U0Bdown

1− Cup

1< (T1 +

U0 − 1)A1 = 0, which is a contradiction.

• If U0 ≤ 0, then we have 0 ≤ W1(ω1) = T1A1 + U0Bup

1− Cup

1< (T1 + U0 +

1)A1 = 0, which is a contradiction.

So there is no arbitrage portfolio.

(c) The contingent claim W1(ω1) = 1, W1(ω2) = W1(ω3) = W1(ω4) = 0 cannot behedged, since we would require

0 = T0A1 + U0Bup

1+ V0C

down

1

= T0A1 + U0Bdown

1+ V0C

down

1= T0A1 + U0B

down

1+ V0C

up

1,

for ω2, ω3, ω4, and these imply T0 = U0 = V0 = 0, but then the fourth equation1 = T0A1 + U0B

up

1+ V0C

up

1fails.

(d) Since the contingent claim does not change as C1 varies, we should considerportfolios of the form (T0, U0, 0). Since the model (A, B) with scenarios “up”

and “down” is complete, the contingent claim ˜W1(up) = W1(ω1), ˜W1(up) =W1(ω3) can be hedged. Specifically,

˜W1(up) = T0A1 + U0Bup

1and ˜W1(down) = T0A1 + U0B

down

1

XXXVI Solutions – MS3b Levy Processes and Finance – Oxford HT 2008

has solution

T0 =˜W1(down)Bup

1−

˜W1(up)Bdown

1

A1(Bup

1− Bdown

1)

and U0 =˜W1(up) − ˜W1(down)

Bup

1− Bdown

1

,

and so we read off from

˜W0 = T0 + U0 =A1 − Bdown

1

A1(Bup

1− Bdown

1)˜W1(up) +

Bup

1− A1

A1(Bup

1− Bdown

1

˜W1(down) (1)

that

qB =A1 − Bdown

1

Bup

1− Bdown

1

∈ (0, 1).

The martingale property is equation (1) for the contingent claim ˜W1(down) =

Bdown

1and ˜W1(up) = Bup

1. The martingale probability qB is unique and does

not depend on ˜W1.

(e) By symmetry, contingent claims of the form W1(ω1) = W1(ω3), W1(ω2) =W1(ω4) can be hedged and priced as e−δ

E(W1), where

qC = P(C1 = Cup

1) =

A1 − Cdown

1

Cup

1− Cdown

1

∈ (0, 1).

The process e−δtCt, t = 0, 1, is a martingale under these probabilities.

(f) In order for both e−δtBt and e−δtCt to be martingales, we need

qB = P(B1 = Bup

1, C1 = Cup

1) + P(B1 = Bup

1, C1 = Cdown

1) = p1 + p2

and

qC = P(B1 = Bup

1, C1 = Cup

1) + P(B1 = Bdown

1, B1 = Bdown

1) = p1 + p3.

Together with the normalisation condition p1 + p2 + p3 + p4, we have threeequations (of rank three) for four unknowns, so there is a one-dimensionalsolution space.

(g) The range of arbitrage-free prices W0 = e−δp1 depends on qB and qC as follows.

• If qB + qC ≤ 1, then p1 can be arbitrarily close to zero, and then W0 willbe arbitrarily close to zero.

• If qB +qC > 1, then qB +qC = 2p1+p2+p3 < p1+1 and so p1 > qB +qC−1and so W0 > e−δ(qB + qC − 1).

• Clearly p1 < minqB, qC and so W0 < e−δ minqB, qC.

So we get eδW0 ∈ (max0, qB + qC − 1, minqB, qC). Note that this range isalways non-empty.

2. (a) The direct proof is to calculate the moment generating function of X(ε)i

E(eγX(ε)i ) = e−γµεe−λε + eγ(1−µε)(1 − e−λε) = e−γµε(1 + (1 − e−λε)(eγ

− 1))

and to see

E(eγS

(ε)[t/ε]) = e−γµε[t/ε]

(

1 +[t/ε](1 − e−λε)(eγ

− 1)

[t/ε]

)[t/ε]

→ e−γµte−λ(eγ−1)

which we recognise as being the moment generating function of Xt = Nt −µt.

Solutions 5 – MS3b Levy Processes and Finance – Oxford HT 2008 XXXVII

(b) This is a special case of the n-step generalisation of the two-asset model (A, S)on two scenarios. Since A0 = S0 = 1, we have no arbitrage if and only ifS1(down) < A1 < Sup

1. Here, this is

e−µε < eδε < e1−µε⇐⇒ −µ < δ < 1/ε − µ.

(c) We need

1 = ˜S(ε)0

= e−δεEq(˜S

(ε)1

) = e−εδ(

e−µε(1 − qε) + e1−µεqε

)

and so

qε =eδε

− e−µε

e−µε(e − 1)=

eµε−δε− 1

e − 1.

(d) This is in complete analogy to (a). We deduce this from the Poisson limit

theorem considering ˜T(ε)n = ˜S

(ε)n − nµε, a Bernoulli random walk with success

probability qε. Noting that

1

εqε =

1

e − 1

eε(δ+µ)− 1

ε→

δ + µ

e − 1as ε ↓ 0,

we obtain ˜T(ε)[t/ε] →

Ñt in distribution, as required. Now, clearly [t/ε]µε → µt,and taking differences in the two limit results completes the argument.

(e) Note from the moment generating function of the Poisson distribution that

E(eeNt) = et δ+µ

e−1(e−1) = eδt+µt

and so Mt = e−δteeNt−µt is a martingale, because for s < t

E(Mt|Fs) = E(e−δseeNs−µse−δ(t−s)e( eNt− eNs)−µ(t−s)

|Fs)

= e−δseeNs−µse−(δ+µ)(t−s)

E(eeNt− eNs) = Ms.

Given Nt = k or Ñt = k, the two processes (eeNs−µs)0≤s≤t and (eNs−µs)0≤s≤t

have the same conditional distribution, since the k jump times of Ñ and Noccur at independent uniform times on [0, t]. Since also P(Nt = k) > 0 if

and only if P( Ñt = k) > 0, the same paths are possible for the two processes.

Since the discounted process e−δteeNt−µt is a martingale, it provides martingale

probabilities for the equivalent process eNt−µt.

(f) (Nt)t≥0 only has jumps of size 1, all other jumps are impossible, and the onlyLevy processes with this property are Poisson processes with drift. If (Yt)t≥0

is a Poisson process with drift −νt, then we have

P((eYs)0≤s≤1 ∈ Dν) = 1.

Since Dν ∩ Dµ = ∅ for µ 6= ν, we must have µ = ν in order that eYt has thesame possible paths as eNt−µt. We can now check that of all intensities λ > 0of Y , only λ = (δ + µ)/(e − 1) is such that Mt = e−δteYt is a martingale:

E(Mt|Fs) = E(e−δseYse−δ(t−s)eYt−Ys|Fs)

= e−δseYse−δ(t−s)E(eYt−Ys) = Mse

−(δ+µ)(t−s)+λ(e−1).

XXXVIII Solutions – MS3b Levy Processes and Finance – Oxford HT 2008

B.6 Time change and subordination

1. (a) First note that

E

[2ny]∑

j=1

(Zj2−n − Z(j−1)2−n)2

=

[2ny]∑

j=1

Var(

(Bf(j2−n) − Bf((j−1)2−n)

)

=

[2ny]∑

j=1

(f(j2−n) − f((j − 1)2−n))

= f([2ny]2−n) − f(0) = f([2ny]2−n) → f(y),

as n → ∞. For L2-convergence we then calculate

E

[2ny]∑

j=1

(Zj2−n − Z(j−1)2−n)2− f(y)

2

≤ Var

[2ny]∑

j=1

(Zj2−n − Z(j−1)2−n)2

+ (f([2ny]2−n) − f(y))2

≤

[2ny]∑

j=1

(f(j2−n) − f((j − 1)2−n))2

Var(B2

1) + (f([2ny]2−n) − f(y))2

→ [f ]yVar(B2

1) = 0,

provided that f is continuous (and increasing). Convergence in L2 impliesconvergence in probability.

(b) Note first that both Z and ˜Z are continuous. For marginal distributions, notethat Zy = Bf(y) ∼ Normal(0, f(y)) and, for yj ≤ y < yj+1,

˜Zy =

j∑

i=1

σi(Wyi− Wyi−1

) + σj+1(Wy − Wyj)

is the sum of independent σi(Wyi− Wyi−1

) ∼ Normal(0, τ 2

i ), where

τ 2

i = σ2

i (yi − yi−1) =

∫ yi

yi−1

f ′(s)ds = f(yi) − f(yi−1),

and these variances add up to f(y), as well. As for joint distributions, Z and˜Z have independent increments: for 0 = u0 < u1 < . . . < un

Zuk− Zuk−1

= Bf(uk) − Bf(uk−1) ∼ Normal(0, f(uk) − f(uk−1)),

are independent as increments of B; similarly, increments ˜Zuk−˜Zuk−1

, forylk−1 < uk−1 ≤ ylk and yrk−1 < uk < yrk

, are independent as linear combina-tions (for lk < rk, just a multiple for lk = rk) of increments of W

˜Zuk−˜Zuk−1

= σlk(Wylk− Wuk−1

) +

rk−1∑

i=lk+1

σi(Wyi− Wyi−1

) + σrk(Wui

− Wyri−1)

∼ Normal(0, f(uk) − f(uk−1)).

Solutions 6 – MS3b Levy Processes and Finance – Oxford HT 2008 XXXIX

(c) Take a Poisson process X of rate λ. Then the process Z = (Xf(y))y≥0 has jumpsof size 1 only, for all continuous functions f : [0,∞) → [0,∞). However, iffor a function as in (b), we have σj 6= 1 for some j ≥ 1, then there is positive

probability that ˜Zy =∫ y

0

√

f ′(s)dXs has jumps of size σj , specifically, therewill be a Poi(λ(yj+1−yj)) number of such jumps in the time interval (yj, yj+1].

2. (a) If Var(X1) < ∞ (and hence Var(Xt) = tVar(X1) and E(Xt) = tE(X1)) andVar(τ1) =

∫∞0

t2gτ (t)dt < ∞, we check the stronger integrability condition

∫ ∞

−∞z2g(z)dz =

∫ ∞

−∞z2

∫ ∞

0

ft(z)gτ (t)dtdz

=

∫ ∞

0

∫ ∞

−∞z2ft(z)dzgτ (t)dt

=

∫ ∞

0

(Var(Xt) + (E(Xt))2)gτ (t)dt

=

∫ ∞

0

(tVar(X1) + t2(E(X1))2)gτ (t)dt < ∞.

(b) If τ is a compound Poisson process, i.e.∫∞0

gτ (t)dt < ∞, then

∫ ∞

−∞g(z)dz =

∫ ∞

0

∫ ∞

−∞ft(z)dzgτ (t)dt =

∫ ∞

0

gτ (t)dt < ∞.

If X is a compound Poisson process with intensity λ and such that P(Xt ∈

(a, b)) =∫ b

aft(x)dx for all (a, b) 6∋ 0 and P(Xt 6= 0) = 1 − e−λt, then

∫

R\0g(z)dz =

∫ ∞

0

∫

R\0ft(z)dzgτ (t)dt =

∫ ∞

0

(1 − e−λt)gτ (t)dt < ∞.

Note that (a) and (b) deal, respectively, with the integrability condition for small

z and large z. The general case, when neither the conditions of (a) nor of (b)

are satisfied, we know that we still obtain a Levy density, from the calculation

of characteristic functions in the lectures, but the integrability condition for Levy

densities is difficult to check directly.

(c) A Levy density is associated with a bounded variation Levy process if andonly if

∫ ∞

−∞(1 ∧ |x|)g(x)dx < ∞

First note that the scaling relation Xt ∼ t1/αX1 implies, by the transformationformula, that

ft(y) = t−1/αf1(t−1/αy), y ∈ R.

XL Solutions – MS3b Levy Processes and Finance – Oxford HT 2008

Therefore, we can check that

∫ ∞

−∞|y|g(y)dy =

∫ ∞

0

∫ ∞

−∞|y|ft(y)dygτ(t)dt

=

∫ ∞

0

∫ ∞

−∞|y|t−1/αf1(t

−1/αy)dygτ(t)dt

=

∫ ∞

0

∫ ∞

−∞|x|t1/αf1(x)dxgτ (t)dt

=

(∫ ∞

0

t1/αgτ (t)dt

)(∫ ∞

−∞|x|f1(x)dx

)

.

The last integral is E(|X1|). To see that it is finite, split X = X(1)+X(2) whereX(2) is the compound Poisson process of big jumps ∆Xt1|∆Xt|≥1, and X(1)

is the martingale of small jumps. Now E(|X1|) ≤ E(|X(1)

1|) + E(|X

(2)

1|) < ∞,

since the moment generating function of X(1) is an entire function, hence allmoments are finite, and

E(|X(2)

1|) ≤ E

(

∑

0≤s≤1

|∆s|

)

=

∫

|x|≥1

|x||x|−α−1 < ∞.

3. Since B ∼ −B, we clearly have for Xy = Bτy

P(Xy ≤ a) =

∫ ∞

0

P(Bt ≤ a)fτy(t)dt =

∫ ∞

0

P(Bt ≥ −a)fτy(t)dt = P(Xy ≥ −a),

and this is enough, since both X and −X have stationary, independent incrementsand right-continuous paths with left limits.

On the other hand, the symmetric process X = P −N for two independent Poissonprocesses of rate λ cannot be obtained as Brownian motion subordinated by an in-

dependent subordinator, since for every random time T with [or without] probabilitydensity function we have

P(BT = 1) =

∫ ∞

0

P(Bt = 1)fT (t)dt = 0

[

or

∫

[0,∞)

P(Bt = 1)P(T ∈ dt) = 0.

]

4. (a) Calculate

E(expγBTs) =

∫ ∞

0

E(expγBt)λαstαs−1

Γ(αs)e−λtdt

=

∫ ∞

0

E(exp1

2tγ2

)λαstαs−1

Γ(αs)e−λtdt

= E(exp1

2γ2Ts) =

(

λ

λ −1

2γ2

)αs

.

and identify this with the formula for Cs − Ds in Exercise A.1.2.(b).

Solutions 6 – MS3b Levy Processes and Finance – Oxford HT 2008 XLI

(b) We condition on St to get

E(eiλRt) = E(eiλBSt ) =

∫ ∞

0

fSt(s)E(eiλBs)ds

=

∫ ∞

0

fSt(s)e−

12λ2sds

= E(e−12λ2St) = e−tΦ(

12λ2

) = e−tΦ(1)(12)α|λ|2α

.

We identify the symmetric 2α-stable distribution.

(c) Now recall Method 3: Let (τy)y≥0 be an increasing process that we can simu-late, and let (Xt)t≥0 be a Levy process with cumulative distribution functionFt of Xt. Fix a time lag δ > 0. Then the process

Z(3,δ)y = S[y/δ], where Sn =

n∑

k=1

Yk and Yk = F−1

τkδ−τ(k−1)δ(Uk)

is the time discretisation of the subordinated process Zy = Xτy.

and Example 84: We can use Method 3 to simulate the Variance Gammaprocess, since we can simulate the Gamma process τ and we can simulate theYk. Actually, we can use the Box-Muller method to generate standard Normalrandom variables Nk and then use

˜Yk ∼

√

τkδ − τ(k−1)δNk, k ≥ 1,

instead of Yk, k ≥ 1.

5. (a) E(Zt) =∫∞0

E(Xs)fSt(s)ds =

∫∞0

µsfSt(ds) = µmt.

(b) E(Z2

t ) =∫∞0

E(Xs)2fSt

(s)ds =∫∞0

(σ2s+µ2s2)fSt(ds) = σ2mt+µ2q2t+µ2m2t2

and then Var(Zt) = σ2mt + µ2q2t + µ2m2t2 − µ2m2t2 = (σ2m + µ2q2)t.

(c) For the Variance Gamma process, we have

Var(Ct − Dt) = Var(Ct) + Var(Dt) = 2αt

√

2λ2

=αt

λ.

On the other hand, for σ2 = 1, µ = 0, m = α/λ

Var(Zt) =α

λt.

Clearly α = λ corresponds to Var(Z1) = 1. Differentiate the moment generat-ing function E(eγBt) = eγ2t/2 four times to get E(B4

t ) = 3t2. Then

E(B4

St) =

∫ ∞

0

E(B4

s )fSt(s)ds =

∫ ∞

0

3s2fSt(s)ds = 3E(S2

t )

Since St ∼ Gamma(α, λ), we get

E(B4

S1) = 3

(

α

λ2+

α2

λ2

)

= 3

(

1 +1

λ

)

∈ (3,∞).

XLII Solutions – MS3b Levy Processes and Finance – Oxford HT 2008

B.7 Level passage events

1. (a) The process ˜X(t) has independent increments: for all 0 = s0 < s1 < . . . <sn = t, we have

˜X(t)sj

−˜X(t)sj−1

= Xt−sj−1− −Xt−sj− = Xtn−j+1− −Xtn−j−, 1 ≤ j ≤ n,

where tn−j+1 = t − sj, and furthermore P(Xtk− = Xtk , 1 ≤ k ≤ n) = 1; nowwe conclude by the independence of increments of X. The same argumentyields the stationarity of increments and identity in distribution with X, sinceany increment of ˜X(t) has been identified with an increment of X of the samelength. Right-continuity and left limits are also deduced, using the followingremarks. First, for a right-continuous function f with left limits the functiong(s) = f(s−) is left-continuous with right limits. Second, for a left-continuousfunction with right limits, the function h(s) = g(t−s) is right-continuous withleft limits.

(b) We condition on τ to get for B = [a, b]

P( ˜X(τ)s1

∈ A1, . . . , ˜X(τ)sm

∈ Am, τ ∈ [sm, sm+1) ∩B)

=

∫ sm+1∧b

sm∨aP( ˜X(τ)

s1 ∈ A1, . . . , ˜X(τ)sm

∈ Am|τ = t)fτ (t)dt

=

∫ sm+1∧b

sm∨aP( ˜X(t)

s1 ∈ A1, . . . , ˜X(t)sm

∈ Am)fτ (t)dt

=

∫ sm+1∧b

sm∨aP(Xs1 ∈ A1, . . . , Xsm

∈ Am)fτ (t)dt

= P(Xs1 ∈ A1, . . . , Xsm∈ Am, τ ∈ [sm, sm+1) ∩B),

where we applied (a) to pass from ˜X(t) to X.

(c) Since ( ˜X(τ)s )0≤s≤τ ∼ (Xs)0≤s≤τ , we deduce from Proposition 86 that the fol-

lowing two random variables are independent:

˜X(τ)

τ = sup0≤s≤τ

(Xτ− −Xτ−s−) = Xτ − inf0≤r≤τ

Xr− = Xτ −Xτ

˜X(τ)

τ −˜X(τ)τ = Xτ −Xτ −Xτ− = −Xτ ,

where we also used that P(Xτ− = Xτ ) = 1.

(d) We calculate from Corollary 89

E(eβXτ ) = E

(

e−β( eX(τ)

τ − eX(τ)τ )

)

=q(Φ(q) − β)

Φ(q)(q − φ(β)).

2. (a) Let us first show that (Tx)x≥0 is a subordinator. See the proof of Proposition90 for the stationarity of increments and the independence of two consecutiveincrements. For 0 = x0 < x1 < . . . < xn we now show that Txk

− Txk−1,

Solutions 7 – MS3b Levy Processes and Finance – Oxford HT 2008 XLIII

k = 1, . . . , n, are independent and proceed by induction on n. Suppose in-dependence holds for n, then the strong Markov property at Txn

shows thatTxn+1 − Txn

is independent of FTxn, where

Txn≤ t ∩ Tx1 ≤ s1, . . . , Txn

≤ sn =n−1⋂

k=1

Txk≤ sk ∧ t ∈ Ft

and so Txn≤ t ∩ Tx1 ≤ s1, . . . , Txn

≤ sn ∈ FTxn. We deduce that

P(Txk≤ sk, 1 ≤ k ≤ n, Txn+1 − Txn

≤ sn+1)

= P(Txk≤ sk, 1 ≤ k ≤ n, ˜Txn+1−xn

≤ sn+1)

= P(Txk≤ sk, 1 ≤ k ≤ n)P(Txn+1−xn

≤ sn+1),

so that (Tx1 , . . . , Txn) (and by linear transformation (Tx1 , . . . , Txn

− Txn−1)) isindependent of Txn+1 −Txn

. By the induction hypothesis, we conclude that allvariables Tx1, . . . , Txn+1 − Txn

are independent.

For the right-continuity let xn = x + δn ↓ x and note that the definitionTx = inft ≥ 0 : Xt > x implies that there is a sequence ε ↓ 0 such thatXTx+εn

> x + δn, but then Tx ≤ Tx+δn ≤ Tx + εn implies that Tx+δn → Tx.The existence of left limits is trivial for increasing x 7→ Tx.

Then we apply Theorem 87 to see that

E(exp−qTx1Tx<∞) = exp−xΦ(q),

where we calculate Φ(q) by inverting q = φ(Φ(q)) = c(Φ(q))α to get Φ(q) =c−1/αq1/α, and we identify the distribution of a stable subordinator with index1/α.

(b) We apply the transformation formula to get

faY (z) =b/a

√

2πz3/a3e−b

2/(2z/a) =

√

ab√

2πz3e−(

√ab)2/(2z) = f√ab(z).

From Exercise 2.(c) we know that fb is the distribution of Tb in the caseα = 1/2, c = 1/2, so we have just shown that aTb ∼ T√ab, or indeed, fora = d2, x =

√

ab that

Tx ∼ d2Tx/d, x ≥ 0.

Since both processes (Tx)x≥0 and (d2Tx/d)x≥0 are subordinators, we have iden-tity of joint distributions and hence another derivation, independent of (a) ofthe fact that (Tx)x≥0 is a stable subordinator of index 1/2.

(c) We now deduce from (b) and (a) respectively that for all γ ≥ 0

∫ ∞

0

e−γxfb(x)dx = E(e−γTb) = e−cb√γ

for some c ∈ (0,∞).

XLIV Solutions – MS3b Levy Processes and Finance – Oxford HT 2008

3. (a) Let us denote the increments of (Nm)m≥0 and of (Rm)m≥0 by Bj = Nj −Nj−1

and Cj = Rj −Rj−1, j ≥ 1. Then we calculate

E(expγC1) = E(expγSB1) =∑

n∈N

P(B1 = n)E(expγSn)

=∑

n∈N

P(B1 = n)M(γ)n = G(M(γ)).

For m ≥ 1, the analogous calculation yields

E(expγ1C1 + . . .+ γmCm)

=∑

n1,...,nm∈N

P(B1 = n1, . . . , Bm = nm)

E(expγ1S1 + . . .+ γm(Sn1+...+nm− Sn1+...+nm−1))

=∑

n1,...,nm∈N

P(B1 = n1) . . .P(Bm = nm)M(γ1)n1 . . .M(γm)nm

= G(M(γ1)) . . . G(M(γm)),

and we can deduce that C1, . . . , Cm are independent and identically distributed,as required.

(b) We apply the same argument as in (a) to first calculate

E(expiλYs) = E(expγXTs) =

∫ ∞

0

fTs(t)E(expiλXt)dt

=

∫ ∞

0

fTs(t) exp−tψ(λ)dt = MTs

(−ψ(λ)),

where we assumed that Ts has a density fTsand that E(eiλX1) = e−ψ(λ). Now,

for r, s ≥ 0,

E(expiλYs + iµ(Ys+r − Ys))

=

∫ ∞

0

∫ ∞

0

fTs,Ts+r−Tr(t, u)E(expiλXt + iµ(Xt+u −Xt))dtdu

=

∫ ∞

0

∫ ∞

0

fTs(t)fTr

(u)e−tψ(λ)e−uψ(µ)dtdu = MTs(−ψ(λ))MTr

(−ψ(µ))

so that we deduce that Ys and Ys+r−Ys are independent, and that Ys+r−Ys ∼Yr. For the right-continuity of paths, note that

limε↓0

Ys+ε = limε↓0

XTs+ε= XTs

= Ys

since Ts + δ := Ts+ε ↓ Ts and therefore XTs+δ → XTs. For left limits, the same

argument applies.

(c) (i) First note that Xt = Bt+bt is also a Levy process, as a linear combinationof two Levy processes. Because of the path continuity of Brownian motion,we have

XTs= s, where Ts = inft ≥ 0 : Xt ∈ (s,∞).

Solutions 7 – MS3b Levy Processes and Finance – Oxford HT 2008 XLV

The strong Markov property of X at the stopping time Ts (first hittingtime of (s,∞)) yields that X(s) = (XTs+u−s)u≥0 is independent of Ts andhas the same distribution as X. Note further that

Ts+r = Ts + T (s)r , where T (s)

r = inft ≥ 0 : X(s)t > r,

since we can wait for X to exceed level s + r by first waiting for level sand then waiting to increase by a further r. Now, we get from the strongMarkov property that

T (s)r ∼ Tr and T

(s)r = Ts+r − Ts is independent of Ts.

This yields the stationarity and independence of increments (n = 2 butcan now be generalised by induction). As an increasing process, we auto-matically get the existence of left and right limits. Assume now that at sthe path s 7→ Ts is not right-continuous. Then there is ε > 0 so that forall δ > 0

Ts+δ − Ts > ε ⇒ XTs+u ≤ s + δ for all 0 ≤ u ≤ ε

but then XTs+u ≤ s for all 0 ≤ u ≤ ε, and this contradicts the definitionof Ts. So, the path s 7→ Ts must be right-continuous at s.

(ii) We check for Et = expγBt −1

2γ2t that

E(Et|Fs) = E(expγBs + γ(Bt − Bs) −1

2γ2t|Fs)

= expγBs −1

2γ2tE(expγ(Bt −NB))

= expγBs −1

2γ2t exp−1

2γ2(t− s) = Es.

Now, applying the Optional Stopping Theorem at Ts (which can be shownto be a stopping time for (Et)t≥0), we get from BTs

= XTs− bTs = s− bTs

that

1 = E(ETs) = E(expγBTs

−1

2γ2Ts) = eγsE(exp−(bγ + 1

2γ2)Ts)

and this yields the claim by choosing ρ = ρ(γ) = −(bγ + 1

2γ2), i.e. γ =

√

b2 − 2ρ− b

4. (a) By the strong Markov property, the post-Ty process ˜Xs = XTy+s− y, s ≥ 0, isa Brownian motion independent of FTy

. Note that we can write

Xt =

Xt t ≤ Tyy + ˜Xt−Ty

t ≥ Ty

and that ˜X ∼ −˜X (by the symmetry of the centred multivariate Normal

distribution), so it remains to replace y + ˜Xt−Ty∼ y −

˜Xt−Ty= 2y − Xt.

Formally, we check for 0 = t0 < t1 < . . . < tn < tn+1 = ∞ that

P(X∗t1∈ A1, . . . , X

∗tn ∈ An)

XLVI Solutions – MS3b Levy Processes and Finance – Oxford HT 2008

=n

∑

k=0

∫ tk+1

tk

P(Xt1 ∈ A1, . . . , Xtk ∈ Ak,

y − ˜Xtk+1−t ∈ Ak+1, . . . , y − ˜Xtn−t ∈ An)fTy(t)dt

=

n∑

k=0

∫ tk+1

tk

P(Xt1 ∈ A1, . . . , Xtk ∈ Ak)

P(y − ˜Xtk+1−t ∈ Ak+1, . . . , y − ˜Xtn−t ∈ An)fTy(t)dt

=

n∑

k=0

∫ tk+1

tk

P(Xt1 ∈ A1, . . . , Xtk ∈ Ak)

P(y + ˜Xtk+1−t ∈ Ak+1, . . . , y + ˜Xtn−t ∈ An)fTy(t)dt

= P(Xt1 ∈ A1, . . . , Xtn ∈ An).

(b) Note that X t > y implies Ty < t and so, if also Xt ≤ x, then X∗t = 2y −Xt ≥

2y− x. Vice versa, if X∗t > 2y− x, then x < y implies that 2y− x > y and so

Ty < t and hence X t > y, but also X∗t = 2y−Xt, so Xt < x. This means that

Xt < x,X t > y = X∗t > 2y−x ⇒ P(Xt < x,X t > y) = P(X∗

t > 2y−x)

We apply (a) to get upon differentiation, first w.r.t. x then y that

fXt,Xt(x, y) =

∂

∂x

1√

2πtexp

−

(2y − x)2

2t

=2(2y − x)√

2πt3exp

−

(2y − x)2

2t

for y ∈ (0,∞), x ∈ (−∞, y).

(c) Since Ty < t if and only if X t > y, we first calculate

fXt(y) =

∫ y

−∞

2(2y − x)√

2πt3exp

−

(2y − x)2

2t

dx =2

√

2πtexp

−

y2

2t

.

and then differentiate the distribution function of Ty

fTy(t) =

d

dtP(Ty ≤ t) =

d

dtP(X t > y) =

d

dt2P(Xt > y)

=d

dt2P(X1 > y/

√

t) =2

√

2πexp

−

y2/t

2

1

yt−3/2

=y

√

2πt3exp

−

y2

2t

,

for all t > 0.

Documents

Part C L´evy Processes and Finance · L´evy processes and indicate some of their applications. By doing so, we will review some results from BS3a Applied Probability and B10 Martingales