Koc5(dba)

Lectures on Lévy Processes and StochasticCalculus (Koc University) Lecture 5: The

Ornstein-Uhlenbeck Process

David Applebaum

School of Mathematics and Statistics, University of Sheffield, UK

9th December 2011

Dave Applebaum (Sheffield UK) Lecture 5 December 2011 1 / 44

Historical Origins

This process was first introduced by Ornstein and Uhlenbeck in the1930s as a more accurate model of the physical phenomenon ofBrownian motion than the Einstein-Smoluchowski-Wiener process.They argued that

Brownian motion = viscous drag of fluid + random molecularbombardment.


Historical Origins




Historical Origins




Let v(t) be the velocity at time t of a particle of mass m executingBrownian motion. By Newton’s second law of motion, the total force

acting on the particle at time t is F (t) = mdv(t)

dt.We then have

mdv(t)

dt= − mkv(t)︸︷︷︸

viscous drag

+ mσdB(t)

dt︸︷︷︸molecular bombardment

,

where k , σ > 0.

Of course,dB(t)

dtdoesn’t exist, but this is a “physicist’s argument”. If

we cancel the ms and multiply both sides by dt then we get alegitimate SDE - the Langevin equation




dt.We then have

mdv(t)

dt= − mkv(t)︸︷︷︸

viscous drag

+ mσdB(t)


,

where k , σ > 0.

Of course,dB(t)






dt.We then have

mdv(t)

dt= − mkv(t)︸︷︷︸

viscous drag

+ mσdB(t)


,

where k , σ > 0.

Of course,dB(t)






dt.We then have

mdv(t)

dt= − mkv(t)︸︷︷︸

viscous drag

+ mσdB(t)


,

where k , σ > 0.

Of course,dB(t)






dt.We then have

mdv(t)

dt= − mkv(t)︸︷︷︸

viscous drag

+ mσdB(t)


,

where k , σ > 0.

Of course,dB(t)




dv(t) = −kv(t)dt + σdB(t) (0.1)

Using the integrating factor ekt we can then easily check that theunique solution to this equation is the Ornstein-Uhlenbeck process(v(t), t ≥ 0) where

v(t) = e−ktv(0) +

∫ t

0e−k(t−s)dB(s).

We are interested in Lévy processes so replace B by a d-dimensionalLévy process X and k by a d × d matrix K . Our Langevin equation is

dY (t) = −KY (t)dt + dX (t) (0.2)

and its unique solution is

Y (t) = e−tK Y0 +

∫ t

0e−(t−s)K dX (s), (0.3)


dv(t) = −kv(t)dt + σdB(t) (0.1)


v(t) = e−ktv(0) +

∫ t

0e−k(t−s)dB(s).


dY (t) = −KY (t)dt + dX (t) (0.2)


Y (t) = e−tK Y0 +

∫ t

0e−(t−s)K dX (s), (0.3)


dv(t) = −kv(t)dt + σdB(t) (0.1)


v(t) = e−ktv(0) +

∫ t

0e−k(t−s)dB(s).


dY (t) = −KY (t)dt + dX (t) (0.2)


Y (t) = e−tK Y0 +

∫ t

0e−(t−s)K dX (s), (0.3)


dv(t) = −kv(t)dt + σdB(t) (0.1)


v(t) = e−ktv(0) +

∫ t

0e−k(t−s)dB(s).


dY (t) = −KY (t)dt + dX (t) (0.2)


Y (t) = e−tK Y0 +

∫ t

0e−(t−s)K dX (s), (0.3)


dv(t) = −kv(t)dt + σdB(t) (0.1)


v(t) = e−ktv(0) +

∫ t

0e−k(t−s)dB(s).


dY (t) = −KY (t)dt + dX (t) (0.2)


Y (t) = e−tK Y0 +

∫ t

0e−(t−s)K dX (s), (0.3)


dv(t) = −kv(t)dt + σdB(t) (0.1)


v(t) = e−ktv(0) +

∫ t

0e−k(t−s)dB(s).


dY (t) = −KY (t)dt + dX (t) (0.2)


Y (t) = e−tK Y0 +

∫ t

0e−(t−s)K dX (s), (0.3)


dv(t) = −kv(t)dt + σdB(t) (0.1)


v(t) = e−ktv(0) +

∫ t

0e−k(t−s)dB(s).


dY (t) = −KY (t)dt + dX (t) (0.2)


Y (t) = e−tK Y0 +

∫ t

0e−(t−s)K dX (s), (0.3)


where Y0 := Y (0) is a fixed F0 measurable random variables. We stillcall the process Y an Ornstein-Uhlenbeck or OU process.Furthermore

Y has càdlàg paths.Y is a Markov process.

The process X is sometimes called the background driving Lévyprocess or BDLP.


















We get a Markov semigroup on Bb(Rd ) called a Mehler semigroup:

Tt f (x) = E(f (Y (t))|Y0 = x)

=

∫Rd

f (e−tK x + y)ρt (dy) (0.4)

where ρt is the law of the stochastic integral∫ t0 e−sK dX (s)

d=∫ t

0 e−(t−s)K dX (s).This generalises the classical Mehler formula (X (t) = B(t),K = kI)

Tt f (x) =1

(2π)d2

∫Rd

f

(e−ktx +

√1− e−2kt

2ky

)e−

y2

2 dy .



Tt f (x) = E(f (Y (t))|Y0 = x)

=

∫Rd

f (e−tK x + y)ρt (dy) (0.4)


d=∫ t


Tt f (x) =1

(2π)d2

∫Rd

f

(e−ktx +

√1− e−2kt

2ky

)e−

y2

2 dy .



Tt f (x) = E(f (Y (t))|Y0 = x)

=

∫Rd

f (e−tK x + y)ρt (dy) (0.4)


d=∫ t


Tt f (x) =1

(2π)d2

∫Rd

f

(e−ktx +

√1− e−2kt

2ky

)e−

y2

2 dy .



Tt f (x) = E(f (Y (t))|Y0 = x)

=

∫Rd

f (e−tK x + y)ρt (dy) (0.4)


d=∫ t


Tt f (x) =1

(2π)d2

∫Rd

f

(e−ktx +

√1− e−2kt

2ky

)e−

y2

2 dy .



Tt f (x) = E(f (Y (t))|Y0 = x)

=

∫Rd

f (e−tK x + y)ρt (dy) (0.4)


d=∫ t


Tt f (x) =1

(2π)d2

∫Rd

f

(e−ktx +

√1− e−2kt

2ky

)e−

y2

2 dy .


In fact (Tt , t ≥ 0) satisfies the Feller property: Tt (C0(Rd )) ⊆ C0(Rd ).We also have the skew-convolution semigroup property:

ρs+t = ρKs ∗ ρt ,

where ρKs (B) = ρs(etK B). Another terminology for this is

measure-valued cocycle.












We get nicer probabilistic properties of our solution if we make thefollowing

Assumption K is strictly positive definite.

OU processes solve simple linear SDEs. They are important inapplications such as volatility modelling, Lévy driven CARMAprocesses, branching processes with immigration.In infinite dimensions they solve the simplest linear SPDE with additivenoise. To develop this theme, let H and K be separable Hilbert spacesand (S(t), t ≥ 0) be a C0-semigroup on H with infinitesimal generatorJ. Let X be a Lévy process on K and C ∈ L(K ,H).


























We have the SPDE

dY (t) = JY (t) + CdX (t),

whose unique solution is

Y (t) = S(t)Y0 +

∫ t

0S(t − s)CdX (s)︸︷︷︸

stochastic convolution

,

and the generalised Mehler semigroup is

Tt f (x) =

∫Rd

f (S(t)x + y)ρt (dy).

From now on we will work in finite dimensions and assume the strictpositive-definiteness of K .


We have the SPDE

dY (t) = JY (t) + CdX (t),


Y (t) = S(t)Y0 +

∫ t

0S(t − s)CdX (s)︸︷︷︸


,


Tt f (x) =

∫Rd




We have the SPDE

dY (t) = JY (t) + CdX (t),


Y (t) = S(t)Y0 +

∫ t

0S(t − s)CdX (s)︸︷︷︸


,


Tt f (x) =

∫Rd




We have the SPDE

dY (t) = JY (t) + CdX (t),


Y (t) = S(t)Y0 +

∫ t

0S(t − s)CdX (s)︸︷︷︸


,


Tt f (x) =

∫Rd




Additive Processes and Wiener-Lévy Integrals

The study of O-U processes focusses attention on Wiener-Lévyintegrals If (t) :=

∫ t0 f (s)dX (s). For simplicity we assume that

f : Rd → Rd is continuous.Recall that Z = (Z (t), t ≥ 0) is an additive process if Z (0) = 0 (a.s.), Zhas independent increments and is stochastically continuous. It followsthat each Z (t) is infinitely divisible.

Theorem(If (t), t ≥ 0) is an additive process.

Proof. (sketch) Independent increments follows from the fact that forr ≤ s ≤ tIf (s)− If (r) =

∫ sr f (u)dX (u) is σ{X (b)− X (s); r ≤ a < b ≤ s} -

measurable,If (t)− If (s) =

∫ ts f (u)dX (u) is σ{X (d)− X (c); s ≤ c < d ≤ t} -

measurable, 2











measurable, 2











measurable, 2











measurable, 2











measurable, 2











measurable, 2


TheoremIf X has Lévy symbol η then for each t ≥ 0,u ∈ Rd ,

E(ei(u,If (t))) = exp{∫ t

0η(f (s)T u)

}.

Proof. (sketch) Define Mf (t) = exp{

i(

u,∫ t

0 f (s)dX (s))}

and use Itô’sformula to show that

Mf (t) = 1 + i(

u,∫ t

0Mf (s−)f (s)dB(s)

)+

∫ t

0

∫Rd−{0}

Mf (s−)(ei(u,f (s)x) − 1)N(ds,dx) +

∫ t

0Mf (s−)η(f (s)T u)ds.

Now take expectations of both sides to get

E(Mf (t)) = 1 +

∫ t

0E(Mf (s))η(f (s)T u)ds,

and the result follows. 2Dave Applebaum (Sheffield UK) Lecture 5 December 2011 11 / 44



0η(f (s)T u)

}.


i(

u,∫ t

0 f (s)dX (s))}


Mf (t) = 1 + i(

u,∫ t


)+

∫ t

0

∫Rd−{0}


∫ t



E(Mf (t)) = 1 +

∫ t





0η(f (s)T u)

}.


i(

u,∫ t

0 f (s)dX (s))}


Mf (t) = 1 + i(

u,∫ t


)+

∫ t

0

∫Rd−{0}


∫ t



E(Mf (t)) = 1 +

∫ t





0η(f (s)T u)

}.


i(

u,∫ t

0 f (s)dX (s))}


Mf (t) = 1 + i(

u,∫ t


)+

∫ t

0

∫Rd−{0}


∫ t



E(Mf (t)) = 1 +

∫ t



If X has characteristics (b,A, ν), it follows that If (t) has characteristics(bf

t ,Aft , ν

ft ) where

bft =

∫ t

0f (s)bds +

∫0

∫Rd−{0}

f (s)x(1B(x)− 1B(f (s)x))ν(dx)ds,

Aft =

∫ t

0f (s)T Af (s)ds,

ν ft (B) =

∫ t

0ν(f (s)−1(B)).

It follows that every OU process Y conditioned on Y0 = y is an additiveprocess. It will have characteristics as above with f (s) = e−sK and bf

ttranslated by e−tK y .



t ,Aft , ν

ft ) where

bft =

∫ t

0f (s)bds +

∫0

∫Rd−{0}


Aft =

∫ t

0f (s)T Af (s)ds,

ν ft (B) =

∫ t

0ν(f (s)−1(B)).





t ,Aft , ν

ft ) where

bft =

∫ t

0f (s)bds +

∫0

∫Rd−{0}


Aft =

∫ t

0f (s)T Af (s)ds,

ν ft (B) =

∫ t

0ν(f (s)−1(B)).





t ,Aft , ν

ft ) where

bft =

∫ t

0f (s)bds +

∫0

∫Rd−{0}


Aft =

∫ t

0f (s)T Af (s)ds,

ν ft (B) =

∫ t

0ν(f (s)−1(B)).





t ,Aft , ν

ft ) where

bft =

∫ t

0f (s)bds +

∫0

∫Rd−{0}


Aft =

∫ t

0f (s)T Af (s)ds,

ν ft (B) =

∫ t

0ν(f (s)−1(B)).





t ,Aft , ν

ft ) where

bft =

∫ t

0f (s)bds +

∫0

∫Rd−{0}


Aft =

∫ t

0f (s)T Af (s)ds,

ν ft (B) =

∫ t

0ν(f (s)−1(B)).




Invariant Measures, Stationary Processes, Ergodicity:General Theory

We want to investigate invariant measures and stationary solutions forOU processes. First a little general theory.First let (Tt , t ≥ 0) be a general Markov semigroup with transitionprobabilities pt (x ,B) = Tt1B(x) so that Tt f (x) =

∫Rd f (y)pt (x ,dy) for

f ∈ Bb(Rd ). We say that a probability measure µ is an invariantmeasure for the semigroup if for all t ≥ 0, f ∈ Bb(Rd ),∫

RdTt f (x)µ(dx) =

∫Rd

f (x)µ(dx) (0.5)






RdTt f (x)µ(dx) =

∫Rd

f (x)µ(dx) (0.5)






RdTt f (x)µ(dx) =

∫Rd

f (x)µ(dx) (0.5)






RdTt f (x)µ(dx) =

∫Rd

f (x)µ(dx) (0.5)






RdTt f (x)µ(dx) =

∫Rd

f (x)µ(dx) (0.5)






RdTt f (x)µ(dx) =

∫Rd

f (x)µ(dx) (0.5)


Equivalently for all Borel sets B∫Rd

pt (x ,B)µ(dx) = µ(B). (0.6)

To see that (0.5)⇒ (0.6) rewrite as∫Rd

∫Rd

f (y)pt (x ,dy)µ(dx) =

∫Rd

f (x)µ(dx),

and put f = 1B. For the converse - approximate f by simple functionsand take limits.



pt (x ,B)µ(dx) = µ(B). (0.6)


∫Rd


∫Rd

f (x)µ(dx),




pt (x ,B)µ(dx) = µ(B). (0.6)


∫Rd


∫Rd

f (x)µ(dx),




pt (x ,B)µ(dx) = µ(B). (0.6)


∫Rd


∫Rd

f (x)µ(dx),



e.g. A Lévy process doesn’t have an invariant probability measure butLebesgue measure is invariant in the sense that for f ∈ L1(Rd )∫

RdTt f (x)dx =

∫Rd

∫Rd

f (x + y)pt (dy)dx =

∫Rd

f (x)dx .

A process Z = (Z (t), t ≥ 0) is (strictly) stationary if for alln ∈ N, t1, . . . , tn,h ∈ R+,

(Z (t1), . . . ,Z (tn))d= (Z (t1 + h), . . . ,Z (tn + h))

TheoremA Markov process Z wherein µ is the law of Z (0) is stationary if andonly if µ is an invariant measure.



RdTt f (x)dx =

∫Rd

∫Rd


∫Rd

f (x)dx .


(Z (t1), . . . ,Z (tn))d= (Z (t1 + h), . . . ,Z (tn + h))




RdTt f (x)dx =

∫Rd

∫Rd


∫Rd

f (x)dx .


(Z (t1), . . . ,Z (tn))d= (Z (t1 + h), . . . ,Z (tn + h))




RdTt f (x)dx =

∫Rd

∫Rd


∫Rd

f (x)dx .


(Z (t1), . . . ,Z (tn))d= (Z (t1 + h), . . . ,Z (tn + h))



Proof. If the process is stationary then µ is invariant since

µ(B) = P(Z (0) ∈ B) = P(Z (t) ∈ B) =

∫Rd

pt (x ,B)µ(dx).

For the converse, its sufficient to prove thatE(f1(Z (t1 + h)) · · · fn(Z (tn + h)))) is independent of h for allf1, . . . fn ∈ Bb(Rd ). Proof is by induction. Case n = 1. Its enough toshow

E(f (Z (t)) = E(E(f (Z (t)|F0)))

= E(Tt f (Z (0)))

=

∫Rd

(Tt f (x))µ(dx)

=

∫Rd

f (x)µ(dx) = E(f (Z (0))).



µ(B) = P(Z (0) ∈ B) = P(Z (t) ∈ B) =

∫Rd

pt (x ,B)µ(dx).


E(f (Z (t)) = E(E(f (Z (t)|F0)))

= E(Tt f (Z (0)))

=

∫Rd

(Tt f (x))µ(dx)

=

∫Rd

f (x)µ(dx) = E(f (Z (0))).



µ(B) = P(Z (0) ∈ B) = P(Z (t) ∈ B) =

∫Rd

pt (x ,B)µ(dx).


E(f (Z (t)) = E(E(f (Z (t)|F0)))

= E(Tt f (Z (0)))

=

∫Rd

(Tt f (x))µ(dx)

=

∫Rd

f (x)µ(dx) = E(f (Z (0))).



µ(B) = P(Z (0) ∈ B) = P(Z (t) ∈ B) =

∫Rd

pt (x ,B)µ(dx).


E(f (Z (t)) = E(E(f (Z (t)|F0)))

= E(Tt f (Z (0)))

=

∫Rd

(Tt f (x))µ(dx)

=

∫Rd

f (x)µ(dx) = E(f (Z (0))).



µ(B) = P(Z (0) ∈ B) = P(Z (t) ∈ B) =

∫Rd

pt (x ,B)µ(dx).


E(f (Z (t)) = E(E(f (Z (t)|F0)))

= E(Tt f (Z (0)))

=

∫Rd

(Tt f (x))µ(dx)

=

∫Rd

f (x)µ(dx) = E(f (Z (0))).



µ(B) = P(Z (0) ∈ B) = P(Z (t) ∈ B) =

∫Rd

pt (x ,B)µ(dx).


E(f (Z (t)) = E(E(f (Z (t)|F0)))

= E(Tt f (Z (0)))

=

∫Rd

(Tt f (x))µ(dx)

=

∫Rd

f (x)µ(dx) = E(f (Z (0))).



µ(B) = P(Z (0) ∈ B) = P(Z (t) ∈ B) =

∫Rd

pt (x ,B)µ(dx).


E(f (Z (t)) = E(E(f (Z (t)|F0)))

= E(Tt f (Z (0)))

=

∫Rd

(Tt f (x))µ(dx)

=

∫Rd

f (x)µ(dx) = E(f (Z (0))).


In general use

E(f1(Z (t1 + h)) · · · fn(Z (tn + h)))

= E(f1(Z (t1 + h)) · · ·E(fn(Z (tn + h))|Ftn−1+h))

= E(f1(Z (t1 + h) · · ·Ttn−tn−1 fn(Z (tn−1 + h)))).

2


In general use

E(f1(Z (t1 + h)) · · · fn(Z (tn + h)))

= E(f1(Z (t1 + h)) · · ·E(fn(Z (tn + h))|Ftn−1+h))

= E(f1(Z (t1 + h) · · ·Ttn−tn−1 fn(Z (tn−1 + h)))).

2


Let µ be an invariant probability measure for a Markov semigroup(Tt , t ≥ 0). µ is ergodic if

Tt1B = 1B (µ a.s.) ⇒ µ(B) = 0 or µ(B) = 1.

If µ is ergodic then “time averages” = “space averages” for thecorresponding stationary Markov process, i.e.

limT→∞

1T

∫ T

0f (Z (s))ds =

∫Rd

f (x)µ(dx) a.s.

Fact: The invariant measures form a convex set and the ergodicmeasures are the extreme points of this set.

It follows that if an invariant measure is unique then it is ergodic.



Tt1B = 1B (µ a.s.) ⇒ µ(B) = 0 or µ(B) = 1.


limT→∞

1T

∫ T

0f (Z (s))ds =

∫Rd

f (x)µ(dx) a.s.





Tt1B = 1B (µ a.s.) ⇒ µ(B) = 0 or µ(B) = 1.


limT→∞

1T

∫ T

0f (Z (s))ds =

∫Rd

f (x)µ(dx) a.s.





Tt1B = 1B (µ a.s.) ⇒ µ(B) = 0 or µ(B) = 1.


limT→∞

1T

∫ T

0f (Z (s))ds =

∫Rd

f (x)µ(dx) a.s.





Tt1B = 1B (µ a.s.) ⇒ µ(B) = 0 or µ(B) = 1.


limT→∞

1T

∫ T

0f (Z (s))ds =

∫Rd

f (x)µ(dx) a.s.





Tt1B = 1B (µ a.s.) ⇒ µ(B) = 0 or µ(B) = 1.


limT→∞

1T

∫ T

0f (Z (s))ds =

∫Rd

f (x)µ(dx) a.s.




The Self-Decomposable Connection

Recall that a random variable Z is self-decomposable if for each0 < a < 1 there exists a random variable Wa that is independent of Zsuch that

Z d= aZ + Wa

or equivalently ρZ = ρaZ ∗ ρWa , where ρa

Z (B) = ρ(a−1B).Now suppose that Y is a stationary Ornstein-Uhlenbeck process on R.Then Y0 is self decomposable with a = e−kt and Wa(t) =

∫ t0 e−ksdX (s)

since

Y (t) = e−ktY0 +

∫ t

0e−(t−s)K dX (s)

and by stationary increments of the process X




Z d= aZ + Wa



∫ t0 e−ksdX (s)

since

Y (t) = e−ktY0 +

∫ t






Z d= aZ + Wa



∫ t0 e−ksdX (s)

since

Y (t) = e−ktY0 +

∫ t






Z d= aZ + Wa



∫ t0 e−ksdX (s)

since

Y (t) = e−ktY0 +

∫ t






Z d= aZ + Wa



∫ t0 e−ksdX (s)

since

Y (t) = e−ktY0 +

∫ t






Z d= aZ + Wa



∫ t0 e−ksdX (s)

since

Y (t) = e−ktY0 +

∫ t




Y (t) d= Y0 and

∫ t

0e−k(t−s)dX (s)

d=

∫ t

0e−ksdX (s)

⇒ Y0d= e−ktY0 + Wa(t).


Y (t) d= Y0 and

∫ t

0e−k(t−s)dX (s)

d=

∫ t

0e−ksdX (s)

⇒ Y0d= e−ktY0 + Wa(t).


Now suppose that µ is self-decomposable - more precisely that

µ = µekt ∗ ρt ,

where ρt is the law of Wa(t). Then

∫R

Tt f (x)µ(dx) =

∫R

∫R

f (e−ktx + y)ρt (dy)µ(dx)

=

∫R

∫R

f (x + y)ρt (dy)µekt(dx)

=

∫R

f (x)(µekt ∗ ρt )(dx)

=

∫R

f (x)µ(dx).

So µ is an invariant measure.





∫R

Tt f (x)µ(dx) =

∫R

∫R


=

∫R

∫R


=

∫R


=

∫R

f (x)µ(dx).






∫R

Tt f (x)µ(dx) =

∫R

∫R


=

∫R

∫R


=

∫R


=

∫R

f (x)µ(dx).






∫R

Tt f (x)µ(dx) =

∫R

∫R


=

∫R

∫R


=

∫R


=

∫R

f (x)µ(dx).






∫R

Tt f (x)µ(dx) =

∫R

∫R


=

∫R

∫R


=

∫R


=

∫R

f (x)µ(dx).






∫R

Tt f (x)µ(dx) =

∫R

∫R


=

∫R

∫R


=

∫R


=

∫R

f (x)µ(dx).



So we have shown that:

TheoremThe following are equivalent for the O-U process Y .

Y is stationary.The law of Y (0) is an invariant measure.

The law of Y (0) is self-decomposable (with Wa(t) =∫ t

0 e−ksdX (s)).






0 e−ksdX (s)).






0 e−ksdX (s)).






0 e−ksdX (s)).


We seek some condition on the Lévy process X which ensures that Yis stationary.

Fact: If Y∞ :=∫∞

0 e−ksdX (s) exists in distribution then it isself-decomposable.

To see this observe that (using stationary increments of X )

∫ ∞0

e−ksdX (s) =

∫ ∞t

e−ksdX (s) +

∫ t

0e−ksdX (s)

d=

∫ ∞0

e−k(t+s)dX (s) +

∫ t

0e−ksdX (s)

= e−kt∫ ∞

0e−ksdX (s) +

∫ t

0e−ksdX (s)






∫ ∞0

e−ksdX (s) =

∫ ∞t

e−ksdX (s) +

∫ t

0e−ksdX (s)

d=

∫ ∞0

e−k(t+s)dX (s) +

∫ t

0e−ksdX (s)

= e−kt∫ ∞

0e−ksdX (s) +

∫ t

0e−ksdX (s)






∫ ∞0

e−ksdX (s) =

∫ ∞t

e−ksdX (s) +

∫ t

0e−ksdX (s)

d=

∫ ∞0

e−k(t+s)dX (s) +

∫ t

0e−ksdX (s)

= e−kt∫ ∞

0e−ksdX (s) +

∫ t

0e−ksdX (s)






∫ ∞0

e−ksdX (s) =

∫ ∞t

e−ksdX (s) +

∫ t

0e−ksdX (s)

d=

∫ ∞0

e−k(t+s)dX (s) +

∫ t

0e−ksdX (s)

= e−kt∫ ∞

0e−ksdX (s) +

∫ t

0e−ksdX (s)






∫ ∞0

e−ksdX (s) =

∫ ∞t

e−ksdX (s) +

∫ t

0e−ksdX (s)

d=

∫ ∞0

e−k(t+s)dX (s) +

∫ t

0e−ksdX (s)

= e−kt∫ ∞

0e−ksdX (s) +

∫ t

0e−ksdX (s)






∫ ∞0

e−ksdX (s) =

∫ ∞t

e−ksdX (s) +

∫ t

0e−ksdX (s)

d=

∫ ∞0

e−k(t+s)dX (s) +

∫ t

0e−ksdX (s)

= e−kt∫ ∞

0e−ksdX (s) +

∫ t

0e−ksdX (s)


When does limt→∞∫ t

0 e−ksdX (s) exist in distribution? Use the Lévy-Itôdecomposition.

X (t) = bt + M(t) +

∫|x |≥1

xN(t ,dx).

It is not difficult to see that limt→∞∫ t

0 e−ksdM(s) exists in L2-sense.

Fact: limt→∞

∫ t

0

∫|x |≥1

e−ksxN(ds,dx) exists in distribution if and only if∫|x |≥1 log(1 + |x |)ν(dx) <∞.




X (t) = bt + M(t) +

∫|x |≥1

xN(t ,dx).



Fact: limt→∞

∫ t

0

∫|x |≥1





X (t) = bt + M(t) +

∫|x |≥1

xN(t ,dx).



Fact: limt→∞

∫ t

0

∫|x |≥1





X (t) = bt + M(t) +

∫|x |≥1

xN(t ,dx).



Fact: limt→∞

∫ t

0

∫|x |≥1





X (t) = bt + M(t) +

∫|x |≥1

xN(t ,dx).



Fact: limt→∞

∫ t

0

∫|x |≥1





X (t) = bt + M(t) +

∫|x |≥1

xN(t ,dx).



Fact: limt→∞

∫ t

0

∫|x |≥1



To prove this you need

1 If (ξn,n ∈ N) are i.i.d. then∑∞

n=1 cnξn converges a.s. (0 < c < 1)if and only if E(log(1 + |ξ1|)) <∞.

2 ∫ n

0

∫|x |≥1

e−ksxN(ds,dx)d=

n−1∑j=0

e−kjMj

where Mj :=∫ j+1

j

∫|x |≥1 e−k(s−j)xN(ds,dx). Note that (Mj , j ∈ N)

are i.i.d.

In this case, Y has characteristics (bf∞,Af

∞, νf∞).

e.g. Brownian motion case. X (t) = B(t). µ ∼ N(0, 1

2k

).





2 ∫ n

0

∫|x |≥1

e−ksxN(ds,dx)d=

n−1∑j=0

e−kjMj

where Mj :=∫ j+1

j


are i.i.d.


∞, νf∞).


2k

).





2 ∫ n

0

∫|x |≥1

e−ksxN(ds,dx)d=

n−1∑j=0

e−kjMj

where Mj :=∫ j+1

j


are i.i.d.


∞, νf∞).


2k

).





2 ∫ n

0

∫|x |≥1

e−ksxN(ds,dx)d=

n−1∑j=0

e−kjMj

where Mj :=∫ j+1

j


are i.i.d.


∞, νf∞).


2k

).





2 ∫ n

0

∫|x |≥1

e−ksxN(ds,dx)d=

n−1∑j=0

e−kjMj

where Mj :=∫ j+1

j


are i.i.d.


∞, νf∞).


2k

).





2 ∫ n

0

∫|x |≥1

e−ksxN(ds,dx)d=

n−1∑j=0

e−kjMj

where Mj :=∫ j+1

j


are i.i.d.


∞, νf∞).


2k

).


In fact, if an invariant measure µ exists then it is unique. For supposethat Y is stationary, then

Y (0)d= e−ktY (0) +

∫0

te−ksdX (s).

Now let ρ be the law of Y (0) and Φρ(u) :=∫R eiuyρ(dy). Then for all

u ∈ R, by independence

Φρ(u) = Φρ(e−ktu) exp{−∫ t

0η(e−ksu)ds

}.

Take limits as t →∞ to get

Φρ(u) = exp{−∫ ∞

0η(e−ksu)ds

}.

So ρ is the law of Y∞.



Y (0)d= e−ktY (0) +

∫0

te−ksdX (s).




0η(e−ksu)ds

}.


Φρ(u) = exp{−∫ ∞

0η(e−ksu)ds

}.




Y (0)d= e−ktY (0) +

∫0

te−ksdX (s).




0η(e−ksu)ds

}.


Φρ(u) = exp{−∫ ∞

0η(e−ksu)ds

}.




Y (0)d= e−ktY (0) +

∫0

te−ksdX (s).




0η(e−ksu)ds

}.


Φρ(u) = exp{−∫ ∞

0η(e−ksu)ds

}.




Y (0)d= e−ktY (0) +

∫0

te−ksdX (s).




0η(e−ksu)ds

}.


Φρ(u) = exp{−∫ ∞

0η(e−ksu)ds

}.




Y (0)d= e−ktY (0) +

∫0

te−ksdX (s).




0η(e−ksu)ds

}.


Φρ(u) = exp{−∫ ∞

0η(e−ksu)ds

}.



Example: Let (X (t), t ≥ 0) be a compound Poisson processX (t) =

∑N(t)i=1 Wi where the Wis are i.i.d. exponential with common

density fW (x) = ae−ax1x>0. Then

η(u) = λa∫ ∞

0(eiux − 1)e−axdx =

λaa− iu

.

You can check that Φρ(u) = (1− ia−1u)−λ as so ρ has a gamma(c, λ)distribution.





η(u) = λa∫ ∞


λaa− iu

.






η(u) = λa∫ ∞


λaa− iu

.



In fact - it is possible to go further. Given any self-decomposabledistribution µ there exists a stationary Ornstein-Uhlenbeck process Ysuch that the law of Y (0) is µ. Let’s sketch the proof of this - due toJurek and Vervaat (1983). Let X be a self-decomposable randomvariable with distribution µ. Then for each t ≥ 0

X d= e−tX + Xt ,

where X and Xt are independent.The key step is the observation that we can construct an additiveprocess (Z (t), t ≥ 0) such that

Z (t) d= Xt and Z (t + h)− Z (t) d

= e−tXh.



X d= e−tX + Xt ,



= e−tXh.



X d= e−tX + Xt ,



= e−tXh.



X d= e−tX + Xt ,



= e−tXh.



X d= e−tX + Xt ,



= e−tXh.



X d= e−tX + Xt ,



= e−tXh.


This follows by Kolmogorov’s theorem since

X d= e−(t+h)X + Xt+hd= e−t (e−hX + Xh) + Xt

⇒ Xt+hd= e−tXh + Xt .

It follows that Y (t) =∫ t

0 esdZ (s) also has independent increments. ButY is a Lévy process since

Y (t + h)− Y (t) =

∫ t+h

tesdZ (s)

=

∫ h

0esetdZ (s + t)

d=

∫ h

0esete−tdZ (s) = Y (h)







Y (t + h)− Y (t) =

∫ t+h

tesdZ (s)

=

∫ h

0esetdZ (s + t)

d=

∫ h








Y (t + h)− Y (t) =

∫ t+h

tesdZ (s)

=

∫ h

0esetdZ (s + t)

d=

∫ h








Y (t + h)− Y (t) =

∫ t+h

tesdZ (s)

=

∫ h

0esetdZ (s + t)

d=

∫ h








Y (t + h)− Y (t) =

∫ t+h

tesdZ (s)

=

∫ h

0esetdZ (s + t)

d=

∫ h








Y (t + h)− Y (t) =

∫ t+h

tesdZ (s)

=

∫ h

0esetdZ (s + t)

d=

∫ h








Y (t + h)− Y (t) =

∫ t+h

tesdZ (s)

=

∫ h

0esetdZ (s + t)

d=

∫ h



We then find that Z (t) =∫ t

0 e−sdY (s) and so

X d= e−tX +

∫ t

0e−sdY (s)

d= e−tX +

∫ t

0e−(t−s)dY (s),

using stationary increments of Y which is extended to an Lévyprocess on the whole of R.In the 1990s, Sato showed that µ is self-decomposable if and only if itis the law of W (1) where (W (t), t ≥ 0) is a self-similar additiveprocess. Recall W self-similar (index H) means for all c ≥ 0

W (ct) d= cHW (t).

So we can embed selfdecomposable distributions into stationary OUprocesses and self-similar additive processes. Is there a connection?




X d= e−tX +

∫ t

0e−sdY (s)

d= e−tX +

∫ t

0e−(t−s)dY (s),


W (ct) d= cHW (t).





X d= e−tX +

∫ t

0e−sdY (s)

d= e−tX +

∫ t

0e−(t−s)dY (s),


W (ct) d= cHW (t).





X d= e−tX +

∫ t

0e−sdY (s)

d= e−tX +

∫ t

0e−(t−s)dY (s),


W (ct) d= cHW (t).





X d= e−tX +

∫ t

0e−sdY (s)

d= e−tX +

∫ t

0e−(t−s)dY (s),


W (ct) d= cHW (t).



To understand the connection between the two “embeddings” of µ weneed the

Lamperti Transform. There is a one-to-one correspondence betweenself-similar processes (W (t), t ≥ 0) and stationary processes(Z (t), t ≥ 0) given by

W (t) = tHZ (log(t)) or equivalently Z (t) = e−tHW (et ).

Indeed if W self-similar

Z (t + h) = e−(t+h)HW (et+h)d= e−tHe−hHehHW (et )

= Z (t).







= Z (t).







= Z (t).







= Z (t).







= Z (t).







= Z (t).


The next step is due to Jeanblanc, Pitman, Yor (SPA 100, 223(2002))

Start with a self-similar additive process (W (t), t ≥ 0). Then we knowthat W (1) is self-decomposable. There exist two independent,identically distributed Lévy processes (X−t , t ≥ 0) and (X +

t , t ≥ 0) suchthat

X−t =

∫ 1

e−t

dW (r)

rH ,X +t =

∫ et

1

dW (r)

rH .

Let (Z (t), t ≥ 0) be the stationary Lamperti transform of W . Then it isan Ornstein-Uhlenbeck process and

Z (t) = e−tHW (1) +

∫ t

0e−(t+s)HdX +

s ,

Z (−t) = e−tHW (1)−∫ t

0e−(t+s)HdX−s .

In the last part of the lecture we’ll briefly look at some recentdevelopments.





X−t =

∫ 1

e−t

dW (r)

rH ,X +t =

∫ et

1

dW (r)

rH .


Z (t) = e−tHW (1) +

∫ t

0e−(t+s)HdX +

s ,

Z (−t) = e−tHW (1)−∫ t

0e−(t+s)HdX−s .






X−t =

∫ 1

e−t

dW (r)

rH ,X +t =

∫ et

1

dW (r)

rH .


Z (t) = e−tHW (1) +

∫ t

0e−(t+s)HdX +

s ,

Z (−t) = e−tHW (1)−∫ t

0e−(t+s)HdX−s .






X−t =

∫ 1

e−t

dW (r)

rH ,X +t =

∫ et

1

dW (r)

rH .


Z (t) = e−tHW (1) +

∫ t

0e−(t+s)HdX +

s ,

Z (−t) = e−tHW (1)−∫ t

0e−(t+s)HdX−s .






X−t =

∫ 1

e−t

dW (r)

rH ,X +t =

∫ et

1

dW (r)

rH .


Z (t) = e−tHW (1) +

∫ t

0e−(t+s)HdX +

s ,

Z (−t) = e−tHW (1)−∫ t

0e−(t+s)HdX−s .






X−t =

∫ 1

e−t

dW (r)

rH ,X +t =

∫ et

1

dW (r)

rH .


Z (t) = e−tHW (1) +

∫ t

0e−(t+s)HdX +

s ,

Z (−t) = e−tHW (1)−∫ t

0e−(t+s)HdX−s .






X−t =

∫ 1

e−t

dW (r)

rH ,X +t =

∫ et

1

dW (r)

rH .


Z (t) = e−tHW (1) +

∫ t

0e−(t+s)HdX +

s ,

Z (−t) = e−tHW (1)−∫ t

0e−(t+s)HdX−s .






X−t =

∫ 1

e−t

dW (r)

rH ,X +t =

∫ et

1

dW (r)

rH .


Z (t) = e−tHW (1) +

∫ t

0e−(t+s)HdX +

s ,

Z (−t) = e−tHW (1)−∫ t

0e−(t+s)HdX−s .






X−t =

∫ 1

e−t

dW (r)

rH ,X +t =

∫ et

1

dW (r)

rH .


Z (t) = e−tHW (1) +

∫ t

0e−(t+s)HdX +

s ,

Z (−t) = e−tHW (1)−∫ t

0e−(t+s)HdX−s .






X−t =

∫ 1

e−t

dW (r)

rH ,X +t =

∫ et

1

dW (r)

rH .


Z (t) = e−tHW (1) +

∫ t

0e−(t+s)HdX +

s ,

Z (−t) = e−tHW (1)−∫ t

0e−(t+s)HdX−s .



Densities of the OU Process

We’ve seen that each Y (t) is infinitely divisible so if the Lévy processX (t) has a Gaussian component then so does Y (t) in which case ithas a density by Fourier inversion.More generally, Priola and Zabczyk (BLMS, 41, 41,(2009)) study

dY (t) = AY (t)dt + BdX (t),

where each Y (t) is Rd -valued but X (t) is Rn-valued (n ≥ d). So A isan n × n matrix and B is an n × d matrix.

















Assume

1 Rank[B,AB, . . . ,An−1B] = n,where [B,AB, . . . ,An−1B] is the matrix of the linear mapping fromRnd to Rn given by

(u0,u1, . . . ,un−1)→ Bu0 + ABu1 + . . .+ An−1Bun−1.

2 The restriction of the Lévy measure ν to Br (0) has a density forsome r > 0.

Then Y (t) has a density for t > 0.


Assume


(u0,u1, . . . ,un−1)→ Bu0 + ABu1 + . . .+ An−1Bun−1.




Assume


(u0,u1, . . . ,un−1)→ Bu0 + ABu1 + . . .+ An−1Bun−1.




Assume


(u0,u1, . . . ,un−1)→ Bu0 + ABu1 + . . .+ An−1Bun−1.




Application - Volatility Modelling

Consider the Black-Scholes model for a stock price

S(t) = S(0) exp{µt + σB(t)},

where µ ∈ R is stock drift and σ > 0 is volatility. By Itô’s formula

dS(t) = σS(t)dB(t) + S(t)(µ+

12σ2)

dt .

In stochastic volatility models the parameter σ2 is replaced by astochastic process (σ2(t), t ≥ 0).




S(t) = S(0) exp{µt + σB(t)},



12σ2)

dt .





S(t) = S(0) exp{µt + σB(t)},



12σ2)

dt .





S(t) = S(0) exp{µt + σB(t)},



12σ2)

dt .



Barndorff-Nielsen and Shephard (JRSS B 63, 167 (2001)) proposedthe OU model

dσ2(t) = −λσ2(t) + dX (λt),

where λ > 0 and X is a subordinator. Then σ2(t) > 0 (a.s.) since

σ2(t) = e−λtσ2(0) +

∫ t

0e−λ(t−s)dX (λt)

= e−λt

σ2(0) +∑

0≤u≤t

e−λu∆X (λu)

.

Assume that∫∞

1 log(1 + x)ν(dx) <∞. Then there is a unique invariantmeasure µ which is self-decomposable and has characteristic function

µ(u) = exp{∫ ∞

0(eiux − 1)

k(x)

xdx},

where k is decreasing.Dave Applebaum (Sheffield UK) Lecture 5 December 2011 36 / 44


dσ2(t) = −λσ2(t) + dX (λt),


σ2(t) = e−λtσ2(0) +

∫ t


= e−λt

σ2(0) +∑

0≤u≤t

e−λu∆X (λu)

.

Assume that∫∞


µ(u) = exp{∫ ∞

0(eiux − 1)

k(x)

xdx},



dσ2(t) = −λσ2(t) + dX (λt),


σ2(t) = e−λtσ2(0) +

∫ t


= e−λt

σ2(0) +∑

0≤u≤t

e−λu∆X (λu)

.

Assume that∫∞


µ(u) = exp{∫ ∞

0(eiux − 1)

k(x)

xdx},



dσ2(t) = −λσ2(t) + dX (λt),


σ2(t) = e−λtσ2(0) +

∫ t


= e−λt

σ2(0) +∑

0≤u≤t

e−λu∆X (λu)

.

Assume that∫∞


µ(u) = exp{∫ ∞

0(eiux − 1)

k(x)

xdx},



dσ2(t) = −λσ2(t) + dX (λt),


σ2(t) = e−λtσ2(0) +

∫ t


= e−λt

σ2(0) +∑

0≤u≤t

e−λu∆X (λu)

.

Assume that∫∞


µ(u) = exp{∫ ∞

0(eiux − 1)

k(x)

xdx},



dσ2(t) = −λσ2(t) + dX (λt),


σ2(t) = e−λtσ2(0) +

∫ t


= e−λt

σ2(0) +∑

0≤u≤t

e−λu∆X (λu)

.

Assume that∫∞


µ(u) = exp{∫ ∞

0(eiux − 1)

k(x)

xdx},



dσ2(t) = −λσ2(t) + dX (λt),


σ2(t) = e−λtσ2(0) +

∫ t


= e−λt

σ2(0) +∑

0≤u≤t

e−λu∆X (λu)

.

Assume that∫∞


µ(u) = exp{∫ ∞

0(eiux − 1)

k(x)

xdx},



dσ2(t) = −λσ2(t) + dX (λt),


σ2(t) = e−λtσ2(0) +

∫ t


= e−λt

σ2(0) +∑

0≤u≤t

e−λu∆X (λu)

.

Assume that∫∞


µ(u) = exp{∫ ∞

0(eiux − 1)

k(x)

xdx},


Problem. Based on discrete-time observationsσ2(0), σ2(∆), σ2((N − 1)∆) find estimates of the parameter λ and k .For a non-parametric approach - see Jongbloed et. al. Bernoulli, 11,759 (2005).


Problem. Based on discrete-time observationsσ2(0), σ2(∆), σ2((N − 1)∆) find estimates of the parameter λ and k .For a non-parametric approach - see Jongbloed et. al. Bernoulli, 11,759 (2005).


Generalised Ornstein-Uhlenbeck Processes

Let X = (X1,X2) be a Lévy process on R2. Then each Xi is areal-valued Lévy process. Let Y0 be independent of X . Thegeneralised Ornstein-Uhlenbeck process is

Y (t) = e−X1(t)(

Y0 +

∫ t

0e−X1(s)dX2(s)

).

The usual OU process is obtained by taking X1(t) = λt (λ > 0).Necessary and sufficient conditions for stationarity solutions werefound by Lindner and Maller (SPA 115, 1701 (2005)). Almost sureconvergence of

∫ t0 e−X1(s)dX2(s) as t →∞ is a sufficient condition but

the general story is more complicated.




Y (t) = e−X1(t)(

Y0 +

∫ t

0e−X1(s)dX2(s)

).







Y (t) = e−X1(t)(

Y0 +

∫ t

0e−X1(s)dX2(s)

).







Y (t) = e−X1(t)(

Y0 +

∫ t

0e−X1(s)dX2(s)

).







Y (t) = e−X1(t)(

Y0 +

∫ t

0e−X1(s)dX2(s)

).







Y (t) = e−X1(t)(

Y0 +

∫ t

0e−X1(s)dX2(s)

).







Y (t) = e−X1(t)(

Y0 +

∫ t

0e−X1(s)dX2(s)

).





In fact a necessary and sufficient condition for stationary solutions isthe almost sure convergence of

∫ t0 e−X1(s)dL(s) as t →∞ where the

one-dimensional Lévy process (L(t), t ≥ 0) is defined by

L(t) := X2(t) +∑

0≤s≤t

(e−∆X1(s) − 1)∆X2(s)− tA1,2,

where A1,2 is the off-diagonal entry of the covariance matrix of theGaussian component of the bivariate Lévy process (X1,L). For furtherwork on generalised O-U processes see Lindner and Sato (AP 37, 250(2009)).





L(t) := X2(t) +∑

0≤s≤t

(e−∆X1(s) − 1)∆X2(s)− tA1,2,






L(t) := X2(t) +∑

0≤s≤t

(e−∆X1(s) − 1)∆X2(s)− tA1,2,






L(t) := X2(t) +∑

0≤s≤t

(e−∆X1(s) − 1)∆X2(s)− tA1,2,



References and Further Reading

These lectures have been broadly based on my recent book:

D.Applebaum Lévy Processes and Stochastic Calculus, CambridgeUniversity Press (second edition) (2009)

and from an earlier course of lectures partly derived from it, whichhave been separately published as

D.Applebaum, Lévy processes in Euclidean spaces and groups inQuantum Independent Increment Processes I: From ClassicalProbability to Quantum Stochastic Calculus, Springer Lecture Notes inMathematics , Vol. 1865 M Schurmann, U. Franz (Eds.) 1-99,(2005)


References and Further Reading

These lectures have been broadly based on my recent book:

D.Applebaum Lévy Processes and Stochastic Calculus, CambridgeUniversity Press (second edition) (2009)

and from an earlier course of lectures partly derived from it, whichhave been separately published as

D.Applebaum, Lévy processes in Euclidean spaces and groups inQuantum Independent Increment Processes I: From ClassicalProbability to Quantum Stochastic Calculus, Springer Lecture Notes inMathematics , Vol. 1865 M Schurmann, U. Franz (Eds.) 1-99,(2005)


A comprehensive account of the structure and properties of Lévyprocesses is:

K-I.Sato, Lévy Processes and Infinite Divisibility, Cambridge UniversityPress (1999)

A shorter account, from the point of view of the French school, whichconcentrates on fluctuation theory and potential theory aspects is

J.Bertoin, Lévy Processes, Cambridge University Press (1996)

From this point of view, you should also look atA.Kyprianou, Introductory Lectures on Fluctuations of Lévy Processeswith Applications, Springer-Verlag (2006)


A comprehensive account of the structure and properties of Lévyprocesses is:

K-I.Sato, Lévy Processes and Infinite Divisibility, Cambridge UniversityPress (1999)

A shorter account, from the point of view of the French school, whichconcentrates on fluctuation theory and potential theory aspects is

J.Bertoin, Lévy Processes, Cambridge University Press (1996)

From this point of view, you should also look atA.Kyprianou, Introductory Lectures on Fluctuations of Lévy Processeswith Applications, Springer-Verlag (2006)


For an insight into the wide range of both theoretical and appliedrecent work wherein Lévy processes play a role, consultO.E.Barndorff-Nielsen,T.Mikosch, S.Resnick (eds), Lévy Processes:Theory and Applications, Birkhäuser, Basel (2001)

For stochastic calculus with jumps, the authoritative treatise isP.Protter, Stochastic Integration and Differential Equations (secondedition), Springer-Verlag, Berlin Heidelberg (2003)


For an insight into the wide range of both theoretical and appliedrecent work wherein Lévy processes play a role, consultO.E.Barndorff-Nielsen,T.Mikosch, S.Resnick (eds), Lévy Processes:Theory and Applications, Birkhäuser, Basel (2001)

For stochastic calculus with jumps, the authoritative treatise isP.Protter, Stochastic Integration and Differential Equations (secondedition), Springer-Verlag, Berlin Heidelberg (2003)


For financial modelling I recommend:

R.Cont, P.Tankov, Financial Modelling with Jump Processes, Chapmanand Hall/CRC (2004)

which is extremely comprehensive and also contains a lot of valuablebackground material on Lévy processes.

W.Schoutens, Lévy Processes in Finance: Pricing FinancialDerivatives, Wiley (2003)

is shorter and aimed at a wider audience than mathematicians andstatisticians.














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