Interest rate modelling with non-Gaussian Ornstein-Uhlenbeck processes

Ornstein-Uhlenbek proesses

by

Marie Kaas Eriksen

THESIS

for the Masters degree of

Modelling and Data Analysis

(Master i modelleringogdataanalyse)

Faulty of Mathematis and Natural Sienes

University of Oslo

June 2008

Detmatematisk-naturvitenskapeligefakultet

Universiteteti Oslo

We study a mean-reverting model for interest rates. The model is an extension of the

Vasiekmodelandisasumofnon-GaussianOrnstein-Uhlenbekproesseswithsubordina-

tors,i.e. Lévyproesseswithonlypositivejumps,givingvariationoftheinterestrate. The

modelhavetheadvantagethatitgivesonlypositiveinterestrates,ontrarytotheVasiek

model. Wealulateexpliitresultsfortheharateristifuntionandtheautoorrelation

funtion oftheinterestrateforbothgeneralsubordinatorsand theasewherethesubor-

dinators are ompound Poisson. We alsond priesof zero-ouponbonds andEuropean

optionswritten on these bonds by applyingFouriermethods. It seems that themodel is

simpleenoughto allowfor analytialpriingof bondsand optionsin addition toapture

theharateristisoftheinterestrate. Intheend wedemonstratein asimulationhowthe

modelbehavewithertainvaluesofthevariables.

I would liketo thankmy patientsupervisor, FredEspen Benth, whoalwayshad timefor

metodisussanyproblem. Thethesisouldnothavebeenwrittenwithoutthehallenging

problemsandveryrelevantfeedbakhegaveme.

I would also like to thank family and friends, espeially my mother and father, who al-

wayssupported meduringmytimeasastudent.

Thankyou!

Oslo, May2008

MarieKaasEriksen

Abstrat i

Aknowledgements iii

1 Introdution 1

2 Some BasiTheory 3

2.1 MeasureTheoryandProbabilityTheory . . . 3

2.2 LévyProesses . . . 4

2.2.1 BrownianMotion . . . 5

2.2.2 LévyProesseswithJumps . . . 6

3 The Vasiek Model 9 3.1 SolutionandDistribution . . . 9

3.2 TheoretialAutoorrelationFuntion. . . 11

3.3 Zero-CouponBondPries . . . 12

4 Extension of the Vasiek Model 15 4.1 Solutionof

dX _k (t)

^and

r(t)

^{. . . . . . . . . . . . . . . . . . . . . . . . . . . 15}

4.2 CharateristiFuntion . . . 16

4.3 Moments . . . 21

4.4 TheoretialAutoorrelationFuntion. . . 25

4.5 ExpliitResultsforCompoundPoisson . . . 28

4.5.1 CharateristiFuntionof

r(t)

^{. . . . . . . . . . . . . . . . . . . . . 28}

4.5.2 TheTheoretialAutoorrelationFuntionof

r(t)

^{. . . . . . . . . . . 35}

5 Zero-Coupon BondPries 37 5.1 BondPries forGeneralSubordinators . . . 37

5.2 BondPries forCompoundPoissonSubordinators . . . 40

6 EuropeanBond Options 43 6.1 BondOptionPrieswith theVasiekModel . . . 43

6.2 BondOptionPrieswith ExtendedVasiekModel . . . 49

6.2.1 BondOptionPrieswithExtendedVasiekModel withGeneral

L _k (t)

⁴⁹ 6.2.2 BondOptionPrieswithExtendedVasiekModelwhen

L k (t)

^isCom- pound Poisson . . . 53

7 Simulation 57

7.1 SimulationoftheInterestRate . . . 57

7.1.1 Simulationof

r(t)

^{intheVasiekModel . . . . . . . . . . . . . . . . 57}

7.1.2 Simulationof

r(t)

^{intheExtendedVasiekModel . . . . . . . . . . 57}

7.2 SimulationofZero-CouponBondPries . . . 60

7.2.1 SimulationofZero-CouponBondPries intheVasiek Model . . . 60

7.2.2 SimulationofZero-CouponBondPriesintheExtendedVasiekModel 60

A MATLAB Files 63

Bibliography 67

Introdution

Tomodelinterestratesandprieinterestratederivativesonbondsaregreatanddemand-

ing areasin mathematialnane. Interest ratederivativesare instrumentswhose payo

dependsontheleveloftheinterestrate. Thevolumeoftradingininterestratederivatives

inreasedinthe1980sand1990s. Thehallengeistondmodelsapturingtheharater-

istis of theinterest rate in a reasonabledegree. It's important that they're analytially

tratableaswell. TheVasiekmodelisoneoftherstmodelsofterm-strutureandisstill

anattrativelassofmodelsbeauseofitsanalytialproperties. Howeverithastheprop-

ertythattheinterestrateanbenegative. OthermodelsderivedaretheCox-Ingersoll-Ross

(CIR) model and theHull and White model, where the latterhas time-dependent oe-

ients. TheCIR model is anextension ofthe Vasiek model, but hasthe advantagethat

it only gives positive values. Some important interest rate derivatives are interest rate

aps/oors,swapoptionsand bond options. Wewill onlyinvestigatebondoptionsinthis

thesis.

In this thesis we disuss two models of the short-term interest rate. Therst oneis

theVasiekmodel,andtheother isanextensionoftheVasiekmodel,proposedin[2℄for

modelling spoteletriitypries. It'smotivatedfrom [1℄. Themodelis asumofOrnstein-

Uhlenbek(OU) proesses,eahwithapurejumpproesswithonlypositivejumps. Itts

well formodellingspot eletriitypries, beause theyareoften dependent oftheseason.

Sine the proess is a sum of OU proesses, it seemsreasonable to takeone of them to

model theseasonality. Inourase,withinterestrates,it hastheadvantagethatitseure

theinterestratetobepositive. Themodelisalsosimpleenough,suhthatoneanalulate

analytialexpressionsforommoninterestratederivatives.

Both of the models we onsider are mean-reverting. That a model is mean-reverting

meansthatitwilleventuallypullbakto someaveragelevel.

Themainpartofthethesisistoexaminethenewlassofmodelsdesribedabove. We

ndanautoorrelationfuntionoftheinterestrategivenbyasumofweightedexponentials

of a onstanttimes the time shift. We want to nd out how easy it is to obtain results

of zero-oupon bond pries and pries of European options written on these bonds. It

turns outthat wean easily deriveexpliit resultsfor the prieof zero-ouponbonds by

looking attheexpressionfortheinterestratediretly. TondpriesofEuropeanoptions

writtenonzero-ouponbondsaremoreompliatedthanndingpriesofEuropeanoptions

written onotherseurities. Thatis beauseinterestratesareused fordisountingaswell

asfor dening the payo from theoption. Wend the priebyapplying inverse Fourier

transform,andonean usefastFouriertransformtehniques toomputeitfurther.

The thesis is organized as follows: In hapter 2we give some well-known denitions

and resultsfrom measure and probabilitytheory. Wealso introdue stohasti proesses

likeBrownianmotionandpurejump proessesand statesomeoftheirproperties. All the

resultsaregivenwithoutproof. Inhapter3weonsidertheVasiekmodelandndpries

ofzero-ouponbondsandbondoptions. Weintroduethenewmodel,theextensionofthe

Vasiek model, in hapter 4. The rest of the thesisis dediated to the extended Vasiek

model. Wend the stationary harateristis, harateristi funtion and theorrelation

funtion. Inhapter5and6wederivepriesofzero-ouponbonds andEuropeanoptions

written on these bonds. Westate allour resultsbothin generaland forthe speial ase

whentheLévyproessesareompoundPoisson. Inhapter7wesimulatetheinterestrate

andpriesofzero-ouponbondswithmaturityinoneyear.

Appendix A ontainsthematlables usedto simulatetheinterestrateand thepries

ofzero-ouponbonds.

Some Basi Theory

Beforewestartlookingatourproblemweneedsomebasitheory. Thetheorystatedinthis

hapteriswell-knownandweskiptheproofs. Moreinformationandproofsanbefoundin

anybook instohastianalysis. Firstweintroduethenotionofa

σ

^-algebra,aprobability measure and a probability spae. We state some well-known and useful theorems from

measuretheory. Intheendwedene astohasti proessandlook atLévyproessesand

theirproperties.

2.1 Measure Theory and Probability Theory

Denition2.1 If

Ω

^{isagiven set,thena}

σ

^-algebra

F

^on

Ω

^isafamily

F

^{of subsetsof}

Ω

with

• ∅ ∈ F

^.

• F ∈ F ⇒ F ^c ∈ F

^,where

F ^c = Ω \ F

^.

• F 1 , F 2 , · · · ∈ F ⇒ F =

∞

S

i=1

F _i ∈ F

^.

Thepair

(Ω, F )

^{isalledameasurable spae.}

Denition2.2 Aprobability measureon

(Ω, F )

^isafuntion

P : F → [0, 1]

^suhthat

• P ( ∅ ) = 0

^,

P (Ω) = 1

^.

•

^If

A 1 , A 2 , · · · ∈ F

^and

{ A _i } ^∞ i=1

^{isdisjoint, i.e.}

A _i ∩ A _j = ∅ , i 6 = j

^,then

P

∞

[

i=1

A i

!

=

∞

X

i=1

P (A i ) .

Thetriple

(Ω, F , P )

^{isalled a}probability spae. Wealsodenethenotionof altration.

Denition2.3 A ltration on a measurable spae

(Ω, F )

^{is an inreasing family of}

σ

^-

algebras

{F ^t } ∈ F

^suhthat

F ^s ⊆ F ^t

^,where

s ≤ t

^.

Altrationisinmathematialnaneusedtodesribetheinformationwegotuptilltoday.

Astimegoesby,weknowmoreandmore. Soithastobeinreasing.

Wemakethefollowingassumptionthroughoutthethesis:

Assumption2.1 All our models are modelled diretly under the risk-neutral probability

measure

Q

^andthe probability spae we're working in, is

(Ω, F , Q)

^.

Thenextthoremwillbeusedtoput thelimitoutsideexpetationswhen omputinghar-

ateristifuntions.

Theorem2.1 Bounded Convergene theorem

Let

µ(Ω) < ∞

^{. If thereexists a}

0 < k < ∞

^{suh that}

| f n | ≤ k µ

^{-a.e. and}

f n → f µ

^-a.e.

then

n→∞ lim Z

f _n dµ = Z

f dµ,

and

n→∞ lim Z

| f n − f | dµ = 0.

Fubini's theoremallowsusto hangetheorderofintegrals.

Theorem2.2 Fubini's theorem

Let

(Ω i , F ⁱ , µ i )

^,

i = 1, 2

^,be

σ

^{-nitemeasurablespaes(i.e. thereexistaountableolletion}

ofsets

A ⁱ ₁ , A ⁱ ₂ , . . . , ∈ F ⁱ

^suhthat

∪ ^n≥1 A ⁱ _n = Ω

^and

µ i (A ⁱ _n ) < ∞

^forall

n ≥ 1

^and

i = 1, 2

⁾

andlet

f ∈ L ¹ (Ω 1 × Ω 2 , F ¹ × F ² , µ 1 × µ 2 )

^{. Then}

Z

Ω 1

Z

Ω 2

f dµ 2

dµ 1 =

Z

Ω 2

Z

Ω 1

f dµ 1

dµ 2 .

Wenowdenethenotionofastohastiproess.

Denition2.4 Astohastiproessisaparametrizedolletionofrandomvariables

{ X t }

^,

t ∈ T

^denedona probability spae

(Ω, F , P )

^{andassuming valuesin}

R ⁿ

^.

2.2 Lévy Proesses

TheLévyproessisanexampleofastohastiproess. It'snamedafterthemathematiian

PaulLévy. ALévyproess

L _t

^{hasthefollowingproperties}

• L 0 = 0.

• L t

^{hasstationaryinrements,i.e. the}probabilitydistributionofanyinrement

L t − L s

^,

dependsonlyonthelength

t − s

^.

• L _t

^hasindependentinrements,i.e. anytwonon-overlappinginrementsareindepen- dentofeah other.

We willuse theharateristi funtion of aLévyproess, givenby theLévy-Khinhin

Theorem2.3 Lévy-Khinhin representation

Let

(X t )

^{beaLévyproesson}

R

^withharateristitriplet

(A, ν, γ)

^{. Then the}harateristi funtion of

(X t )

^is

E e ^{izX t}

= e ^ψ(z)t , z ∈ R ^d ,

where

ψ(z) = − 1

2 z.Az + iγ.z + Z

R

e ^izx − 1 − izx

¹

_|x|≤1 ν(x)dx.

A

^{istheovarianematrixofaBrownianmotion,}

ν

^{istheLévymeasureand}

γ

^isthedrift.

2.2.1 Brownian Motion

A Brownian motion

B t

^{is an example of aLévy proess. It wasrst studied by Robert}

Brownin 1827. He wasstudying pollenpartiles oating in water under themirosope.

Brownianmotion is often usedbeause itmakesomputations simple, notbeause of its

auray. ItisaLévyproess,soitsatisesallthepropertiesabove,butitalsosatises

• B t − B s ∼ N (0, t − s)

^.

ForaBrownianmotion,theharateristitripletis

(1, 0, 0)

^,so,fromLévy-Khinhinrepre- sentation, theharateristifuntion ofaBrownianmotion

B(t)

^beomes

E h e ^izB(t) i

= e ^{− 1 2 z 2 t}

^(2.1)

A simple, but useful tool for solving stohasti dierential equations (SDE) is the It

formula.

Theorem2.4 The one - dimensional It formula.

Let

X t

^{bean Itproess wherethe dynamis isgiven by}

dX t = udt + vdB t .

Let

f (t, x) ∈ C ² ([0, ∞ ] × R)

^{, i.e.}

f

^{is twotimes}ontinuously dierentiable on

[0, ∞ ] × R

^.

Then

Y t = f (t, X t )

isagainan It proess,and

dY _t = f _t (t, X t )dt + f _x (t, X t )dX t + 1

2 f _xx (t, X t )(dX t ) ² ,

where

f _t (t, X t ) = ∂

∂t f (t, X t ), f _x (t, X t ) = ∂

∂x f (t, X t ), f _xx (t, X t ) = ∂ ²

∂x ² F (t, X t ),

and

(dX t ) ²

^{isomputedaording tothe rules}

dt · dt = dt · dB _t = dB _t · dt = 0, dB _t · dB _t = dt.

Proof. For proof,lookin Øksendal[4℄.

Anotherusefulproperty,whihwewillusetondthevarianeoftheVasiekmodel,is

Lemma2.1 It isometry.

E



 Z T

S

f (t, ω)dB t

! 2 

 = E

"

Z T S

f ² (t, ω)dt

# ,

for all

f

^{in the lassof funtions}

g(t, ω) : [0, ∞ ) × Ω → R

suhthat

• (t, ω) → g(t, ω)

^is

B × F

-measurable,where

B

^{isthe Borel}

σ

^-algebraon

[0, ∞ )

^.

• g(t, ω)

^is

F t

^{-adapted,i.e.}

g(t, ω)

^is

F t

-measurablefor all

t

^.

• E h R T

S g ² (t, ω)dt i

< ∞

^.

In the setion disussion the priing of European bond options we make use of the

Girsanovtheoremto hangemeasure.

Theorem2.5 Girsanov's theorem

Let

B(t)

^{be astandardbrownian motion on a}probability spae

(Ω, F , P)

^{. Supposethat}

γ u

isameasurableproess suhthat

E _P h

e ^{R 0 T γ u dB(u)− 1 2 R 0 T |γ u | 2 du} i

= 1.

^(2.2)

Deneaprobability measure

P ˜

^on

(Ω, F T )

^equivalentto

P

^bymeansofthe Radon-Nikodým derivative

d P ˜

dP = e ^{R 0 T γ u dB(u)− 1 2 R 0 T |γ u | 2 du} , P − a.s.

Then the proess

B(t) ˜

^{given bythe formula}

B(t) = ˜ B(t) −

Z t 0

γ u du,

for all

t ∈ [0, T ]

^{,follows astandardBrownian motionon the spae}

(Ω, F , ˜ P)

^.

Asuientonditionfor(2.2)toholdis theNovikovondition:

E _P h

e ^{1 2 R 0 T |γ u | 2 du} i

< ∞ .

2.2.2 Lévy Proesses with Jumps

To ndan expliitrepresentationof

r(t)

^{in the extended Vasiekmodelwewill need the}

ItformulaforsalarLévyproesses.

Proposition 2.1 It formula for salar Lévy proesses

Let

(X t ) t≥0

^{be a Lévy proess and}

f : R → R

^a

C ²

^{-funtion(i.e.}

2

^timesdierentiable withontinuousderivatives). Then

f (X t ) = f (0) + Z t

0

σ ² 2

d ²

dx ² f (X s )ds + Z t

0

d

dx f (X s− )dX s

+ X

0≤s≤t

∆X s 6=0

f (X s− + ∆X s ) − f (X s− ) − ∆X s

d

dx f (X s− )

.

Wearegoing tousethegeneralizedasewhere

f

^{alsodependsontime. Let}

(X t ) t≥0

^bea

Lévyproessandlet

f : [0, T ] × R → R

^bea

C ^1,2

^-funtion(i.e

1

^timedierentiablein the rstvariableand

2

^timesdierentiablein theseond). Then

f (t, X t ) = f (0, X 0 ) + Z t

0

∂f

∂x (s, X s− )dX s + Z t

0

∂f

∂s (s, X s ) + σ ² 2

∂ ² f

∂x ² (s, X s )

ds

+ X

0≤s≤t

∆X s 6=0

f (s, X s− + ∆X s ) − f (s, X s− ) − ∆X s

∂f

∂x (s, X s− )

.

TheLévyproessesintheextendedVasiekmodelaresubordinators. Thatis,theyare

jump proesseswithonlypositivejumps. Theharateristitripletis then

(0, ν, 0)

^{. From}

Lévy-Khinhinrepresentationtheharateristifuntion ofsuhproessesis

E h e ^izL(t) i

= e ^ψ(z)t ,

^(2.3)

where

ψ(z) = R

R e ^izx − 1 − izx

¹

|x|≤1

ν (x)dx

^.

AnexampleofasubordinatoristheompoundPoissonproessandisdenedasfollows

Denition2.5 AompoundPoisson proesswith intensity

λ > 0

^{andjump sizedistribu-}

tion

f

^{isastohasti proess}

X t

^denedas

X t =

N t

X

i=1

Y i ,

where the jump sizes

Y _i

^areindependent and idential distributed(i.i.d.) with distribution

f

^,and

(N t )

^{isaPoisson proess withintensity}

λ

^,independent from

(Y i ) i≥1

^.

The Vasiek Model

ThemodelanalysedbyVasiekin1977isoneoftherstmodelsoftermstruture. It has

somequalitiesthatmakesitattrative. Itislinearandanthereforebesolvedexpliitly. Its

distributionisGaussian,andzero-ouponbondsandotherderivativesareeasily obtained.

However,ahugedrawbakisthatitallowstheinterestratetobenegative.

TheVasiekmodeltakestheform

dr(t) = (µ − αr(t))dt + σdB(t).

^(3.1)

It's a mean-reverting Ornstein-Uhlenbek proess where

B(t)

^{is a Brownian motion and}

where

µ, α

^and

σ

^{arestritlypositiveonstants. Thattheproessis}mean-revertingmeans thatitwilleventuallypullbaktowardssomelong-runaveragelevel. Thatis,iftheinterest-

rateis higherthan theexpeted, itwill tendtoderease, and ifitislower,itwill tendto

inrease.

3.1 Solution and Distribution

Thesolutionofthestohastidierentialequationaboveisgivenbythefollowingproposi-

tion.

Proposition 3.1 The solutionof (3.1)isgiven by

r(t) = r(s)e ^−α(t−s) + µ α

1 − e ^−α(t−s) + σ

Z t s

e ^−α(t−u) dB(u),

^(3.2)

wherethe proess startsattime

s ≤ t

^.

Proof. Toprovethepropositionwehaveto usetheItformula (Theorem2.4). If weuse

It'sformulaon

e ^αt r(t)

^{andinsertthedynamisof}

r(t)

^{from(3.1)weget}

d(e ^αt r(t)) = αe ^αt r(t)dt + e ^αt dr(t) = e ^αt [µdt + σdB(t)].

So,

e ^αt r(t) − e ^αs r(s) = µ Z t

s

e ^αu du + σ Z t

s

e ^αu dB(u).

Andnallywegetthesolutionof

r(t)

^:

r(t) = r(s)e ^−α(t−s) + µ

α

1 − e ^−α(t−s) + σ

Z t s

e ^−α(t−u) dB(u)

Weannowndthedistributionof

r(t)

^.

Proposition 3.2 Theproess

r(t)

^{,givenby (3.2),isGaussian}distributedwithexpetation

E [r(t)] = r(s)e ^−α(t−s) + µ α

1 − e ^−α(t−s)

andvariane

Var [r(t)] = σ ² 2α

1 − e ^{−2α(t−s)} .

Andwhentimegoestoinnityweget

t→∞ lim E [r(t)] = µ α

and

t→∞ lim Var [r(t)] = σ ² 2α .

Proof. Sine

e ^−α(t−u)

^isdeterministi,wegetfromthepropertiesofBrownianmotionand It isometry that

σ R t

s e ^−α(t−u) dB(u)

^{is Gaussian with expeted value zero and variane}

givenby

Var

σ Z t

s

e ^−α(t−u) dB(u)

= E

"

σ Z t

s

e ^−α(t−u) dB(u) ² #

− E

σ Z t

s

e ^−α(t−u) dB(u) ²

= σ ² Z t

s

e ^{−2α(t−u)} du = σ ² 2α

1 − e ^{−2α(t−s)} .

Itfollowsthattheproess

r(t)

^isGaussiandistributedwith

E [r(t)] = r(s)e ^−α(t−s) + µ α

1 − e ^−α(t−s)

and

Var [r(t)] = σ ² 2α

1 − e ^{−2α(t−s)} .

Findingthepropertiesof

r(t)

^{whentimegoestoinnityisstraightforward,}

t→∞ lim E [r(t)] = lim

t→∞

h r(s)e ^−α(t−s) + µ α

1 − e ^−α(t−s) i

= µ α

and

t→∞ lim Var [r(t)] = lim

t→∞

σ ² 2α

1 − e ^{−2α(t−s)}

= σ ² 2α .

Andtheproofisomplete.

3.2 The Theoretial Autoorrelation Funtion of

r ( t )

Wewanttoalulatethetheoretialautoorrelationfuntionfor

r(t)

^{. The}autoorrelation funtion says something about the degree of similarity between

r(t)

^{and a time shifted}

versionof itself. It an takevaluesin theinterval

[ − 1, 1]

^,where

1

^{means perfet positive}

orrelation,and

− 1

^{meansperfet negative}orrelation.

Theorrelationfuntionof r(t)isdenedby

corr (r(t), r(t + τ)) = E [(r(t) − E [r(t)]) (r(t + τ) − E [r(t + τ)])]

p Var [r(t)] Var [r(t + τ)]

= E [r(t)r(t + τ)] − E [r(t)] E [r(t + τ )]

p Var [r(t)] Var [r(t + τ)] .

Weomputethepartsseperately. Firstlookat

E [r(t)] E [r(t + τ )]

^{. Weusetheexpetation}

derivedin theprevioussetioninProposition (3.2).

E [r(t)] E [r(t + τ)]

=

r(s)e ^−α(t−s) + µ

α (1 − e ^−α(t−s) ) r(s)e ^{−α(t+τ−s)} + µ

α (1 − e ^{−α(t+τ−s)} )

= B.

Thenwelookat

E [r(t)r(t + τ)]

^{. Toalulatethispart,notiethatwefromtheproperties}

ofBrownianmotionhavethat

E h R t

s e ^−α(t−u) dB(u) i

iszero,sine

e ^−α(t−u)

^isdeterministi.

Rememberthat

r(t)

^{isgivenby(3.2).}

E [r(t)r(t + τ)] = E r(s)e ^−α(t−s) + µ

α (1 − e ^−α(t−s) ) + σ Z t

s

e ^−α(t−u) dB(u)

×

r(s)e ^{−α(t+τ−s)} + µ

α (1 − e ^{−α(t+τ−s)} ) + σ Z t+τ

s

e ^{−α(t+τ−u)} dB(u)

= B + σ ² E Z t

s

e ^−α(t−u) dB(u) Z t+τ

s

e ^{−α(t+τ−u)} dB(u)

.

Theexpetationin thelast termoftheaboveequation anbeomputedas

E Z t

s

e ^−α(t−u) dB(u) Z t+τ

s

e ^{−α(t+τ−u)} dB(u)

= E Z t

s

e ^−α(t−u) dB(u) Z t

s

e ^{−α(t+τ−u)} dB(u) + Z t+τ

t

e ^{−α(t+τ−u)} dB(u)

= E

"

e ^−ατ Z t

s

e ^−α(t−u) dB(u) ² #

+ E Z t

s

e ^−α(t−u) dB(u)

E Z t+τ

t

e ^{−α(t+τ−u)} dB(u)

= e ^−ατ E Z t

s

e ^{−2α(t−u)} du

= 1

2α e ^−ατ (1 − e ^{−2α(t−s)} ).

Hereweusedthe fat that

R t

s e ^−α(t−u) dB(u)

^and

R t+τ

t e ^{−α(t+τ−u)} dB(u)

^are independent

Let'sputitalltogether. Thenweget

E [r(t)r(t + τ)] − E [r(t)] E [r(t + τ)] = B + σ ² 2α e ^−ατ

1 − e ^{−2α(t−s)}

− B

= σ ² 2α e ^−ατ

1 − e ^{−2α(t−s)} .

Let'slookat

Var [r(t)] Var [r(t + τ)]

^{. Weusethevarianederivedintheprevioussetionin}

Proposition3.2

Var [r(t)] Var [r(t + τ)] = σ ² 2α

1 − e ^{−2α(t−s)} σ ² 2α

1 − e ^{−2α(t+τ−s)}

= σ ⁴ 4α ²

1 − e ^{−2α(t−s)} 1 − e ^{−2α(t+τ−s)} .

Proposition 3.3 The orrelation funtionof

r(t)

^is

corr (r(t), r(t + τ)) = e ^−ατ 1 − e ^{−2α(t−s)} q

1 − e ^{−2α(t−s)}

1 − e ^{−2α(t+τ−s)} .

Whentime goes toinnitythe orrelationfuntion of

r(t)

^tendsto

t→∞ lim corr (r(t), r(t + τ)) = e ^−ατ .

3.3 Zero-Coupon Bond Pries

Weareinterestedinndingthepriesofzero-ouponbonds,whereweassumethat

r(t)

^is

modelled diretlyunder therisk-neutralprobability measure

Q

^{. A zero-ouponbondis a}

bondpaying

1

^{urrenyat afuture time}

T

^{withnoouponspaidinbetween.}

Denition3.1 Theprieof azero-ouponbondattime

t ≤ T

^is

P(t, T ) = E Q

h

e ^{− R t T r(s)ds} | F ^t i .

Tondtheprieweneedtoevaluate

− R T

t r(s)ds

^{. Asin (3.2),}

r(s)

^isgivenby

r(s) = r(t)e ^−α(s−t) + µ

α

1 − e ^−α(s−t) + σ

Z s t

e ^−α(s−u) dB(u),

wheretheproessstartsattime

t ≤ s

^.

− Z T

t

r(s)ds = − Z T

t

r(t)e ^−α(s−t) ds − µ α

Z T t

1 − e ^−α(s−t) ds

− σ Z T

t

Z s t

e ^−α(s−u) dB(u)ds

= − I 1 − I 2 − I 3 .

Tomakeiteasier,weomputetheintegralsseperately. Westartwith

I 1

^and

I 2

^{. It'seasy}

toseethat

I 1 = r(t) Z T

t

e ^−α(s−t) ds = r(t) 1 α

1 − e ^{−α(T −t)}

(3.3)

I 2 = µ α

Z T t

1 − e ^−α(s−t) ds = µ

α (T − t) − µ α ²

1 − e ^{−α(T −t)}

.

^(3.4)

Thenlookat thelast integral

I 3

^.

I 3 = σ Z T

t

Z s t

e ^−α(s−u) dB(u) ds = σ Z T

t

e ^αu Z T

u

e ^−αs ds dB(u)

= σ Z T

t

1 α

1 − e ^{−α(T −u)} dB(u).

Hereweapplied Fubini'stheoremto hangetheorderoftheintegrals. Denenow

n(t, T ) = 1 α

1 − e ^−α(T−t) .

Thenweget

− Z T

t

r(s)ds = − r(t)n(t, T ) − µ 1

α (T − t) − 1 α n(t, T )

− σ Z T

t

n(u, T )dB(u).

Nowalulate

Z T t

n(u, T )du = Z T

t

1 α

1 − e ^{−α(T −u)} du = 1

α (T − t) − 1 α ²

1 − e ^{−α(T −t)}

= 1

α (T − t) − 1

α n(t, T ),

sowegetthat

− Z T

t

r(s)ds = − r(t)n(t, T ) − µ Z T

t

n(u, T )du − σ Z T

t

n(u, T )dB(u).

Tomakethenotationeasier,let

ξ _T = − R T

t r(s)ds

^.

n(t, T )

^isdeterministi,sowehavethat

σ R T

t n(u, T )dB(u)

^{isGaussianwithexpetationzeroandvariane}

σ ² R T

t n ² (u, T )du

^byIt

isometry,Lemma2.1. Alsonotiethat

R T

t n(u, T )dB(u)

^isindependentof

F ^t

^andthat

r(t)

is

F ^t

-measurable. Wethereforehave

P (t, T ) = E Q e ^{ξ T} |F ^t

= E Q e ^{ξ T}

= e ^{−r(t)n(t,T )−µ R t T n(u,T)du} E Q

h e ^{−σ R t T} n(u,T)dB(u) i

= e ^{−r(t)n(t,T )−µ R t T n(u,T)du+ 1 2 σ 2 R t T n 2 (u,T)du} .

Proposition 3.4 The zero-ouponbondprie, when

r(t)

^{isgiven by (3.2), is}

P(t, T ) = e ^m(t,T )−n(t,T)r(t) ,

where

n(t, T ) = 1 α

1 − e ^{−α(T −t)}

and

m(t, T ) = 1 2 σ ²

Z T t

n ² (u, T )du − µ Z T

t

n(u, T )du.

Extension of the Vasiek Model

with Subordinators

Inthis hapter,and therest ofthethesis, wearegoing toinvestigatean extensionof the

Vasiekmodeloftheform

r(t) =

n

X

k=1

w k X k (t),

^(4.1)

where

dX _k (t) = − α _k X _k (t)dt + dL _k (t),

^(4.2)

and

X _k (0) = r(0)

w 1

if

k = 1 0

^if

k ≥ 2 .

Here,

L k (t), k = 1, 2, . . . , n

^are subordinators. Havingseveral

X k

^{'s, rather thanjust one,}

gives an opportunityto apture dierent fators with inuene onthe interest rate

r(t)

^.

The model havethe advantagethat it always givespositiveinterest rates,something the

Vasiek model fails to do. It is, as mentioned before, proposed to model spot eletriity

priesin [2℄.

4.1 Solution of

dX _k ( t )

^and

r ( t )

To nd an expliit solution of (4.2) we use the It formula for salar Lévy proesses,

Proposition2.1. ApplyingIt'sformulaon

f (t, X k (t)) = e ^{α k t} X _k (t)

^weget

e ^{α k t} X _k (t) − e ^{α k s} X _k (s)

= Z t

s

e ^{α k u} dX _k (u) + Z t

s

α _k e ^{α k u} X _k (u)du

+ X

s≤u≤t

∆X k (u)6=0

[e ^{α k u} (X k (u − ) + ∆X k (u)) − e ^{α k u} X k (u − ) − ∆X k (u)e ^{α k u} ]

= Z t

s

e ^{α k u} [ − α _k X _k (u)du + dL _k (u)] + Z t

s

α _k e ^{α k u} X _k (u)du

= Z t

s

e ^{α k u} dL k (u).

Soanexpliitsolutionof (4.2)beomes

X _k (t) = X _k (s)e ^{−α k (t−s)} + Z t

s

e ^{−α k (t−u)} dL _k (u),

^(4.3)

wheretheproessstartsatageneraltime

s ≤ t

^{. Wewant}

r(t)

^{tostarttoday,so set}

s = 0

^.

Ifweput

X k (t)

^{intotheexpressionfor}

r(t)

^weget

r(t) =

n

X

k=1

w _k

X _k (0)e ^{−α k t} + Z t

0

e ^{−α k (t−u)} dL _k (u)

= w 1

r(0) w 1

e ^{−α 1 t} +

n

X

k=1

w k

Z t 0

e ^{−α k (t−u)} dL k (u)

= r(0)e ^{−α 1 t} +

n

X

k=1

w k

Z t 0

e ^{−α k (t−u)} dL k (u).

Proposition 4.1 An expliitsolution of

r(t)

^{startingattime}

0

^,is

r(t) = r(0)e ^{−α 1 t} +

n

X

k=1

w k

Z t 0

e ^{−α k (t−u)} dL k (u)

^(4.4)

4.2 The Charateristi Funtion of

r ( t )

We want to nd the harateristi funtion of

r(t)

^{. The} harateristi funtion denes ompletelythe distributionof anyrandomvariable. Generally,theharateristifuntion

ofarandomvariable

X

^isgivenby

ϕ X (z) = E e ^izX

.

Sofor

r(t)

^{wehavetoompute}

E e ^izr(t)

;

E h e ^izr(t) i

= e ^{izr(0)e −α 1 t} E h

e ^{iz P n k=1 w k R 0 t e −αk (t−u) dL k (u)} i .

First wetakealookat

E h

e ^{iz P n k=1 w k R 0 t e −αk (t−u) dL k (u)} i

. Todoso,let

f _k (u) = e ^{−α k (t−u)}

^.

The

L k

^'sareindependentofeahother, so

E h

e ^{iz P n k=1 w k R 0 t f k (u)dL k (u)} i

=

n

Y

k=1

E h

e ^{izw k R 0 t f k (u)dL k (u)} i .

Let

{ u j } ^m j=1

^{be any partition of the interval [}

0, t

^{℄ with}

max j | u j+1 − u j | < ǫ

^{. Then the}

integralanbewrittenas

Z t 0

f _k (u)dL k (u) = lim

ǫ→0 m

X

j=1

f _k (u j )∆L k (u j ),

where

∆L k (u j ) = L k (u j+1 ) − L k (u j )

^and,sine

e ^g(t)

^{isaontinuousfuntion,weget}

n

Y

k=1

E h

e ^{izw k R 0 t f k (u)dL k (u)} i

=

n

Y

k=1

E h

ǫ→0 lim e ^{izw k P m j=1 f k (u j )∆L k (u j )} i .

The BoundedConvergene Theorem is applied to takethe limit outsidethe expetation.

Notiethat

∆L k (u j )

^areindependentof

∆L k (u j+1 )

^forall

j

^{sinetheLévyproesses,}

L _k (u)

^,

haveindependentinrements. Thus

n

Y

k=1

E h

ǫ→0 lim e ^{izw k P m j=1 f k (u j )∆L k (u j )} i

=

n

Y

k=1 ǫ→0 lim E h

e ^{izw k P m j=1 f k (u j )∆L k (u j )} i

=

n

Y

k=1 ǫ→0 lim

m

Y

j=1

E h

e ^{izw k f k (u j )∆L k (u j )} i .

Generally,wehavethatforaLévyproess

L(t)

^,theharateristifuntionis

E e ^izL(t)

= e ^ψ(z)t

^{, by} Lévy-Khinhin representation (Theorem 2.3). It follows that

E

e ^iz∆L(t)

= e ^ψ(z)∆u

^{, where}

ψ(z) = R

R e ^izx − 1 − izx

¹

|x|≤1

ν(x)dx

^{, sine the proesses havehara-}

teristitriplet

(0, ν, 0)

^{. Finally}

n

Y

k=1 ǫ→0 lim

m

Y

j=1

E h

e ^{izw k f k (u j )∆L k (u j )} i

=

n

Y

k=1 ǫ→0 lim

m

Y

j=1

e ^{ψ(zw k f k (u j ))∆u j}

=

n

Y

k=1

ǫ→0 lim e ^{P m j=1 ψ(zw k f k (u j ))∆u j}

=

n

Y

k=1

e ^{R 0 t ψ(zw k f k (u))du}

= e ^{P n k=1 R 0 t ψ(zw k f k (u))du} .

Weputitalltogetherinaproposition.

Proposition 4.2 The harateristifuntion of

r(t)

^{isgiven by}

E h

e ^izr(t) i

= e ^{izr(0)e −α 1 t} e ^{P n k=1 R 0 t ψ(zw k f k (u))du} ,

^(4.5)

where

ψ(zw k f k (u)) = Z

R

e ^{izw k f k (u)x} − 1 − izw k f k (u)x

¹

|x|≤1

ν (x)dx

and

f k (u) = e ^{−α k (t−u)}

^.

Next we nd the expetation and variane of

r(t)

^{. We state the result in a} proposition beforeweproveit.

Proposition 4.3 The expetationandvarianeof

r(t)

^{aregiven by}

E [r(t)] = r(0)e ^{−α 1 t} +

n

X

k=1

w k

α _k 1 − e ^{−α k t} Z

R

x − x

¹

|x|≤1

ν(x)dx,

^(4.6)

and

Var [r(t)] =

n

X

k=1

w _k ² 2α k

1 − e ^{−2α k t} Z

R

x ² ν(x)dx.

^(4.7)

Whentimegoestoinnitytheybeome

t→∞ lim E [r(t)] =

n

X

k=1

w _k α _k

Z

R

x − x

¹

_|x|≤1 ν (x)dx

and

t→∞ lim Var [r(t)] =

n

X

k=1

w _k ² 2α k

Z

R

x ² ν(x)dx.

Intheproof wewill makeuseof thefollowingorollary. We stateit withoutproofbefore

provingProposition4.3.

Corollary4.1 If

X

^{isarandomvariablewith}harateristifuntion

ϕ X (z) = E[e ^izX ]

^one

annd it's

n

^{'thmoment byusing the formula}

E[X ⁿ ] = (i) ⁻ⁿ d ⁿ

dz ⁿ E [ϕ X (z)]

_z=0 .

Proof of Proposition4.3. Westartwiththeexpetation. Notiethat

L k

^isindependentof

L j

^when

k 6 = j

^{. Rememberthat}

r(t)

^{isgivenby(4.4).}

E [r(t)] = E

"

r(0)e ^{−α 1 t} +

n

X

k=1

w _k Z t

0

e ^{−α k (t−u)} dL _k (u)

#

= r(0)e ^{−α 1 t} +

n

X

k=1

w _k E Z t

0

e ^{−α k (t−u)} dL _k (u)

.

Wehavetond

E h R t

0 e ^{−α k (t−u)} dL _k (u) i

. Todothenotationeasier,let

f _k (u) = e ^{−α k (t−u)}

^.

From the alulations leading up to Proposition 4.2, we know that the harateristi

funtion of

R t

0 f _k (u)dL k (u)

^{, with}

ψ(zf k (u)) = R

R e ^{izf k (u)x} − 1 − izf _k (u)x

¹

|x|≤1

ν (x)dx

^,

is

e ^{R 0 t ψ(zf k (u))du}

^{. SobyCorollary4.1}

E

Z t 0

f _k (u)dL k (u)

= − i d dz E h

e ^{iz R 0 t f k (u)dL k (u)} i z=0

= − i d

dz e ^{R 0 t ψ(zf k (u))du} z=0

= − i d

dz Z t

0

ψ(zf k (u))du

e ^{R 0 t ψ(zf k (u))du} _z=0

= − i Z t

0

d

dz ψ(zf k (u))du e ^{R 0 t ψ(zf k (u))du} z=0

,

with

ψ(zf k (u))

^{asabove. Wehavethat}

ψ(0) = 0

^,so

e ^{R 0 t ψ(zf k (u))du} | ^z=0 = 1

^{. Further}

E

Z t 0

f _k (u)dL k (u)

= − i Z t

0

d

dz ψ(zf k (u))du _z=0

(4.8)

Wethenneedtoompute

d

dz ψ(zf k (u)) | ^z=0

^.

d

dz ψ(zf k (u)) z=0

= Z

R

if _k (u)xe ^{izf k (u)x} − ixf _k (u)

¹

|x|≤1

ν(x)dx z=0

= Z

R

if _k (u)x − ixf _k (u)

¹

|x|≤1

ν(x)dx.

Weget

E Z t

0

f _k (u)dL k (u)

= 1

α _k 1 − e ^{−α k t} Z

R

x − x

¹

_|x|≤1

ν (x)dx,

^(4.9)

whih isaresultofthefollowingomputation

E Z t

0

f _k (u)dL k (u)

= − i Z t

0

Z

R

if _k (u)x − ixf _k (u)

¹

|x|≤1

ν (x)dx

du

= Z t

0

f k (u) Z

R

x − x

¹

|x|≤1

ν(x)dx

du

= Z

R

x − x

¹

|x|≤1 ν (x)dx

Z t 0

f k (u)du

= Z

R

x − x

¹

_|x|≤1 ν (x)dx

Z t 0

e ^{−α k (t−u)} du

= Z

R

x − x

¹

_|x|≤1

ν (x)dx 1 α k

1 − e ^{−α k t} .

Ifweputitalltogether,wegetthat

E [r(t)] = r(0)e ^{−α 1 t} +

n

X

k=1

w _k α k

1 − e ^{−α k t} Z

R

x − x

¹

_|x|≤1

ν(x)dx.

Thisproves(4.6). Lettingtimegoto innity,

lim t→∞ e ^{−α k t} = 0

^gives

t→∞ lim E [r(t)] =

n

X

k=1

w k

α k

Z

R

x − x

¹

|x|≤1

ν (x)dx.

Itremainstoshowthatthevarianeisgivenby(4.7). Let

f k (u)

^{stillbegivenby}

e ^{−α k (t−u)}

^.

Notiethat

L k

^isindependentof

L j

^when

k 6 = j

^{. Wethennd}

Var [r(t)] = Var

"

r(0)e ^{−α 1 t} +

n

X

k=1

w k

Z t 0

f k (u)dL k (u)

#

= Var

" _n X

k=1

w k

Z t 0

f k (u)dL k (u)

#

=

n

X

k=1

w ² _k Var Z t

0

f k (u)dL k (u)

.

We want to ompute

Var h R t

0 f _k (u)dL k (u) i

. Generally, we have that

Var[X ] = E[X ² ] − E[X ] ²

^{. So}

Var h

R t

0 f k (u)dL k (u) i

= E

R t

0 f k (u)dL k (u) 2

− E h R t

0 f k (u)dL k (u) i 2

. Compute

the

n

^{'thmomentofarandomvariable}

X

^{byusingtheformula}

E[X ⁿ ] = (i) ^{−n d} _dz ⁿ n ϕ X (z) | ^z=0

^,

where

ϕ X (z)

^istheharateristifuntionof

X

^{. Wehavealreadyalulatedtheharater-}

istifuntionof

R t

0 f k (u)dL k (u)

^{,whendealingwiththe}harateristifuntionof

r(t)

^before

Proposition4.2. Hene

E

" Z t 0

f k (u)dL k (u) ² #

= (i) ⁻² d ² dz ² E h

e ^{iz R 0 t f k (u)dL k (u)} i _z=0

= − d ²

dz ² e ^{R 0 t ψ(zf k (u))du} _z=0

= − d dz

Z t 0

d

dz ψ(zf k (u))du e ^{R 0 t ψ(zf k (u))du} _z=0

= −

"

Z t 0

d

dz ψ(zf k (u))du ²

e ^{R 0 t ψ(zf k (u))du}

+ Z t

0

d ²

dz ² ψ(zf k (u))du e ^{R 0 t ψ(zf k (u))du}

# _z=0

= −

" Z t 0

d

dz ψ(zf k (u))du ²

+ Z t

0

d ²

dz ² ψ(zf k (u))du

# z=0

,

where weused that

e ⁰ = 1

^{. We havethat}

R t 0

d

dz ψ(zf k (u))du

z=0 = i E h R t

0 f k (u)dL k (u) i

from(4.8). Rememberthat

E h R t

0 f _k (u)dL k (u) i

isgivenby(4.9). Wethenget

Z t 0

d

dz ψ(zf k (u))du z=0

²

= − 1

α ² _k 1 − e ^{−α k t} 2 Z

R

x − x

¹

|x|≤1

ν (x)dx 2

Soit remains to ompute

R t 0

d ²

dz ² ψ(zf k (u))du | ^z=0

^{. Werst omputethe expression inside}

theintegral;

d ²

dz ² ψ(zf k (u)) z=0

= d ² dz ²

Z

R

e ^{izf k (u)x} − 1 − ixzf _k (u)

¹

|x|≤1

ν(x)dx

z=0

= Z

R

d dz

h

if k (u)xe ^{izf k (u)x} − ixf k (u)

¹

|x|≤1

ν (x) i

dx z=0

= − Z

R

f _k ² (u)x ² ν (x)dx.

Rememberthat

f k (u) = e ^{−α k (t−u)}

^{. It followsthat}

Z t

0

d ²

dz ² ψ(zf k (u))du z=0

= − Z t

0

Z

R

f _k ² (u)x ² ν(x)dx

du

= − Z

R

x ² ν(x)dx Z t

0

f _k ² (u)du

= − Z

R

x ² ν(x)dx Z t

0

e ^{−2α k (t−u)} du

= − Z

R

x ² ν(x)dx 1 2α k

1 − e ^{−2α k t}

Goingbaktoourexpressionfor

E

R t

0 f k (u)dL k (u) 2

,andusewhatwehavefound,we

get

E

"

Z t 0

f k (u)dL k (u) ² #

= 1

α ² _k 1 − e ^{−α k t} 2 Z

R

x − x

¹

|x|≤1

ν (x)dx 2

+ 1 2α k

1 − e ^{−2α k t} Z

R

x ² ν(x)dx.

From(4.9)itfollowsdiretlythat

E Z t

0

f _k (u)dL k (u) ²

= 1

α ² _k 1 − e ^{−α k t} 2 Z

R

x − x

¹

_|x|≤1 ν (x)dx

2

.

Tomaketheproofof (4.7)omplete;

Var [r(t)] =

n

X

k=1

w _k ² Var Z t

0

f k (u)dL k (u)

=

n

X

k=1

w _k ² E

" Z t 0

f _k (u)dL k (u) ² #

− E Z t

0

f _k (u)dL k (u) ² !

=

n

X

k=1

w _k ²

"

1

α ² _k 1 − e ^{−α k t} 2 Z

R

x − x

¹

_|x|≤1 ν(x)dx

2

+ 1 2α k

1 − e ^{−2α k t} Z

R

x ² ν(x)dx

#

−

n

X

k=1

w _k ²

α ² _k 1 − e ^{−α k t} 2 Z

R

x − x

¹

_|x|≤1 ν(x)dx

2

=

n

X

k=1

w _k ² 2α k

1 − e ^{−2α k t} Z

R

x ² ν (x)dx.

Wealsohavetohekwhathappenswhentimegoestoinnity;

t→∞ lim Var [r(t)] =

n

X

k=1

w ² _k 2α k

Z

R

x ² ν(x)dx.

Andourproofis omplete.

4.3 Moments of

r ( t )

We wantto nd the rsttwo momentsof

r(t)

^{. As mentionedbefore, wean dothat by}

usingCorollary4.1. Therstmomentisalreadyomputedandgivenby(4.6). Westateit

Proposition 4.4 The rstmoment of

r(t)

^is

E [r(t)] = r(0)e ^{−α 1 t} +

n

X

k=1

w k

α k

1 − e ^{−α k t} Z

R

x − x

¹

|x|≤1

ν(x)dx.

Tondtheseondmomentwelookat

E r ² (t)

= (i) ⁻² d ² dz ² E h

e ^izr(t) i z=0

= (i) ⁻² d dz

d dz E h

e ^izr(t) i z=0

.

Let

f _k (u) = e ^{−α k (t−u)}

^{asbefore. First,letusompute}

_dz ^{d 2} 2 E e ^izr(t)

. Weusetheexpression

for theharateristifuntion of

r(t)

^{given by} Proposition 4.2. Let

b(t) = r(0)e ^{−α 1 t}

^and

rememberthat

ψ(zw k f k (u)) = Z

R

e ^{izw k f k (u)x} − 1 − izw k f k (u)x

¹

|x|≤1

ν(x)dx.

^(4.10)

Thenweget

d dz E h

e ^izr(t) i

= d dz

h

e ^izb(t) e ^{P n k=1 R 0 t ψ(zw k f k (u))du} i

= ib(t)e ^izb(t)

n

Y

k=1

e ^{R 0 t ψ(zw k f k (u))du} + e ^izb(t) d dz

n

Y

k=1

e ^{R 0 t ψ(zw k f k (u))du} ,

and

d ² dz ² E h

e ^izr(t) i

= − b ² (t)e ^izb(t)

n

Y

k=1

e ^{R 0 t ψ(zw k f k (u))du} + ib(t)e ^izb(t) d dz

n

Y

k=1

e ^{R 0 t ψ(zw k f k (u))du}

+ib(t)e ^izb(t) d dz

n

Y

k=1

e ^{R 0 t ψ(zw k f k (u))du} + e ^izb(t) d ² dz ²

n

Y

k=1

e ^{R 0 t ψ(zw k f k (u))du} .

Now,let

D 1 = − b ² (t)e ^izb(t)

n

Y

k=1

e ^{R 0 t ψ(zw k f k (u))du}

_z=0

^(4.11)

D 2 = ib(t)e ^izb(t) d dz

n

Y

k=1

e ^{R 0 t ψ(zw k f k (u))du}

_z=0

^(4.12)

D 3 = e ^izb(t) d ² dz ²

n

Y

k=1

e ^{R 0 t ψ(zw k f k (u))du}

z=0 .

^(4.13)

Notiethat

E r ² (t)

= (i) ⁻² [D 1 + 2D 2 + D 3 ] .

^(4.14)

Weomputethepartsseparately,but notierstthat

e ^{R 0 t ψ(zw k f k (u))du}

z=0 = 1,

^(4.15)

sine

ψ(zw k f _k (u))

z=0 = 0

^{. Westartbyevaluating}

D 1

^{. It'sfairlyeasytosee that}

D 1 = − b ² (t)e ^izb(t)

n

Y

k=1

e ^{R 0 t ψ(zw k f k (u))du} z=0

= − b ² (t).

^(4.16)