ORNSTEIN-UHLENBECK PROCESSES: limit theorems and …ajacquie/WorkshopLDP/ICWorkshopLDP2013 -...

ORNSTEIN-UHLENBECK PROCESSES:limit theorems and Markov modulation

Michel Mandjes1,2,3

1Korteweg-de Vries Institute for Mathematics, University of Amsterdam2CWI, Amsterdam

3Eurandom, Eindhoven

Imperial College, London, April 2013

Workshop on Large Deviations & Asymptotic Methods in Finance

OVERVIEW

I Introduction on Ornstein-Uhlenbeck processes, relation withinfinite-server queues

I Part I: Reflection (joint work with Gang Huang and PeterSpreij)

I Part II: CLT under Markov modulation (joint work with DaveAnderson, Joke Blom, Offer Kella, Peter Spreij, HalldoraThorsdottir, Koen de Turck)

I Part III: Large deviations under Markov modulation (jointwork with Joke Blom, Koen de Turck)

OVERVIEW

Hence I’ll (slightly) deviate from the abstract I submitted.There I announced I was also going to cover work with Kosinski onmultidimensional ruin probabilities.

Doesn’t mean that paper is not interesting... , — see arXiv.

Combines multidimensional aspect (as in Collamore’s work) withnon-standard scaling (as in Duffield/O’Connell and Duffie et al.).

OVERVIEW

Doesn’t mean that paper is not interesting... ,

— see arXiv.

OVERVIEW

PAPER WITH KOSINSKI

I Covering strong correlation structures, e.g. fractionalBrownian motion.

I One way is to use (generalized version of) ‘Schilder/Azencott’.Is needed for ‘involved’ probabilities as arise in queueing: longbusy period (M./Mannersalo/Norros/van Uitert, Stoch. Proc.Appl. 2006), downstream queue in tandem (M./van Uitert,Ann. Appl. Probab. 2005), convergence to stationarity(M./Norros/Glynn, Ann. Appl. Probab. 2009).Difficulty: no explicit expression for rate function of givenpath.

PAPER WITH KOSINSKI

I This is not needed for ruin or overflow in single queue.‘Principle of largest term’. See also predecessor of this paperfor Gaussian case (Debicki/Kosinski/M./Rolski, Stoch. Proc.Appl., 2011).

I As in Duffield/O’Connell (single dimension): most likelyepoch of ruin determines large deviations. Non-linear scalefunctions, so as to cover long-range dependence.

I Paper with Kosinski: extension to multidimensional risk.

PAPER WITH KOSINSKI

ORNSTEIN AND UHLENBECK

INTRODUCTION

I Stochastic differential equation

dMt = (α− γMt)dt + σdBt , Y0 = x > 0,

where α, γ, σ > 0, Bt is a standard Brownian motion.

I Similarity with the infinite-server queue. There jobs aregenerated according to a Poisson process of rate λ. Theyremain in system exp(µ) time; they don’t “see” each other, sodeparture rate is µ multiplied by number of jobs present.

INTRODUCTION

where α, γ, σ > 0, Bt is a standard Brownian motion.

I Similarity with the infinite-server queue. There jobs aregenerated according to a Poisson process of rate λ. Theyremain in system exp(µ) time; they don’t “see” each other, sodeparture rate is µ multiplied by number of jobs present.

INTRODUCTION

can be solved trivially, for instance as follows.

I F (Mt , t)) := Mteγt . Ito’s lemma:

dF (Mt , t) = eγt(γMtdt + dMt) = eγt(αdt + σdBt).

I Integrating:

Mteγt = M0 +

0αeγsds +

0σeγsdBs .

I Hence

Mt = M0e−γt +

γ(1− e−γt) + σ

0e−γ(t−s)dBs .

INTRODUCTION

I Integrating:

Mteγt = M0 +

0αeγsds +

0σeγsdBs .

I Hence

Mt = M0e−γt +

γ(1− e−γt) + σ

0e−γ(t−s)dBs .

INTRODUCTION

I Integrating:

Mteγt = M0 +

0αeγsds +

0σeγsdBs .

I Hence

Mt = M0e−γt +

γ(1− e−γt) + σ

0e−γ(t−s)dBs .

INTRODUCTION

I Integrating:

Mteγt = M0 +

0αeγsds +

0σeγsdBs .

I Hence

Mt = M0e−γt +

γ(1− e−γt) + σ

0e−γ(t−s)dBs .

INTRODUCTION

I From

Mt = M0e−γt +

γ(1− e−γt) + σ

0e−γ(t−s)dBs

we observe

I mean:EMt = EM0e

−γt +α

γ(1− e−γt),

I variance:

VarMt = Var(σ

e−γ(t−s)dBs

2γ(1− e−2γt).

I In fact, Mt has a Normal distribution with this mean andvariance.

INTRODUCTION

I From

Mt = M0e−γt +

γ(1− e−γt) + σ

0e−γ(t−s)dBs

we observe

I mean:EMt = EM0e

−γt +α

γ(1− e−γt),

I variance:

VarMt = Var(σ

e−γ(t−s)dBs

2γ(1− e−2γt).

INTRODUCTION

I From

Mt = M0e−γt +

γ(1− e−γt) + σ

0e−γ(t−s)dBs

we observe

I mean:EMt = EM0e

−γt +α

γ(1− e−γt),

I variance:

VarMt = Var(σ

e−γ(t−s)dBs

2γ(1− e−2γt).

INTRODUCTION

I From

Mt = M0e−γt +

γ(1− e−γt) + σ

0e−γ(t−s)dBs

we observe

I mean:EMt = EM0e

−γt +α

γ(1− e−γt),

I variance:

VarMt = Var(σ

e−γ(t−s)dBs

2γ(1− e−2γt).

INTRODUCTION

I OU is a Markovian, Gaussian process, that is mean-reverting(towards the limiting mean α/γ).

INTRODUCTION

I OU is a Markovian, Gaussian process, that is mean-reverting(towards the limiting mean α/γ).

INTRODUCTION

In this talk we consider two complications:

I Reflection at 0 (from above), or double reflection at 0 (fromabove) and d > 0 (from below).

I Markov modulation: parameters α, γ, σ > 0 have valuesαi , γi , σi > 0 when independent background Markov chain isin state i .

INTRODUCTION

In this talk we consider two complications:

I Reflection at 0 (from above), or double reflection at 0 (fromabove) and d > 0 (from below).

I Markov modulation: parameters α, γ, σ > 0 have valuesαi , γi , σi > 0 when independent background Markov chain isin state i .

Part IREFLECTED ORNSTEIN-UHLENBECK (ROU)

REFLECTED ORNSTEIN-UHLENBECK (ROU)

I The reflected Ornstein-Uhlenbeck process (ROU) withreflection at 0 is defined as the unique strong solution to theSDE

dYt = (α− γYt)dt + σdBt + dLt , Y0 = x > 0,

where α, γ, σ > 0, Bt is a standard Brownian motion and Lt isthe minimal nondecreasing process which makes Yt ≥ 0. Wehave ∫

[0,T ]1(Yt > 0)dLt = 0, ∀T > 0.

I Idea (in queueing lingo): Lt can be interpreted as ‘cumulativeidle time’ (i.e., this can only grow if Yt is positive).

I The reflected Ornstein-Uhlenbeck process (ROU) withreflection at 0 is defined as the unique strong solution to theSDE

dYt = (α− γYt)dt + σdBt + dLt , Y0 = x > 0,

where α, γ, σ > 0, Bt is a standard Brownian motion and Lt isthe minimal nondecreasing process which makes Yt ≥ 0. Wehave ∫

[0,T ]1(Yt > 0)dLt = 0, ∀T > 0.

I Idea (in queueing lingo): Lt can be interpreted as ‘cumulativeidle time’ (i.e., this can only grow if Yt is positive).

I The existence and uniqueness of the solution is guaranteed byb(x) = α− γx is uniformly Lipschitz continuous and grows nofaster than linearly, and σ is a bounded constant.

I In Ward and Glynn’s papers [3, 5], ROU is used toapproximate the number-in-system processes in M/M/1 andGI/GI/1 queues both with reneging.

I The existence and uniqueness of the solution is guaranteed byb(x) = α− γx is uniformly Lipschitz continuous and grows nofaster than linearly, and σ is a bounded constant.

I In Ward and Glynn’s papers [3, 5], ROU is used toapproximate the number-in-system processes in M/M/1 andGI/GI/1 queues both with reneging.

I The number-in-system process in a GI/M/n loss model can beapproximated by ROU, see e.g. Srikant and Whitt [2].

I Useful in finance as well, if certain quantity cannot crossspecific boundaries (for instance if it has to stay positive).

I The scaled ROU (with small perturbation)

dY εt = (α− γY ε

t )dt +√εσdBt + dLεt , Y ε

0 = x > 0.

I The first question to be addressed is: given T > 0 andb > E(Y ε

T | Y ε0 = x), what is the probability that the process

Y εt exceeds b at time T . I.e., compute

limε→0

ε logP(Y εT > b | Y ε

0 = x). (1)

I Shwartz and Weiss [1] calculate this blocking probability ofM/M/n queue (or overflow probability in correspondingM/M/∞) analytically by large deviations techniques.

0 = x > 0.

limε→0

0 = x). (1)

0 = x > 0.

limε→0

0 = x). (1)

0 = x > 0.

limε→0

0 = x). (1)

0 = x > 0.

limε→0

0 = x). (1)

STRATEGY:

I Step 1: Derive Large Deviations Principle for Y εt with an

explicit rate function. Observe that rate function hassameexpression as the one of OU but operates on smaller functionspace.

I Step 2: Minimize rate function of OU over all continuouspaths f such that f (0) = x and f (T ) > b. Observe thatoptimizing path f ? does not hit level 0 between 0 and T .

I Step 3: Prove that the optimizing path f ? also solvesvariational problem of minimizing rate function of ROU overall positive continuous paths f such that f (0) = x andf (T ) > b. Hence the decay rate (1) is obtained.

STRATEGY:

STEP 1: Identification of ROU’s rate function

Proposition 1

dY εt = b(Y ε

0 = x > 0;

recall that in our case b(x) = α− γx . Define:

{f : f (t) = x +

0φ(s)ds, φ ∈ L2([0,T ])

I When b(0) > 0, Y εt satisfies LDP with rate function

I+(h) =1

(h′t − b(ht)

if h ∈ Hx and ∞ else.I When b(0) < 0, Y ε

t satisfies LDP with rate function

I+(h) =1

(h′t − b(ht)

)2dt− 1

01{0}(ht)(b(0))2dt.

if h ∈ Hx and ∞ else.

Observe: coincides with LDP for OU, adapted at boundary 0.

Proposition 1

dY εt = b(Y ε

0 = x > 0;

{f : f (t) = x +

0φ(s)ds, φ ∈ L2([0,T ])

I+(h) =1

(h′t − b(ht)

I When b(0) < 0, Y εt satisfies LDP with rate function

I+(h) =1

(h′t − b(ht)

)2dt− 1

01{0}(ht)(b(0))2dt.

Proposition 1

dY εt = b(Y ε

0 = x > 0;

{f : f (t) = x +

0φ(s)ds, φ ∈ L2([0,T ])

I+(h) =1

(h′t − b(ht)

I+(h) =1

(h′t − b(ht)

)2dt− 1

01{0}(ht)(b(0))2dt.

Proposition 1

dY εt = b(Y ε

0 = x > 0;

{f : f (t) = x +

0φ(s)ds, φ ∈ L2([0,T ])

I+(h) =1

(h′t − b(ht)

I+(h) =1

(h′t − b(ht)

)2dt− 1

01{0}(ht)(b(0))2dt.

STEP 2: Solving a variational problem

I Observe that we can write P(Y εT > b | X ε

0 = x) = P(Y εt ∈ S),

S :=⋃a>b

Sa, Sa :={f ∈ C[0,T ](R) : f (0) = x , f (T ) = a

I Hence,

limε→0

0 = x) = − inff ∈S

I+(f ).

I Observe that we can write P(Y εT > b | X ε

0 = x) = P(Y εt ∈ S),

S :=⋃a>b

Sa, Sa :={f ∈ C[0,T ](R) : f (0) = x , f (T ) = a

I Hence,

limε→0

0 = x) = − inff ∈S

I+(f ).

I Mεt , scaled OU. E(Mε

T | Mε0 = x) = α

γ +(x − α

)e−γT . It is

greater than 0 when the starting point x > 0. Define the

crossing level b > αγ +

(x − α

)e−γT for both OU and ROU.

I Proposition 2 Let a > b. We have, with Ix(f ) the ratefunction for OU,

inff ∈Sa

Ix(f ) =[a− (αγ + (x − α

γ )e−γT )]2

[1− e−2γT ](σ2/γ).

The optimizing path is given by

f ?(t) =(x − α

γ )(eγT−γt − e−γT+γt) + (a− αγ )(eγt − e−γt)

eγT − e−γT+α

I Moreover, f ?(t) > 0 on t ∈ [0,∞) when the starting pointx > 0.

I f ?(t) ∈ [0, d ] on t ∈ [0,T ] when the starting point x ∈ [0, d ],a ∈ [0, d ] and α

γ < d .

T | Mε0 = x) = α

γ +(x − α

)e−γT . It is

(x − α

inff ∈Sa

Ix(f ) =[a− (αγ + (x − α

γ )e−γT )]2

[1− e−2γT ](σ2/γ).

f ?(t) =(x − α

eγT − e−γT+α

γ < d .

T | Mε0 = x) = α

γ +(x − α

)e−γT . It is

(x − α

inff ∈Sa

Ix(f ) =[a− (αγ + (x − α

γ )e−γT )]2

[1− e−2γT ](σ2/γ).

f ?(t) =(x − α

eγT − e−γT+α

γ < d .

T | Mε0 = x) = α

γ +(x − α

)e−γT . It is

(x − α

inff ∈Sa

Ix(f ) =[a− (αγ + (x − α

γ )e−γT )]2

[1− e−2γT ](σ2/γ).

f ?(t) =(x − α

eγT − e−γT+α

γ < d .

STEP 3: Computation of the decay rate

I Theorem 1 Let b > αγ +

(x − α

)e−γT . Then

limε→0

0 = x) = −[b − (αγ + (x − α

γ )e−γT )]2

[1− e−2γT ](σ2/γ).

Moreover, the optimizing path is the one given in Prop. 2(with a replaced by b).

TRANSIENT ASYMPOTOTICS FOR DROU

I Doubly reflected Ornstein-Uhlenbeck process (DROU) on[0, d ] is defined as the strong solution to

dZt = (α− γZt)dt + σdBt + dLt − dUt , Z0 = x ∈ [0, d ],

where Ut is the minimal nondecreasing process which makesZ (t) ≤ d , i.e., we have

∫[0,T ] 1(Zt > 0)dLt = 0 as well as∫

[0,T ] 1(Zt < d)dUt = 0.

I In context of queues with finite-capacity, Ut is the continuousanalog to the cumulative amount of loss over [0, t] — oftenreferred to (queueing lingo!) as ‘loss process’.

I In addition, we assume that the upper boundary d > αγ . It

means that asymptotic mean αγ lies between two reflecting

boundaries.

∫[0,T ] 1(Zt > 0)dLt = 0 as well as∫

[0,T ] 1(Zt < d)dUt = 0.

boundaries.

∫[0,T ] 1(Zt > 0)dLt = 0 as well as∫

[0,T ] 1(Zt < d)dUt = 0.

boundaries.

I Recall in Prop. 2, let f ?(t) be the solution to the variationalproblem inff ∈Sa Ix(f ). Then, importantly, f ?(t) ∈ [0, d ] ont ∈ [0,T ] when the starting point x ∈ [0, d ], a ∈ [0, d ] andαγ < d .

I Theorem 2Let d > b > α

γ +(x − α

)e−γT . Then

limε→0

ε logP(Z εT > b | Z ε0 = x) = −[b − (αγ + (x − α

γ )e−γT )]2

[1− e−2γT ](σ2/γ).

I Recall in Prop. 2, let f ?(t) be the solution to the variationalproblem inff ∈Sa Ix(f ). Then, importantly, f ?(t) ∈ [0, d ] ont ∈ [0,T ] when the starting point x ∈ [0, d ], a ∈ [0, d ] andαγ < d .

I Theorem 2Let d > b > α

γ +(x − α

)e−γT . Then

limε→0

ε logP(Z εT > b | Z ε0 = x) = −[b − (αγ + (x − α

γ )e−γT )]2

[1− e−2γT ](σ2/γ).

CENTRAL LIMIT THEOREM OF LOSS PROCESSES

I CLT of Ut (Lt) of DROU

dZt = (α− γZt)dt + σdBt + dLt − dUt , Z0 = x ∈ [0, d ].

I Zhang & Glynn [6] solve the same problem for doublyreflected Brownian motion by a martingale approach.

I Let h be a twice continuously differentiable function on R,and Zt be the DROU process. By Ito’s formula, we have:

dh(Zt) =

((α− γZt)h

′(Zt) +σ2

2h′′(Zt)

+ σh′(Zt)dBt + h′(Zt)dLt − h′(Zt)dUt .

dh(Zt) =

((α− γZt)h

′(Zt) +σ2

2h′′(Zt)

dh(Zt) =

((α− γZt)h

′(Zt) +σ2

2h′′(Zt)

I Since Lt ,Ut are local times at 0 and d respectively

dh(Zt) = (Lh)(Zt)dt+σh′(Zt)dBt +h′(0)dLt−h′(d)dUt (2)

where the operator L is defined through

L := (α− γx)d

dx+σ2

I The ODE(Lh)(x) = q, 0 6 x 6 d ,

such that h(0) = 0, h′(0) = 0, and h′(d) = 1, has a uniquesolution pair (h(x), qU).

I h(0) = 0, h′(0) = −1, and h′(d) = 0, has a unique solutionpair (h1(x), qL).

L := (α− γx)d

dx+σ2

I The ODE(Lh)(x) = q, 0 6 x 6 d ,

L := (α− γx)d

dx+σ2

I The ODE(Lh)(x) = q, 0 6 x 6 d ,

I Theorem 3 The loss process Ut satisfies the central limittheorem, with η2

U defined below in (3),

Ut − qUt√t

⇒ N (0, η2U), as t →∞.

I Proof: Insert the unique solution h(x) into (2). Sinceh′(0) = 0, h′(d) = 1, and (Lh)(Zt) = qU ,

Ut − qUt + h(Zt)− h(Z0) = σ

0h′(Zs)dBs .

I Mt := Ut − qUt + h(Zt)− h(Z0) is zero-mean squareintegrable martingale;〈Mt〉: quadratic variation process of Mt .

I By the ergodic theorem,

〈Mt〉t

0h′(Zs)2ds

P→ σ2

0h′(x)2π(dx) =: η2

U , (3)

where π is the stationary distribution corresponding Zt . [4]

Ut − qUt√t

⇒ N (0, η2U), as t →∞.

0h′(Zs)dBs .

〈Mt〉t

0h′(Zs)2ds

P→ σ2

0h′(x)2π(dx) =: η2

U , (3)

Ut − qUt√t

⇒ N (0, η2U), as t →∞.

0h′(Zs)dBs .

〈Mt〉t

0h′(Zs)2ds

P→ σ2

0h′(x)2π(dx) =: η2

U , (3)

Ut − qUt√t

⇒ N (0, η2U), as t →∞.

0h′(Zs)dBs .

〈Mt〉t

0h′(Zs)2ds

P→ σ2

0h′(x)2π(dx) =: η2

U , (3)

I By the martingale central limit theorem,

Ut − qUt + h(Zt)− h(Z0)√t

⇒ N (0, η2U)

as t →∞.

I Since Zt ∈ [0, d ] and h is continuous, h(Zt) is bounded. So,

h(Zt)− h(Z0)√t

a.s. as t →∞.

I Hence,Ut − qUt√

t⇒ N (0, η2

U), as t →∞.

⇒ N (0, η2U)

as t →∞.I Since Zt ∈ [0, d ] and h is continuous, h(Zt) is bounded. So,

h(Zt)− h(Z0)√t

a.s. as t →∞.

t⇒ N (0, η2

U), as t →∞.

⇒ N (0, η2U)

as t →∞.I Since Zt ∈ [0, d ] and h is continuous, h(Zt) is bounded. So,

h(Zt)− h(Z0)√t

a.s. as t →∞.

t⇒ N (0, η2

U), as t →∞.

So far:

I Tail probabilities for DROU;

I CLT Ut and Lt for DROU;

I To be done: LD Ut and Lt for DROU. Tricky!

So far:

Bibliography of Part I

Adam Shwartz and Alan Weiss.Large deviations for performance analysis.Stochastic Modeling Series. Chapman & Hall, London, 1995.Queues, communications, and computing, With an appendix by Robert J.Vanderbei.

Rayadurgam Srikant and Ward Whitt.Simulation run lengths to estimate blocking probabilities.ACM Trans. Model. Comput. Simul., 6(1):7–52, January 1996.

Amy R. Ward and Peter W. Glynn.A diffusion approximation for a Markovian queue with reneging.Queueing Syst., 43(1-2):103–128, 2003.

Amy R. Ward and Peter W. Glynn.Properties of the reflected Ornstein-Uhlenbeck process.Queueing Syst., 44(2):109–123, 2003.

Amy R. Ward and Peter W. Glynn.A diffusion approximation for a GI/GI/1 queue with balking or reneging.Queueing Syst., 50(4):371–400, 2005.

Xiaowei Zhang and Peter W. Glynn.On the dynamics of a finite buffer queue conditioned on the amount of loss.Queueing Syst., 67(2):91–110, 2011.

Part IIMARKOV MODULATED ORNSTEIN-UHLENBECK (MMOU)

Central Limit Theorems

MODEL: MMOU

I (X (t))t>0: irreducible, Markov process on {1, . . . , d}.

I Transition rates: Q = (qij)di ,j=1, (unique) invariant

distribution: π.

I Now we suppose that the process X (·) modulates anOrnstein-Uhlenbeck process: while X (·) in state i , the process(M(t))t>0 behaves as an Ornstein-Uhlenbeck process Ui (·)with parameters αi , γi , and σi , independently of the‘background process’ X (·).

I Hence, M(·) obeys the following SDE:

dM(t) = (αX (t) − γX (t)M(t))dt + σX (t) dB(t);

(B(t))t>0 standard BM independent of (X (t))t>0.

I Queueing: Markov modulation — Finance: regime switching.

MODEL: MMOU

I (X (t))t>0: irreducible, Markov process on {1, . . . , d}.I Transition rates: Q = (qij)

di ,j=1, (unique) invariant

distribution: π.

MODEL: MMOU

distribution: π.

MODEL: MMOU

distribution: π.

MODEL: MMOU

distribution: π.

First: deterministic modulationI First consider OU with deterministically changing parameters:

dM(t) = (α(t)− γ(t)M(t))dt + σ(t)dB(t),

with B(t) a standard Brownian motion, and α(t), γ(t), andσ(t) arbitrary positive, deterministic functions.

I Solve SDE. Define F (M(t), t) := M(t)eΓ(t), withΓ(t) :=

∫ t0 γ(s)ds. Then, by virtue of Ito’s lemma,

dF (M(t), t) = eΓ(t) (γ(t)M(t)dt + dM(t))

= eΓ(t) (α(t)dt + σ(t) dB(t)) .

I Now integrating yields

M(t)eΓ(t) = M(0) +

0eΓ(s)α(s)ds +

0eΓ(s)σ(s) dB(s),

so that, trivially, M(t) equals

M(0)e−Γ(t)+

0e−(Γ(t)−Γ(s))α(s)ds+

0e−(Γ(t)−Γ(s))σ(s)dB(s).

dF (M(t), t) = eΓ(t) (γ(t)M(t) dt + dM(t))

= eΓ(t) (α(t) dt + σ(t) dB(t)) .

M(t)eΓ(t) = M(0) +

0eΓ(s)α(s)ds +

0eΓ(s)σ(s) dB(s),

M(0)e−Γ(t)+

0e−(Γ(t)−Γ(s))σ(s)dB(s).

dF (M(t), t) = eΓ(t) (γ(t)M(t) dt + dM(t))

= eΓ(t) (α(t) dt + σ(t) dB(t)) .

M(t)eΓ(t) = M(0) +

0eΓ(s)α(s) ds +

0eΓ(s)σ(s) dB(s),

M(0)e−Γ(t)+

0e−(Γ(t)−Γ(s))σ(s) dB(s).

Deterministic modulation, ctd.

I From now on: M(0) equals the constant m0.

I Now the random variable M(t) necessarily has a Normaldistribution, with mean

µt = m0e−Γ(t) +

0e−(Γ(t)−Γ(s))α(s)ds,

and variance

vt = Var(∫ t

0e−(Γ(t)−Γ(s))σ(s)dB(s)

0e−2(Γ(t)−Γ(s))σ2(s)ds.

Deterministic modulation, ctd.

I From now on: M(0) equals the constant m0.

I Now the random variable M(t) necessarily has a Normaldistribution, with mean

µt = m0e−Γ(t) +

0e−(Γ(t)−Γ(s))α(s)ds,

and variance

vt = Var(∫ t

0e−(Γ(t)−Γ(s))σ(s)dB(s)

0e−2(Γ(t)−Γ(s))σ2(s)ds.

MMOU: conditional mean and variance

I In MMOU α(t) = αi , γ(t) = γi , and σ2(t) = σ2i if X (t) = i .

We have found the following result.

I Denote by X the path (X (s), s ∈ [0, t]). (M(t) |X ) has aNormal distribution with random parameters m and s given by

m := E(M(t) |X )

= m0 exp

(−∫ t

0γX (s)ds

(−∫ t

sγX (r)dr

)αX (s) ds

s := Var(M(t) |X ) =

sγX (r)dr

)σ2X (s) ds.

I Similarity with corresponding result for Markov modulatedinfinite-server queue by D’Auria: there number of jobs insystem has a Poisson distribution with random parameter.

m := E(M(t) |X )

= m0 exp

(−∫ t

0γX (s)ds

(−∫ t

sγX (r)dr

)αX (s) ds

s := Var(M(t) |X ) =

sγX (r)dr

)σ2X (s) ds.

m := E(M(t) |X )

= m0 exp

(−∫ t

0γX (s)ds

(−∫ t

sγX (r)dr

)αX (s) ds

s := Var(M(t) |X ) =

sγX (r)dr

)σ2X (s) ds.

MMOU: unconditional mean and variance

I Now: expressions for mean and variance.

I Special cases: γi equal, t →∞, certain scalings.

I Let Z (t) ∈ {0, 1}d such that Zi (t) = 1 if X (t) = i and 0 else.

I Directly from the definitions

dµt = E(αTZ (t)− γTY (t))dt,

with Y (t) := Z (t)M(t). Let νt := EY (t) and

pt := (P(X (t) = 1), . . . ,P(X (t) = d))T.

I Conclude thatµ′t = αTpt − γTνt .

I Clearly, dZ (t) = QTZ (t) dt + dK (t), for a d-dimensionalmartingale K (t). With Ito’s rule,

dY (t) = M(t)(QTZ (t)dt + dK (t)

)+Z (t)

(αTZ (t)− γTY (t)

)dt + σTZ (t)dB(t).

pt := (P(X (t) = 1), . . . ,P(X (t) = d))T.

dY (t) = M(t)(QTZ (t)dt + dK (t)

)+Z (t)

pt := (P(X (t) = 1), . . . ,P(X (t) = d))T.

dY (t) = M(t)(QTZ (t) dt + dK (t)

)+Z (t)

I Taking expectations of both sides, we obtain, with

Qγ := QT − diag{γ},

ν ′t = Qγνt + diag{α}pt .

I From this non-homogeneous linear system of differentialequations, we know that νt is given by

νt = eQγ tν0 +

0eQγ(t−s)diag{α}psds;

and then µt = 1Tνt .

I Taking expectations of both sides, we obtain, with

Qγ := QT − diag{γ},

ν ′t = Qγνt + diag{α}pt .I From this non-homogeneous linear system of differential

equations, we know that νt is given by

νt = eQγ tν0 +

0eQγ(t−s)diag{α}psds;

and then µt = 1Tνt .

I Equations simplify drastically if background process starts offin equilibrium at time 0; then pt = π for all t ≥ 0. As a result,

νt = eQγ tν0 − Q−1γ (I − eQγ t)diag{α}π.

I It follows that ν∞ = −Q−1γ diag{α}π, and

µ∞ = 1Tν∞ = −1TQ−1γ diag{α}π.

I We further note that γ = −(Q − diag{γ})1, and hence

γTQ−1γ = −1T,

so that γTν∞ = πTα.

γTQ−1γ = −1T,

I Variance can be found similarly. Define Y (t) := Z (t)M2(t),and w t := EY (t).

I Starting point is the relation

d(M(t)− µt) =(αT(Z (t)− pt)− γT(Y (t)− νt)

+σTZ (t)dB(t),

so that

d(M(t)− µt)2 = 2(M(t)− µt)(αT(Z (t)− pt)− γT(Y (t)− νt)

+ 2(M(t)− µt)σTZ (t)dB(t) + σTdiag{Z (t)}σ dt.

I Taking expectations of both sides,

v ′t = 2αTνt − 2µtαTpt − 2γTw t + 2µtγ

Tνt +σTdiag{pt}σ.

+σTZ (t)dB(t),

so that

+σTZ (t)dB(t),

so that

I Clearly, to evaluate this expression, we first need to identifyw t . To this end, we set up and equation for dY (t) as before,take expectations, so as to obtain

w ′t = Q2γw t + 2diag{α}νt + diag{σ2}pt .

I This leads to

w t = eQ2γ tw0+

0eQ2γ(t−s)

(2diag{α}νs + diag{σ2}ps

so that vt = 1Tw t − µ2t .

I Clearly, to evaluate this expression, we first need to identifyw t . To this end, we set up and equation for dY (t) as before,take expectations, so as to obtain

w ′t = Q2γw t + 2diag{α}νt + diag{σ2}pt .

I This leads to

w t = eQ2γ tw0+

0eQ2γ(t−s)

(2diag{α}νs + diag{σ2}ps

so that vt = 1Tw t − µ2t .

I Again simplifications can be made if p0 = π (and hencept = π for all t). In that case, we had already found anexpression for νs above, and as a result w t can be explicitlyevaluated.

I For the stationary situation (t →∞, that is) we obtain

w∞ = −Q−12γ

(2diag{α}ν∞ + diag{σ2}π

and v∞ = 1Tw∞ − µ2∞.

I Again simplifications can be made if p0 = π (and hencept = π for all t). In that case, we had already found anexpression for νs above, and as a result w t can be explicitlyevaluated.

I For the stationary situation (t →∞, that is) we obtain

w∞ = −Q−12γ

(2diag{α}ν∞ + diag{σ2}π

and v∞ = 1Tw∞ − µ2∞.

MMOU: unconditional mean and variance, γi equal

I It is directly seen that µ∞ = πTα/γ.

I Now γTQ−1γ = −1T implies 1TQ−1

δ1 = −δ−11T for any δ > 0,so that

v∞ = 1Tw∞ − µ2∞ =

1Tdiag{α}ν∞γ

+πTσ2

2γ−(πTα

= −1Tdiag{α}Q−1

γ1 diag{α}πγ

+πTσ2

2γ−(πTα

I It is directly seen that µ∞ = πTα/γ.

I Now γTQ−1γ = −1T implies 1TQ−1

δ1 = −δ−11T for any δ > 0,so that

v∞ = 1Tw∞ − µ2∞ =

1Tdiag{α}ν∞γ

+πTσ2

2γ−(πTα

= −1Tdiag{α}Q−1

γ1 diag{α}πγ

+πTσ2

2γ−(πTα

MMOU: unconditional mean and variance, γi equalI Let Dij(γ) :=

∫∞0 pij(v)e−γvdv . Integration by parts:

QD(γ) =

∫ ∞0

QP(v)e−γvdv =

∫ ∞0

P ′(v)e−γvdv

= −I +

∫ ∞0

γP(v)e−γvdv = −I + γD(γ).

I As a consequence, −(Q − γI )D(γ) = I , so that

v∞ =πTσ2

γαTdiag{π}D(γ)α−

(πTα

where Dij(γ) :=∫∞

0 (pij(v)− πj)e−γvdv = Dij(γ)− πj/γ.I We find:

v∞ =πTσ2

γαTdiag{π}D(γ)α.

I Mean and variance can also be found by elementary, insightfularguments, however!

QD(γ) =

∫ ∞0

QP(v)e−γvdv =

∫ ∞0

P ′(v)e−γvdv

= −I +

∫ ∞0

v∞ =πTσ2

(πTα

0 (pij(v)− πj)e−γvdv = Dij(γ)− πj/γ.

I We find:

v∞ =πTσ2

QD(γ) =

∫ ∞0

QP(v)e−γvdv =

∫ ∞0

P ′(v)e−γvdv

= −I +

∫ ∞0

v∞ =πTσ2

(πTα

v∞ =πTσ2

QD(γ) =

∫ ∞0

QP(v)e−γvdv =

∫ ∞0

P ′(v)e−γvdv

= −I +

∫ ∞0

v∞ =πTσ2

(πTα

v∞ =πTσ2

I First: mean µt . It is immediate that µt is a convex mixture ofm0 and πTα/γ:

µt = m0e−γt + e−γt

0eγsds

πiαi

= m0e−γt +

γ(1− e−γt),

which converges, as t →∞, to πTα/γ, as expected.

I Then: variance vt . Use law of total variance. Denote again byX the path (X (s), s ∈ [0, t]). We now compute

VarM(t) = E (Var(M(t) |X )) + Var (E(M(t) |X )) .

I First term: use result for conditional mean. We get:

E (Var(M(t) |X )) = E(∫ t

0e−2γ(t−s)σ2

X (s) ds

0e−2γ(t−s)E

(σ2X (s)

πiσ2i

(1− e−2γt

I Then: variance vt . Use law of total variance. Denote again byX the path (X (s), s ∈ [0, t]). We now compute

VarM(t) = E (Var(M(t) |X )) + Var (E(M(t) |X )) .

I First term: use result for conditional mean. We get:

E (Var(M(t) |X )) = E(∫ t

0e−2γ(t−s)σ2

X (s) ds

0e−2γ(t−s)E

(σ2X (s)

πiσ2i

(1− e−2γt

I Also,

Var (E(M(t) |X )) = Var(∫ t

0e−γ(t−s)αX (s) ds

(e−γ(t−s)αX (s), e

−γ(t−u)αX (u)

)du ds

= e−2γt

0eγ(s+u)Cov

(αX (s), αX (u)

)du ds.

I The latter integral expression can be evaluated as

2e−2γt

0eγ(s+u)Cov

(αX (s), αX (u)

)du ds = . . . =

d∑i=1

d∑j=1

αiαj

(e−γv − e−γ(2t−v)

)πi (pij(v)− πj)dv

(change order of integration and use elementary calculus).

I Also,

Var (E(M(t) |X )) = Var(∫ t

0e−γ(t−s)αX (s) ds

(e−γ(t−s)αX (s), e

−γ(t−u)αX (u)

)du ds

= e−2γt

0eγ(s+u)Cov

(αX (s), αX (u)

)du ds.

I The latter integral expression can be evaluated as

2e−2γt

0eγ(s+u)Cov

(αX (s), αX (u)

)du ds = . . . =

d∑i=1

d∑j=1

αiαj

)πi (pij(v)− πj)dv

(change order of integration and use elementary calculus).

I Specializing to the situation that t →∞, we obtain

VarM(∞) =πTσ2

d∑i=1

d∑j=1

αiαjπiDij(γ)

=πTσ2

γαTdiag{π}D(γ)α,

in accordance with the expression we found before.

MMOU: unconditional mean and variance, γi equalI Scale α 7→ Nα, σ2 7→ Nσ2, and Q 7→ N fQ for some f > 0.

VarM(N)(t) = Nd∑

πiσ2i

(1− e−2γt

+N2d∑

d∑j=1

αiαj

)πi (pij(vN

f )− πj)dv ,

which for N large behaves as, with D := D(0),(1− e−2γt

)Nd∑

πiσ2i + 2N2−f

d∑i=1

d∑j=1

αiαjπiDij

(1− e−2γt

)(NπTσ2 + 2N2−fαTdiag{π}Dα

I Dichotomy: for f > 1 the variance is essentially linear in N,while for f < 1 it behaves superlinearly (more specifically,proportionally to N2−f ).

I The matrix D is referred to as the deviation matrix.

VarM(N)(t) = Nd∑

πiσ2i

(1− e−2γt

+N2d∑

d∑j=1

αiαj

)πi (pij(vN

f )− πj)dv ,

)Nd∑

πiσ2i + 2N2−f

d∑i=1

d∑j=1

αiαjπiDij

(1− e−2γt

VarM(N)(t) = Nd∑

πiσ2i

(1− e−2γt

+N2d∑

d∑j=1

αiαj

)πi (pij(vN

f )− πj)dv ,

)Nd∑

πiσ2i + 2N2−f

d∑i=1

d∑j=1

αiαjπiDij

(1− e−2γt

MMOU: Central Limit Theorem

I Result:

(M(N)(t)

N−mt

)converges to a Normal distribution with zero mean andvariance σ2(t).

I Here β := min{f , 1}, and

σ2(t) :=

(1− e−2γt

)(πTσ21{f≥1} + 2αTdiag{π}Dα1{f≤1}

I Future work: weak convergence to OU process withappropriate parameters.

I Result:

(M(N)(t)

N−mt

σ2(t) :=

(1− e−2γt

I Result:

(M(N)(t)

N−mt

σ2(t) :=

(1− e−2γt

Part IIIMARKOV MODULATED ORNSTEIN-UHLENBECK (MMOU)

Large Deviations

MMOU: Large deviations

Two regimes!

I First regime: α 7→ Nα, σ2 7→ Nσ2, Q 7→ N fQ with f > 1.

I Idea: Markov chain moves faster than OU processes. Hence:we see effectively OU with parameters Nα∞ := N πTα,Nσ2∞ := N πTσ2, γ∞ := πTγ.

limN→∞

NlogP(M(N)(t) ≥ Na) = −1

(a−m∞(t))2

s∞(t),

m∞(t) =α∞γ∞

(1− e−γ∞t),

s∞(t) =σ2∞

2γ∞(1− e−2γ∞t).

Two regimes!

limN→∞

(a−m∞(t))2

s∞(t),

m∞(t) =α∞γ∞

(1− e−γ∞t),

s∞(t) =σ2∞

2γ∞(1− e−2γ∞t).

Two regimes!

limN→∞

(a−m∞(t))2

s∞(t),

m∞(t) =α∞γ∞

(1− e−γ∞t),

s∞(t) =σ2∞

2γ∞(1− e−2γ∞t).

Proof technique:

I Construct lower bound by considering specific scenario.

I Split interval in subintervals of length t/√N.

I Within each interval consider scenario that backgroundprocess is close to π, viz. in δ-environment.

I Find lower bound on mean and upper bound on variance ofthe Normally distribution M(N)(t).

I Let δ ↓ 0.

Proof technique:

I Let δ ↓ 0.

Proof technique:

I Let δ ↓ 0.

Proof technique:

I Let δ ↓ 0.

Proof technique:

I Let δ ↓ 0.

Proof technique, ctd.:

I Construct upper bound by showing all other scenarios are lesslikely, as follows.

I Split interval in Nε/2 subintervals of length t/Nε/2.

I For any event E ,

P(M(N)(t) ≥ Na) ≤ P(M(N)(t) ≥ Na,E ) + P(E c).

Let E be the event of being close to π (i.e., δ-environment).

I Second term decays superexponentially.

I Find upper bound on mean and lower bound on variance ofthe Normally distribution M(N)(t) on E .

I Let δ ↓ 0.

I For any event E ,

I Let δ ↓ 0.

I For any event E ,

I Let δ ↓ 0.

I For any event E ,

I Let δ ↓ 0.

I For any event E ,

I Let δ ↓ 0.

I For any event E ,

I Let δ ↓ 0.

Two regimes!

I Second regime: α 7→ Nα, σ2 7→ Nσ2, Q unchanged.

I A single path f (t) of X (t) determines asymptotics.

I mf (t) = E(M(t) |X = f ) and sf (t) = Var(M(t) |X = f )

minf :f (t)∈{1,...,d}

(a−mf (t))2

sf (t).

Two regimes!

minf :f (t)∈{1,...,d}

(a−mf (t))2

sf (t).

Two regimes!

minf :f (t)∈{1,...,d}

(a−mf (t))2

sf (t).

Goal: estimate P(M(t) ≥ a) for large a (rare event).

A few thoughts on rare-event simulation by importance sampling:

I Two sources of randomness: in background process X (·) andin individual OU processes Ui (·).

I Change-of-measure can be constructed? (cf. work Pham)

I ‘Hybrid’ idea: sample background process, and then computeprobability.

CONCLUSION

I OU models allow fairly explicit analysis;

I Features such as reflection and Markov modulation can bebrought in;

I ... and there is still a lot of work to be done.

CONCLUSION

ORNSTEIN-UHLENBECK PROCESSES: limit theorems and …ajacquie/WorkshopLDP/ICWorkshopLDP2013 -...

Documents

Transcript of ORNSTEIN-UHLENBECK PROCESSES: limit theorems and …ajacquie/WorkshopLDP/ICWorkshopLDP2013 -...