Karhunen-Loève Expansions of Lévy Processes

Karhunen-Loève Expansions of Lévy Processes Daniel Hackmann June 2016, Barcelona supported by the Austrian Science Fund (FWF), Project F5509-N26 Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 0 / 40

Outline 1 Introduction 2 Lévy Processes and Infinitely Divisible Random Vectors 3 Main Results KLE Components Simulation 4 Examples Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 0 / 40

The main idea We want to expand a continuous time stochastic process X in a stochastic Fourier series on the interval [0, T ]: X t = k 1 Y k φ k (t), where {φ k } k 1 is an orthonormal basis of L 2 ([0, T ], R), and our stochastic Fourier coefficients are given by Y k := T 0 X t φ k (t)dt. Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 1 / 40

The main idea Things to think about: In what sense does X t = k 1 Y kφ k (t)? Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 2 / 40

The main idea Things to think about: In what sense does X t = k 1 Y kφ k (t)? How should we choose {φ k } k 1? Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 2 / 40

The main idea Things to think about: In what sense does X t = k 1 Y kφ k (t)? How should we choose {φ k } k 1? Distribution of, and dependence among {Y k } k 1? Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 2 / 40

The main idea Assumptions: (a) E[X t ] = 0, (b) E[X 2 t ] <, and (c) Cov(X s, X t ) is continuous Basis: The eigenfunctions {e k } k 1 corresponding to the non-zero eigenvalues {λ k } k 1 of the operator K : L 2 ([0, T ]) L 2 ([0, T ]), (Kf)(s) := T constitute a basis for L 2 ([0, T ]). We define Z k := 0 T 0 Cov(X s, X t )f(t)dt X t e k (t)dt and determine the order of {e k } k 1, {Z k } k 1, and {λ k } k 1 according to λ 1 λ 2 λ 3.... Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 3 / 40

The main idea Theorem (The Karhunen-Loève Theorem (KLT)) (i) ( ) 2 d E X t Z k e k (t) 0, as d k=1 uniformly for t [0, T ]. Additionally, the {Z k } k 1 are uncorrelated and satisfy E[Z k ] = 0 and E[Z 2 k ] = λ k. (ii) For any other basis {φ k } k 1 with corresponding Fourier coefficients {Y k } k 1, and any d N, we have T 0 [ E (ε d (t)) 2] T [ dt E ( ε d (t)) 2] dt, 0 where ε d and ε d are the remainders ε d (t) := d+1 Z ke k (t) and ε d (t) := d+1 Y kφ k (t). Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 4 / 40

The main idea Proof in: Ghanem, R. G.. and Spanos, P.D.. (1991). Stochastic finite elements: A spectral approach. Springer Verlag, New York Berlin Heidelberg. They credit: Kac, M. and Siegert, A. (1947). An explicit representation of a stationary Gaussian process. Ann. Math. Stat. 18, 438 442. Karhunen, K. (1947). Über lineare Methoden in der Wahrscheinlichkeitsrechnung. Amer. Acad. Sc. Fennicade, Ser. A, I 37, 3 79. Loéve, M. (1948). Fonctions aleatoires du second ordre. In Processus stochastic et mouvement Brownien. ed. P. Lévy. Gauthier Villars, Paris. Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 5 / 40

The main idea In order to use the KLT for a chosen process X in any meaningful way we see that we need to determine: The eigenvalues {λ k } k 1 and eigenfunctions {e k } k 1 of the operator K. I.e. solve: T Cov(X s, X t )e k (s)ds = λ k e k (t), 0 a homogeneous Fredholm integral equation of the second kind The distribution of the Fourier coefficients {Z k } k 1 If we want to simulate an approximate path of X via a Karhunen Loève expansion (KLE) we also need to know how to simulate the first d Fourier coefficients, i.e. simulate the random vector Z (d) = (Z 1,..., Z d ). Note: The components of this vector need not be independent! Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 6 / 40

The main idea We don t have full knowledge for many processes, and of those, we know primarily about Gaussian processes: Brownian motion Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 7 / 40

The main idea We don t have full knowledge for many processes, and of those, we know primarily about Gaussian processes: Brownian motion Brownian bridge Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 7 / 40

The main idea We don t have full knowledge for many processes, and of those, we know primarily about Gaussian processes: Brownian motion Brownian bridge Anderson-Darling Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 7 / 40

The main idea We don t have full knowledge for many processes, and of those, we know primarily about Gaussian processes: Brownian motion Brownian bridge Anderson-Darling More Gaussian references and interesting result for generalized BB in: Barczy, M. and Lovas, R.L.. (2016). Karhunen Loève expansion for a generalization of the Wiener bridge. preprint: arxiv:1602.05084v2. Few complete results (any?) for Non-Gaussian process. A numerical example in: Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 7 / 40

The main idea We don t have full knowledge for many processes, and of those, we know primarily about Gaussian processes: Brownian motion Brownian bridge Anderson-Darling More Gaussian references and interesting result for generalized BB in: Barczy, M. and Lovas, R.L.. (2016). Karhunen Loève expansion for a generalization of the Wiener bridge. preprint: arxiv:1602.05084v2. Few complete results (any?) for Non-Gaussian process. A numerical example in: Phoon, K.K.., Huang, H.W.. and Quek, S.T.. (2005). Simulation of strongly non-gaussian processes using Karhunen Loeve expansion. Probabilistic Engineering Mechanics 20, 188 198. Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 7 / 40

The main idea Goal: For a one-dimensional Lévy process X satisfying E[X t ] = 0 and E[X 2 t ] < determine {λ k } k 1, {e k } k 1 Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 8 / 40

The main idea Goal: For a one-dimensional Lévy process X satisfying E[X t ] = 0 and E[X 2 t ] < determine {λ k } k 1, {e k } k 1 The distribution of {Z k } k 1 Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 8 / 40

The main idea Goal: For a one-dimensional Lévy process X satisfying E[X t ] = 0 and E[X 2 t ] < determine {λ k } k 1, {e k } k 1 The distribution of {Z k } k 1 The distribution and dependence structure of Z (d) = (Z 1,..., Z d ) Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 8 / 40

The main idea Goal: For a one-dimensional Lévy process X satisfying E[X t ] = 0 and E[X 2 t ] < determine {λ k } k 1, {e k } k 1 The distribution of {Z k } k 1 The distribution and dependence structure of Z (d) = (Z 1,..., Z d ) How to simluate Z (d) Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 8 / 40

Outline 1 Introduction 2 Lévy Processes and Infinitely Divisible Random Vectors 3 Main Results KLE Components Simulation 4 Examples Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 8 / 40

Quick review The distribution of a d-dimensional Lévy process X or infinitely divisible vector ξ is completely determined by the characteristic exponent: Ψ X (z) := 1 t log E[ei z,xt ], or Ψ ξ (z) := log E[e i z,ξ ]. Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 9 / 40

Quick review The distribution of a d-dimensional Lévy process X or infinitely divisible vector ξ is completely determined by the characteristic exponent: Ψ X (z) := 1 t log E[ei z,xt ], or Ψ ξ (z) := log E[e i z,ξ ]. The characteristic exponent always has the form Ψ(z) = 1 2 zt Qz i a, z e i z,x 1 i z, x h(x)ν(dx). R n \{0} Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 9 / 40

BV Assumption So Ψ and therefore the distribution are determined by the generating triple (a, Q, ν) h where If a R d (a R), Q R d d is positive semi definite (σ 2 0) (Gaussian component), and ν is a measure (Lévy measure) h a cut-off function needed in general to make the integral converge x <1 x ν(dx) < then we can set h 0. (X has bounded variation when Q = 0 or σ 2 = 0.) Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 10 / 40

Two examples A scaled Brownian motion with drift with E[X t ] = µt and Var(X t ) = σ 2 t has characteristic exponent Ψ(z) = σ2 2 z2 iµz. Since ν 0 there are no jumps and we have generating triple (µ, σ 2, 0) Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 11 / 40

Two examples For c, ρ, ˆρ > 0 set Then Ψ(z) = eˆρx ν(dx) = I(x < 0)c dx + I(x < 0)ce ρx x x dx. R\{0} ( ) ( ( e izx 1 ν(dx) = c log 1 iz ) ( + log 1 + izˆρ )) ρ is the characteristic exponent of a variance gamma process with generating triple (0, 0, ν). Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 12 / 40

Outline 1 Introduction 2 Lévy Processes and Infinitely Divisible Random Vectors 3 Main Results KLE Components Simulation 4 Examples Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 12 / 40

Covariance and the operator K Under the KLT assumption it is straight-forward to show that Var(X t ) = E[X 2 t ] = αt, and Cov(X s, X t ) = α min(s, t), where α = Ψ (0). For example, for our Brownian motion we have α = σ 2 and for our variance gamma process α = c(ρ 2 + ˆρ 2 ). But this shows that the eigenvalues/functions of the operator K for a Lévy process can be derived in exactly the same manner as those of a Brownian motion. And so we get our first result essentially for free... Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 13 / 40

Basis functions Proposition The eigenvalues and associated eigenfunctions of the operator K defined with respect to a Lévy process X are given by αt 2 λ k = ) 2, and e k (t) = π (k 2 1 2 for k N and t [0, T ]. 2 T sin ( π T ( k 1 ) ) t, 2 Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 14 / 40

Basis functions Idea of the Proof: Differentiate both sides of T α min(s, t)e k (s)ds = λ k e k (t), 0 twice with respect to t to reduce to an ODE. Details for Brownian Motion case in Ash, R.B.. and Gardner, M.F.. (1975). Topics in stochastic process. Academic Press, New York San Francisco London. Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 15 / 40

Total Variance Define the total variance of X on [0, T ] as v(t ) := T 0 E[X 2 t ]dt = αt 2 But we also have v(t ) = k 1 λ k (in general). Therefore, the total variance explained by a d-term approximation is 2 dk=1 λ k v(t ) = 2 d 1 π 2 ( ) 2. k=1 k 1 2 Computation yields: the first 2, 5 and 21 terms already explain 90%, 95%, and 99% of the total variance of the process. Holds independently of X, α or T. Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 16 / 40

Gaussian processes A key property of Gaussian processes, is that the coefficients {Z k } k 1 are again Gaussian. For example, ) B t = 2 sin (π(k 1 2 Z t) k ( ) k 1 π k 1 2 where B is a standard Brownian Motion on [0, 1] and the {Z k } k 1 are i.i.d. random variables with common distribution N (0, 1). Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 17 / 40

Key Lemma Lemma Let X be a Lévy process and {f k } d k=1 be a collection of functions which are in L 1 ([0, T ]). Then the vector ξ consisting of elements ξ k = T 0 X t f k (s)ds, k {1, 2,..., d}, has an ID distribution with characteristic exponent Ψ ξ (z) = T 0 Ψ X ( z, u(t) ) dt, z R d, where u : [0, T ] R d is the function with k-th component u k (t) := T t f k (s)ds, k {1, 2,..., d}. Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 18 / 40

Main Theorem 1 Theorem If X is a Lévy process with generating triple (a, σ 2, ν) that satisfies the KLT and BV assumptions then Z (d) has generating triple (a, Q, Π) where a is the vector with entries a k := a ( 1)k+1 2T 3 2 ) 2, k {1, 2,..., d}, π (k 2 1 2 Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 19 / 40

Main Theorem 1 Theorem (cont.) Q is a diagonal d d matrix with entries and Π is the measure, Π(B) := q k,k := σ2 2 T 2 π 2 ( k 1 2 ) 2, k {1, 2,..., d}, I(f(v) B)(ν λ)(dv), B B R d \{0}, R\{0} [0,T ] where λ is the Lebesgue measure on [0, T ] and f : R [0, T ] R d is the function ( 2T x cos ( ( ) ) π T 1 1 2 t (x, t) ( ),..., cos ( ( ) )) T π T d 1 2 t ( ). π 1 1 d 1 2 2 Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 20 / 40

Main Theorem 1 Idea of Proof: From Lemma: Ψ Z (d)(z) = T 0 Ψ X( z, u(t) )dt Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 21 / 40

Main Theorem 1 Idea of Proof: From Lemma: Ψ Z (d)(z) = T 0 Ψ X( z, u(t) )dt Evaluate integral in the components of u u k (t) = 2 T T t ( π sin T ( ( k 1 ) ) s ds = 2T cos π T 2 ( ) ) k 1 2 t π(k 1 2 ) Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 21 / 40

Main Theorem 1 Idea of Proof: From Lemma: Ψ Z (d)(z) = T 0 Ψ X( z, u(t) )dt Evaluate integral in the components of u u k (t) = 2 T T t ( π sin T ( ( k 1 ) ) s ds = 2T cos π T 2 ( ) ) k 1 2 t π(k 1 2 ) Evaluation of integral over [0, T ] and the orthogonality of {u k } 1 k d give a and Q. Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 21 / 40

Main Theorem 1 Idea of Proof: From Lemma: Ψ Z (d)(z) = T 0 Ψ X( z, u(t) )dt Evaluate integral in the components of u u k (t) = 2 T T t ( π sin T ( ( k 1 ) ) s ds = 2T cos π T 2 ( ) ) k 1 2 t π(k 1 2 ) Evaluation of integral over [0, T ] and the orthogonality of {u k } 1 k d give a and Q. Fubini s theorem and a change of variables yields Π. Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 21 / 40

Dependence Corollary Z (d) has independent entries if, and only if, ν is the zero measure. Idea of Proof: ( ) known, or because Q is diagonal ( ) we show that Π is not supported by the coordinate axes Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 22 / 40

Outline 1 Introduction 2 Lévy Processes and Infinitely Divisible Random Vectors 3 Main Results KLE Components Simulation 4 Examples Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 22 / 40

The problem How can we simulate a random vector with dependent entries with only knowledge of the characteristic function? In general, this seems to be a difficult problem. Infinite divisbility makes it possible! Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 23 / 40

Shot noise Idea: Write Z (d) as a(n) (infinite) sum of simpler random vectors. Random sequences {V i } i 1 and {Γ i } i 1 which are independent of each other and defined on a common probability space. Γ i sum of i i.i.d exponential r.v. s with mean 1. {V i } i 1 are independent and take values in some measurable space D with common dist. F Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 24 / 40

Shot noise Idea: Write Z (d) as a(n) (infinite) sum of simpler random vectors. Random sequences {V i } i 1 and {Γ i } i 1 which are independent of each other and defined on a common probability space. Γ i sum of i i.i.d exponential r.v. s with mean 1. {V i } i 1 are independent and take values in some measurable space D with common dist. F H : (0, ) D R d, measurable, and n S n := H(Γ i, V i ), n N, i=1 and µ(b) := 0 D I(H(r, v) B)F (dv)dr, B B R d \{0}. Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 24 / 40

Shot noise Theorem If µ is a Lévy measure satifying the BV assumption, then S n converges almost surely to an ID random vector with generating triple (0, 0, µ) as n. Theorems 3.1, 3.2, and 3.4 in Rosiński, J. (1990). On series representations of infinitely divisible random vectors. The Annals of Probability 18, 405 430. Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 25 / 40

Shot noise Idea: Show that Π has a disintegrated form like µ. Simplifying assumptions: X has no Gaussian component X has only positive jumps d X t = X + t Xt + B t Z (d) X = Z (d) X Z (d) + X + Z (d) B Necessary assumption: ν has a strictly positive density π. d We want g(x) := x π(s)ds to have a well defined, non-increasing inverse g 1. Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 26 / 40

Main Result 2 Theorem If X is a Lévy process with only positive jumps, satisfying the KLT and BV assumptions, such that X has triple (a, 0, π), where π is strictly positive, then Z (d) d = a + H(Γ i, U i ), where {U i } i 1 is an i.i.d. sequence of uniform random variables on [0, 1], and i 1 H(r, v) := f(g 1 (r/t ), T v), where f is the function defined previously, i.e. ( ( 2T x cos π ( ) ) T 1 1 2 t f(x, t) = ( ) π 1 1,..., cos ( ( ) )) T π T d 1 2 t ( ) 2 d 1. 2 Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 27 / 40

Discussion Nice/Unique features: S (d) 2 t := T d k=1 ( ( π Z k sin k 1 ) ) t T 2 d Z k = ak + 2T g 1 (Γ i /T ) cos ( π ( ) ) k 1 2 Ui ( ) π i 1 k 1, 2 Z (d) is independent of t Potentially difficult: Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 28 / 40

Discussion Nice/Unique features: S (d) 2 t := T d k=1 ( ( π Z k sin k 1 ) ) t T 2 d Z k = ak + 2T g 1 (Γ i /T ) cos ( π ( ) ) k 1 2 Ui ( ) π i 1 k 1, 2 Z (d) is independent of t We can increase d incrementally very easily (without starting fresh simulation) S (d) has smooth paths Potentially difficult: Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 28 / 40

Discussion Nice/Unique features: S (d) 2 t := T d k=1 ( ( π Z k sin k 1 ) ) t T 2 d Z k = ak + 2T g 1 (Γ i /T ) cos ( π ( ) ) k 1 2 Ui ( ) π i 1 k 1, 2 Z (d) is independent of t We can increase d incrementally very easily (without starting fresh simulation) S (d) has smooth paths Potentially difficult: Each summand of i 1 H(Γ i, U i ) requires the evaluation of d cosine functions Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 28 / 40

Discussion Nice/Unique features: S (d) 2 t := T d k=1 ( ( π Z k sin k 1 ) ) t T 2 d Z k = ak + 2T g 1 (Γ i /T ) cos ( π ( ) ) k 1 2 Ui ( ) π i 1 k 1, 2 Z (d) is independent of t We can increase d incrementally very easily (without starting fresh simulation) S (d) has smooth paths Potentially difficult: Each summand of i 1 H(Γ i, U i ) requires the evaluation of d cosine functions Inverting g Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 28 / 40

Discussion Nice/Unique features: S (d) 2 t := T d k=1 ( ( π Z k sin k 1 ) ) t T 2 d Z k = ak + 2T g 1 (Γ i /T ) cos ( π ( ) ) k 1 2 Ui ( ) π i 1 k 1, 2 Z (d) is independent of t We can increase d incrementally very easily (without starting fresh simulation) S (d) has smooth paths Potentially difficult: Each summand of i 1 H(Γ i, U i ) requires the evaluation of d cosine functions Inverting g Convergence of sum depends on decay of g 1 Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 28 / 40

Outline 1 Introduction 2 Lévy Processes and Infinitely Divisible Random Vectors 3 Main Results KLE Components Simulation 4 Examples Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 28 / 40

Example 1 Consider a VG process with c = 1, ρ = 2, ˆρ = 5, and T = 3. The process X + has Lévy measure ν(dx) = e ρx x dx so that we have e ρx g(x) = c x x dx = ce 1(ρx), and g 1 where E 1 is the exponential integral function. ( ) Γi T = 1 ρ E 1 1 ( ) Γi, T c We truncate the random series when Γ i /(T c) > 46 since at this point g 1 (Γ i /T ) < ρ 1 10 19. We expect to generate 46T c = 138 random pairs (U i, Γ i ) per path of X +. Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 29 / 40

Example 1 Sample paths of S (d) with d {5, 10, 15, 20, 25, 100, 250, 500, 3000} Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 30 / 40

Sample paths of S (d) A closer look at d = 3000 shows...... the Gibbs phenomenon... Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 31 / 40

Sample paths of S (d)... but we can remove this if we want. For example if we form the Cesàro sums C (d) dk=1 t S (k) t the Gibbs phenomenon disappears. := 1 d Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 32 / 40

Sample paths of C (d) Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 33 / 40

Monte Carlo test Suppose we want to compute E[e Xt ] = e Ψ( i) = (5/3) t on the interval [1, 2] by simulating 10 6 paths of S (d). The errors: Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 34 / 40

Discussion This method could be useful if we have to evaluate E[f(X t )] at many different points in the interval [0, T ], or at previously unknown points, say if we wanted to find the minimum in t. If, for example, d 100 is deemed good enough, then 100 10 6 single precision floating point numbers (realizations of the {Z k } 1 k d ) can be stored in memory: 4 Bytes 10 8 0.37 GB. Even d = 3000 is not impossible: 4 Bytes 3 10 9 11.2 GB. If we want to compute E[f(X t )] at points t 1 t 2... t d in [0, T ] then we can also use a fractional FFT algorithm to compute the realizations of {S (d) t k } 1 k d with O(d log(d)) operations. Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 35 / 40

Financial Applications Consider a VG process X with c = 5, ρ = 21.8735, ˆρ = 56.4414, and T = 3. We will also add a drift µ whose value is to be determined. Suppose we want to compute [ e rτ E (A 0 exp (X τ ) K) +], or [ (A0 τ ) ] + e rτ E exp (X t ) dt K, τ 0 where r, A 0, K > 0. Then µ is determined by the risk neutral condition Ψ( i) = r, and the expressions represent the price of a European and Asian call option respectively. Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 36 / 40

European option Replace X by S (d) and do Monte Carlo with 10 6 iterations. Errors for τ [1, 2] compared to a benchmark computed using a Fourier transform technique. Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 37 / 40

Asian option Replace X by C (d) and do Monte Carlo with 10 6 iterations. Errors for τ [1, 2] compared to a benchmark computed using a Fourier transform technique. Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 38 / 40

Smooth paths Let c (d) be a realization of C (d) on [0, T ]. Then t exp(c (d) t ) is smooth function of t. Applying the right change of variables cos(θ) = 2 ( t τ 1) transforms τ 0 ( exp c (d) t ) dt into the integral of a periodic function over [0, π]. We expect exponential convergence of the trapezoidal rule for such a function. Numerical evaluation is then possible via the Clenshaw-Curtis quadrature. Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 39 / 40

Some ideas for the future Can we speed the convergence of E[f(X t )] or E[F({X t : t [0, T ]})] by: Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 40 / 40

Some ideas for the future Can we speed the convergence of E[f(X t )] or E[F({X t : t [0, T ]})] by: Expanding X in a different orthonormal basis Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 40 / 40

Some ideas for the future Can we speed the convergence of E[f(X t )] or E[F({X t : t [0, T ]})] by: Expanding X in a different orthonormal basis Expanding X in a wavelet basis Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 40 / 40

Some ideas for the future Can we speed the convergence of E[f(X t )] or E[F({X t : t [0, T ]})] by: Expanding X in a different orthonormal basis Expanding X in a wavelet basis Unser, M. and Tafti, P.D.. (2014). An introduction to sparse stochastic processes. Cambridge University Press, Cambridge. Determining the KLE of Y, Y t := f(x t ) directly Other Lévy generalizations of Gaussian processes, e.g. Lévy Bridge In the discrete case, orthogonal transforms seem to speed rate of Quasi-Monte Carlo convergence when generating Brownian paths. Is something similar true here as well? Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 40 / 40

Some ideas for the future Can we speed the convergence of E[f(X t )] or E[F({X t : t [0, T ]})] by: Expanding X in a different orthonormal basis Expanding X in a wavelet basis Unser, M. and Tafti, P.D.. (2014). An introduction to sparse stochastic processes. Cambridge University Press, Cambridge. Determining the KLE of Y, Y t := f(x t ) directly Other Lévy generalizations of Gaussian processes, e.g. Lévy Bridge In the discrete case, orthogonal transforms seem to speed rate of Quasi-Monte Carlo convergence when generating Brownian paths. Is something similar true here as well? Hackmann, D. (2016). Karhunen Loève expansions of Lévy processes. preprint: arxiv:1603.00677. Daniel Hackmann (JKU Linz) KLE s of Lévy Processes 40 / 40