Modeling Jump Dependence using L evy copulas Andrea Krajina, - - PowerPoint PPT Presentation

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Modeling Jump Dependence using L´ evy copulas

Andrea Krajina, Institut f¨ ur Matematische Stochastik, G¨

• ttingen

Joint work with Roger Laeven, Tilburg University and EURANDOM EURANDOM, Eindhoven, August, 2011.

Outline

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Introduction

– L´

evy process

– L´

evy copula

M-Estimator

– Definition. Asymptotic properties – Simulation study: Clayton copula

Example: MSCI equity index data (Europe)

L´ evy process

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bivariate L´

evy process: a stochastic process Y = (Y1, Y2) such that

– Y0 = (0, 0); – independent and stationary increments; – stochastically continuous.

by L´

evy -Khintchine representation: E

• eiz,Y

= eφ(z), where φ(z) = i γ, z−1 2 z, Az+

• R2\(0,0)
• eiz,y − 1 − i z, y 1{|y| ≤ 1}
• ν(dy),

any L´ evy process: (γ, A, ν), with γ ∈ R2, A ∈ R2×2 and a positive sigma-finite measure ν on R2 \ {(0, 0)}.

L´ evy copula

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denote R := (−∞, ∞] a function F : S → R, S ⊆ R

2 is 2-increasing if

F(a1, a2) + F(b1, b2) − F(a1, b2) − F(b1, a2) ≥ 0, for all (a1, a2), (b1, b2) ∈ S with a1 ≤ b1, a2 ≤ b2

a function F : R

2 → R is a L´

evy copula if

– F(u1, u2) = ∞, if (u1, u2) = (∞, ∞), – F(u1, u2) = 0, if ui = 0, for at least one i = 1, 2, – F is 2-increasing – F i(u) = u, i = 1, 2, u ∈ R

L´ evy copula

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a function U : R2 \ {(0, 0)} → R, S ⊆ R

2 is the tail integral of a L´

evy process with the L´ evy measure ν if U(x1, x2) = sgn(x1)sgn(x2)ν (I(x1) × I(x2)) , where I(x1) =

• (x1, ∞),

if x1 ≥ 0, (−∞, x1], if x1 < 0.

L´

evy copula vs. copula

L´ evy copula

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Theorem (Kallsen and Tankov)

Let F be a bivariate L´ evy copula and U1 and U2 the tail integrals of its marginal processes. Then there exists a bivariate L´ evy process Y whose components have tail integrals U1 and U1, and such that U I((xi)i∈I) = F I((Ui(xi))i∈I) for any non-empty I ⊆ {1, 2}. The L´ evy measure ν is uniquely determined by U1, U2 and F.

L´ evy copula. Limit relation

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Theorem (Kallsen and Tankov)

Let X = (X1, X2) be a bivariate L´ evy process with L´ evy copula F and tail integrals U1 and U2. Denote by C(α1,α2)

t

: [0, 1]2 → [0, 1] a copula of (−α1X1, −α2X2), for t ∈ (0, ∞)2 and α1, α2 ∈ {−1, 1}. Then F(u1, u2) = lim

t→0 C(sgnu1,sgnu2) t

(t|u1|, t|u2|)sgnu1sgnu2, for any (u1, u2) ∈ Ran(U1) × Ran(U2)

M-Estimator: Basic Assumptions

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bivariate L´

evy process Y observed at n distinct times separated by ∆n: Y∆n, Y2∆n, . . . , Yn∆n

n increments: Y∆n, Y2∆n − Y∆n, . . . , Yn∆n − Y(n−1)∆n sample X1, . . . , Xn, where Xi = Yi∆n − Y(i−1)∆n, with distribution

function H and marginals H1 and H2

assume that its L´

evy copula F belongs to a parametric family F ∈ {Fθ : θ ∈ Θ ⊆ Rp},

M-Estimator: Nonparametric Estimator

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nonparametric estimator of F (Laeven)

ˆ F(u1, u2) := sgn(u1)sgn(u2)1 k

n

• i=1

1

• R1

i > n + 1 − ku1,

if u1 ≥ 0 R1

i ≤ k|u1|,

if u1 < 0

• ,
• R2

i > n + 1 − ku2,

if u2 ≥ 0 R2

i ≤ k|u2|,

if u2 < 0

• – Rj

i is the rank of Xij among X1j, . . . , Xnj, j = 1, 2

– k ∈ {1, 2, . . . , n}

k = kn → ∞ and k/n → 0 when n → ∞ ∆n → 0 and n∆n → ∞ when n → ∞

M-Estimator: Definition

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let T > 0; denote ST := [−T, T]2 ∩ Ran(U1) × Ran(U2) let g ≡ (g1, . . . , gq)T : T → Rq, q ≥ p, be an integrable function such

that ϕ : Θ → Rq defined by ϕ(θ) :=

• T

g(u)F(u; θ) du is a homeomorphism between Θ and its image ϕ(Θ)

M-estimator ˆ

θn of θ0 is defined as the minimizer of the criterion function Qk,n(θ) =

q

• m=1
• T

gm(x)

• ˆ

Fn(x) − F (x; θ)

• dx

2

M-Estimator: Asymptotic Properties

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existence, uniqueness and consistency asymptotic normality conditions:

– there exists α ≥ 1 such that k/n − ∆n = o (k/n)α; – F has continuous first derivatives F 1 and F 2, – there exist α > 0 and c > 0 such that, as t → 0,

t−1C(sgn(u1),sgn(u2))

t

(t|u1|, t|u2|)sgn(u1)sgn(u2)−F(u1, u2) = O(tα) uniformly on {(u1, u2) ∈ Ran(U1) × Ran(U2): u2

1 + u2 2 = c},

– for the α from (AN2), k/n − ∆n = O((k/n)1+α) and

k = o(n2α/(1+2α));

M-Estimator: Consistency

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Theorem 1 (Consistency) If the following conditions are satisfied, (C1) Θ is open, (C2) ϕ is a homeomorphism from Θ to ϕ(Θ), (C3) ϕ is a twice continuously differentiable, (C4) ˙ ϕ(θ0) is of full rank; (C5) k = kn is an intermediate sequence: k → ∞ and k/n → 0, as n → ∞, (C6) ∆n is such that ∆n → 0 and n∆n → ∞, (C7) there exists α ≥ 1 such that k/n − ∆n = o (k/n)α; then with probability tending to one the criterion function Qk,n has a unique minimizer ˆ θn, and ˆ θn

P

→ θ0, as n → ∞.

M-Estimator: Asymptotic Normality

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Theorem 2 (Asymptotic normality) If in addition to the assumptions (C1)-(C6) of Theorem 1 the following assumptions hold, (AN1) F has continuous first derivatives F (1) and F (2), (AN2) there exist α > 0 and c > 0 such that, as t → 0, t−1C(sgn(u1),sgn(u2))

t

(t|u1|, t|u2|)sgn(u1)sgn(u2)−F(u1, u2) = O(tα) uniformly on {(u1, u2) ∈ Ran(U1) × Ran(U2): u2

1 + u2 2 = c},

(AN3) for the α from (AN2), k/n − ∆n = O((k/n)1+α) and k = o(n2α/(1+2α)); then as n → ∞, √ k(ˆ θn − θ0)

d

→ N((0, 0)T , M(θ0)). (2)

Simulation Study: Clayton L´ evy Copula

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Clayton L´

evy copula is given by F(u1, u2) =

• |u1|−ζ + |u2|−ζ−1/ζ

(η1{u1u2 ≥ 0} − (1 − η)1{u1u2 < 0}) , with parameters ζ > 0 and η ∈ [0, 1]

ζ > 0 unknown, η = 1 fixed L´

evy process: bivariate Poisson jump diffusion dXt = µdt + σdWt + JdNt

Simulation Study: Clayton L´ evy Copula

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ν(dx) = λ1f1(x1; θ1)dx1 +λ2f2(x2; θ2)dx2 +λ12f12(x1, x2; θ12)dx1dx2, model parameters:

µ = (0, 0), λ+

1 + λ+ 12 = λ+ 2 + λ+ 12 = 6, θ1 = θ2 = 1/30,

σ = (0.1, 0.1), ρ = 0.4, three different values for the L´ evy copula parameter ζ ∈ {0.3, 1, 3}

simulations: 50 samples of size n = 7500, corresponding to Tn = 30 and

∆n = 1/250 (e.g. daily returns over 30 years)

Simulation Study: Clayton L´ evy Copula

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0.00 0.05 0.10 0.15 0.20 0.25 0.00 0.05 0.10 0.15 0.20 0.25

ζ=0.3, with Brownian noise

0.00 0.05 0.10 0.15 0.00 0.05 0.10 0.15 0.20

ζ=1, with Brownian noise

0.00 0.05 0.10 0.15 0.00 0.05 0.10 0.15

ζ=3, with Brownian noise

Simulation Study: Clayton L´ evy Copula

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20 40 60 80 100 0.30 0.32 0.34 0.36 0.38 0.40 0.42 k ζ ^

Clayton family with ζ=0.3, η=1; M−est. of ζ, rep=50

T=5 T=10 20 40 60 80 100 0.00 0.02 0.04 0.06 0.08 0.10 0.12 k RMSE

Clayton family with ζ=0.3, η=1; M−est. of ζ, rep=50

T=5 T=10

Figure 1: Estimation of ζ = 0.3, for different values of k and T.

Simulation Study: Clayton L´ evy Copula

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20 40 60 80 100 0.90 0.95 1.00 1.05 k ζ ^

Clayton family with ζ=1, η=1; M−est. of ζ, rep =50

T=5 T=10 20 40 60 80 100 0.00 0.05 0.10 0.15 0.20 k RMSE

Clayton family with ζ=1, η=1; M−est. of ζ, rep =50

T=5 T=10

Figure 2: Estimation of ζ = 1, for different values of k and T.

Simulation Study: Clayton L´ evy Copula

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20 40 60 80 100 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 k ζ ^

Clayton family with ζ=3, η=1; M−est. of ζ, rep =50

T=5 T=10 20 40 60 80 100 0.0 0.2 0.4 0.6 0.8 1.0 k RMSE

Clayton family with ζ=3, η=1; M−est. of ζ, rep =50

T=5 T=10

Figure 3: Estimation of ζ = 3, for different values of k and T.

Simulation Study: Clayton L´ evy Copula

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jump dependence coefficient: F(1, 1) = 2−1/0.3 ≈ 0.099

20 40 60 80 100 0.08 0.10 0.12 0.14 0.16 0.18 0.20 k F(1, 1) ^

Clayton(0.3,1); M− and nonpar. est. of F(1,1)=0.099, rep=50

T=5 T=10 non.par. 20 40 60 80 100 0.00 0.02 0.04 0.06 0.08 0.10 k RMSE

Clayton(0.3,1); M− and nonpar. est. of F(1,1)=0.099, rep=50

T=5 T=10 non.par.

Figure 4: Estimation of F(1, 1) = 2−1/0.3 ≈ 0.099, for different values of k and T.

Simulation Study: Clayton L´ evy Copula

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jump dependence coefficient: F(1, 1) = 2−1/1 = 0.5

20 40 60 80 100 0.46 0.47 0.48 0.49 0.50 0.51 0.52 k F(1, 1) ^

Clayton(1,1); M− and nonpar. est. of F(1,1)=0.5, rep=50

T=5 T=10 non.par. 20 40 60 80 100 0.00 0.05 0.10 0.15 k RMSE

Clayton(1,1); M− and nonpar. est. of F(1,1)=0.5, rep=50

T=5 T=10 non.par.

Figure 5: Estimation of F(1, 1) = 0.5, for different values of k and T.

Simulation Study: Clayton L´ evy Copula

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jump dependence coefficient: F(1, 1) = 2−1/3 ≈ 0.794

20 40 60 80 100 0.74 0.75 0.76 0.77 0.78 0.79 k F(1, 1) ^

Clayton(3,1); M− and nonpar. est. of F(1,1)=0.794, rep=50

T=5 T=10 non.par. 20 40 60 80 100 0.00 0.05 0.10 0.15 k RMSE

Clayton(3,1); M− and nonpar. est. of F(1,1)=0.794, rep=50

T=5 T=10 non.par.

Figure 6: Estimation of F(1, 1) = 2−1/3 ≈ 0.794, for different values of k and T.

Example: MSCI equity data; Europe

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we consider log-returns from the price index for the following pairs of

countries:

– UK-Germany, time period Jan 2, 1980 - Jun 30, 2011, n = 8127 – UK-Portugal, time period Jan 4, 1988 - June 30, 2011, n = 6129 – UK-Greece, time period Jan 4, 1988 - June 30, 2011, n = 6129 – summary statistics: mean≈ 0, std.dev 0.012 − 0.02

ˆ ζ, k = 30 F(1, 1) = 2−1/ˆ

ζ

ˆ F(1, 1) UK-Germany 0.885 0.457 0.5 UK-Portugal 0.545 0.280 0.33 UK-Greece 0.439 0.206 0.23

Example: MSCI equity data; Europe

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−0.15 −0.10 −0.05 0.00 0.05 0.10 −0.10 −0.05 0.00 0.05 0.10 UK Germany

UK − Germany

−0.10 −0.05 0.00 0.05 0.10 −0.10 −0.05 0.00 0.05 0.10 UK Portugal

UK − Portugal

−0.10 −0.05 0.00 0.05 0.10 −0.10 −0.05 0.00 0.05 0.10 0.15 UK Greece

UK − Greece

Example: MSCI equity data; Europe

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20 40 60 80 100 0.0 0.2 0.4 0.6 0.8 k ζ ^

UK − Germany

20 40 60 80 100 0.0 0.2 0.4 0.6 0.8 k ζ ^

UK − Portugal

20 40 60 80 100 0.0 0.1 0.2 0.3 0.4 0.5 k ζ ^

UK − Greece