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A Strategy for the Use of the Cross Recurrence Quantification Analysis

  • Teresa Aparicio , Eduardo F. Pozo EMAIL logo and Dulce Saura
Published/Copyright: April 23, 2020

Abstract

In this work, our goal is to analyze the use of the Cross Recurrence Plot (CRP) and its quantification (CRQA) as tools to detect the possible existence of a relationship between two systems. To do that, we define three tests that are a bivariate extension of those proposed by Aparicio et al. (Aparicio, T., E. Pozo, and D. Saura. 2008. “Detecting Determinism Using Recurrence Quantification Analysis: Three Test Procedures.” Journal of Economic Behavior & Organization 65: 768–787, Aparicio, T., E. F. Pozo, and D. Saura. 2011. “Detecting Determinism Using Recurrence Quantification Analysis: A Solution to the Problem of Embedding.” Studies in Nonlinear Dynamics and Econometrics 15: 1–10) within the context of the Recurrence Quantification Analysis. These tests, based on the diagonal lines of the CRP, are applied to a large number of simulated pairs of series. The results obtained are not always satisfactory, with problems being detected specifically when the series have a high degree of laminarity. We study the identified problems and we implement a strategy that we consider adequate for the use of these tools. Finally, as an example, we apply this strategy to several economic series.

Appendix: A brief description of the dynamic systems employed

This Appendix provides details of the equations that generate the dynamic systems used in our study.

Lorenz System:

= 10·(y – x)

= 28·(x – y – x·z)

ż = x·y – 8 3 x·z

Initial conditions:

x = 0.1, y = 0.1, z = 0.1

Time step (fixed) = 0.015.

Rossler System:

= – yz

= x + 0.2 · y

ż = 0.4 + z · (x – 5.7)

Initial conditions:

x = 8.6274, y = −1.8956,

z = 0.9850.

Time step (fixed) = 0.105.

Henon System:

xt+1 = 1 – 1.4·xt 2 + yt

yt+1 = 0.3·xt

Initial conditions:

x = 1, y = 0.

Martin System:

xt +1 = yt – sin(t)

yt +1 = 3.14 – xt

Initial conditions:

x = 0, y = 0.

Ikeda System:

t = 0.4 6 1 + x t 2 + y t 2

xt +1 = 1 + 0.9·[xt ·cos(t) – yt ·sin(t)]

yt +1 = 0.9·[xt ·sin(t) + yt ·cos(t)]

Initial conditions:

x = 1, y = 0.

Lu Chen System

= 36·(y – x)

= xx·z + 20·y

ż = x·y – 3·z

Initial conditions:

x = 0.1, y = 0.3, z = − 0.6

Time step (fixed) = 0.005.

Pickover System:

xt +1 = sin(2.87·yt ) + 0.76·sin(2.87·xt )

yt +1 = sin(−0.96·xt ) + 0.74·sin(−0.96·yt )

Initial conditions:

x = 1, y = 1.

Duffing System:

= y

= x – x 3 – 0.25·y + 0.3·sin(t)

Initial conditions:

x = 1, y = 1, t = 0

Time step (fixed) = 0.05.

Peter de Young System:

xt +1 = sin(1.76·yt ) – sin(1.67·xt )

yt +1 = sin(−0.85·xt ) – cos(2.1·yt )

Initial conditions:

x = 0, y = 0.5.

Hopalong System:

xt +1 = yt – [abs(0.5·xt – 9.25)]·sign(xt – 1)

yt +1 = −0.85 – xt

Initial conditions:

x = 0, y = 0.

Van der Pol – Duffing System:

= y

= −0.1·y – [x 3 + 12·cos(2·π·t)]

Initial conditions:

x = 2, y = 0.1, t = 0

Time step (fixed) = 0.02.

Ueda System:

= y

= −0.05·y – x 3 + 0.75·sin(t)

Initial conditions:

x = 2.5, y = 0, t = 0

Time step (fixed) = 0.02.

Rayleigh System:

= y

= – x + y – y 3

Initial conditions:

x = 1, y = 1

Time step (fixed) = 0.5.

Tinkerbell System:

x t+1 = xt 2yt 2 + 0.9·xt – 0.6013·yt

y t+1 = 2·xt ·yt + 2·xt + 0.5·yt

Initial conditions:

x = −0.72, y = −0.64.

Chua modified System

f = 0.11 sin [ π x 2.6 ]

= 10.814·(yf)

= x – y + z

ż = −14.286·y

Initial conditions:

x = 1, y = 1, z = 0

Time step (fixed) = 0.1.

Marwan et al. (2007) System:

xt : variable x of the Lorenz System

y t+1 = 0.86·yt + 0.2·xt 2 + 0.5·zt

zt : Gaussian N(0,1) signal.

Initial conditions:

x = 0.1, y = 0.

Bilinear model:

x t+1 = e t+1 + 0.3·xt ·et

et : Gaussian N(0,1) signal.

Initial condition: x = 0.1.

ARCH1: ARCH(1) model with constant = 0.05 and ARCH coefficient = 0.2.

ARCH2: ARCH(1) model with constant = 0.05 and ARCH coefficient = 0.7.

GARCH1: GARCH(1,1) model with constant = 0.05, ARCH coefficient = 0.2 and GARCH coefficient = 0.7.

GARCH2: GARCH(1,1) model with constant = 0.05, ARCH coefficient = 0.7 and GARCH coefficient = 0.2.

EGARCH1: EGARCH(1,1) model with constant = 0.05, ARCH coefficient = 0.2, GARCH coefficient = 0.7 and leverage coefficient = −0.03.

EGARCH2: EGARCH(1,1) model with constant = 0.05, ARCH coefficient = 0.7, GARCH coefficient = 0.2 and leverage coefficient = −0.03.

Bivariate VAR(1): Stationary bivariate VAR(1) model, with Constant = [0.05; 0], AR coefficients = [0.5, 0; 0.1, 0.3]} and Covariance matrix = I(2).

BEKK-GARCH(1,1,1): Bivariate scalar BEKK-GARCH(1,1,1), with CCp = [1, 0.5; 0.5, 4], A2 = 0.05, G2 = 0.1 and B2 = 0.88.

CCC-GARCH(1,0,1): Multivariate 3 × 3 CCC-GARCH(1,0,1) model: GARCH_coefficients = [0.1, 0.1, 0.8] and constant conditional correlation = [1, 0.2, 0.5; 0.2, 1, 0.3; 0.5, 0.3, 1].

Vect-GARCH(1,1,1),a: Bivariate vector-GARCH(1,1,1) model: GARCH parameters = (0.2, 0.4, 0.4) and Intercept = [1, 0.8; 0.8, 1].

Vect-GARCH(1,1,1),b: Bivariate vector-GARCH(1,1,1) model: GARCH parameters = (0.1, 0.1, 0.1) and Intercept = [1, 0.8; 0.8, 1].

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Supplementary Material

The online version of this article offers supplementary material (DOI: https://doi.org/10.1515/snde-2018-0103).


Published Online: 2020-04-23

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