A two-sample test for comparison of long memory parameters F. - PowerPoint PPT Presentation

Introduction Test Statistic Consistency bi-fBm bi-linear models In practice Simu A two-sample test for comparison of long memory parameters F. Lavancier 1 , A. Philippe 1 , D. Surgailis 2 1 Laboratoire Jean Leray, Université de Nantes 2 Institute of Mathematics and Informatics, Vilnius JSTAR 2008

Introduction Test Statistic Consistency bi-fBm bi-linear models In practice Simu 1 Introduction 2 Test Statistic 3 Consistency of the test 4 The bivariate fractional Brownian motion 5 The case of bivariate linear models 6 Practical implementation of the test 7 Some simulations

Introduction Test Statistic Consistency bi-fBm bi-linear models In practice Simu 1 Introduction

Introduction Test Statistic Consistency bi-fBm bi-linear models In practice Simu Introduction Let X ( t ) , t ∈ Z , be a second order, stationary time series and let γ ( h ) be its covariance function, i.e. γ ( h ) = cov( X ( t ) , X ( t + h )) . We say that X exhibit long memory when its covariance function is not summable : � | γ ( h ) | = ∞ . h ∈ Z Exemple : FARIMA(p,d,q) processes, defined for p ∈ N , q ∈ N , d ∈ ( − 1 / 2 , 1 / 2) , verify γ ( h ) ∼ h 2 d − 1 when h → ∞ and long memory occurs when 0 < d < 1 / 2 .

Introduction Test Statistic Consistency bi-fBm bi-linear models In practice Simu Let X 1 a FARIMA ( p 1 , d 1 , q 1 ) with 0 ≤ d 1 < 1 / 2 , X 2 a FARIMA ( p 2 , d 2 , q 2 ) with 0 ≤ d 2 < 1 / 2 . We want to test the null hypothesis H 0 : d 1 = d 2

Introduction Test Statistic Consistency bi-fBm bi-linear models In practice Simu Let X 1 a FARIMA ( p 1 , d 1 , q 1 ) with 0 ≤ d 1 < 1 / 2 , X 2 a FARIMA ( p 2 , d 2 , q 2 ) with 0 ≤ d 2 < 1 / 2 . We want to test the null hypothesis H 0 : d 1 = d 2 Our framework : X 1 and X 2 may not be independent. We do not restrict to FARIMA models (see the assumptions later)

Introduction Test Statistic Consistency bi-fBm bi-linear models In practice Simu 2 Test Statistic

Introduction Test Statistic Consistency bi-fBm bi-linear models In practice Simu H 0 : d 1 = d 2 First idea : estimate d 1 and d 2 by ˆ d 1 and ˆ d 2 (different estimators are available : log-periodogram, Whittle, GPH, FEXP, etc.) evaluate ( ˆ d 1 − ˆ d 2 ) to conclude

Introduction Test Statistic Consistency bi-fBm bi-linear models In practice Simu H 0 : d 1 = d 2 First idea : estimate d 1 and d 2 by ˆ d 1 and ˆ d 2 (different estimators are available : log-periodogram, Whittle, GPH, FEXP, etc.) evaluate ( ˆ d 1 − ˆ d 2 ) to conclude Drawbacks : the joint probability law of ˆ d 1 and ˆ d 2 in the dependent case is not known. the behavior of ( ˆ d 1 − ˆ d 2 ) is strongly sensitive to the short-memory part of the induced processes X 1 and X 2 (e.g. the ARMA part of a FARIMA), leading to a bad size of the test.

Introduction Test Statistic Consistency bi-fBm bi-linear models In practice Simu Our approach : If X exhibits long memory, � [ nτ ] S n ( τ ) = t =1 ( X ( t ) − EX ( t )) does not have a standard asymptotic behavior.

Introduction Test Statistic Consistency bi-fBm bi-linear models In practice Simu Our approach : If X exhibits long memory, � [ nτ ] S n ( τ ) = t =1 ( X ( t ) − EX ( t )) does not have a standard asymptotic behavior. For testing H 0 : d = 0 (short memory) vs H 1 : d � = 0 (long memory) several procedures rely on the variations of S n . R/S (Lo, 1991) : based on the range of S n , KPSS (Kwiatkowski et al. , 1992) : based on E ( S 2 n ) , V/S (Giraitis et al. , 2003) : based on V ar ( S n ) .

Introduction Test Statistic Consistency bi-fBm bi-linear models In practice Simu Our approach : If X exhibits long memory, � [ nτ ] S n ( τ ) = t =1 ( X ( t ) − EX ( t )) does not have a standard asymptotic behavior. For testing H 0 : d = 0 (short memory) vs H 1 : d � = 0 (long memory) several procedures rely on the variations of S n . R/S (Lo, 1991) : based on the range of S n , KPSS (Kwiatkowski et al. , 1992) : based on E ( S 2 n ) , V/S (Giraitis et al. , 2003) : based on V ar ( S n ) . In the same spirit, for testing H 0 : d 1 = d 2 our statistic is V 1 /S 1 ,q + V 2 /S 2 ,q T n,q = , V 2 /S 2 ,q V 1 /S 1 ,q where V 1 /S 1 ,q is the standard V/S statistic for X 1 , V 2 /S 2 ,q is the standard V/S statistic for X 2 .

Introduction Test Statistic Consistency bi-fBm bi-linear models In practice Simu More precisely V 1 /S 1 ,q + V 2 /S 2 ,q T n,q = . V 2 /S 2 ,q V 1 /S 1 ,q For i=1,2, X i denotes the sample mean of X i ˆ γ i ( h ) the empirical covariance function of X i . � k � n � 2 � 2 n k � � � � n − 2 − n − 3 V i = ( X i ( t ) − X i ) ( X i ( t ) − X i ) t =1 t =1 k =1 k =1 ( V i is the empirical variance of the partial sums of X i ) q q +1 � | h | � 1 � � S i,q = 1 − γ i ( h ) = ˆ γ i ( h − ℓ ) . ˆ q + 1 q + 1 h = − q h,ℓ =1 ( S i,q estimates the variance of the limiting law of the partial sums of X i )

Introduction Test Statistic Consistency bi-fBm bi-linear models In practice Simu The dependent case Consider the cross-covariance estimator q q +1 � | h | � 1 � � S 12 ,q = 1 − γ 12 ( h ) = ˆ γ 12 ( h − ℓ ) ˆ q + 1 q + 1 h = − q h,ℓ =1 γ 12 ( h ) = n − 1 � n − h where, ˆ t =1 ( X 1 ( t ) − X 1 )( X 2 ( t + h ) − X 2 ) , h > 0 .

Introduction Test Statistic Consistency bi-fBm bi-linear models In practice Simu The dependent case Consider the cross-covariance estimator q q +1 � | h | � 1 � � S 12 ,q = 1 − ˆ γ 12 ( h ) = γ 12 ( h − ℓ ) ˆ q + 1 q + 1 h = − q h,ℓ =1 γ 12 ( h ) = n − 1 � n − h where, ˆ t =1 ( X 1 ( t ) − X 1 )( X 2 ( t + h ) − X 2 ) , h > 0 . When X 1 and X 2 are dependent, we introduce ˜ X 1 ( t ) = X 1 ( t ) − ( S 12 ,q /S 2 ,q ) X 2 ( t ) , t = 1 , . . . , n. so that the partial sums of ˜ X 1 and X 2 are uncorrelated.

Introduction Test Statistic Consistency bi-fBm bi-linear models In practice Simu The dependent case Consider the cross-covariance estimator q q +1 � | h | � 1 � � S 12 ,q = 1 − ˆ γ 12 ( h ) = γ 12 ( h − ℓ ) ˆ q + 1 q + 1 h = − q h,ℓ =1 γ 12 ( h ) = n − 1 � n − h where, ˆ t =1 ( X 1 ( t ) − X 1 )( X 2 ( t + h ) − X 2 ) , h > 0 . When X 1 and X 2 are dependent, we introduce ˜ X 1 ( t ) = X 1 ( t ) − ( S 12 ,q /S 2 ,q ) X 2 ( t ) , t = 1 , . . . , n. so that the partial sums of ˜ X 1 and X 2 are uncorrelated. Then we consider V 1 / ˜ ˜ S 1 ,q + V 2 /S 2 ,q ˜ T n = , V 1 / ˜ ˜ V 2 /S 2 ,q S 1 ,q where ˜ V 1 and ˜ S 1 ,q are the same as before but with respect to ˜ X 1 .

Introduction Test Statistic Consistency bi-fBm bi-linear models In practice Simu 3 Consistency of the test

Introduction Test Statistic Consistency bi-fBm bi-linear models In practice Simu Assumptions Assumption A ( d 1 , d 2 ) There exist d i ∈ [0 , 1 / 2) , i = 1 , 2 such that for any i, j = 1 , 2 the following limits exist n 1 � 1) c ij = lim γ ij ( t − s ) . n 1+ d i + d j n →∞ t,s =1 Moreover, when q → ∞ , n → ∞ , n/q → ∞ , � q k,l =1 ˆ γ ij ( k − l ) 2) → p 1 � q k,l =1 γ ij ( k − l ) This assumption claims that 1) the second moment of the partial sums of X i converge with the proper normalization, 2) the natural estimation of this second moment is consistent.

Introduction Test Statistic Consistency bi-fBm bi-linear models In practice Simu Assumptions Assumption B ( d 1 , d 2 ) The partial sums of X 1 and X 2 � � n − d 1 − (1 / 2) � [ nτ ] t =1 ( X 1 ( t ) − EX 1 ( t )) , n − d 2 − (1 / 2) � [ nτ ] t =1 ( X 2 ( t ) − EX 2 ( t )) converge (jointly) in finite dimensional distribution to ( √ c 11 B 1 ,d 1 ( τ ) , √ c 22 B 2 ,d 2 ( τ )) , where ( B 1 ,d 1 ( τ ) , B 2 ,d 2 ( τ )) is a bivariate fractional Brownian motion with parameters d 1 , d 2 and the correlation coefficient ρ = c 12 / √ c 11 c 22 . Some questions : What is a bivariate fractional Brownian motion ? see later. Is this assumption restrictive (especially the joint convergence) ? → We will show that it holds for linear processes.

Introduction Test Statistic Consistency bi-fBm bi-linear models In practice Simu Recall that H 0 : d 1 = d 2 and the test statistic is V 1 / ˜ ˜ V 2 /S 2 ,q + V 2 /S 2 ,q S 1 ,q ˜ T n = , V 1 / ˜ ˜ S 1 ,q

Introduction Test Statistic Consistency bi-fBm bi-linear models In practice Simu Recall that H 0 : d 1 = d 2 and the test statistic is V 1 / ˜ ˜ V 2 /S 2 ,q + V 2 /S 2 ,q S 1 ,q ˜ T n = , V 1 / ˜ ˜ S 1 ,q Proposition (Consistency of the test) (i) Let Assumptions A ( d 1 , d 2 ) and B ( d 1 , d 2 ) be satisfied with some d 1 = d 2 = d ∈ [0 , 1 / 2) . Then, as n, q, n/q → ∞ , T n → law T = U 1 U 2 + U 2 ˜ U 1 , where Z 1 „Z 1 « 2 ( B 0 i,d ( τ )) 2 d τ − B 0 U i = i,d ( τ )d τ ( i = 1 , 2) 0 0 and where B 0 1 ,d ( τ ) , B 0 2 ,d ( τ ) are mutually independent fractional bridges with the same parameter d .