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Introduction Test Statistic In practice Simulations 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 10th International Vilnius Conference on Probability and Mathematical Statistics

Introduction Test Statistic In practice Simulations 1 Introduction

Introduction Test Statistic In practice Simulations The statistical problem ( X 1 ( t )) t ∈ Z is a stationary process with long memory parameter d 1 ≥ 0 ( X 2 ( t )) t ∈ Z is a stationary process with long memory parameter d 2 ≥ 0 X 1 and X 2 may be dependent. We want to test the null hypothesis H 0 : d 1 = d 2

Introduction Test Statistic In practice Simulations 2 The Testing procedure and its consistency

Introduction Test Statistic In practice Simulations First idea : Testing from the estimation of ( d 1 , d 2 ) H 0 : d 1 = d 2 Procedure 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 In practice Simulations First idea : Testing from the estimation of ( d 1 , d 2 ) H 0 : d 1 = d 2 Procedure 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 Drawback 1 The joint probability law of ˆ d 1 and ˆ d 2 in the dependent case is not known. 2 The behavior of | ˆ d 1 − ˆ d 2 | is strongly sensitive to the short-memory part of 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 In practice Simulations Our testing procedure 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 In practice Simulations Our testing procedure If X exhibits long memory, � [ nτ ] S n ( τ ) = t =1 ( X ( t ) − EX ( t )) does not have a standard asymptotic behavior. Example for F ( d ) 1 D [0 , 1] n d +1 / 2 S n ( τ ) − − − − → W d ( τ )

Introduction Test Statistic In practice Simulations Our testing procedure If X exhibits long memory, � [ nτ ] S n ( τ ) = t =1 ( X ( t ) − EX ( t )) does not have a standard asymptotic behavior. Example for F ( d ) 1 D [0 , 1] n d +1 / 2 S n ( τ ) − − − − → W d ( τ ) In particular, typically, V ar ( S n ( τ )) = O ( n 2 d +1 L ( n )) , where L is a slowly varying function. = ⇒ The idea is to base the test on the variations of S n .

Introduction Test Statistic In practice Simulations Univariate time series For univariate time series, the variations of S n are used to test : H 0 : d = 0 (short memory) vs H 1 : d � = 0 (long memory) The most standard statistics 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 In practice Simulations Univariate time series For univariate time series, the variations of S n are used to test : H 0 : d = 0 (short memory) vs H 1 : d � = 0 (long memory) The most standard statistics 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 In practice Simulations 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 ) k =1 t =1 k =1 t =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)

Introduction Test Statistic In practice Simulations 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 ) k =1 t =1 k =1 t =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) Basically, when n, q, n/q → ∞ , � − 2 d i � n V i /S i,q = ⇒ U ( d i ) q

Introduction Test Statistic In practice Simulations 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 ) Remark. This assumption claims that 1) the second moment of the partial sums of ( X 1 , X 2 ) converge with the proper normalization, 2) the natural estimation of this second moment is consistent.

Introduction Test Statistic In practice Simulations Assumptions (cont.) 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 nonanticipative bivariate fractional Brownian motion with parameters d 1 , d 2 and the correlation coefficient ρ = corr( B 1 (1) , B 2 (1)) = c 12 / √ c 11 c 22 Remark. This is fulfilled for bivariate (super)linear processes under mild assumptions.

Introduction Test Statistic In practice Simulations Asymptotic behavior of T n Proposition Let Assumptions A ( d 1 , d 2 ) and B ( d 1 , d 2 ) be satisfied with some d 1 , d 2 ∈ [0 , 1 / 2) and ρ ∈ [ − 1 , 1] , and let n, q, n/q → ∞ . (i) If d 1 = d 2 = d then T = U 1 U 2 + U 2 T n → law U 1 , (1) where Z 1 „Z 1 « 2 ( B 0 i ( τ )) 2 d τ − B 0 U i = i ( τ )d τ , i = 1 , 2 , (2) 0 0 where ( B 0 i ( τ ) = B i ( τ ) − τB i (1) , τ ∈ [0 , 1]) , i = 1 , 2 are fractional Brownian bridges obtained from bivariate fBm (( B 1 ( τ ) , B 2 ( τ )) , τ ∈ R ) with the same memory parameters d 1 = d 2 = d and correlation coefficient ρ . (ii) If d 1 � = d 2 then T n → p ∞ . (3)

Introduction Test Statistic In practice Simulations The dependent case When X 1 and X 2 are dependent, we introduce X 1 ( t ) = X 1 ( t ) − S 12 ,q ˜ X 2 ( t ) , t = 1 , . . . , n. S 2 ,q where q +1 1 X S 12 ,q = γ 12 ( h − ℓ ) ˆ q + 1 h,ℓ =1 γ 12 ( h ) = n − 1 P n − h and ˆ t =1 ( X 1 ( t ) − X 1 )( X 2 ( t + h ) − X 2 ) , h > 0 . ⇒ 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 In practice Simulations Proposition (Consistency of the test) Let Assumptions A ( d 1 , d 2 ) and B ( d 1 , d 2 ) be satisfied with some d 1 , d 2 ∈ [0 , 1 / 2) and ρ ∈ [ − 1 , 1] , and let n, q, n/q → ∞ . (i) If d 1 = d 2 = d ∈ [0 , 1 / 2) , then 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 .

Introduction Test Statistic In practice Simulations Proposition (Consistency of the test) Let Assumptions A ( d 1 , d 2 ) and B ( d 1 , d 2 ) be satisfied with some d 1 , d 2 ∈ [0 , 1 / 2) and ρ ∈ [ − 1 , 1] , and let n, q, n/q → ∞ . (i) If d 1 = d 2 = d ∈ [0 , 1 / 2) , then 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 . (ii) If d 1 > d 2 , then ˜ T n → p ∞ .

Introduction Test Statistic In practice Simulations Proposition (Consistency of the test) Let Assumptions A ( d 1 , d 2 ) and B ( d 1 , d 2 ) be satisfied with some d 1 , d 2 ∈ [0 , 1 / 2) and ρ ∈ [ − 1 , 1] , and let n, q, n/q → ∞ . (i) If d 1 = d 2 = d ∈ [0 , 1 / 2) , then 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 . (ii) If d 1 > d 2 , then ˜ T n → p ∞ . Remark. T n → p ρ 2 + ρ − 2 . ˜ When d 1 < d 2 ,

Introduction Test Statistic In practice Simulations 3 Practical implementation of the test