CUSUM TEST FOR PARAMETER CHANGE IN TIME SERIES MODELS. Sangyeol Lee

Transcription

1 CUSUM TEST FOR PARAMETER CHANGE IN TIME SERIES MODELS Sangyeol Lee 1

2 Contents 1. Introduction of the CUSUM test 2. Test for variance change in AR(p) model 3. Test for Parameter Change in Regression Models with ARCH Errors 4. Cusum test for general parameter case 5. Test for Parameter Change based on the MDPDE 2

3 Introduction of the CUSUM test The cusum test is one of the most frequently used method to detect change points. It started in quality control (Page, 1955) and moved to time series analysis since time series data suffer from changes due to the change of governmental policy and critical social events. It is easy to understand and implement in actual usage, and can be used for both testing and estimating the locations of changes. The cusum test proposed by Inclán and Tiao (1994). Suppose that {X t } is a series of independent r.v. s with (0, σ 2 t ) and one wishes to test H 0 : σ 2 t is constant over X 1,..., X n vs. H 1 : not H 0 3

4 Test statistic: IT n := n/2 max 1 k n D k, where D k = k X2 t n X2 t k n If X 1,..., X n are i.i.d. N(0, σ 2 ) under H 0, then w n/2 Dk sup W o (s), 0 s 1 sup 1 k n where W o is a Brownian bridge: Gaussian process with zero mean and Cov(W o (s), W o (t)) = s t for all s, t in [0,1] Estimate of the change point : k = arg max k D k. This procedure can be implemented for detecting multiple changes. 4

5 The background for the weak convergence is Donsker s invariance principle. Note that n/2dk = 1 k ( ) n k Xt 2 X 2 t n 2 n 1 n 1 n. X2 t ( n n 1 Xt 2 σ 2 and V ar(x1 2) = ) 2σ 2 So, n/2d[ns] W o (s), 0 s 1. sup n/2 Dk = sup n/2 D[ns] 1 k n 0 s 1 d sup W o (s). 0 s 1 Reject H 0 if sup 1 k n n/2 Dk is large. 5

6 For a general i.i.d. case, we can construct the cusum test : T n = 1 k ( ) n k max X 2 nˆτ t X 2 t 1 k n n, where ˆτ 2 = n 1 n X 4 t ( n 1 n Xt 2 ) 2 τ 2 = V ar(x 2 1 ). This can be extended to stationary time series and the residuals from AR models with unit roots and GARCH models 6

7 Test for variance change in AR(p) model Unstable AR(p) model: X t φ 1 X t 1 φ p X t p = ɛ t, (1) where ɛ t are iid r.v. s with Eɛ 1 = 0, Eɛ 2 1 = σ 2 and Eɛ 4 1 < and the corresponding characteristic polynomial φ(z) has a decomposition φ(z) = 1 φ 1 z φ p z p l = (1 z) a (1 + z) b (1 2 cos θ m z + z 2 ) d m ψ(z), m=1 where a, b, l, d m are nonnegative integers, θ m belongs to (0, π) and ψ(z) is the polynomial of order q = p (a + b + 2d d l ) that has no zeros on the unit disk in the complex plane. 7

8 Our goal is to test the following hypotheses: H 0 : the ɛ t have the same variance σ 2 vs. H 1 : not H 0. In order to perform a test, we employ the cusum of squares test statistic : T n = 1 k max ɛ 2 nκ t k n ɛ 2 t 1 k n n, (2) where κ 2 = V ar(ɛ 2 1). 8

9 Since ɛ t are unobservable, we use the cusum of squares test T n1 based on the residuals: T n1 = 1 k max ˆɛ 2 t k n ˆɛ 2 t nˆκn 1 k n n, (3) where ˆɛ t = X t ˆφ n X t 1, t = 1,..., n, X t = (X t,..., X t p+1 ) with X t = 0, t 0, ( n ) 1 n ˆφ n = X t 1 X [ t 1 X t 1 X t LSE of φ = (φ1,..., φ p ) ], n ( ˆκ 2 n = n 1 ˆɛ 4 t n 1 n ˆɛ 2 t ) 2 [ estimator of V ar(ɛ 2 1) ]. 9

10 Theorem 1 Under H 0, as n, T n1 w sup W (s), (4) 0 s 1 where W o is a Brownian bridge. We reject H 0 if T n1 is large. Unit roots do not affect the limiting distribution of the test unlike the empirical process case (cf. Lee and Wei, 1999). 10

11 Test for Parameter Change in Regression Models with ARCH Errors Let us consider the model y t = β z t + ɛ t, (5) ɛ t = h t ξ t, h 2 t = a(θ) + b j (θ)ɛ 2 t j, where ξ t are iid r.v. s with zero mean and unit variance, {z t } is a p-dim l strictly stationary process, and θ a(θ) and θ b(θ) are nonnegative continuous real functions defined on a subset N in R d with a(θ) > 0 and j=1 b j(θ) < for all θ N. We assume that y s, z s, s < t are independent of ξ u, u t, and {(ɛ t, h t, z t )} is strong mixing. j=1 11

12 The Model (5) covers a broad class of important models in the financial time series context including GARCH models. In particular, it becomes a GARCH(1,1) model if we put z t = 0, θ = (ω, α 1, α 2 ), ω, α 1, α 2 > 0, α 1 + α 2 < 1, a(θ) = ω/(1 α 1 ) and b j (θ) = α 1 α j 1 2. In this case, {(ɛ t, h t, z t )} is geometrically strong mixing (cf. Carrasco and Chen (2002)). 12

13 The objective here is to test the hypotheses H 0 : η = (β, θ ) remains the same for the whole series H 1 : Not H 0. vs. For a test, one may construct a cusum test based on {ˆε t := y t β z t } as in Inclán and Tiao (1994) and Kim, Cho and Lee (2000). However, as observed in the simulation study, the test in GARCH(1,1) models is unstable and produces low powers. Thus one has to develop a better test which is not much affected by the GARCH parameters. As a candidate, one can naturally consider the cusum test based on { ξ 2 t }, say, T n := 1 nτ max 1 k n k ξt 2 ( ) k n ξt 2 n, (6) where τ 2 = V ar ( ξ 2 1), since Tn is free from the GARCH parameters. 13

14 In this case, however, one may speculate whether T n can detect any changes since T n itself has no information about the GARCH parameters. But since ξ t are not observable, one should replace ξt 2 s by the residuals ξ 2 t, which are obtained via estimating the unknown parameters. Those estimators play an important role to detect changes in the parameters in the presence of changes: the iid property of the true errors still remains when there are no changes. From this reasoning, one can anticipate that the residual cusum test should be more stable and produce better powers. 14

15 Now, we construct the residual cusum test. To this end, we assume that (A1) E z 1 4+δ 1 <, E ɛ 1 4+δ 1 < and E ξ 1 4+δ 1 < for some δ 1 > 0. (A2) There exists δ 2 > 0 such that sup θ θ δ 2,θ N ȧ(θ) < and j=1 sup θ θ δ 2,θ N b j (θ) <, where ȧ(θ) and ḃj(θ) denote the gradient vectors of a and b j at θ. (A3) There exists a sequence of positive integers with q, q/ n 0 and n j=q+1 b j(θ) 0 as n. (A4) {(ɛ t, h t, z t )} is strong mixing with order γ(h) satisfying h=1 γ(h) δ 1 4+δ 1 <. 15

16 Observe that the last condition in (A3) is satisfied if b j (θ) are geometrically bounded (as in GARCH models), and q = [(log n)] ζ, ζ > 1. Also, if z t are identically zero and {y t } is a GARCH process, {(y t, h t )} is geometrically strong mixing (cf. Carrasco and Chen (2002)), so that (A4) is satisfied. Now, we construct the residual cusum test. In analogy of h 2 t, we define ĥ 2 t = a(ˆθ) + q b j (ˆθ)ˆɛ 2 t j, j=1 ˆɛ t = y t ˆθ z t and ˆξ t = ˆɛ t /ĥt, where ˆη = (ˆβ, ˆθ ) is an estimator of η with n(ˆη η) = O P (1). 16

17 Then, we have the following result. Theorem 2 Assume that (A1)-(A4) hold. Set ˆT n := 1 k ( ) max ˆξ 2 k n t nˆτ q+1 k n n t=q+1 t=q+1 ˆξ 2 t where ˆτ 2 = 1 n q n t=q+1 ˆξ 4 t ( 1 n q n t=q+1 ˆξ 2 t ) 2. Then, under H 0, ˆT n d sup B o (u), n, 0 u 1 where B o is a Brownian bridge. 17

18 GARCH(1,1) case: we demonstrate that a modification of the test, free from a choice of q, can be constructed for the models with h 2 t satisfying a specific equation. Here, considering its extreme popularity in the financial time series context, we concentrate ourselves on the case of GARCH(1,1) errors: y t = β z t + ε t, (7) ε t = h t ξ t, h 2 t = ω + α 1 ε 2 t 1 + α 2 h 2 t 1 with ω > 0, α 1, α 2 0 and α 1 + α 2 < 1. In this case, we can write h 2 t = a + α 1 j=1 α j 1 2 ε 2 t j (8) 18

19 with a = ω/(1 α 1 ), and its estimate is ĥ 2 t = â + ˆα 1 q j=1 ˆα j 1 2 ˆε 2 t j, (9) ˆε t = y t ˆβ z t, ˆβ, â, ˆα 1, ˆα 2 are the estimators for β, a, α 1 and α 2 satisfying n (ˆβ β ) = O P (1), n (â a) = O P (1), n (ˆα1 α 1 ) = O P (1) and n (ˆα 2 α 2 ) = O P (1), and q is a sequence of positive integers with q, q/ n 0 and nα q 2 0, which ensures (A3). 19

20 Note that the estimate of the conditional variance can be obtained recursively from the equation h 2 t = ˆω + ˆα 1ˆε 2 t 1 + ˆα 2 h2 t 1, (10) in sofar as initial values ˆε 2 0 and h 2 0 are provided. In fact, we can see that for t 2, h 2 t can be used instead of ĥ2 t. Therefore, we have the following. Theorem 3 Let h 2 t be the one in (10), and set ξ t 2 = ˆε 2 t / h 2 t. Let T n := max T n,k := 1 k ( ) n max ξ 2 k t ξ 2 t 1 t n n τ 1 k n n, where τ 2 = 1 n n ξ 4 t ( 1 n n ξ t ) 2. Then if (A1) holds, under H 0, T d n sup B o (u), n. 0 u 1 20

21 Cusum test for General parameter case Suppose that X 1, X 2,... are i.i.d. r.v. s and one wishes to test for certain parameter change in θ R based on an estimator ˆθ n. Suppose that n (ˆθn θ) = 1 n n l t (θ) + n, where n is negligible asymptotically and l 1 (θ),..., l n (θ) are i.i.d. r.v. s with El t (θ) = 0, El 2 t (θ) <. If l t (θ) are observable, as in the mean and variance case, one may construct the cusum test based on [ns] 1 V n (s) = l n (El 2 1 (θ)) 1/2 t (θ) [ns] n l t (θ) W o (s). n 21

22 However, in many situations, those are unobservable. But we can still construct a cusum test as a functional of the estimators. Since k one can write T n (s) = ) l t (θ) = k (ˆθk θ k k, 1 k n, [ns] (ˆθ[ns] ˆθ ) n (El 2 1 (θ)) 1/2 n = V n (s) + ( El 2 1(θ) ) 1/2 ( ) [ns] [ns] [ns] n n n, So, W o (s) if max 1 k n k n k = o P (1). sup T n (s) 2 d 0 s 1 sup W o (s) 2. 0 s 1 22

23 CUSUM test : Lee, Ha, Na and Na (2003) Let {X t } be stationary time series. θ = (θ 1,..., θ p ) is the parameter vector which will be examined for constancy. H 0 : θ does not change for X 1,..., X n vs. H 1 : not H 0 Suppose that ˆθ k satisfies the following k (ˆθk θ) = 1 k k l t (θ) + k, where l t (θ) forms stationary martingale differences w.r.t. {F t } and Γ = V ar(l t (θ)) is nonsingular. 23

24 If under H 0, and [ns] 1 Γ 1/2 n max k l t (θ) w W p (s) k n k = o P (1), as n, the test statistic T n := max p k n k 2 n (ˆθk ˆθ n ) ˆΓ 1 (ˆθk ˆθ n ) where ˆθ k = (ˆθ k1,..., ˆθ kp ) be the estimator of θ based on X 1,..., X k, k = 1,..., n, satisfies ˆT n w sup 0 s 1 p ( W o j (s) ) 2 j=1 24

25 Example 1 : Test for RCA(1) model Let {x t ; t = 0, ±1, ±2,...} be the time series of the RCA(1) model x t = (φ + b t )x t 1 + ɛ t, (11) where b t ɛ t iid 0 0, ω2 0 0 σ 2. Nicholls & Quinn (1982) showed that a sufficient condition for the strict stationarity and ergodicity of {x t } in (11) is φ 2 + ω 2 < 1. We assume that Eɛ 2k t < and E(φ + b t ) 2k < 1 for some k that will be specified later, which immediately yields Ex 2k t < (cf. Feigin & Tweedie, 1985). 25

26 Now, we consider the problem of testing for a change of the parameter vector θ = (φ, ω 2, σ 2 ) based on a (conditional) LSE ˆθ. Suppose that x 1,..., x n are a sample from (11) and assume x 0 = 0. We intend to test the following hypotheses H 0 : θ = (φ, ω 2, σ 2 ) is constant over x 1,..., x n. vs. H 1 : not H 0. 26

27 In order to construct a cusum test, consider the estimators ˆθ k = ( ˆφ k, ˆω 2 k, ˆσ 2 k) based on x 1,..., x k, k = 1,..., n. The estimator ˆφ k of φ is defined as the minimizer of k (x t φx t 1 ) 2, and the estimators ˆω 2 k and ˆσ 2 k are defined as the minimizers of k (û2 k,t ω 2 x 2 t 1 σ 2 ) 2, where û k,t = x t ˆφ k x t 1, by noticing the equation E(u 2 t F t 1 ) = ω 2 x 2 t 1 + σ 2 under H 0, where u t = x t φx t 1 and F t = σ(ɛ s, b s ; s t). Then, ˆθ k is written as ˆθ k = ˆφ k ˆω 2 k ˆσ 2 k = where m 2,k = k 1 k x2 t 1. k x t 1x t k x2 t 1 k (x2 t 1 m 2,k )û 2 k,t k (x2 t 1 m 2,k) 2 k 1 27 k û 2 k,t ˆω km 2 2,k,

28 Let Γ be the 3 3 symmetric matrix whose entries are Γ 11 = ω2 Ex σ 2 Ex 2 1 (Ex 2 1 )2, Γ 22 = ( Ex 4 1 (Ex 2 1) 2) 2 ( (Eb 4 1 ω 4 )(Ex 8 1 2Ex 2 1Ex (Ex 2 1) 2 Ex 4 1) + 4ω 2 σ 2 (Ex 6 1 2Ex 2 1Ex (Ex 2 1) 3 ) + (Eɛ 4 1 σ 4 )(Ex 4 1 (Ex 2 1) 2 ) ), Γ 33 = ( Eb 4 1 ω 4) ( Ex 4 1 2Ex2 1(Ex 6 1 Ex 2 1Ex 4 ) 1) Ex 4 1 (Ex2 1 )2 4ω 2 σ 2 Ex Eɛ 4 1 σ 4 + (Ex 2 1) 2 Γ 22, Γ 12 = Eb3 1Ex 6 1 Eb 3 1Ex 2 1Ex Eɛ 3 1Ex 3 1 Ex 2 1 Ex4 1, (Ex2 1 ( )3 ) Γ 13 = Eb3 1Ex 2 1Ex Eb 3 1 Ex Eɛ 3 1Ex 2 1Ex 3 1 Ex 2 1 Ex4 1, (Ex2 1 )3 Γ 23 = (Eb4 1 ω 4 )(Ex 6 1 Ex 2 1Ex 4 1) Ex 4 1 (Ex2 1 )2 + 4ω 2 σ 2 Ex 2 1Γ

29 Theorem 4 Suppose that Eɛ 16 1 < and E(φ + b 1 ) 16 < 1. Then under H 0, as n, [ns] n Γ 1/2 (ˆθ [ns] ˆθ n ) w W 3(s). In view of the result of Theorem 4, we can construct the test statistic T n = max 1 k n k 2 n (ˆθ k ˆθ n ) ˆΓ 1 (ˆθ k ˆθ n ), where ˆΓ is a consistent estimator of Γ. We reject H 0 at the level α if T n C α, where C α is the (1 α)-quantile point of sup 0 s 1 3 j=1 ( W j (s) )2. 29

30 Example 2 : Test for Autocovariance Function Let {x t ; t = 0, ±1, ±2,...} be a stationary linear process of the form x t = a i ɛ t i, (12) i=0 where the real sequence {a i } satisfies the summability condition i=0 i a i < and ɛ t are iid random variables with mean 0, variance σ 2 ɛ, and E ɛ 1 4λ < for some λ > 1. Assume that x 1,..., x n are observed, and denote the autocovariance at lag h by γ(h). As an estimate of γ(h), we use ˆγ n (h) = 1 n n h (x t x n )(x t+h x n ), x n = 1 n n x t. 30

31 Theorem 5 Let ( [ns] Λ n (s) = (ˆγ[ns] (0) ˆγ n (0) ),..., [ns] (ˆγ[ns] (m) ˆγ n (m) )), n n 0 s 1. Then under H 0, where no changes are assumed to occur in the autocovariance function, we have Λ n (s) Γ 1 Λ n (s) w m ( W j (s) )2, where Γ is the (m + 1) (m + 1) matrix whose (i, j)th entry is j=0 Γ ij = κ 4 γ(i)γ(j) + r= (γ(i + r)γ(j + r) + γ(i r)γ(j + r)), for i, j = 0,..., m, and κ 4 is the kurtosis of ɛ 1. 31

32 Since Γ is unknown, we should replace it by a consistent estimator ˆΓ. Now we assume that {x t } in (12) can be rewritten as x t = π j x t j + ɛ t, (13) j=1 where π(z) := 1 j=1 π jz j is analytic on an open set containing the unit disk in the complex plane, and have no zeros in the unit disk. Notice that {x t } can be rewritten as in (12), and covers stationary and invertible ARMA processes. We introduce a sequence of positive integers {h n }, such that as n, h n and h n = O(n β ) for some β (0, (λ 1)/2λ). 32

33 Then if ˆκ 4 is a consistent estimator of κ 4, we have ˆΓ ij P Γ ij, (14) where ˆΓ ij = ˆκ 4ˆγ n (i)ˆγ n (j) + h n r= h n (ˆγ n (i + r)ˆγ n (j + r) + ˆγ n (i r)ˆγ n (j + r)). Note that a consistent ˆκ 4 can be obtained by calculating residuals ˆɛ t via fitting a long AR(q) model to observations, viz., ˆκ 4 = n 1 n ˆɛ4 t /(n 1 n ˆɛ2 t ) 2 3. A typical example of q is (log n) 2. 33

34 From Theorem 5 and (14), we obtain the following result. Theorem 6 Under H 0, Λ n (s) ˆΓ 1 Λ n (s) w m ( W j (s) )2. j=0 Theorem 6 ensures the Brownian bridge result for the cusum test statistic. The test statistic is defined as Λ n := max Λ n(k/n) ˆΓ 1 Λ n (k/n). m+1 k n 34

35 Test for Parameter Change based on the MDPDE Suppose that the time series are contaminated by outliers, and we observe X t = (1 p t )X o,t + p t X c,t, where p t are iid Bernoulli r.v. s with succes probability p, {X o,t } is a strong mixing process with the marginal density f θ, the contaminating process {X c,t } is a sequence of iid r.v. s, and {p t }, {X o,t } and {X c,t } are all independent. Assume that one wishes to test H 0 : θ does not change over X 1,..., X n vs. H 1 : not H 0. 35

36 Minimum Density Power Divergence Estimator (MDPDE) Basu, Harris, Hjort and Jones (1998) Let X 1,..., X n be iid r.v. s with distribution G having density g We want to choose a model that fits the data in a reasonable way from the parametric family of models {F θ } indexed by the unknown parameter θ Θ R p, possessing densities {f θ }. Then the best fitting parameter is defined by θ α := T α (G) = arg min θ Θ d α (g, f θ ), where d α, α 0 is a density power divergence defined by { f 1+α (z) ( ) α g(z)f α (z) + 1 α d α (g, f) := g1+α (z) } dz, α > 0 g(z) (log g(z) log f(z)) dz, α = 0. 36

37 Note that θ α = arg min θ Θ d α (g, f θ ) = arg min θ Θ E GV α (θ; X), where X G and V α (θ; x) := f 1+α θ (z)dz ( α ) f α θ (x), α > 0 log f θ (x), α = 0, so the MDPDE ˆθ α,n of θ α is defined by where H α,n (θ) = n 1 n V α(θ; X t ). If α = 0, the MDPDE is nothing but MLE. ˆθ α,n = arg min θ Θ H α,n(θ), (15) 37

38 Basu et al. (1998) showed that the MDPDE with α > 0 has robust features against outliers but still possesses the efficiency that the MLE has when the true density belongs to the parametric family {F θ }. Under certain regularity conditions, ˆθ α,n is weakly consistent for θ α and n(ˆθ α,n θ α ) is asymptotically normal with mean zero vector. 38

39 Main result Our objective here is to test the constancy of the unknown parameter θ α through the cusum test, based on the MDPDE, introduced by Lee et al. (2003). For this task, we set up the null and alternative hypotheses: H 0 : θ α does not change over X 1,..., X n vs. H 1 : not H 0. In order to achieve the asymptotic distribution of the cusum test statistic, we need the strong consistency and the functional central limit theorem for the estimator. 39

40 (B1) The distribution F θ and G have common support, so that the set X on which the densities are greater than zero is independent of θ. (B2) There is an open set ϑ of the parameter space Θ containing θ α such that for all x X, and all θ ϑ, the density f θ (x) has continuous partial derivatives of order r( 0) with respect to θ and E j f θ (X) θ i1 θ ij <, 0 j r. (B3) The integral f 1+α θ (z)dz can be differentiated r-times (r 0) with respect to θ, and the derivative can be taken under the integral sign. (B4) For each 1 i 1,, i r p, there exist functions M α,i1 i r (x) with EM α,i1 i r (X) < such that r V α (θ; x) θ i1 θ ir M α,i 1 i r (x) for all θ ϑ and x X. 40

41 (B5) There exists a nonsingular matrix J α, defined by J α := 1 ( ) α E V α (θ α ; X) θ 2. (B6) {X t } is ergodic and strictly stationary. In what follows, we assume that θ α exists and is unique, and keep the same definition for the estimator ˆθ α,n as in the iid case. In fact, the estimator is obtained by solving the estimating equations U α,n (θ) = (1 + α) 1 H α,n(θ) θ = 0. 41

42 Theorem 7 (Strong consistency) Suppose that H 0 holds and Conditions (B1) - (B4) and (B6) hold for some nonnegative integer r. Then there exists a sequence {ˆθ α,n }, such that } P {ˆθα,n θ α, as n = 1. (16) Also, if V α (θ; x) is differentiable w.r.t. θ, then we have U α,n (ˆθ α,n ) = 0 (17) for sufficiently large n. 42

43 Theorem 8 (functional central limit theorem) Assume that Conditions (B1)-(B6) hold with r=3. In addition, suppose that (C1) {X t } is strong mixing with mixing order β( ) of size γ/(γ 2) for γ > 2, i.e., n=1 β(n)1 2/γ <. (C2) E V α (θ α ; X)/ θ i γ < for i = 1,..., m. (C3) nk α,n K α for some positive definite and symmetric matrix K α, where K α,n is the covariance matrix of U α,n (θ α ). Then, under H 0, we have [ns] ) (ˆθα,[ns] θ α n w J 1 α Kα 1/2 W p (s). 43

44 Utilizing Theorem 8, we can have the following. Theorem 9 Define T 0 n3(α) := max p k n k 2 n (ˆθα,k ˆθ α,n ) Jα K 1 α J α (ˆθα,k ˆθ α,n ). (18) Suppose that the conditions of Theorem 8 hold. Then, under H 0, Tn3(α) 0 w W o p (s) 2. We reject H 0 if T 0 n3(α) is large. sup 0 s 1 44

45 Since J α and K α are unknown, we should replace them by consistent estimators Ĵα and ˆK α. Note that J α = u θα (z)u θα (z) f 1+α θ α (z)dz + (iθα(z) αu θα(z)u θα(z) ) (g(z) f θα(z)) f α θ α (z)dz, where u θ (z) = log f θ (z)/ θ and i θ (z) = u θ (z)/ θ. Therefore, if we put Ĵ α = { (1 + α)uˆθα,n (z)uˆθα,n (z) Iˆθα,n (z) } + 1 n n { f 1+α ˆθ α,n (z)dz Iˆθα,n (X t ) αuˆθα,n (X t )uˆθα,n (X t ) } f αˆθα,n (X t ), then one can show that Ĵα converges to J α almost surely. 45

46 Under the assumptions of Theorem 8, ( 1 K α = (1 + α) Cov Vα (θ α, X 0 ) 2 θ k= exists due to Theorem 1.5 in Bosq (1996, page 32). Assuming that (D1) n=1 β(n)1/3 < ; (D2) E V α (θ α ; X)/ θ i 6 < for i = 1,..., m;, V ) α(θ α, X k ) θ (D3) there exists a function M α (x) with EM α (X) 2 < such that p i,j=1 2 V α (θ; x)/ θ i θ j M α (x) for all θ Θ and x X, one can show that ˆK α K α in probability, where ˆK α = h n k= h n 1 n(1 + α) 2 n k V α (ˆθ α,n ; X t ) θ and {h n } is a sequence of positive integers such that V α(ˆθ α,n ; X t+k ) θ h n and h n / n 0. (19) 46

47 Then we have the cusum test. Theorem 10 Define the test statistic T n3 (α) by T n3 (α) := max p k n k 2 n (ˆθα,k ˆθ ) α,n Ĵα ˆK 1 α Ĵα (ˆθα,k ˆθ ) α,n. (20) Suppose that the conditions of Theorem 8 and (D1)-(D3) hold. Then, under H 0, T n3 (α) w We reject H 0 if T n3 (α) is large. sup W o p (s) 2. 0 s 1 47