LAD Estimation for Time Series Models With Finite and Infinite Variance

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1 LAD Estimatio for Time Series Moels With Fiite a Ifiite Variace Richar A. Davis Colorao State Uiversity William Dusmuir Uiversity of New South Wales 1

2 LAD Estimatio for ARMA Moels fiite variace ifiite variace A Real Data Example LAD estimatio uit root problem 2

3 Αsymptotics i No-staar Settigs (Whe Taylor series expasios o ot work.) Applicatios to: LAD estimatio (Pollar `91, Davis a Dusmuir `95) M-estimatio with ifiite variace (Davis, Kight a Liu `92, Davis `95) Uit root problems (AR + MA) (Davis a Dusmuir `95, Davis, Che a Dusmuir `95) 3

4 Εxample (The meia). Data. Z 1,..., Z, IID meia 0 a pf f(0) > 0. Meia. m = meia(z 1,..., Z ) Asymptotics: Note: m (assume m 0 = populatio meia = 0) m is AN(0,??) miimizes T (m) = ( Z t m Z t ) 4

5 Set u= 1/2 m a put The S (u ) = T (u -1/2 ) u = 1/2 m. Usig the key ietity, we have t = 1 = ( Z t u -1/2 Z t ). z-y - z = -y sg(z) + 2(y-z) (I(0<z<y) - I(y<z<0)), S (u) = u -1/2 sg(z t ) + 2 ( -1/2 u-z t ){ I(0< Z t < -1/2 u)- I( -1/2 u<z t <0)} 5

6 S (u) = u -1/2 sg(z t ) + 2 ( -1/2 u-z t ){ I(0< Z t < -1/2 u) I( -1/2 u<z t <0)} = : A +B Results: A -un, N ~ N(0,1) (CLT) EB = 2E( -1/2 u-z t ) I(0< Z t < -1/2 u) (for u>0) -1/2 u = 2 ( -1/2 u z) F(z) 0-1/2 u ~ 2 ( -1/2 u z) f(0)z = u 2 f(0). 0 6

7 Coclue : A B P -un u 2 f(0) S (u) S(u):= un + u 2 f(0) [o C( R) ] u = 1/2 m u := miimizer of S(u). Solve S'(u) = N + 2 u f(0) = 0, we obtai u = N/(2f(0)) ~ N(0,1/(4f 2 (0))) or 1/2 m N(0,1/(4f 2 (0))) 7

8 The Paraigm: Objective fuctio to be miimize: T (θ) Reparameterize by settig u = a (θ θ 0 ) θ 0 = true value, a = scalig Form ew objective fuctio: S (u) = T (θ 0 +u/a ) Establish weak covergece of S (u) to S(u) o C(R) R). Show u = a (θ θ 0 ) u:= argmi S(u) 8

9 8 Theorem. Let {Y t } be the liear process Y t = c j Z t-j, c j <, where {Z t }~IID(0,σ 2 ), meia(z 1 )=0, a f(0)>0. The S := ( Z t - -1/2 Y t-1 - Z t ) j= 0 j= 0 γ f(0) + N, where N ~ N(0, γ) (γ = Var(Y t )). Key ietity: z-y - z = -y sg(z) + 2(y-z) (I(0<z<y) - I(y<z<0)) 9

10 Usig the key ietity, z-y - z = -y sg(z) + 2(y-z) (I(0<z<y) - I(y<z<0)), we have S = -1/2 Y t-1 sg(z t ) + 2 ( -1/2 Y t-1 -Z t ){ I(0< Z t < -1/2 Y t-1 )- I( -1/2 Y t-1 <Z t <0)} = : A + B Result: A N (MG CLT) B P γ f(0) (Ergoic Theorem) 10

11 Results : A un 1 v N 2, (N 1,N 2 ) T ~ N( )., 0 1 EX 1 0 EX 1 EX 1 2 EB B P EY 1 2 f(0) = (u,v) (u,v) T f(0). (u,v) (u,v) T f(0). Coclue : S (u,v) un 1 v N 2 + (u,v) (u,v) T f(0). (u,v ) T -1 (N 1,N 2 ) T / 2f(0) ~ N(0, -1 / 4f 2 (0)) ( 1/2 (µ µ 0 ), 1/2 (φ φ 0 )) T N(0, -1 / 4f 2 (0)) 11

12 Εxample (AR(1)). Data. X 1,..., X Moel. X t = µ 0 + φ 0 X t-1 +Z t, LAD estimatio: Miimize φ 0 < 1, {Z t } ~ IID(0,σ 2 ), f(0)>0. T (µ,φ) = ( X t µ φx t-1 Z t ) = ( Z t (µ µ 0 ) (φ φ 0 )X t-1 Z t ) Set u= 1/2 (µ µ 0 ), v= 1/2 (φ φ 0 ), S (u,v) = ( Z t u -1/2 v -1/2 X t-1 Z t ) 12

13 S (u,v) = ( Z t u -1/2 v -1/2 X t-1 Z t ) = ( Z t -1/2 Y t-1 Z t ) ( -1/2 Y t-1 = u -1/2 + v -1/2 X t-1 ) = - -1/2 Y t-1 sg(z t ) + 2 ( -1/2 Y t-1 -Z t ){ I(0< Z t < -1/2 Y t-1 )- I( -1/2 Y t-1 <Z t <0)} = : A + B 13

14 Results : A un 1 v N 2, (N 1,N 2 ) T ~ N( )., 0 1 EX 1 0 EX 1 EX 1 2 EB B P EY 1 2 f(0) = (u,v) (u,v) T f(0). (u,v) (u,v) T f(0). Coclue : S (u,v) un 1 v N 2 + (u,v) (u,v) T f(0). (u,v ) T -1 (N 1,N 2 ) T / 2f(0) ~ N(0, -1 / 4f 2 (0)) ( 1/2 (µ µ 0 ), 1/2 (φ φ 0 )) T N(0, -1 / 4f 2 (0)) 14

15 Εxample (MA(1)). Data. X 1,..., X Moel. X t = Z t + θ Z t-1, LAD estimatio: θ 0 < 1, {Z t } ~ IID(0,σ 2 ), f(0)>0. Miimize T (θ) = ( Z t (θ) - Z t (θ 0 ) ) Set u= 1/2 (θ θ 0 ), = ( X t θx t-1 +θ 2 X t (-θ) t-1 X 1 - Z t (θ 0 ) ) 15

16 S (u) = T (θ 0 + u -1/2 ) = ( Z t (θ 0 + u -1/2 ) - Z t (θ 0 ) ) (Not a covex fuctio of u!) Liearize Z t (θ 0 + u -1/2 ) to get S (u) ( Z t (θ 0 ) + u -1/2 Z t ' (θ 0 ) - Z t (θ 0 ) ) where -Z t ' (θ 0 ) is the AR(1) process Y t = θ 0 Y t-1 +Z t. Result : Same limit result as i the AR(1) case, i.e. 1/2 (θ LAD θ 0 ) N / (Var(Y t )2f(0)) ~ N(0, (1-θ 2 ) / (σ 2 4f 2 (0))) 16

17 Liearize Versio : Iitial estimate : θ 0 = θ 0 + O p ( -1/2 ) Objective Fuctio: The T (θ) = ( Z t (θ 0 ) + Z' t (θ 0 )(θ θ 0 ) - Z t (θ 0 ) ). 1/2 (θ L θ 0 ) N(0, (1-θ 2 ) / (4f 2 (0))), where θ L = argmi T (θ) 17

18 Extesios : ARMA Moel : φ(b)x t = θ(b)z t, {Z t } ~ IID(0,σ 2 ), f(0)>0. Set β = (φ 1,..., φ p, θ 1,..., θ q ) T a The v = 1/2 (β β 0 ) (i) S (v ) := ( Z t (β 0 + v -1/2 ) - Z t (β 0 ) ) f(0)v T Γ Q -1/2 v + v T N, N~N( 0, Γ Q ), (i C(R p+q )) (ii) v LAD = 1/2 (β LAD β 0 ) Γ -1 Q N/(2f(0)) ~ N(0, Γ Q -1 /(4f 2 (0))) Note: Γ Q -1 σ 2 is the limitig covariace matrix i Gaussia case. 18

19 ARMA Moel With Stable Noise: (Davis, Kight a Liu `92 for AR case, Davis `95 for ARMA.) Moel: φ(b)x t = θ(b)z t, {Z t } ~ IID symmetric stable(α), 0<α < 2. 1/α (β LAD β 0 ) W I this case both A a B have raom limits! Least squares estimates: ( / l ) 1/α (β LS β 0 ) V Simulatio results: Cauchy oise LS LAD AR(1) φ=.4.395(.041).399(.015) MA(1) θ=.8.795(.049).794(.036) ARMA(1,1) φ=.4.399(.053).399(.026) θ=.8.781(.046).781(.033) 19

20 Liear Regressio with ARMA Errors : Moel : Y t = A T t α + X t, where {X t } follows a ARMA process φ(b)x t = θ(b)z t, {Z t } ~ IID(0,σ 2 ), f(0)>0. Assume A T t = (a 1t,..., a rt ) T satisfies Greaer s coitios: (a) a j /b j 0 (b) (c) B -1 A t A t+j T B -1 Γ A (j) b j P P8 where b j 2 = A t 2 a B = iag(b 1t,..., b rt ). P 20

21 Estimatio: Let X t (α) = 0, if t < 1, Y t A t T α, if t > 0, a for τ Τ =( α Τ, β Τ ), efie Z t (τ) =... 0, if t < 1, φ(b)x t (α) θ 1 Z t-1 (τ) θ q Z t-q (τ), if t > 0, Miimize S ( u, v) = ( Z t (τ 0 + (u,v)) - Z t (τ 0 ) ) where (u,v) = ( (B -1 u) T, -1/2 v T ) T. 21

22 Result : S ( u, v) S ( u, v) o C(R p+q+r ) B (α LAD α 0 ), 1/2 (β LAD β 0 ) Γ 1 N /(2f(0)) ~ N(0, Γ -1 /(4f 2 (0))), where a Γ = Γ C 0 0 T Γ Q j, k Γ C = π j π k Γ A (j-k), π(b) = θ 0 (B) / φ 0 (B). 22

23 A Example : (overshorts Y 1,..., Y 57 storage tak.) from uergrou (gallos) ACF Lag 23

24 Moel. Y t = µ + Z t +θz t-1 Problem. Estimate µ a costruct a C.I.? (Is µ < -5 gallos/ay?) Estimatio. Estimates Asymptotic Var µ MLE = 4.87 (1+θ) 2 σ 2 / = (1.408) 2 θ MLE =.849 (1 θ 2 )/ = (.070) 2 µ LAD = (1+θ) 2 /(4f 2 (0)) = (2.236) 2 θ LAD =.673 (1 θ 2 ) / (σ 2 4(f 2 (0))) = (.106) 2 24

25 Uit Root. If θ=1, the µ MLE is AN(µ 0, 12σ 2 / 3 ) (if θ is ot estimate) µ MLE is asymptotically o-ormal (if θ is estimate) Asymptotic istributio of µ LAD a θ LAD whe θ = 1? 25

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