A representation theory for a class of vector autoregressive models for fractional processes

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1 A representation theory for a class of vector autoregressive moels for fractional processes Søren Johansen Department of Applie Mathematics an Statistics, University of Copenhagen November 2006 Abstract Base on an iea of Granger (1986), we analyze a new vector autoregressive moel e ne from the fractional lag operator 1 (1 L) : We rst erive conitions in terms of the coe cients for the moel to generate processes which are fractional of orer zero. We then show that if there is a unit root, the moel generates a fractional process X t of orer ; > 0; for which there are vectors so that 0 X t is fractional of orer b; 0 < b : We n a representation of the solution which emonstrates the fractional properties. Finally we suggest a moel that allows for a polynomial fractional vector, that is, the process X t is fractional of orer ; 0 X t is fractional of orer b an a linear combination of 0 X t an b X t is fractional of orer 2b: The representations an conitions are analogous to the well known conitions for I(0), I(1) an I(2) variables. 1 Introuction an motivation Since the e nition of fractional processes by Granger an Joyeux (1980) an Hosking (1981), attention has been given to the estimation of the fractional orer of a time series, see for instance Beran (1994), with minimal assumptions on the processes. The problem consiere in this paper is i erent. We propose a vector autoregressive moel that allows for a fractional process as solution, but also allows for moelling cofractional relations an ajustments to these, so that economic hypotheses can be formulate within a moel, that can be I woul like to thank the Danish Social Sciences Research Council for continuing support, an the members of the ESF-EMM network for iscussions. As usual the insight an help of Søren Tolver Jensen is greatly appreciate in connection with Lemma 10. I am grateful for comments from an anonymous referee which helpe clarify some explanations, an to Bent Nielsen who helpe with the proof of Lemma 11. Finally Massimo Franchi ha some goo comments to the almost nal version. 1

2 Fractional autoregressive processes 2 teste against the ata. In this way we have a platform for making moel base inference on coe cients, relations an fractional orer. The contribution of the present paper is to stuy the solution of this autoregressive moel in orer to n conitions uner which the process is fractional of orer zero an conitions for the solution to be fractional of orer > 0; but allowing for linear combinations 0 X t ; that are fractional of orer b; 0 < b : We exten the results to a moel that further allows polynomial fractionality, that is, a linear combination of b X t an 0 X t ; which is fractional of orer 2b. Such moels are well known for I(0); I(1); an I(2) variables, see for example Johansen (1996), an we formulate the moels so that results for the usual cointegrating moels can be carrie over to the new framework. 1.1 Granger s moel for fractional processes Granger (1986) suggeste an autoregressive moel A (L) X t = (1 b ) b 0 X t 1 + (L)" t ; (1) where " t is inepenent ientically istribute with mean zero an positive e nite variance > 0; enote i.i.. (0; ): We have use the matrix 0 instea of 0 from Granger (1986). He note that the lag operator L b = 1 (1 L) b ; (2) plays the role of the usual lag operator (b = 1; L 1 = L) in moels for fractional processes. This moel has been analyze by Lyhagen (1998) from the point of view of ning the properties of the solution, but unfortunately the results an their proof are not correct. 1 Dittmann (2004) attempts to erive moel (1) from a moving average form, assuming the processes are fractional, using the results of Engle an Granger (1987). However, the results are not correctly prove. 2 This type of moel can for instance be erive as follows. Let us assume that X t is fractional of orer ; an that there are r linear combinations, ; that are fractional of orer b: Let be p (p r) so that (; ) has rank p; then the assumptions are 0 X t = u 1t ; 0 b X t = u 2t ; 1 The proof is attempte by the usual transformation from X t to 0 X t an 0?X t ; but for fractional processes a i erent metho is neee. This becomes apparent when it is state that the matrix D(z) = iag((1 z) i (1 z)) is of full rank for z = 1: 2 A counter example to Lemma 1 in the paper Engle an Granger (1987) is given by taking G() = iag(1 + ; 2 ; 2 ): What is missing in Lemma 1 is a conition corresponing to the I(1) conition of cointegration (see Johansen 1996, Theorem 4.5.)

3 Fractional autoregressive processes 3 where for simplicity u t = (u 0 1t; u 0 2t) 0 is i.i.. (0; ). This formulation was use by Breitung an Hassler (2002) an allows for moelling an estimating both the cofraction vectors,, an the common trens vectors, : For a p m matrix we e ne a? to be a p (p m) matrix of rank p m; for which a 0 a? = 0: When (; ) has full rank p; the ientity shows that? ( 0? ) 1 0 +? ( 0? ) 1 0 = I p ; X t =? ( 0? ) 1 u 1t +? ( 0? ) 1 b u 2t (3) =? ( 0? ) 1 u 1t +? ( 0? ) 1 u 2t? ( 0? ) 1 (1 b )u 2t = (1 b ) 0 b X t + " t ; where " t =? ( 0? ) 1 u 1t +? ( 0? ) 1 u 2t ; is i.i.. The ajustment matrix =? ( 0? ) 1 satis es 0 = I r ; an (3) is a special case of moel (1) with A (z) = 1; apart from the lagge X t : It is now a natural iea to make the moel more exible by aing a lag structure, an Granger suggeste to a lags of X t : Example 1. As a simple example of moel (1) consier the univariate moel with one lag an b = ; X t = (1 ) 1 X t X t 1 + " t : (4) The characteristic function, which is not a polynomial unless is an integer, is (z) = (1 z) 1 (1 (1 z) )z 2 (1 z) z: (5) The process is stationary if the roots of (z) = 0 are greater than one in absolute value. This criterion involves solving an unpleasant transcenental equation. Fining the roots of this equation is not a stanar problem. Thus moel (1) is an autoregressive moel for fractional processes, but its lag structure is inconvenient to analyze in the sense that the stochastic properties of the solution generate by the equations are not easily re ecte in properties of the coe cients. Therefore we propose a slightly i erent moel with i erent choice of lag structure, for which we get a feasible algebraic analysis, in the sense that we get a veri able criterion for the solution of the moel equations to be fractional of various orers. 1.2 An alternative autoregressive moel for fractional processes We propose an autoregressive moel, V AR ;b (k); k = 0; 1; : : : ; 0 < b ; of the form A(L b ) X t = (1 b ) b 0 X t + " t ;

4 Fractional autoregressive processes 4 or, if A(z) = I p P k i=1 iz i ; the moel is X t = b 0 L b X t + kx i=1 i L i bx t + " t : (6) We assume throughout that " t is i.i.. (0; ) in p imensions. This moel preserves the main structure of (1) in that it allows for moeling of cofractionality an ajustment. We have roppe the lag on X t ; which seems super uous because L b is alreay a lag operator. The essential i erence is, that we have replace the usual lag operator in the polynomial A (L) by the new lag operator, see (2). Note that the moel (6) is not a fractional ARIMA moel, that is, it is not of the form D(L) X t = B(L)" t ; where D(L) an B(L) are nite orer lag polynomials. It is, however, a fractional ARIMA moel in the new lag operator, because it can be expresse as (L b ) b X t = " t ; for (u) = (1 u)i p 0 P u k i=1 i(1 u)u i. Example 1. (continue) The simple univariate example from (4) is change into X t = 1 L X t + 2 L X t + " t : The characteristic function is (z) = (1 z) 1 (1 (1 z) ) 2 (1 z) (1 (1 z) ) = 1 u 1 u 2 (1 u)u = (u); u = 1 (1 z) : 7 Note that (:) is not a polynomial but (:) is a secon egree polynomial in the variable u = 1 (1 z) : Hence (z) = 0 is equivalent to (u) = 0; which is the usual polynomial equation, which is well unerstoo. We just have to unerstan the mapping z! 1 (1 z) an its range in the complex plane C; see Appenix 6.3. Thus the iea, to get a tractable theory, is to replace A (L) in (1) by a polynomial in the lag operator L b ; or equivalently, replace in the usual cointegrate VAR moel the i erence operator by b : This trick allows us to carry over the result from the I(0)=I(1)=I(2) theory an evelop the analogous theory for fractional processes. This paper oes not eal with inference, but the moels (1) an (6) have the avantage that for known fractional orers, it is easy maximize the Gaussian likelihoo applying the usual reuce rank techniques, which are well known from the I(1) moel. This leaves a function of only one or two variables (b; ) to be optimize. For moel (1) this was note by Lyhagen (1998) an Lasak (2005). Lyhagen erive the asymptotic istribution of the test for no cointegration, when the fractional orer is known, an Lasak erive the limit istribution of this test in case the fractional orer b is unknown an = 1. Moel (1) has been applie by Davison (2002) an Breitung an Hassler (2002).

5 Fractional autoregressive processes An overview of the paper In the next section we give the basic properties of the fractional operator an e ne the class of fractional processes F() an the notion of a cofractional vector. In section 3 we analyze the solution of V AR ;b (k) an start by ning a criterion for the solution of the univariate AR (1) moel to be fractional of orer zero in terms of the roots of a characteristic polynomial. Next we give the theory for the cofractional moel V AR ;b (k) an show that the moel allows for fractional processes of orer ; for which 0 X t is fractional of orer b: Finally in section 4 we e ne an analyze a moel for polynomially cofractional processes, with the property that X t is fractional of orer ; 0 X t fractional of orer b an nally that a linear combination of 0 X t an b X t is fractional of orer 2b: In the Appenix we have collecte some mathematical tools: The notion of a regular (or holomorphic) function, an a few results on some special complex functions. We suggest a convenient formalism for expressing the solution of i erence equations, which involve fractional i erences, as a function of (in nitely many) initial values an the errors of the equations. We en the Appenix with a result from Grenaner an Rosenblatt (1956) which is the key tool in the paper an a trigonometric inequality. 2 Fractional processes We let C n be a sequence of p p matrices for which P 1 jjc njj 2 < 1; where jjc n jj 2 = tr(cnc 0 n ). We e ne C(z) = P 1 zn C n ; jzj < 1 an let " t be a sequence of p imensional i.i.. variables with E(" t ) = 0 an V ar(" t ) = > 0. This allows us to e ne the stationary linear process X t = P 1 C n" t n with mean zero an nite variance P 1 k=0 C kck 0 : In orer to have an expression for the spectral ensity, we assume that C(z) can be extene to the bounary jzj = 1 by continuity so that the spectral ensity is f X () = 1 2 C(e i )C(e i ) 0 : If P 1 P jjc njj < 1 we can e ne I(0); or F(0); by the conition that C(1) = 1 C n 6= 0; see Johansen (1996), but for the processes stuie here, the sum in P the sense of lim N N!1 C n nee not P be e ne. Instea, when C(z) is continuous for jzj 1; we have C(1) = lim 1 r!1 rn C n ; which we use in the e nition of F(0). We rst e ne an F(0) process, that is, a process which is fractional of orer zero. De nition 1 If P 1 jjc njj 2 < 1 an C(z) = P 1 C nz n ; jzj < 1 can be extene to a continuous function on the bounary jzj = 1; we call the stationary linear process X t = P 1 k=0 C n" t n fractional of orer zero, F(0); if the spectrum at zero f X (0) = 1 2 C(1)C(1)0 6= 0:

6 Fractional autoregressive processes 6 For such processes we enote F(0) + the class of asymptotically stationary processes of the form X + t = C(L) + " t = C(L)" t 1ft 1g = P t 1 C n" t n t = 1; 2; : : : 0 t = 0; 1; : : : (7) Note that we o not assume in De nition 1, that f X (0) > 0, so that any linear combination is also fractional of orer zero, an we o not assume that iag(f X (0)) > 0, which woul imply that all components of X t were fractional of orer zero. The class F(0) will be use to e ne fractional processes of other orers an examples will be iscusse below. We next turn to fractional processes of orer, see for example Brockwell an Davis (1991, Chapter 13.2) or Beran (1994). The binomial expansion 1X (1 z) = ( 1) n z n ; jzj < 1; 2 R; n e nes the coe cients ( 1) n n = ( + 1) : : : ( + n 1) n! = ( + n) () (n + 1) ; which are O(n 1 ): This shows that P 1 2 n < 1 for < 1=2; so that for a sequence of i.i.. variables " t with mean zero an nite variance we can e ne 1X " t = (1 L) " t = ( 1) n " t n ; < 1 n 2 ; (8) as a stationary process with nite variance. For 1=2 the in nite sum oes not exist, but we can e ne a non-stationary process by the operator + ; see Appenix 6.4, + " t = Xt 1 ( 1) n n " t n = " t + " t ( 1) t 1 " 1 ; t = 1; : : : ; T; t 1 see for instance Robinson an Marinucci (2001) who use the notation " t 1ft 1g: De nition 2 We say that X t is fractional of orer ; an write X t 2 F() if conitionally on the past fx s ; s 0g; +X t t 2 F(0) + for some function t of the past. Note that we o not assume that all components of X t have the same orer of fractionality. This is in line with the e nition that a process is I(1) if the i erence is I(0): This allows the components to be either I(1) or I(1) in the cointegrate VAR. Note that fractional processes, as e ne here for < 1 nee not be stationary, 2 because nothing is assume about the values before time 1, but they are asymptotically stationary. The term t is neee because when solving i erence equations,

7 Fractional autoregressive processes 7 we get a contribution from initial values, see Appenix 6.5. If we conition on these, we get a eterministic term. Example 2 Let " t be i.i.. in p imensions with mean zero an nite variance. The equations X t = " t ; t = 1; : : : ; T (9) have the interpretation E t 1 X t = (1 )X t = L X t ; V ar t 1 X t = ; where the subscript inicates that we take conitional mean an variance given the past fx s ; s t 1g: Note that for the equations to make sense we must P assume that the initial values of X s ; s 0 are such that the expression X t = 1 ( 1)n Xt n n converges. Obviously this woul be the case if all initial values were zero, but for applications that is probably not so useful. The equations can be solve, see Appenix 6.5, by applying + to both sies of the equations. We then n, using + + = 1 + ; X t = t + + " t ; t = 1; : : : ; T; where t = + X t = E 0 X t is a function of initial values fx s ; s 0g: When < 1=2; the stationary process Xt = " t ; see (8), with spectrum f X () = 1 2 j1 e i j 2 ; is a solution of (9). We note that Xt an X t are fractional of orer ; an that the spectral ensity of Xt has a pole of orer 2 at = 0 for > 0; in the sense that 2 f X ()! 1 > 0. 2 De nition 3 If X t 2 F() an there exists a vector so that 0 X t 2 F( some b; 0 < b ; we call X t cofractional with cofraction vector : b) for Note that the zero vector is not a cofraction vector by this e nition, because the process X t = 0 is not fractional of any orer. The linear combination coul be the unit vector (1; 0; 0 : : : ; 0) 0 in which case the rst component of X t is F( b). The e nition of fractional an cofractional process use here are chosen with the following point of view in min. When we t the autoregressive fractional moel we rst check that the moel type as e ne by ; b; k; an the error structure is aequate. Base on the moel we then make inference about the rank of 0 an the roots of the characteristic function. If we n r < p; we infer that there are unit roots an that there are linear combinations of the ata that are fractional of orer b: Only then o we start the ienti cation of the cofractional vectors ; which of course may contain a unit vector.

8 Fractional autoregressive processes 8 A process Y t 2 F() is also calle fractionally integrate, or just integrate, of orer, an fractionally cointegrate, or just cointegrate, CI(; b), but integration gives connotations of nonstationarity an it is therefore perhaps better to use fractional an cofractional. The linear combination 0 X t oes not eliminate nonstationarity but reuces the orer of fractionality, that is, the pole at zero frequency of the spectrum. Example 3 As a simple example of fractional an cofractional processes we consier the process X t = (X 1t ; X 2t ; X 3t ) 0 e ne by its moving average representation for t = 1; : : : ; T X 1t = + 0:4 " 1t + 0:2 " 2t + " 3t ; X 2t = + 0:4 " 1t + + 0:2 " 2t + " 3t ; X 3t = + 0:2 " 1t + " 2t + " 3t : where " t is i.i.. (0; I 3 ): The transfer function for 0:4 X t at = 0 is C(1) A ; an the spectrum is C(1) 0 C(1) 6= 0, so that 0:4 X t is fractional of orer zero accoring to De nition 1, an X t is fractional of orer 0:4; accoring to De nition 2. The vectors = e ne the bivariate process Y t = 0 X t which is fractional of orer 0.2, so that is a cofraction vector, see De nition 3. Note that X 3t is fractional of a lower orer than X t ; an that is expresse by containing the unit vector (0; 0; 1): Finally note that there is another type of fractionality in this example because if (L) 0 = 0: :2 1 then (L) 0 X t is fractional of orer 0, so we have polynomial cofractionality between levels X t an i erences 0:2 X t. We conclue this section with a e nition of polynomial cofractionality, to which we shall return in section 4. De nition 4 If X t 2 F() an 0 X t 2 F( b) for some b, 0 < b ; an we can n 1 6= 0 an 2 so that X t b X t 2 F(c) for some 0 c < b; then we say that X t has a polynomial cofraction vector b : Note that if X t 2 F(); then of course b X t 2 F( b); but this is not calle polynomial cofractionality, because the levels have to enter a cofractional relation. The moels we propose only allow for fractional processes of orer an b; an polynomial cofractional vectors of orer 2b: This allows for a simpler representation theory an a simpler statistical theory. ;

9 Fractional autoregressive processes 9 3 Autoregressive moel for cofractional processes In this section we rst analyze the univariate case with one lag for b = ; which we call AR (1); an n a criterion for the solution to be F(0): Next we analyze the moel V AR ;b (k) an n criteria for the solution to be a cofractional process. 3.1 The AR (1) moel an a conition for F(0) As the simplest case of (6) we take b = ; k = 0; an p = 1, an the moel is or X t = (1 )X t + " t ; t = 1; : : : ; T: X t = L X t + " t = (1 (1 L) )X t + " t ; t = 1; : : : ; T; (10) with = 1 + : We assume that " t is univariate i.i.. (0; 2 ): Note that for = 1 we get a moel with unit root of fractional multiplicity: X t = " t : First, however, we want to n the set of for which (10) has a solution which is fractional of orer zero. We e ne the characteristic function, which is a polynomial only if is a nonnegative integer, (z) = 1 (1 (1 z) ): The equations (10) etermine the value of X t as a function of " 1 ; : : : ; " t ; an the in nitely many initial values by successive substitution an the solution is given in Theorem 5. We allow any complex value of so X t may be complex. In orer to iscuss the orer of fractionality of (10), we nee the set C e ne as the image of the unit isk uner the mapping z 7! 1 (1 z) ; see Appenix 6.3 an Figure 1. The main result in Theorem 5 generalizes the well known case of = 1; where the solution of X t = X t 1 +" t is given by X t = t X 0 + P t 1 n " t n for any ; an if jj < 1 the process Xt = P 1 n " t n is a stationary solution, an if = 1 we get the ranom walk P t i=1 " i +X 0. For the fractional process in Theorem 5, the stationarity conition jj < 1 is replace by 1 =2 C ; which reuces to (1 2 ) 1 < < 1 if is real. Note that the set of parameter values that give an F(0) process epens on : Theorem 5 1. For any 2 C, there exists a > 0; so that (z) 1 = P 1 nz n ; for jzj < : The solution of (10) has the representation X t = + (L) 1 " t + t = Xt 1 n" t n + t ; t = 1; 2; : : : (11) where t = + 1 (L) (L)X t is a function of in nitely many initial values X 0 ; X 1 ; : : : 2. If 1 =2 C ; see Appenix 6.3, (z) 1 is boune an continuous on the close unit isk, an has the expansion (z) 1 = P 1 P nz n for jzj < 1; where 1 j nj 2 < 1: It follows that X t = (L) 1 " t = 1X n" t n (12)

10 Fractional autoregressive processes 10 is stationary with mean zero an nite variance. The process Xt (10) for all t with spectrum so that f X (0) > 0; an X t 2 F(0). 3. If = 1; then (1) = 0 an where t = is a solution of f X () = j1 (1 (1 e i ) )j ; (13) 2 X t = + " t + t ; t = 1; 2; : : : (14) + X t is a function of initial values, an X t 2 F(). Proof of Theorem 5. Because (0) 6= 0; we can expan aroun z = 0 in a neighbourhoo, jzj < ; so that (z) = P 1 nz n ; jzj < : We introuce the operators xt ; t = 1; 2; : : : 1 + x t = x t 1fT 1g = 0; t = : : : ; 1; 0 an 1 = ; + (L) = (L)1 + ; an (L) = (L)1 ; see Appenix 6.4. The equations are (L)X t = + (L)X t + (L)X t = " t ; t = 1; : : : ; T: Now apply + (L) 1 on both sies, using + (L) 1 + (L) = 1 + ; an we n, with X + t = X t 1 + ; X + t + + (L) 1 (L)X t = + (L) 1 " t = Xt 1 n" t n : Finally e ne t = + (L) 1 (L)X t ; which is a function of initial values. This completes the proof of (11). If 1 62 C ; the equation (z) = 0 (or 1 = 1 (1 z) ) has no solution for jzj 1; an (z) is regular for jzj < 1 an continuous for jzj 1; when > 0: Thus the expansion of (z) is vali for jzj < 1; an by Lemma 10, the continuity on jzj 1 implies that P 1 j nj 2 < 1: We use the coe cients n to construct the process 1X Xt = (L) 1 " t = n" t n ; which is a stationary process with mean zero, nite variance, an spectrum given by (13). The value for = 0; is 2 j1 j 2 =2 > 0; so that Xt is F(0): Because the coe cients n are the coe cients in (z) 1 ; it follows that Xt solves (10) for all t. Finally, (14) is a special case of (11) using (1 z) 1 = P 1 zn ; jzj < 1: Equation (14) shows that X t 2 F(). Example 4 Consier the special case with = 0:4; that is, the moel X t = (1 (1 L) 0:4 )X t + " t ; t = 1; : : : ; T:

11 Fractional autoregressive processes 11 The bounary of the set C 0:4 intersects the negative axis at the point 1 2 0:4 = 0:32: If we take = 2, then 1 = 0:5 is outsie C 0:4 ; see Figure 7, an the process X t is F(0). Note that there are values of arbitrarily close to z = 1; an insie the unit circle, for which the process (10) is F(0) an hence stationary. Finally for = 1, we n 0:4 X t = " t which amits a solution Xt = P 1 0:4 k=0 ( 1)k "t with spectrum k 2 j1 e i j 0:8 _ 0:8 ;! 0; so that although X 2 t is stationary it is fractional of orer 0.4. Theorem 5 can easily be extene to the moel for the p imensional process given by A(L )X t = " t : We formulate the result in Corollary 6 Let X t be given by A(L )X t = P k i=0 A il i X t = " t ; t = 1; : : : ; T: If the roots of j P k i=0 A iu i j = 0 are outsie C ; then X t is F(0). Proof of corollary 6. The stacke process X ~ t = (Xt; 0 L Xt; 0 : : : ; L k 1 Xt) 0 0 is a V AR (1) process with coe cient matrix A; ~ say. By a linear transformation we can reuce the system to its Joran form, where it is seen, using Theorem 5, that the conition on the (possibly complex) roots means that all components are F(0). The proof of this result can also be given as the proof of Theorem 8 where we stuy the cofractional moel V AR ;b (k). We conclue this section by giving a criterion for checking the conition that 1 62 C : The bounary of C is escribe by the curves x( ) = 1 (2 cos ) cos( ); y( ) = (2 cos ) sin( ); for 0 min( ; 2 ): Let 1 = x + iy: Because of the symmetry of C ; there is no loss of generality in assuming that y > 0: Lemma 7 The function g( ) = x( ) y( ) = 1 (2 cos ) cos( ) ; 0 < < min( (2 cos ) sin( ) ; 2 ): is strictly increasing an has range is R. Thus for any (x; y) with y > 0, we can etermine u so that g( u ) = x=y: Then 1 62 C if an only if y( u ) < y: Proof Note that for 0 < < min( ; ); y( ) > 0; see Figure 7, so that g( ) is 2 well e ne. It is seen that! 0 implies g( )! 1; an! min( ; ) implies 2 g( )! 1; so the range is R. Next we n the erivatives _x( ) = x( ) = 2 (cos ) 1 sin((1 + ) ); _y( ) = y( ) = 2 (cos ) 1 cos((1 + ) );

12 Fractional autoregressive processes 12 an _g( ) = (y( ) _x( ) _y( )x( ))=y 2 = [(2 cos ) sin( )2 (cos ) 1 sin((1 + ) ) (1 (2 cos ) cos( ))2 (cos ) 1 cos((1 + ) )]=y 2 = 2 (cos ) 1 [2 (cos ) +1 cos((1 + ) )]=y 2 : This is positive by the trigonometric inequality in Lemma The cofractional V AR ;b (k) moel We next turn to the moel V AR ;b (k) an want to n the conitions uner which the solution is fractional of orer an 0 X t fractional of orer b: The moel is X t = 0 b L b X t + kx i=1 i L i bx t + " t ; t = 1; : : : ; T; where 0 < b : We e ne the characteristic polynomial (z) = (1 z) I p 0 (1 (1 z) b )(1 z) b kx i=1 i(1 (1 z) b ) i (1 z) an the polynomial (u) = (1 u)i p 0 u kx i=1 i(1 u)u i : (15) These are relate via the transformation u = 1 (1 z) so that (z) = (1 z) b (1 (1 z) ); jzj 1: Note that an have a unit root if r < p because (1) = (1) = 0 has eterminant zero, an that (u) is the characteristic polynomial P for the usual cointegration moel ( = b = 1; L = L). We e ne = I k i=1 i an formulate the main result, where the conitions are given on the roots of the polynomial j (u)j = 0; an the coe cients of the moel. Theorem 8 Assume that j (u)j = 0 implies that either u = 1 or u =2 C b ; that an have rank r < p; an that j 0??j 6= 0; then ( b) X t = C+ " t + + Y t + t ; t = 1; : : : ; T; where C =? ( 0?? ) 1 0?: (16)

13 Fractional autoregressive processes 13 The process Y t is fractional of orer zero, an has continuous spectrum which at zero frequency is given by C C 0 =2 6= 0; where C = ( 0 + C C + C( 0 + kx i i )C); (17) i=1 an t = + 1 (L) (L)X t epens on initial values. Thus X t is fractional of orer, while b X t an 0 X t are fractional of orer b: There is no 0 X t which is fractional of lower orer then b; an even though 0 X t an b X t are both F( b) it hols that if X t b X t is fractional of lower orer than b; then 1 = 0 an 2 = for some ; so that X t oes not allow for cofractionality. Proof of Theorem 8. From Theorem 3 in Johansen (2005) it follows that because j 0? it hols that u (u)j u=1? j = j 0?( I p 0 + kx i=1 i)? j = ( 1) p r j 0?? j 6= 0; (u) 1 = C 1 1 u + C + (1 u)h(u); 0 < ju 1j < ; (18) for some > 0; where C an C are given by (16) an (17). function H (u) e ne by Furthermore, the H (u) = C + (1 u)h(u) is regular in C with poles at the roots where j (u)j = 0; but no singularity at u = 1. The function f(z) = 1 (1 z) b is regular for jzj < 1; an continuous for jzj 1; when b > 0: If the remaining roots are outsie C b ; then there is no z in the close unit isk for which 1 (1 z) b = u i : Hence the compoun function F (z) = H (f(z)) = H (1 (1 z) b ); jzj 1 (19) is continuous for jzj 1 an regular without singularities on the open unit isk, jzj < 1: It therefore has an expansion F (z) = P 1 F nz n ; jzj < 1; where the coe cients satisfy P 1 jjf njj 2 < 1; see Lemma 10. We apply these coe cients to e ne Y t = F (L)" t = P 1 F n" t n as a stationary process with mean zero, nite variance, an continuous spectral ensity f Y () = 1 2 F (e i )F (e i ) 0 = 1 2 H (1 (1 e i ) b )H (1 (1 e i ) b ) 0 : For = 0 we get 1 2 F (1)F (1)0 = 1 2 H (1)H (1) 0 = 1 2 C C 0 :

14 Fractional autoregressive processes 14 The inequality ( 0 1 ) 1 0 =? ( 0?? ) 1 0? 0 (20) shows that 0 C C 0 > 0 C ( 0 1 ) 1 0 C 0 = ( 0 1 ) 1 > 0; (21) because 0 C = I r ; see (17). This shows that f Y (0) 6= 0; so that Y t 2 F(0): We n from (18) that 1 (z) = C(1 z) + (1 z) +b F (z); an we next apply 1 + (L) to the equation (L)X t = " t an n the solution X + t ( b) = C+ " t + + Y t + 1 (L) (L)X t : This shows that X t is F() because C 6= 0; an + b 0 X t = 0 Y t + is in F(0) + ; because Y t 2 F(0): Next consier a linear combination 0 X t : For this to be fractional of orer less that ; it must satisfy 0 C = 0; that is, = for some : Then we n ( b) 0 X t = 0 0 X t = Y t t : The stationary process 0 0 Y t has spectrum at frequency zero which is boune below by 0 ( 0 1 ) 1 ; see (21). Thus the spectrum is only zero if = 0; which means that no linear combination has lower fractional orer than b: Finally consier Z t = X t b X t ( b) = + [ Y t C" + t + b + 0 2Y t + ] + ( b ) t b X t : The spectral ensity of U t = Y t + 0 2C" t + b 0 2Y t at zero frequency is f U (0) = ( C + 0 2C)(C C 0 1 ): Because of the inequality (20), this is boune below by ( C + 0 2C)( 0 1 ) 1 0 (C C 0 1 ) = 0 1( 0 1 ) 1 1 because C = 0; an 0 C = I r ; see (17). If Z t were fractional of lower orer than b; then f U (0) = 0; which implies that 1 = 0 an hence 0 2C = 0; so that 2 = for some. Note that the last result implies that moel (6) oes not allow for polynomial fractional cointegration, see Engste an Johansen (1999) for the I(1) case. A moel that allows this possibility is analyze in the next section. We coul easily inclue the case r = p in the formulation an we woul get C = 0; an C = 0 = ( 0 ) 1 = (1) 1 : In this sense, the result of Theorem 8 contains Corollary 6, as a special case for r = p.

15 Fractional autoregressive processes 15 4 A moel for polynomial cofractional processes We consier the V AR ;b (k) moel (6) reparametrize as X t = 2b ( 0 L b X t b L b X t ) + kx i=1 i L i bx t + " t ; (22) where we assume that an have rank r < p; an that 0?? = 0 of rank s < p r; an that 0 < 2b : This is an analogue of the moel for I(2) variables, which we get for = 2; b = 1; see Johansen We e ne the irections (; 1 ; 2 ) = (;? ;?? ); (; 1 ; 2 ) = (;? ;?? ): (23) Note that (; 1 ; 2 ) an (; 1 ; 2 ) are orthogonal ecompositions of R p : From (22) we n the characteristic function kx (z) = (1 z) I p (1 (1 z) b )(1 z) 2b ( 0 (1 z) b ) i(1 z) (1 (1 z) b ) i e ne for jzj 1; an the polynomial (u); see (15), which is reparametrize as kx (u) = (1 u) 2 I p u 0 + u(1 u) i(1 u) 2 u i ; an relate to (z) by i=1 i=1 (z) = (1 z) 2b (u); u = 1 (1 z) b : Theorem 9 Assume that j (u)j = 0 implies that either u = 1 or u =2 C b ; an that an have rank r < p: If 0?? = 0 has rank s < p r an if the conition j j 6= 0 hols, where = I p kx i=1 then X t = C 2 + ( b) " t + C 1 + " t + The matrices C 1 an C 2 are i ; ( 2b) + Y + t + t : (24) C 2 = 2 ( ) 1 0 2; (25) C 1 = [ ]C 2 + C 2 [ ] (26) C 2 [ kx i i]c 2 an t = + (L) 1 (L)X t epens on initial values. The process Y t is stationary with continuous spectrum, an X t is fractional of orer ; (; 1 ) 0 X t is fractional of orer b; an 0 X t 0 b X t is fractional of orer 2b. i=1

16 Fractional autoregressive processes 16 Proof of Theorem 9. From Theorem 5 of Johansen (2005) it follows that implies that j 0 2( u (u)j u=1 0 u (u)j u= u 2 (u)j u=1 ) 2 j = j j 6= 0 (u) 1 1 = C 2 (1 u) + C u + C 0 + (1 u)h(u); 0 < ju 1j < ; (27) where C 2 an C 1 are given by (25) an (26). No complete expression was foun for C 0 ; but it was shown that Furthermore, the function 0 C 0 = I r + 0 C 2 : (28) H (u) = C 0 + (1 u)h(u) is regular without a pole at u = 1: Uner the conition that the remaining roots of j (u)j = 0 are outsie C b ; H (u) is regular without singularities insie C b an continuous on C b : We e ne F (z) = H (1 (1 z) b ) for jzj 1: By Lemma 10 this is regular for jzj < 1 an continuous on the bounary when > 0, so that Y t = P 1 F n" t n is a stationary process with continuous spectrum, where F (z) = P 1 F nz n ; jzj < 1. We then n 1 (z) = (1 z) ( 2b) (u) 1 = C 2 (1 z) + C 1 (1 z) ( b) + (1 z) ( 2b) F (z) We construct the solution of the equations (L)X t = " t by applying + 1 (L) an n X t + = C 2 + ( b) ( 2b) " t + C 1 + " t + + Y t + + (L) 1 (L)X t ; which proves (24). It is seen that X t is F() because C 2 6= 0; that (; 1 ) 0 X t is F( b) because (; 1 ) 0 C 2 = 0; an (; 1 ) 0 C 1 1 = (0; I s ) 6= 0; see (25) an (26). We next consier polynomial cofractionality an want to show that 0 X t 0 b X t 2 F( 2b); an have to show that the process + 2b ( 0 X t 0 b X t ) = + 2b ( 0 X t + 0 b +X t + 0 b X t ) is in F(0) + : We n from (24) that 2b + ( 0 X t + 0 b +X t + 0 b X t ) = ( 0 C 1 0 C 2 ) b +" t + ( 0 Y t + 0 C 1 " + t 0 b +Y + +( + 2b b ) t + 2b 0 b X t : From (25) an (26) it follows that 0 C 1 0 C 2 = 0; so that the rst parenthesis is zero. In orer to see that + 2b ( 0 X t 0 b X t ) is in F(0) + ; we n the spectrum at zero frequency for the process 0 Y t 0 C 1 " t 0 b Y t ; which is proportional to ( 0 C 0 0 C 1 )( 0 C 0 0 C 1 ) 0 : t )

17 Fractional autoregressive processes 17 By the inequality (20), this is boune below by ( 0 C 0 0 C 1 )( 0 1 ) 1 0 (C0 0 C1 0 0 ): From (28) an (26) it follows that 0 C 0 0 C 1 = I r + 0 C 2 0 C 2 = Ir ; so that 0 Y t 0 C 1 " t 0 b Y t is F(0). Note that Y t is stationary with spectrum at zero frequency given by C 0 C0=2; 0 an hence probably non-zero so that Y t is probably F(0). We o not have a complete expression for C 0 ; only an expression for 0 C 0 which is enough to prove what we nee, namely that 0 X t 0 b X t is F( 2b). 5 Conclusion We have e ne a class of vector autoregressive moels, V AR ;b (k); that generate fractional, cofractional, an polynomial cofractional processes. The moels have the avantage that one can give simple criteria in terms of the parameters for fractionality, cofractionality, an polynomial cofractionality. In this way we have a class of moels for cofractional processes an their ajustment to equilibrium relations, so that economic hypotheses can be formulate within the moel, which can be teste against the ata. Thus we have a platform for making moel base inference on coe cients, relations, an fractional orer. Inference for these moel has still to be worke out. 6 Appenix In this Appenix we rst iscuss regular functions in particular the function f(z) = 1 (1 z) ; an the image of the unit circle uner the mapping f(z). We introuce some useful operators for solving i erence equations involving fractional i erences. Finally we give a result from Grenaner an Rosenblatt (1956), which is the key to unerstaning the structure solution of the moel equations. 6.1 Regular functions A regular (or holomorphic) function, see Phillips (1958) is e ne as a complex value i erentiable function on an open (an arc connecte) set D of C. We give two examples. Consier rst the function g(z) = 1=(1 3z): This is e ne for z 2 Cnf1=3g; an jg(z)j! 1; z! 1=3: Note that (1 3z)g(z); z 6= 1=3; can be extene by continuity to the point z = 1=3; an the result is a regular function e ne on C. Such a point is calle an isolate singularity an (1 3z)g(z) has a removable singularity at z = 1=3, whereas g(z) has a pole of orer one. The regular function g(z) can be

18 Fractional autoregressive processes 18 expane in a Taylor s series at all other points except z = 1=3: For instance, aroun z = 0 we get the series g(z) = P 1 3n z n ; jzj < 1 ; with exponentially increasing 3 coe cients, P whereas aroun z = 2; the function can be represente by the series g(z) = ( 3 5 )n (2 z) n ; j2 zj < 5 ; where the coe cients are exponentially 3 ecreasing. Thus g is regular on Cnf1=3g: Another example is the function f(z) = 1 (1 z) 0:8 : In orer to get a single value function we remove from the complex plane the line L = fim(z) = 0; Re(z) 1g: Then f(z) is well e ne an regular on D = CnL. Moreover it can be extene by continuity to the point z = 1; but the extene function is not i erentiable at z = 1: The point is not an isolate singularity. The main result about a regular function is that at any point of the omain of e nition it can be expane in a Taylor s series which converges in the largest open isk that oes not contain any singularity. Notice that a polynomial p(z) = P k i=0 a iz i is a regular function on C, so that in a sense the regular functions correspon to in nite orer polynomials. Another funamental result is that for two regular functions the composition f(g(z)) is again regular provie the range of g is in the omain of f. If (z) is a matrix polynomial, then 1 (z) is a regular function with isolate singularities (poles) at the roots of j(z)j = 0: The representation (18) shows that (1 u) (u) 1 has a removable singularity at u = 1; or that by subtracting C(1 u) 1 from (u) 1 we get a function with removable singularity. If there are no further poles in the unit isk, then 1 (z) C(1 u) has an expansion aroun zero with exponentially ecreasing coe cients. For fractional processes, see (6), the function 1 (z) is regular but with a i erent type of singularity at z = 1, because we cannot move aroun the pole insie the omain of e nition. The main trick in this paper is to e ne (z) as (1 (1 z) ); so that by removing the poles from 1 (u) we remove the cause for the singularity from 1 (u): If we remove these singularities, the function is regular without singularities in the open unit isk an continuous on the whole isk, the result given in Section 6.6 shows that this is enough to iscuss the processes in the sense of L 2 : 6.2 The function z For > 0; the function h(z) = z ; is multi-value with a singularity at z = 0. For any n = 0; 1; : : : ; the representation z = re i = re i+2ni ; shows that z = r e i+2ni ; n = 0; 1; 2; : : : has many i erent values unless is an integer. In orer to get a unique value we associate with a complex number z the main argument Arg(z), which is the argument for which z = re i ; < : Then Arg(z) is iscontinuous at the negative real line, because for z! 1 with Im(z) > 0; we have Arg(z)! ; but for z! 1; with Im(z) < 0; we have

19 Fractional autoregressive processes 19 Arg(z)! : For > 0; we can e ne the power function z uniquely if we avoi the negative real axis. Thus for z = re i ; < < we e ne z = r e i = jzj e iarg(z) : Notice that we have z a z b = z a+b ; but in general (z a ) b 6= z ab ; as the following example shows. Example 5 Let z = e i 3 4 ; then with the above e nition z 2 = e i 2 ; so that (z 2 ) 1=2 = e i 4 ; whereas z 1=2 = e i 3 8 implies that (z 1=2 ) 2 = e i 3 4 6= (z 2 ) 1=2 : A consequence is that the function 1 (1 z) has singularities at the line L = fim(z) = 0; Re(z) 1g; but it is regular on CnL an we can expan it aroun z = 0; an n 1X 1 (1 z) = 1 ( 1) n z n ; jzj < 1: n 6.3 The image C of the unit isk uner f(z) = 1 (1 z) We investigate here the image of the unit isk uner the mapping f(z) = 1 (1 z) where 0 < < 1: We have for 0 an z = e i ; 1 z = 1 e i = 1 cos() i sin() = 2 sin 2 2i sin 2 2 cos 2 = 2 sin 2 (sin i cos 2 2 ) = 2(cos )(cos i sin ) = 2(cos )(cos i sin ) = 2(cos )e i ; = 1 2 ( ) 2 [0; 1 2 ]: For 0 we get the same result but with replace by : We n for i erent the curves given in Figure 7. A point on the curve is (x( ); y( )) given by x( ) = 1 (2 cos ) cos( ); y( ) = (2 cos ) sin( ): They intersect the real axis for = 0 an = min( ; ): Note that y( ) is positive 2 for 0 < < min( ; ): The operators h + (L) an h (L) Let the complex function h be given by a convergent power series 1X h(z) = h n z n ; jzj < 1;

20 Fractional autoregressive processes 20 for which h(0) = 1; so that h(z) 6= 0; jzj <. The function g(z) = 1=h(z) is given by a power series 1X g(z) = g n z n ; jzj < ; where the coe cients satisfy X n+m=k g n h m = 1 fk=0g : (29) If P 1 h nx n t is well e ne we e ne the operators P t 1 h + (L)x t = h(l)(x t 1ft 1g) = h nx t n ; t = 1; 2; : : : 0; t = : : : ; 1; 0 ; h (L) = h(l) h + (L): Note that the operator h + (L) is e ne for any sequence because only nitely many terms occur. Similarly we e ne g + (L) for any sequence an notice that it follows from (29) that h + (L)g + (L)x t = 1 + x t = g + (L)h + (L)x t for any sequence, whereas h (L)g + (L)x t = 0: (30) As an example we consier the function h(z) = 1 z an h(l) = 1 L =. The function is inverte as g(z) = 1=(1 z) = P 1 zn ; jzj < 1; so that g n = 1; n = 0; 1; : : : Thus 1 x t is only e ne if P 1 P x t n is well e ne, but + 1 x t = t i=1 x i; t = 1; 2; : : : is always e ne. 6.5 Solving fractional i erence equations We can use the operators h + an h to solve i erence equations involving fractional i erences. Thus for instance we can solve the i erence equation X t = " t ; t = 1; : : : ; T as function of " 1 ; : : : ; " T an initial values fx s ; s 0g: We cannot apply the operator on both sies because it may not be e ne on the sequence " t : We instea apply the operator + an use the relation + + = 1 + = 1ft 1g; to get + X t = + ( +X t + X t ) = X t + + X t = + " t ; t = 1; : : : ; T; which gives X t = + " t + t = Xt 1 ( 1) n n " t n + t ; t = 1; : : : ; T;

21 where t = Fractional autoregressive processes 21 + X t = Xt 1 X 1 ( 1) n ( 1) k n k k=t X t n k is a function of the in nitely many initial values X 0 ; X 1 ; : : : In the special case = 1; we n 8 8 < x t x t 1 ; t = 2; 3; : : : < 0; t = 2; 3; : : : + x t = x 1 ; t = 1 ; x : t = x 0 ; t = 1 : 0; t = : : : ; 1; 0 x t x t 1 ; t = : : : ; 1; 0 so that t = 1 + x t = x ; which gives the well known solution x t = x 0 + tx " i : i=1 6.6 A result from spectral theory We can infer the properties of the coe cients of a linear process by analyzing the transfer function, as the next lemma shows. The result is taken from Grenaner an Rosenblatt (1956, page 288) an shows how continuity of the transfer function on the bounary of the unit isk can be use to prove the existence of a stationary process with nite variance, see also Brockwell an Davis (1991, Theorem ). Lemma 10 Let be Lebesgue measure on the interval [0; 2]: Consier the regular matrix value function F (z) which has no singularities in the open unit isk, so that F (z) = P 1 F nz n ; jzj < 1: Then F can be extene by L 2 () continuity to the bounary jzj = 1 if an only if P 1 jjf njj 2 < 1: In this case X t = P 1 k=0 F k" t k is a stationary process with mean zero an variance P 1 k=0 F kfk 0: If in particular F (z) is continuous for jzj 1; it is L 2 () continuous, an the spectral ensity is continuous an given by f() = 1 2 F (e i )F (e i ) 0 : Proof of Lemma 10. For r < 1; we n, using the regular function F; Z 1 F (re i )F (re i ) 0 = 2 1X n;m=0 r n+m 1 Z F n F 2 me 0 i(m n) = 1X F n Fnr 0 2n (31) First assume that F has been extene by L 2 () continuity to the bounary, so R that F (e i ) = lim r!1 F (re i ); where the limit is in the sense of L 2 (); that is, jf (ei ) F (re i )j 2! 0; r! 1: Then the left han sie of (31) converges R towars the nite limit 1 F (e i )F (e i ) 0. Hence the limit of the right han 2 sie, P 1 F nfn; 0 is nite.

22 Fractional autoregressive processes 22 Next, assume that P 1 F nf 0 n is nite, then k () = F (r k e i ); r k! 1 is a Cauchy sequence in L 2 (); because by (31) Z ( k () m ())( k ( ) m ( )) 0 = 1X F n Fn(r 0 k n rm) n 2! 0 as r k an r m ten to one. By completeness of L 2 () there is an L 2 limit (), an we therefore exten F by L 2 () continuity by e ning F (e i ) = (): P Let P enote the probability measure unerlying the ranom sequence " t : If 1 F nfn 0 < 1; then P 1 F nrk n" t n; k = 1; 2; : : : is an L 2 (P) Cauchy sequence for r k! 1; because E( 1X F n (rk n rm)" n t n ) 2 = n=m 1X F n Fn(r 0 k n rm) n 2! 0; r k ; r m! 1: n=m Thus X t exists as an element of L 2 (P) an the variance is as given. If nally F (z) is continuous for jzj 1; it is boune an therefore L 2 () continuous. This gives the value of the continuous spectral ensity. 6.7 A trigonometric inequality Lemma 11 (cos ) +1 cos((1 + ) ); 0 min( 2 ; ); > 1: (32) Proof We apply a prouct representation of cos ; see Grashteyn an Ryzhik (1971, # ), 1Y 4 2 cos = (1 (2k + 1) 2 ); 2 an we therefore want to prove k=0 1Y (1 k=0 4 2 (2k + 1) 2 2 )+1 > 1Y (1 k=0 4 2 ( + 1) 2 ): (33) (2k + 1) 2 2 The function f(z) = (1 u) +1 (1 ( + 1) 2 u) is convex on [0; 1[; f(0) = 0; an erivative f(0) _ = ( + 1) > 0; for > 0: Therefore f is strictly increasing an positive on the interval u 2 [0; 1[: For each of the factors we then have (1 4 2 (2k + 1) 2 2 )+1 > ( + 1) 2 ; k = 0; 1; : : : (2k + 1) 2 2

23 Fractional autoregressive processes 23 The inequality (33) woul follow if the factors were positive. The right han sie is positive for k = 1; 2; : : : because ( + 1) 2 (2k + 1) ( + 1) ( + 1) min(1; 2 )2 : For 0 < 2 we n min(1; 2) = 1; an 1 (+1) 2 0; an for 2; we n 3 2 min(1; 2=) = 2= an 1 4( + 1) 2 =(3 2 2 ) 0: Thus the result (32) hols when the factor with k = 0 on the right han sie is positive, but when it is negative the inequality hols trivially. 7 References 1. Anerson, T.W. (1971), The Statistical Analysis of Time Series, New York, Wiley. 2. Beran, J. (1994). Statistics for Long-memory Processes. Chapman & Hall, New York. 3. Breitung, J. an Hassler, U. (2002). Inference on the Cointegration Rank in Fractionally Integrate Processes, Journal of Econometrics, 110, Brockwell, P.J. an Davis, R.A. (1991). Time Series: Theory an Methos. (2. e.) Springer, New York. 5. Davison, J. (2002). A moel of fractional cointegration, an tests for cointegration using the bootstrap. Journal of Econometrics 110, Dittmann, I. (2004). Error correction moels for fractionally cointegrate time series. Journal of Time Series Analysis 25, Engle, R.F. an Granger, C.W.J. (1987). Co-integration an error correction: representation, estimation an testing. Econometrica 55, Engste, T. an Johansen, S. (1999). Granger s representation theorem an multicointegration. In Engle, R.F. an White, H. (es.) Cointegration, Causality an Forecasting. A Festschrift in Honour of Clive W.J. Granger, Oxfor University Press, Oxfor. 9. Grashteyn, I.S., an Ryzhik, I.M. (1971). Tables of Integrals, Sums, Series an Proucts (6th En.), es A. Je rey an Zwillinger, Acaemic Press, New York. 10. Granger, C.W.J. (1986). Developments in the stuy of cointegrate economic variables. Oxfor Bulletin of Economics an Statistics 48,

24 Fractional autoregressive processes Granger, C.W.J. an Joyeux, R. (1980). An introuction to long memory time series moels an fractional i erencing. Journal of Time Series Analysis 1, Grenaner, U. an Rosenblatt M. (1956). Statistical Analysis of Stationary Time Series. Almqvist an Wiksell, Stockholm. 13. Hosking, J.R.M. (1981). Fractional i erencing. Biometrika Johansen, S. (1996). Likelihoo Base Inference on Cointegration in the Vector Autoregressive Moel. Oxfor University Press, Oxfor. 15. Johansen, S. (1997). Likelihoo analysis of the I(2) moel. Scaninavian Journal of Statistics 24, Johansen, S. (2005). Representation of cointegrate autoregressive processes with application to fractional processes. Preprint 7, University of Copenhagen. 17. Lasak, K. (2005). Likelihoo base testing for fractional cointegration. Discussion paper, University of Barcelona. 18. Lyhagen, J. (1998). Maximum likelihoo estimation of the multivariate fractional cointegrating moel. Working paper, Stockholm School of Economics. 19. Phillips, E.G. (1958). Functions of a Complex Variable with Applications. Oliver an Boy, Lonon. 20. Robinson, P.M. an Marinucci, D. (2001). Semiparametric fractional cointegration analysis. Journal of Econometrics 105,

25 Fractional autoregressive processes = = = = Figure 1 The images C of the unit isk uner the mapping z! 1 (1 z) ; for four values of : The curves are e ne for 0 =2 by x( ) = 1 (2 cos ) cos( ); y( ) = (2 cos ) sin( ); an intersect the real axis for = 0 an = min( 2 ; ):