Random coefficient autoregression, regime switching and long memory

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1 Random coefficient autoregression, regime switching and long memory Remigijus Leipus 1 and Donatas Surgailis 2 1 Vilnius University and 2 Vilnius Institute of Mathematics and Informatics September 13, 2002 Abstract We discuss long memory properties of AR(1) process X t with random coefficient a t taking independent values A j [0, 1] on consecutive intervals of a stationary renewal process with a power-law interrenewal distribution. In the case when the distribution of generic A j has either an atom at the point a = 1, or a beta-type probability density in a neighborhood of a =1,weshow that the covariance function of X t decays hyperbolically with exponent between 0 and 1, and that a suitably normalized partial sums process of X t weakly converges to a symmetric Lévy process. 0 Introduction The first order autoregressive equation X t = ax t 1 + ε t (0.1) is fundamental to the theory and applications of time series. In the most simple case of (0.1), a is a scalar nonrandom coefficient taking values in the interval ( 1, 1) and ε t,t Z 1

2 are independent normally distributed random variables with zero mean. A natural generalization of (0.1) is to assume the coefficient random and/or time-dependent. Vervaat (1979), Lewis and Lawrence (1981), Tjøstheim (1986), among others, studied the simple model X t = a t X t 1 + ε t, (0.2) {a t } and {ε t } are two random processes, which are usually assumed to be mutually independent. The so-called Random Coefficient AutoRegressive (RCAR) model (see e.g. Nicholls and Quinn (1982)) usually refers to the situation when the coefficient sequence {a t } is i.i.d. Dependence properties of the RCAR model with independent coefficients are broadly similar to those of the standard AR(1) process (0.1). The other extreme case of (0.2) when a t is a random constant corresponds to a nonergodic process X t whose covariance need not tend to zero (see e.g. Robinson (1978)). Granger (1980) showed that aggregation of random coefficient AR(1) can lead to Gaussian long memory processes whose spectral density has power-law singularity at the origin. Pourahmadi (1988) was probably the first to observe that the model (0.2) with random time-dependent a t (taking only two distinct values 0 and 1) may exibit long memory, in the sense that the covariance of X t is nonsummable and decays hyperbolically. In his case, the values 0 and 1 can be taken on consecutive renewal intervals with a power-law interrenewal distribution. In our paper we discuss a similar (but more general) situation when the process a t is the so-called renewal-reward process: a t := A j, S j 1 <t S j, j Z (0.3)...<S 1 < 0 S 0 <S 1 <... (0.4) is a strictly stationary renewal process on Z and {A i,i Z} is a sequence of i.i.d. r.v. s independent of the renewal process (0.4). We assume that the interrenewal times U i := S i S i 1, i Z are independent {1, 2,...} valued random variables having a common distribution U with finite expectation µ := EU <. The above conditions imply strict stationarity of the renewal-reward process a t. We also assume that the sequences {S j,a j,j Z} and {ε j,j Z} are independent and ε j,j Z are i.i.d. r. v. s with zero mean and unit variance. We do not impose gaussianity or any particular distributional assumptions on ε j s. On the other hand, the assumptions on the distributions of U and generic A = A i given below play an important role in the proof of our results, although they probably can be relaxed (in particular, the requirement that A takes values in the interval [0, 1]). 2

3 Assumption U(α). There exist constants c U > 0andα>1 such that P [U = u] c U u α 1, u. Assumption A(q). P [0 A 1] = 1 and (i) If q = 0, then A has an atom at 1: 0 <f 1 := P [A =1]< 1. (ii) If q>0, then A has a probability density f(a) in some neighborhood of a = 1 such that f(a) =f 1 (a)(1 a) q 1, f 1 (a) is a continuous function such that f 1 := f 1 (1) > 0. Let us describe the main results of the paper. In Sect. 1 we find sufficient and necessary conditions on U and A for the existence of a covariance stationary solution X t of (0.2) given by i 1 X t = ε t i a t p, (0.5) i=0 with the convention 1 p=0 a t p := 1. These conditions are satisfied under Assumptions U(α) and A(q). The main results of the paper are Theorems 1 and 2 below. They refer to X t of (0.5), with a t defined by (0.3). p=0 Theorem 1 Let Assumptions U(α) anda(q) be satisfied, 2 <α+ q<3. Then r t = EX 0 X t c 2 t 2 α q, t. (0.6) The explicit form of the asymptotic constant c 2 > 0 (depending on α and q) is given in Sect. 2. Theorem 1 is a particular case of Theorem 2.1 Assumption A(q) is replaced by regular decay condition ν u c A u q of moments ν u := EA u,asu. Theorem 1 implies that for 2 <α+ q<3, the stationary solution X t (0.5) has long memory. The fact that the intensity of long memory, or the exponent α + q 2 (0, 1) in (0.6), depends both on the tail parameter α of the interrenewal time distribution, as well as on the parameter q characterizing the average closeness of (0.2) to the unit root a = 1 of the standard AR(1) model, is very natural. Theorem 1 refers to second-order properties of X t only. Theorem 2 discusses distributional properties of partial sums process of X t for large n. Put λ := 2(α + q)/3. Write = for weak convergence of finite dimensional distributions. 3

4 Theorem 2 Let Assumptions U(α) anda(q) be satisfied, 2 <α+q <3, andlete ε 3 <. Then n 1/λ X s = Z λ (t), (0.7) 0 s<[nt] Z λ (t) is a λ stable symmetric Lévy process whose characteristic function is given in (3.19) below. The question of the functional convergence in (0.7) (e.g., in the Skorohod topology) is open; see also Pipiras, Taqqu and Levy (2002, p.4). Theorem 2 might appear surprising as the summands X t have finite variance (but infinite fourth moment, see Proposition 2.3) and their marginal distribution is not heavy-tailed. However, stable limits of sums of (stationary) r.v. s may occur under long memory even when the summands are bounded. One of the simplest models of such kind is the renewal-reward model (0.3) itself which was recently discussed in several papers in connection with aggregation, see Mikosch, Resnick, Rootzén and Stegeman (2002), Pipiras, Levy and Taqqu (2002) and the references therein. It is easy to show that the process a t of (0.3) exibits long memory if EA 2 < and the distribution of U satisfies Assumption U(α) with1 < α < 2. Taqqu and Levy (1986) discuss the convergence of partial sums process of a t to a stable Lévy process under more general assumptions on A and U. A far reaching generalization of the renewal-reward model is the stochastic regime model of Davidson and Sibbertsen (2002), in which the rewards are random short memory processes which fluctuate around a local mean (regime). One of the principal results of the last paper is the fact that suitably normalized sums of their stochastic regime model converge to a stable Lévy process. A different class of long memory processes with finite variance whose partial sums converge to a Lévy process is discussed in Surgailis (2002a, 2002b). The random coefficient autoregressive process of (0.2)-(0.3) can be also regarded as a stochastic regime model, each regime corresponds to a standard AR(1) process (0.1) with fixed coefficient a. As a varies between 0 and 1, the class of possible regimes ranges from I(0) (i.i.d.) to I(1) (random walk) behavior. In the most simple case when a t, or A j, assume values 0 and 1 only, the solution alternates between I(0) and I(1) regimes: { εt, if a t =0, X t = ε t + ε t ε S 0 (t), if a t =1, S 0 (t) := max{s j : S j < t,a j = 0} = max{s : s < t,a s = 0}, see (0.5), also Pourahmadi (1988). Several other models involving structural changes, stochastic regime switching and long memory were recently discussed in the econometric literature, see Parke (1999), Granger and Hyung (1999), Diebold and Inoue (2001), Liu (2001), Gourieroux and Jasiak (2001), Leipus and Viano (2002). 4

5 1 Existence of stationary solution Conditions for the existence of covariance stationary solution (0.5) of the random coefficient autoregressive equation (0.2) with general a t were discussed in Pourahmadi (1988). Brandt (1986) studied the existence and uniqueness of strictly stationary solution of (0.2) without any moments. In this section we discuss the existence of covariance stationary solution in the case of a t given by (0.3), but without imposing specific Assumptions U(α) anda(q). In the sequel, C stands for generic constant which may change from line to line. Recall the distribution of the first arrival time S 0 0 and the last arrival time S 1 before t = 0 in a strictly stationary renewal process (0.4) satisfy P [S 0 = u] = P [S 1 = u 1] = µ 1 P [U u +1], u =0, 1,... (1.1) Put ν k := EA k, k =1, 2,... Theorem 1.1 Equation (0.2) with a t as in (0.3) admits a covariance stationary solution X t given by (0.5) if and only if Eν 2U < 1 (1.2) and E U (ν ν 2u ) <. (1.3) Proof. According to Pourahmadi (1988), the existence of covariance stationary solution of (0.2) is equivalent to u J := E[a a 2 0a ]=E a 2 p <. We have J = E u=0 p=0 u a 2 pi(s 1 < u)+e p=0 By (1.1) and the independence of A 0 and S 1, J 1 = EA 2u 0 I(S 1 u) = and ν 2u v=u = µ 1 J 2 = E v=1 u=0 p=0 u a 2 pi( u S 1 < 0) =: J 1 + J 2. ν 2u P [S 1 u] P [U v] =µ 1 E A 2v 0 I(S 1 = v) = µ 1 ν 2v P [U v]j 0 v=1 5 u U (ν ν 2u ), (1.4) a 2 p u=v p=v = J 0 µ 1 E U ν 2v, (1.5) v=1

6 J 0 := E u u=0 p=0 a 2 0, p and a 2 0,t is the renewal-reward process corresponding to the renewal process... < S 0, 1 < S 0,0 =0<S 0,1 <... with the same distribution U of interrenewal times and fixed renewal point S 0,0 = 0. Similarly as above, for J 0 one obtains the equation J 0 = = u=0 ( EA 2(u+1) 0 I(S 0, 1 < u)+ u v=1 ν 2u P [U u]+ ν 2v P [U = v]j 0 v=1 = E(ν ν 2U )+J 0 Eν 2U. Clearly, J 0 < if (1.2) and (1.3) hold in which case ) EA 2v 0 I(S 0, 1 = v)a 2 v...a 2 u J 0 = E(ν ν 2U ) 1 Eν 2U. (1.6) From (1.4), (1.5), (1.6) one obtains that under conditions (1.2) and (1.3), J = µ 1 E = µ 1 E U (ν ν 2u )+J 0 µ 1 E(ν ν 2U ) U (ν ν 2u )+ (E(ν ν 2U )) 2 < µ(1 Eν 2U ) and therefore covariance stationary solution (0.5) is well-defined. Moreover, the above argument shows J = if either (1.2) or (1.3) is violated. This concludes the proof of Theorem 1.1. Corollary 1.2 Assume P [ A 1] = 1 and µ = EU <. Then: (i) Covariance stationary solution (0.5) exists if P [ A < 1] > 0 and µ 2 := EU 2 <. (ii) If P [ A =1]> 0, then the sufficient conditions of (i) are also necessary for the existence of covariance stationary solution (0.5). Proof. (i) Note P [ A < 1] > 0 implies ν 2u < 1,u 1 and hence (1.2). Condition (1.3) is also satisfied as E U (ν ν 2u ) <EU 2 <. (ii) The necessity of P [ A =1]< 1 for (1.2) is obvious. Let us check the necessity of µ 2 <. Since ν 2u P [ A =1]> 0, together with (1.3) this yields >E U (ν ν 2u ) P [ A =1]E U u P [ A =1]EU2 /2, or µ 2 <. Corollary 1.3 Let Assumptions U(α) anda(q) besatisfied,α + q>2. Then covariance stationary solution (0.5) exists. 6

7 Proof. Note Assumption A(q) implies ν u = EA u = { f1 + o(1), q =0, (c A + o(1))u q, q > 0, (1.7) c A = f 1 0 e x x q 1 dx. If q = 0 then α>2 implying µ 2 < and the existence of X t (0.5) by Corollary 1.2. Let q>0. Then (1.2) easily follows we need to check condition (1.3) only. We have E U (ν ν 2u )=µe(ν ν 2U0 ) U 0 =1, 2,... is distributed according to P [U 0 = u] =µ 1 P [U u] cu α. By (1.7), v v 1, q > 1, ν 2u C u q C log v, q =1, v 1 q, q < 1. Hence, if q<1, E U (ν ν 2u ) CEU 1 q 0 C 1 u 1 q u α du = C 1 u 1 α q du <, implying (1.3). If q 1, (1.3) follows similarly by α > 1. Hence, by Theorem 1.1, the covariance stationary solution (0.5) exists. 2 Decay of covariance Theorem 2.1 Let Assumption U(α) besatisfied.letp [0 A 1] = 1, P[A =1]< 1, and let there exists q 0, 2 <α+ q<3 and a constant c A > 0 such that Then ν u c A u q, u. (2.1) r t = EX 0 X t c 2 t 2 α q, t, (2.2) c 2 := c A c U (µα(α 1)) 1 0 (1 + 2x) q (1 + x) 1 α dx. Proof. Note first check that the covariance stationary solution X t exists under conditions of the theorem, which follows from (2.1) and the proof of Corollary 1.3. Let us prove (2.2). According to Pourahmadi (1988), the covariance of X t (0.5) equals r t = E j 1 a t...a 1 j=0 p=0 a 2 p = E j 1 a 0...a 1 t j=0 p=0 a 2 t p. 7

8 Let us split r t = r t + r t into two parts r t and r t, corresponding to S 1 t j and t j<s 1 1, respectively. Then Ea 0...a 1 t a 2 t...a 2 t j+1i(s 1 t j) = EA t+2j 0 P [S 1 t j]. Hence by (1.1), (2.1) and Assumption U(α), r t = = j 1 Ea 0...a 1 t a 2 t pi(s 1 t j) j=0 p=0 ν t+2j P [S 1 t j] c 2 t 2 α q. (2.3) j=0 Let us prove that the term r t = j 1 Ea 0...a 1 t a 2 t pi( t j<s 1 1) j=0 s t+1 p=0 Ea 0...a s I(s S 1 1) =: ρ t is negligible w.r.t. r t. By definition of a t, ρ t = π 0 (u 1 )π(u 1 u 2 )...π(u k 1 u k )π (u k s), (2.4) k=1 s t s<u k <...<u 1 <0 π 0 (u) :=ν u P [S 1 = u], π(u) :=ν u P [U = u], π (u) :=ν u P [U u], u 1. Note π := π(u) =Eν U EA < 1, π(u) Cu q α 1,π 0 (u) Cu q α,π (u) Cu q α. According to Giraitis, Robinson and Surgailis (2000, Lemma 4.2), π(u 1 )π(u 2 u 1 )...π(t u k 1 ) Ck 3 π k t q α 1, t,k 1. 0<u 1 <...<u k 1 <t Therefore ( ρ t C k 3 π k) π 0 (u 1 ) u 1 u k q α 1 π (u k s) k=1 C s t s<u k <u 1 <0 s t s<u k <u 1 <0 C s q α Ct 1 q α. s t u 1 q α u 1 u k q α 1 u k s q α Therefore r t = o(r t), which concludes the proof of the theorem. 8

9 From the proof of the above theorem, it is easy to observe that long memory behavior of X t is due to the past of the autoregressive process until the first change of the coefficient, while the remaining part of the process is short memory. More precisely, let S (t) be the last renewal time before time t: S (t) :=max{s j : S j <t}. As it follows from the definition (0.3), the process a s is constant on the interval [S (t)+1,t]: a s = A j(t) on s [S (t)+1,t], j(t) Z is defined by S (t) =S j(t) 1. By stationarity of the renewal process, the backward time t S (t) has the same distribution as S 1, i.e. P [t S (t) =u] =µ 1 P [U u], u =1, 2,... Then X t = X 0 t + X 1 t, (2.5) Xt 0 := ε t + A j(t) ε t 1 + A 2 j(t) ε t A t S (t) 1 j(t) Xt 1 :=: a t a t 1...a s+1 ε s. s S (t) ε S (t)+1, Clearly, both X 0 t and X 1 t are stationary processes. Put r 0 t := EX 0 0 X0 t,r 1 t := EX 1 0 X1 t. Corollary 2.2 Assume conditions of Theorem 2.1. Then (2.5) represents the decomposition of X t of (0.5) into the long memory part Xt 0 and the short memory part Xt 1,inthesense that rt 0 r t c 2 t 2 α q, rt 1 = O(t 1 q α ). (2.6) In particular, t=0 r1 t <. Proof. Let us show the second relation of (2.6). We have Xt 1 = a t...a t s ε s, and therefore for t 0 s S (t) r 1 t := EX 1 0X 1 t = s S ( t) Ea 0...a 1 t a 2 t...a 2 t s. Clearly under conditions of Theorem 2.1, as S ( t) S 1 (= S (0)) a.s., rt 1 r t, r t defined in the proof of Theorem 2.1 and satisfies rt 1 = O(t 1 q α ). The first relation of (2.6) also follows from the proof of Theorem

10 Proposition 2.3 Assume conditions of Theorem 2.1. Then EX 4 0 =. Proof. As X 0 = X0 0 + X1 0 and X0 0 and X1 0 are conditionally independent given A j,s j,j Z, it suffices to show E(X0 0)4 =. Then E[(X 0 0) 4 A j,s j,j Z] C ( S 1 1 i=0 ) 2 A 2i 0 with some constant C>0 depending on Eε 2i,i=1, 2 only. Hence E(X 0 0) 4 CE ( S 1 1 i=0 Hence by (2.1), (1.1) and Assumption U(α), E(X 0 0) 4 C C i,j=1 u α ) 2 A 2i 0 = C u (i + j) q i,j=1 u i+j u 1 i,j=0 (i + j) q u α C ν 2(i+j) P [S 1 = u]. (i + j) 1 q α = i,j=1 for q + α<3. 3 Convergence to a Lévy process (proof of Theorem 2) By Corollary 2.2, it suffices to show (0.7) with X t replaced by the long memory part Xt 0, i.e. n 1/λ Xs 0 = Z λ (t), (3.1) 0 s<[nt] as 0 s<n X1 t = O P (n 1/2 )and1/λ > 1/2. Note Xs 0 = Y0 0 + Y i + YN 0 [nt] 1 +1 (3.2) 1 i N [nt] 1 0 s<[nt] N t := max{j 0:S j t} is the number of renewal points in the interval [0,t], and Y i is the sum of the autoregressive process with parameter A i in the renewal interval [S i +1,S i ], more precisely ( ) Y i := ε s + A i ε s A s S i 1 1 i ε Si 1 +1, i =1, 2,... (3.3) 0 s S 0 S i 1 <s S i The two extreme terms on the r.h.s. of (3.2) are appropriate modifications of (3.3): Y0 0 := ( ( ε s +A 0 ε s A s 0ε 0 ), YN 0 [nt] 1 +1 := ε s +A 0 ε s A s 0ε 0 ), 10 S ([nt])<s<[nt]

11 and can be easily seen negligible in the limit (3.1), which then follows from N [nt] n 1/λ i=1 Y i = Z λ (t). (3.4) Observe the Y i s are conditionally independent given S j,a j,j Z and Law(Y i S j,a j,j Z) =Law(T (A i,u i )), (3.5) U i = S i S i 1 is the length of the renewal interval, as usual, and T (a, n) := n s=1 (ε s + aε s a s 1 ε 1 ). (3.6) As (A i,u i ),i 1 are independent, this implies that Y i,i 1 are independent (and identically distributed) r.v. s. Hence the l.h.s. of (3.4) is a sum of a random number N [nt] [nt]/µ of i.i.d. r.v. s. Our next aim is to show that the distribution of generic Y = Y i belongs to the domain of attraction of a symmetric λ stable law. To that end, we first study the tail behavior of the distribution of the conditional variance W i := E[Yi 2 A j,s j,j Z]. We have W i =Φ(A i,u i ), Φ(a, n) := ET 2 (a, n) = (a t s + a t s a t+s 2 ). (3.7) Put 1 t,s n 1 Θ(v) :=v 2 (e vτ e vτ )(e vτ e v )dτ, v > 0. Lemma 3.1 Under conditions of Theorem 2, 0 P [Φ(A, U) >x] c V x λ/2, x, (3.8) c V := { cu f 1 3 α/3 α 1, q =0, ( ) c U f 1 0 y 3 Θ(v) > 1, q > 0. dy y 1+α+q 0 dv v 1 q I (3.9) Proof. For simplicity of notation, assume c U = f 1 =1. Put Θ n (v) := 1 n 3 Φ ( 1 v n,n ), v 0. Observe by the Lebesgue dominated convergence theorem, Θ n (v) = 1 {( n 3 1 v ) t s ( + 1 v ) t s +2 ( v ) t+s 2 } n n n = Θ(v), 1 t,s n { t+s t s } e vτ dτ dtds 11

12 as n, the convergence being uniform on each compact interval v [0,K]. We note the bound Θ n (v) C/(1 + v) 2. (3.10) For v>1, this follows by writing Φ(a, n) = n k=1 (1 + a ak 1 ) 2 Cn/(1 a) 2, a =1 v/n. Then Φ(a, n) Cn 3 /v 2, implying (3.10). For v 1, (3.10) is obvious by Φ(a, n) n 3. To prove (3.8), consider first the case q>0. Then P [Φ(A, U) >x]=p 0 (x)+p 1 (x), p 0 (x) := P [Φ(A, U) >x,1 ɛ<a<1, U>n 0 ], p 1 (x) P [Φ(A, U) >x,a<1 ɛ]+p [Φ(A, U) >x,u n 0 ]. Here, n 0,ɛ 1 > 0 are large enough and p 0 (x) = l n n 1+α n>n ɛ da f 1 (a) I(Φ(a, n) >x), (1 a) 1 q l n c U =1(n ), f 1 (a) f 1 =1(a 1). By choosing n 0,ɛ appropriately, we may assume above l n = f 1 (a) 1. Then with c V (x) := p 0 (x) = n>n = n 0 n 1+α 1 dz [z] 1+α+q da I(Φ(a, n) >x) (1 a) 1 q dv ( 1 ( v 1 q I [z] 3 Φ 1 v ) [z], [z] > x ) [z] 3 1 ɛ ɛ[z] = c V (x)x (α+q)/3 = c V (x)x λ/2, 0 dy y 1+α+q ω dv 1(y; x) 0 v 1 q ω 2(v; x, y)i(y 3 Θ [x 1/3 y] (v) >ω 3(v; x, y)), ω 1 (y; x) := (x 1/3 y/[x 1/3 y]) 1+α+q I(x 1/3 y>n 0 ), ω 2 (v; x, y) := I(0 <v<ɛ[x 1/3 y]), ω 3 (v; x, y) := (x 1/3 y/[x 1/3 y]) 3, and ω 1 (y; x) 1,ω 2 (v; x, y) 1,ω 3 (v; x, y) 1 for any v, y > 0asx. Then c V (x) c V (x ) follows by the dominated convergence theorem. To justify the use of the latter, note ω 1,ω 2 C, ω 3 1 uniformly in all arguments. Then, using (3.10), c V (x) c V := C 0 dy y 1+α+q 12 0 dv v 1 q I(Cy3 /(1 + v) 2 > 1),

13 the last integral converges by 1 + α + q>1, 2(α + q)/3 >q.thus, p 0 (x) c V x λ/2, x. To finish the proof of (3.8) in the case q>0, we need to estimate the term p 1 (x), namely, to show p 1 (x) =o(x λ/2 ), x. Note Φ(a, n) Cn on a<1 ɛ and therefore P [Φ(A, U) >x,a<1 ɛ] P [U >C 1 x] Cx α = o(x (α+q)/3 ) as α>1 > (α + q)/3 =λ/2. A similar bound on the set U n 0 follows from the fact that Φ(a, n) is bounded for n n 0 and any a [0, 1]. If q = 0, then P [Φ(A, U) >x]=f 1 P [Φ(1,U) >x]+p [Φ(A, U) >x,a<1]. As Φ(1,n)=n(n + 1)(2n +1)/6 n 3 /3so P [Φ(1,U) >x] P [U 3 > 3x] =P [U >(3x) 1/3 ] (c V /f 1 )x α/3, x, with c V given in (3.9). It remains to show lim sup x α/3 P [Φ(A, U) >x,a<1] = 0. (3.11) x For any δ > 0 one can find δ > 0 such that P [1 δ < A < 1] < δ. Then using Φ(a, n) Φ(1,n) P [Φ(A, U) >x,1 δ <A<1] δp[φ(1,u) >x] Cδx α/3 according to the argument above. Finally, for any fixed δ > 0, sup 0 a 1 δ Φ(a, n) Cn/(1 δ ) 2 Cn and therefore P [Φ(A, U) >x,a 1 δ ] P [CU > x] Cx 1 EU = o(x α/3 ) as α>3. As δ>0 is arbitrary, this proves (3.11) and the lemma. If the conditional law (3.5) is Gaussian (i.e., the errors ε i s are Gaussian), Lemma 3.1 easily implies the regular tail behavior of Y i with exponent λ and the corresponding statement about the domain of attraction. However we do not assume conditional gaussianity of our random coefficient autoregressive process. In the general case, this tail behavior will follow from Lemmas 3.2 and 3.3 below. 13

14 Lemma 3.2 Let be given a r.v. Ỹ and a σ algebra F such that E[Ỹ F] =0, E[Ỹ 2 F] =: W >0, a.s., P [ W >u] c 0 u λ/2 (u ) (3.12) for some c 0 > 0, 1 <λ<2. Moreover, let the normalized r.v. Z( W ):=Ỹ W 1/2 satisfy the following asymptotic normality condition: there exists a nonrandom function δ(u),u > 0 with lim u δ(u) =0, such that sup P [Z( W ) x F] G(x) δ( W ), a.s. (3.13) x R G(x) =P [Z x] is the c.d.f. of standard normal Z N(0, 1). Then P [Ỹ > x] c 1x λ (x ), P[Ỹ x] c 1 x λ (x ), (3.14) c 1 := (c 0 /2)E Z λ. Proof. By definition, Ỹ = W 1/2 Z( W ) and therefore P [Ỹ > x]=e{p [Z( W ) >xu 1/2 F] u= W }. Let X := W 1/2 Z, Z N(0, 1) is independent of W. Then the lemma follows from P [ X >x] c 1 x λ (x ), P[ X x] c 1 x λ (x ). (3.15) and P [Ỹ > x] P [ X >x] = o(x λ ) (x ), (3.16) P [Ỹ x] P [ X x] = o( x λ ) (x ). (3.17) Relation (3.15) is well-known, see e.g. Breiman (1965). Let us show (3.16)-(3.17). We have P [ X >x]=e{p [Z >xu 1/2 ] u= W } and therefore P [Ỹ > x] P [ X >x] 3 d i (x), i=1 d 1 (x) := E{ P [Z( W ) >xu 1/2 F] u= W P [Z >xu 1/2 ] u= W I(x W 1/2 K)}, d 2 (x) := E{P [Z( W ) >xu 1/2 F] u= W I(x W 1/2 >K)}, d 3 (x) := E{P [Z >xu 1/2 ] u= W I(x W 1/2 >K)}. 14

15 Consider the last term. As P [Z >xu 1/2 ] u/x 2 so (x/k) 2 d 3 (x) x 2 E{ WI( W (x/k) 2 )} = x 2 wdp[ W >w] (x/k) 2 = x 2 (x/k) 2 P [ W >(x/k) 2 ]+x 2 P [ W >w]dw (C/K 2 λ )x λ. A similar bound for d 2 (x) is immediate by P [Z( W ) >xu 1/2 F] (u/x 2 )E[Z 2 ( W ) F] = u/x 2. Therefore, sup x>0 x λ (d 2 (x)+d 3 (x)) can be made arbitrary small by choosing K>0 large enough. Let us estimate d 1 (x). As lim u δ(u) 0, for any K<,δ 0 > 0 there exists u 0 = u 0 (K, δ 0 ) > 0 such that δ(u) <δ 0 /K λ for all u>u 0. Then by (3.13), sup P [Z( W ) >v F] P [Z >v] δ(u) δ 0 /K λ, a.s. on W >u0. v R Then for all x>u 1/2 0 K 2 large enough, d 1 (x) (δ 0 /K λ )P [ W (x/k) 2 ] Cδ 0/K λ (x/k) λ Cδ 0 x λ. Hence lim sup x x λ d 1 (x) Cδ 0, thereby proving (3.16). Relation (3.17) follows similarly. The following Lemma 3.3 is needed to verify condition (3.13) of Lemma 3.2 for Ỹ = Y = T (A, U), W = W =Φ(A, U) andf = σ{a j,s j,j Z}, witht (a, n), Φ(a, n) defined in (3.6), (3.7) respectively. 0 0 Lemma 3.3 sup sup P a [0,1] x R [ T (a, n) Φ 1/2 (a, n) x ] G(x) 0. (3.18) Proof. We use the elegant bound due to Berry and Esséen (see e.g. Petrov (1975)). Observe T (a, n) is a sum of independent r.v. s: T (a, n) =ε n + ε n 1 (1 + a)+...+ ε 1 (1 + a a n 1 ). Then the Berry -Esséen theorem yields [ T (a, n) ] sup P x R Φ 1/2 (a, n) x G(x) KE ε 3 n i=1 (1 + a ai 1 ) 3 Φ 3/2, (a, n) K is an absolute constant. The square of the r.h.s. clearly does not exceed (up to a constant (KE ε 3 ) 2 ) the quantity Γ n (a) := (13 +(1+a) (1+a a n 1 ) 3 ) 2 (1 2 +(1+a) (1+a a n 1 ) 2 ) 3. 15

16 Hence the lemma follows from sup Γ n (a) 0. a [0,1] According to Lemma 4.1 below, the function Γ n is strictly increasing in a for any n 1. Therefore sup Γ n (a) = Γ n (1) = ( n 3 ) 2 27n(n +1) a [0,1] ( n 2 3 = ) 2(2n +1) 3 0. The lemma is proved. From Lemmas 3.2 and 3.3 and the trivial bound n Φ(a, n) n 3 ( a [0, 1]) one obtains Corollary 3.4 Let Y be a copy of Y i (3.3). Then P [Y > x] c Y x λ (x ), P[Y x] c Y x λ (x ), c Y := (c V /2)E Z λ, with Z N(0, 1) and c V given in (3.9). Proof of Theorem 2 (continued). one has the weak convergence From Corollary 3.4 and the classical central limit theorem, [nt]/µ n 1/λ i=1 Y i = Z λ (t), (3.19) Z λ (t) is a symmetric Lévy process, with the characteristic function Ee iuz λ(1) = e c u λ, u R, (3.20) c := Γ(2 λ) cos(πλ/2)c V E Z λ /(λ 1), Z N(0, 1) and c V is given in (3.9). Clearly, relation (3.4) and the theorem follow from (3.18) and N [nt] [nt]/µ Q n (t) := Y i Y i = o P (n 1/λ ). (3.21) i=1 Let us prove (3.20). Assume µ = t = 1 for simplicity. Put Q n := Q n (1), F := σ{s j,a j,j Z}. WehaveP [ Q n >δn 1/λ ]=EP[ Q n >δn 1/λ F] and (3.20) follows from i=1 E[Q 2 n F] =o P (n 2/λ ). (3.22) As Y i,i 1 are conditionally uncorrelated given F and E[Yi 2 F] =W i =Φ(A i,u i ), so (3.21) follows from N n n R n := W i W i = o P (n 2/λ ). (3.23) i=1 i=1 16

17 By the renewal theorem, N n /n µ 1, n, a.s. (3.24) For any δ, δ 1 > 0, one has P [ R n >δn 2/λ ] = P [ R n >δn 2/λ, N n n >δ 1 n] + P [ R n >δn 2/λ,n(1 δ 1 ) N n (1 + δ 1 )n] =:β + β. As β P [ 2nδ 1 i=1 W i >δn 2/λ ], by Lemma 3.1 and the central limit theorem, for any δ, δ 2 > 0 one can find δ 1 > 0 sufficiently small so that β <δ 2 for all n large enough. Next, by (3.23), for any δ 1,δ 2 > 0 one can find n 0 > 0 such that β P [ N n n >δ 1 n] <δ 2 for any n>n 0 and all δ. This proves (3.22) and Theorem 2 as well. 4 Appendix Consider the function (1 3 +(1+a 1 ) (1+a 1 + a 1 a a 1 a 2...a n ) 3 ) 2 Γ n (a 1,...,a n ) := (1 2 +(1+a 1 ) (1+a 1 + a 1 a a 1 a 2...a n ) 2 ) 3, in real variables a i [0, 1], i=1,...,n. Lemma 4.1 The function Γ n (a 1,...,a n ) is strictly increasing in each a i [0, 1], i = 1,...,n. Proof. The lemma follows from Γ n (a 1,...,a n )/ a j > 0, (4.1) for each j =1,...,n. Fix j and let x := a j. Then 1 3 +(1+a 1 ) (1+a 1 + a 1 a a 1 a 2...a n ) 3 { = A 3 0 B +1+(1+A 1 x) 3 +(1+A 2 x) 3 +(1+A 3 x) (1+A n j+1 x) 3}, B := ( 1+(1+a 1 ) (1+a a 1...a j 2 ) 3) /A 3 0, A 0 := 1 + a 1 + a 1 a a 1...a j 1, A k := a 1...a j 1 (1 + a j a j+1...a j+k 1 )/A 0, k =1,...,n j

18 Similarly, 1 2 +(1+a 1 ) (1+a 1 + a 1 a a 1 a 2...a n ) 2 { = A 2 0 C +1+(1+A 1 x) 2 +(1+A 2 x) 2 +(1+A 3 x) (1+A n j+1 x) 2}, C := ( 1+(1+a 1 ) (1+a a 1...a j 2 ) 2) /A 2 0. Put also Then Σ i := n j+1 k=1 A i k, i =1, 2, 3. Γ n (a 1,...,a n ) = (B +2+n j +3Σ 1x +3Σ 2 x 2 +Σ 3 x 3 ) 2 (C +2+n j +2Σ 1 x +Σ 2 x 2 ) 3. We obtain Γ n / x (Σ 1 +2Σ 2 x +Σ 3 x 2 )(C +2+n j +2Σ 1 x +Σ 2 x 2 ) (Σ 1 +Σ 2 x)(b +2+n j +3Σ 1 x +3Σ 2 x 2 +Σ 3 x 3 ), the proportionality factor is strictly positive. Finally, Γ n / x [C B]Σ 1 + [Σ 2 (2C B +1)+(n j +1)Σ 2 Σ 2 1]x + [Σ 3 (C +1)+(n j +1)Σ 3 Σ 1 Σ 2 ]x 2 + [Σ 1 Σ 3 Σ 2 2]x 3. Now (4.1) follows from B C, (4.2) Σ 2 1 (n j +1)Σ 2, (4.3) Σ 1 Σ 2 (n j +1)Σ 3, (4.4) Σ 2 2 Σ 1 Σ 3. (4.5) Here, inequality (4.2), or 1+(1+a 1 ) (1+a a 1...a j 2 ) 3 (1+(1+a 1 ) (1+a a 1...a j 2 ) 2)( ) 1+a 1 + a 1 a a 1...a j 1 is obvious for any a i 0,i =1,...,j 1. The remaining inequalities (4.3) -(4.5) easily follow from the Cauchy -Schwarz and Hölder inequalities. The lemma is proved. 18

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