On Least Absolute Deviation Estimators For One Dimensional Chirp Model

Transcription

1 On Least Absolute Deviation Estimators For One Dimensional Chirp Model Ananya Lahiri & Debasis Kundu, & Amit Mitra Abstract It is well known that the least absolute deviation (LAD) estimators are more robust than the least squares estimators particularly in presence of heavy tail errors. We consider the LAD estimators of the unknown parameters of one dimensional chirp signal model under independent and identically distributed error structure. The proposed estimators are strongly consistent and it is observed that the asymptotic distribution of the LAD estimators are normally distributed. We perform some simulation studies to verify the asymptotic theory for small sample sizes and the performance are quite satisfactory. Key Words and Phrases: Chirp signals; least absolute deviation estimators; strong consistency, asymptotic distribution. Department of Mathematics and Statistics, Indian Institute of Technology Kanpur, Pin 0806, India. Corresponding author, Phone: , Fax:

2 Introduction Let us consider the following chirp signal model; y(n) = A 0 cos(α 0 n + β 0 n ) + B 0 sin(α 0 n + β 0 n ) + X(n); n =,,. () Here y(n) is the real valued signal observed at n =,,. A 0, B 0 are amplitudes, and α 0 and β 0 are frequency and frequency rate respectively. The additive error {X(n)} is a sequence of independent and identically distributed (i.i.d.) random variables with mean zero and finite second moment. The explicit assumptions on X(n)s will be provided later. In signal processing literature, chirp signal models are used to detect an object with respect to a fixed receiver. Such models are typically one-dimensional chirp model as described in (), where the dimension is usually the time. In this model, frequency varies with time in a non-linear fashion like a quadratic function and it is this property that has been exploited for measuring the distance of an object from a fixed receiver. In various areas of science and engineering, for example in sonar, radar and communications systems, such models are used. Oceanography and geology are some other areas where this model has been used quite extensively. On this model () or on its variations, extensive work has been done by several authors, see for example Abatzoglou (986), Kumaresan and Verma (987), Djuric and Kay (990), Gini, Montanari and Verrazani (000),Saha and Kay (00), andi and Kundu (004), Kundu and andi (008) and the references cited therein. andi and Kundu (004) first established the consistency and asymptotic normality property of the LSE of the one dimensional (D) chirp signal model for i.i.d. errors. The authors, see Kundu and andi (008), extended the results when X(n) s are obtained from a linear stationary processes. But there is no discussion about any method like least absolute deviation (LAD) estimation which is well known to be more robust than the LSEs, particularly in presence of outliers.

3 Unfortunately, the model does not satisfy the assumption B5 of Oberhofer (98) and therefore the strong consistency of the LAD estimators in this case is not immediate. It may be mentioned that even the ordinary sinusoidal model does not satisfy the assumption B5 of Oberhofer (98), and in that case Kim et al. (000) provided the consistency and asymptotic normality results of the LAD estimators. The main aim of this paper is to provide the consistency and asymptotic normality properties of the LAD estimators of the unknown parameters of model (). It is known that the LSE of α 0 has the convergence rate O p ( 3/ ), whereas the LSE of β 0 has the convergence rate O p ( 5/ ), see andi and Kundu (004). Here z = O p ( δ ) means z δ is bounded in probability. In this paper it is observed that the LAD estimators of α 0 and β 0 have the same rates of convergence as the corresponding LSEs. But it is observed that asymptotic efficiency of LAD estimators relative to LSE is 4f(0) σ, here f( ) is the probability density function (PDF) of the error random variable X(n). Therefore it is clear that LAD estimators are more efficient than LSEs for heavy tailed error distributions. We perform some extensive simulation experiments to study the effectiveness of the LAD estimators for finite samples, and it is observed that the performances of the LAD estimators are quite satisfactory. The rest of the paper is organized as follows. In Section, we mainly provide the model assumptions and methodology. In Section 3 the strong consistency and asymptotic normality of LAD estimators are provided. umerical results are presented in Section 4, and finally we conclude the paper in Section 5. Model Assumptions and Preliminary Results. Model Assumptions We make the following assumptions on the error random variables. 3

4 Assumption : The error random variable X(n) satisfies the following conditions; {X(n)} is a sequence of i.i.d. absolute continuous random variables with mean zero, variance σ, and has the PDF f( ). It is further assumed that f( ) is symmetric and differentiable in (0,ǫ) and ( ǫ, 0) for some ǫ > 0 and f(0) > 0. We use the following notations; F( ) the cumulative distribution function corresponds to f( ). The parameter vector θ = (A,B,α,β), the true parameter vector θ 0 = (A 0,B 0,α 0,β 0 ), and the parameter space Θ = [ M,M [ M,M [0,π [0,π. Assumption : It is assumed that θ 0 is an interior point of Θ.. Least Absolute Deviation Estimation Procedure In this section we propose the LAD estimation procedure to estimate the unknown parameters of the model (). The LAD estimators are obtained by minimizing Q(θ), with respect to θ, where, Q(θ) = we note that y(n) ( A cos(αn + βn ) + B sin(αn + βn ) ) () n= Q(A,B,α,β) > Q(Â(α,β), B(α,β),α,β) > Q(Â( α, β), B( α, β), α, β) where Â(α,β), B(α,β) are the minimizer of Q(A,B,α,β) for known α,β and Â( α, β), B( α, β) are the minimizer of Q(A,B, α, β). ow ( α, β) = arg minq(â(α,β), B(α,β),α,β). So, LAD estimators of θ 0 will be θ = (Â( α, β), B( α, β), α, β) = (Â, B, α, β). 4

5 3 Asymptotic Properties of Least Absolute Deviation Estimators 3. Strong Consistency ow we will provide the consistency results for the proposed estimators. Theorem. If the Assumptions - are satisfied then (Â, B, α, β) is a strongly consistent estimator of (A 0,B 0,α 0,β 0 ). We need the following results to prove Theorem. Lemma. If (θ,θ ) in (0,π) (0,π), t = 0,, then except for countable number of points the followings are true. (i) (ii) lim n= lim lim lim cos(θ n + θ n ) = lim t+ t+ t+ sin(θ n + θ n ) = 0. (3) n= n t cos (θ n + θ n ) = n= n t sin (θ n + θ n ) = n= (t + ) (4) (t + ). (5) n t sin(θ n + θ n ) cos(θ n + θ n ) = 0. (6) n= Proof: Using the result of Vinogradov (954) Lemma can be easily established. Lemma. If, D(θ) = Q(θ) Q(θ 0 ), then D(θ) lim E[ D(θ) 0 a.s.uniformly θ Θ. Proof: Let us denote W n (θ) = h n (θ)+x(n) X(n). Then D(θ) = W n (θ). n= We note that W n (θ) = h n (θ) + X(n) X(n) h n (θ) 4M, as the parameter 5

6 space is compact. Also W n (θ)s are independent and non identically distributed random variables with E[W n (θ) < and V [W n (θ) <. It may be easily seen similarly as in Oberhofer (98) that these bounds do not depend on n. Since Θ is a compact set, there exists Θ,, Θ K, such that Θ = K i=θ i and on each Θ i, sup W n (θ) inf W n (θ) < ǫ 4 n. a.s. ow for θ Θ i, = D(θ) lim E[ D(θ) [ [ W n (θ) E sup W n (θ) + n= = A(θ) + B(θ) n= where, A(θ) = W n (θ) n= n= n= sup[ W n (θ) n= n= sup W n (θ) n= E sup W n (θ) lim E[ D(θ) E sup W n (θ) n= E sup W n (θ) E sup W n (θ). ote that sup W n (θ) s are independent and non identically distributed random variables with finite mean and variance, and the variance is bounded by a quantity not depending on n. Applying Kolmogorov s strong law of large numbers, choose 0i large enough, so that for 0i, A(θ) < ǫ 3 a.s., uniformly θ Θ i. ow B(θ) = = n= n= = C(θ) + D(θ), E sup W n (θ) lim E[ D(θ) E sup W n (θ) E lim [ D(θ) using DCT where C(θ) = n= E sup W n (θ) E lim sup W n (θ), and DCT stands for dominated convergence theorem. We take U (θ) = n= n= E sup W n (θ) and we want to apply DCT to pass the limit inside the expectation and we get i 6

7 such that C(θ) < ǫ. Further, note that 3 D(θ) = E lim = E lim E lim lim n= sup W n (θ) E lim [ D(θ) sup W n (θ) E lim W n (θ) n= n= sup W n (θ) E lim n= ǫ 4 n = 0. n= n= inf W n (θ) Combining we get D(θ) lim E[ D(θ) 0 a.s.uniformly θ Θ. Lemma 3. The global minimum of lim E[ D(θ) is attained at θ0. Proof: At θ 0 the value of lim E[ D(θ) is zero, and for θ θ0, if we can show lim E[ D(θ) > 0 then we are through. To achieve that we verify the assumptions B7, B8, B9 of Lemma 4 by Oberhofer (98). For convenience we reproduce the assumptions B7, B8, B6 as A, A, A3 respectively, below. A: For every closed set Θ 0 not containing θ 0, there exist numbers c > 0, d > 0, 0 > 0 such that for all θ Θ 0 and all 0, {n : n 0, h n (θ) c} / d > 0. A: For every c > 0, there exists a real number d > 0, such that for all n min[f n (c) /, / F n ( c) d > 0 A3: There exists e > 0 and 0 such that for all 0, Q = inf Θ 0 h n (θ) min[f n (c) /, / F n ( c) e > 0 n= Lemma 4 of Oberhofer (98) states that A3 is fulfilled if A and A holds. ote that Lemma of Oberhofer (98) gives D(θ) Q. Then it is enough to show lim E[ D(θ) lim Q > 0. ow, lim Q > 0 condition is same as A3. Using Lemma 4 of Oberhofer (98) instead of A3 we try to show A and A. If f(0) > 0 7

8 then A is automatically satisfied. It remains to show that A is satisfied in our case. If there exists c > 0 such that inf h n (θ) c > 0 for all 0 then Θ 0 n= A will be satisfied. Let us consider Θ 0 = S c = {θ : θ θ 0 3c > 0} S A c S B c S (α,β) c (7) where S A c = {θ : A A 0 c > 0} {θ : A A 0 c, (α,β) = (α 0,β 0 )} {θ : A A 0 c, (α,β) (α 0,β 0 )}, S B c = {θ : B B 0 c > 0} {θ : B B 0 c, (α,β) = (α 0,β 0 )} {θ : B B 0 c, (α,β) (α 0,β 0 )}, S (α,β) c = {θ : (α,β) (α 0,β 0 ) c > 0} ow on the set {θ : A A 0 c, (α,β) = (α 0,β 0 )} Case-, if B B 0 = 0, then h n (θ) = (A A 0 ) cos(α 0 n + β 0 n ) and, lim inf n= h n (θ) = A A 0 lim inf A A 0 lim cos(α 0 n + β 0 n ) n= (cos(α 0 n + β 0 n )) n= = A A 0 c > 0 Case-, if B B 0 0, then h n (θ) = (A A 0 ) cos(α 0 n + β 0 n ) + (B B 0 ) sin(α 0 n + β 0 n ) = r cos(ω) cos(α 0 n + β 0 n ) + r sin(ω) sin(α 0 n + β 0 n ) for some r > 0, ω = r cos(α 0 n + β 0 n ω) 8

9 So, lim inf n= h n (θ) = r lim inf r lim = r > 0 cos(α 0 n + β 0 n ω) n= (cos(α 0 n + β 0 n ω)) n= On the set {θ : A A 0 c, (α,β) (α 0,β 0 )} h n (θ) = A cos(αn + βn ) + B sin(αn + βn ) A 0 cos(α 0 n + β 0 n ) B 0 sin(α 0 n + β 0 n ) = r cos(αn + βn ω) r 0 cos(α 0 n + β 0 n ω 0 ) for some r,r 0 > 0, ω,ω 0 We recall that h n (θ) 4M. Then 0 and r0 4M = R0 > 0. Then lim inf 4M lim inf n= = 4M lim = 4M R + R 0 ( ) hn (θ) 4M h n (θ) = 4M lim inf n= ( ) hn (θ) 4M h n (θ) 4M n= <. We denote r 4M = R > h n(θ) 4M [R cos(αn + βn ω) R 0 cos(α 0 n + β 0 n ω 0 ) n= > 0 Similarly on other sets lim inf θ θ 0. n= h n (θ) > 0. So, we get lim E[ D(θ) > 0 for Proof of Theorem : ow to prove the strong consistency of the LAD estimators, first let us observe that the minimizer of Q(θ) will be same as the minimizer of D(θ) = Q(θ) Q(θ 0 ). So we develop our result based on minimizer of D(θ) instead of Q(θ). ote that Q(θ) = y(n) ( A cos(αn + βn ) + B sin(αn + βn ) ) = h n (θ) + X(n) n= n= 9 (8)

10 where h n (θ) = A 0 cos(α 0 n+β 0 n )+B 0 sin(α 0 n+β 0 n ) A cos(αn+βn ) B sin(αn+βn ) and note that h n (θ) 4M for θ Θ and Q(θ 0 ) = shown that X(n). In Lemma we have D(θ) lim E[ D(θ) 0 a.s. uniformly θ Θ. [ and in Lemma 3 we have shown that θ 0 is the global minimizer of lim E D(θ). Therefore, by Lemma of Jennrich (969) or by Lemma. of White (980) we can conclude that minimizer of D(θ) is a strong consistent estimators of θ 0. n= 3. Asymptotic ormality ow we want to show that the estimators obtained have the following asymptotic ( ) normality result. Let us take D = diag,,, Theorem. If the Assumptions - are satisfied then here, Σ = A 0 + B 0 ) ( θ θ 0 )D d 4 (0, f(0) Σ (9) ( ) A 0 + 9B 0 4A 0 B 0 8B 0 5B 0 ( ) 4A 0 B 0 9A 0 + B 0 8A 0 5A 0 8B 0 8A , (0) 5B 0 5A d means converges in distribution, Proof: We recall that Q(θ) is not a differentiable function, to find the asymptotic distribution of θ, we want to approximate Q(θ) by Q(θ) with some nice property (differentiability). For that purpose we need to approximate x by some nice function ρ (x) near zero, such that lim ρ (x) = x. Let us consider the interval near 0

11 zero as ( γ, γ ) where γ is an increasing function of satisfying lim = 0. γ Let us approximate x by a polynomial. We want to approximate x separately in ( γ, 0) and (0, γ ). In each of these intervals we observe that the degree of the polynomial has to be at least 3 to make the approximating function twice continuously differentiable. If possible the degree of the polynomial is less than 3, say and it is P(x) = Ax + Bx + C. Then P ( γ ) = A should match with the second derivative of x at boundary point γ, which is zero. In that case A = 0 makes polynomial degree, if not then there will be a jump discontinuity at γ for the function P (x). So, let the approximating polynomial is P(x) = Ax 3 + Bx + Cx+D in (0, γ ). As x is symmetric about zero the approximating polynomial in ( γ, 0) will be P(x) = Ax 3 + Bx Cx + D. ow to find the coefficients of the polynomial we match the function value and its derivatives at the joining points. P( γ ) = γ gives P ( γ ) = gives P ( γ ) = 0 gives A γ 3 + B γ 3A γ and P (0) agrees from both parts of the polynomial giving + C γ + D = γ () + B γ + C = () 6A γ + B = 0 (3) C = 0. (4) Solving previous four equations we get the suitable cubic spline as ρ (x) = [ 3 γ x 3 + γ x + I + xi 3γ 0<x x> γ γ ρ ( x) = ρ (x) which is symmetric, twice continuously differentiable and γ is an increasing function of satisfying some extra conditions, = o(γ 3 ), γ = o() and <, γ =

12 which we will be needing later. After getting the nice function ρ (x) we now define Q(θ) = ρ (h n (θ) + X(n)) (5) n= and note that Q(θ 0 ) = n= ρ (X(n)). ow we want to prove the following two results (Lemma 4 and Lemma 5) which when combined will give the required asymptotic normality result. P Lemma 4. If the Assumptions - are satisfied then ( θ θ)d means convergence in probability. 0 where P Lemma 5. If the Assumptions - are satisfied, then θ, the minimizer of Q(θ) has the following asymptotic distribution ( θ θ 0 )D d 4 (0, f(0) Σ) To prove Lemma 4 and Lemma 5 we need some more lemmas. Lemma 6. sup( Q(θ) Q(θ)) = o P () and sup θ Θ θ Θ Q(θ) Q(θ) 0 a.s. where o P () means converges to zero in probability. Proof: To calculate the following quantity Q(θ) Q(θ). we write explicitly the function ρ (x) x. ρ (x) x = [ 3 γ x 3 + γ x x + 3γ [ + 3 γ x 3 + γ x + x + 3γ I 0<x γ I x 0 γ

13 and we note that ρ (x) x C γ. ow P( Q(θ) Q(θ) > ǫ) C γ = C γ = C γ = C γ E Q(θ) Q(θ) ǫ n= n= n= n= = C γ C 3 γ EI 0< h n(θ)+x(n) γ ( P 0 < h n (θ) + X(n) γ P( h n (θ) γ X(n) h n (θ) + γ ) F( h n (θ) + γ ) F( h n (θ) γ ) f( h n (θ)) using mean value theorem n= 0 as. So, we get Q(θ) Q(θ) = o P () and hence sup( Q(θ) Q(θ)) = o P () as Θ is compact. θ Θ Also we note that P( Q(θ) Q(θ) > ǫ) C 3 γ and ) P( Q(θ) Q(θ) > ǫ) C γ = = < implies Q(θ) Q(θ) 0 a.s. and hence sup θ Θ Q(θ) Q(θ) 0 a.s. Lemma 7. θ, the minimizer of Q(θ) is strong consistent estimator of θ 0. Proof: We take W n (θ) = ρ (h n (θ)+x(n)) ρ (X(n)). Then D(θ) = W n (θ) and D(θ) = Q(θ) Q(θ 0 ). As before θ is also minimizer of D(θ). Then we proceed with exactly same technique as that used for proving strong consistency of ˆθ and we finally get D(θ) lim E[ D(θ) 0 a.s.uniformly θ Θ. ow at θ 0 the value n= 3

14 of lim E[ D(θ) is zero. And for θ θ 0, lim E[ D(θ) = lim E[ D(θ) D(θ) + lim E[ D(θ) = lim E[ Q(θ) Q(θ) lim E[ Q(θ 0 ) Q(θ0 ) + lim E[ D(θ) The first two terms converge to zero, using Lemma 6 we get lim E[ D(θ) > 0 for θ θ 0. So, θ is strong consistent estimator of θ 0. Let us denote Q (θ) as the 4 first derivative vector and Q (θ) as the 4 4 second derivative matrix of Q(θ). To get explicit expressions of Q (θ) and Q (θ), let us write explicitly the functions ρ (x), ρ (x) and ρ (x). ρ (x) = + [ 3 γ x 3 + γ x + 3γ [ 3 γ x 3 + γ x + 3γ I + xi 0<x x> γ γ I xi x 0 x< γ γ ρ (x) = [ γx + γ x I + I 0<x x> γ γ + [ γx + γ x I I x 0 x< γ γ ρ (x) = [ γx + γ I + [ γ 0<x x + γ I x 0 γ γ Lemma 8. D Q (θ 0 )D converges to f(0)σ which is a positive definite matrix, in probability. Proof: First we note that Q (θ 0 ) depends on. Step- Let us calculate the quantity Eρ (X(n)). We want to show E[ ρ (X(n)) = f(0) + o() n= Also we recall that X(n) s are i.i.d. and have symmetric density function f with f(0) <. ow f is differentiable in (0, γ ) and ( γ, 0) for sufficiently large. 4

15 ote that in that case f is bounded in (0, γ ), say less than M. E[ρ (X(n)) = γ 0 γ = = 4 0 γ 0 [ γ x + γ f(x)dx + 0 [ γ x + γ f(x)dx [ γ x + γ f(x)dx Integration by parts gives this is equal to, [ 4 f(x) ( γx + γ )dx f (x)[ γ γ [ γ x + γ f(x)dx x + γ γ xdx 0 = 4[ f( γ ) R = f(0) + (f( γ ) f(0)) R where, [ R = 4 4M [ = 4M [γ f (x)[γ [γ x γ γ xdx x γ xdx x 3 6 γ 4M [ 6γ + γ γ x 0 0 γ 0 0 and (f( γ ) f(0)) 0 as. Step- ext we want to show V [ ρ (X(n)) = o(), using Step-, which is equivalent to [ E ρ (X(n)) = 4f (0) + o(). n= For variance calculation let us write the expression for [ρ (x). n= [ρ (x) = [ 4γ 4 x 8γ 3 x + 4γ I 0<x γ + [ 4γx 4 + 8γx 3 + 4γ I x 0 γ 5

16 which is an even function. Then, = E [ρ (X(n)) γ 0 γ = 8 0 γ 0 C 0 γ. [ 0 [ 4γ 4 x 8γx 3 + 4γ f(x)dx + 4γ 4 x + 8γx 3 + 4γ f(x)dx γ [ 4γ 4 x 8γx 3 + 4γ f(x)dx γ 4 x + γ dx Therefore, using the above and independence of X(n) we get E[ ρ (X(n)) = 4f (0) + o() n= Step-3: Let us consider the (,)-th element of lim D Q (θ 0 )D, which is lim ρ (X(n)) cos (α 0 n + β 0 n ), n= Using Step- and Step-, and by applying Chebychev s inequality we get B 0 0 B 0 3 D Q (θ 0 )D = P 0 A0 A0 3 f(0) B 0 A 0 +B 0 A 0 +B 0 + o P() = f(0)σ + o P () A0 B 0 A A 0 +B A 0 +B 0 5 Then D Q (θ 0 )D converges to f(0)σ, which is a positive definite matrix, in probability. Lemma 9. If θ is a function of X(),,X(), such that θ θ 0 a.s. as then, D[ Q ( θ) Q (θ 0 )D 0 a.s. Proof: To calculate D[ Q ( θ) Q (θ 0 ) let us consider [ρ (h n ( θ) + X(n)) ρ (X(n)). n= ow note that as, θ θ 0 a.s. This implies h n ( θ) 0 a.s. n as h n (θ) is a continuous function of θ which implies for fixed n, lim k P( n=k h n ( θ) < ǫ ) = ǫ. 6

17 ow h n ( θ) + X(n) X(n) a.s. n and ρ (.) is a continuous function, then given ǫ > 0 ǫ > 0 such that for fixed n, h n ( θ) < ǫ ρ (h n ( θ) + X(n)) ρ (X(n)) < ǫ. which implies, ρ (h n( θ) + X(n)) ρ (X(n)) 0 a.s. And using this fact we get, D[ Q ( θ) Q (θ 0 )D 0 a.s. Proof of Lemma 4 Step- By definition of θ, Q( θ) Q( θ) > 0. Adding to both sides, Q( θ) Q( θ), which is again > 0 by definition of θ, we get, Q( θ) Q( θ) + Q( θ) Q( θ) > Q( θ) Q( θ) (6) By Lemma 6 the left hand side of (6) is o P (). So is the right hand side. i.e. Q( θ) Q( θ) = o P () Step- ow by Taylor series expansion of Q around θ Q( θ) Q( θ) = ( θ θ) Q ( θ) + ( θ θ) Q (θ )( θ θ) T. By definition of θ, Q ( θ) = 0, So, Q( θ) Q( θ) = ( θ θ)d [D Q (θ )DD ( θ θ) T (7) where, θ is a point on line joining θ and θ. We note that ˆθ θ 0 a.s. and θ θ 0 a.s. Then θ θ 0 a.s. So, using Lemma 9 lim D Q (θ )D converges in probability to a positive definite matrix and that implies its minimum eigen value, say λ is strictly positive. By using step-, the left hand side of (7) is o P (). Then P which implies ( θ θ)d 0. ( θ θ)d D ( θ θ) T < o P() λ Lemma 0. [ Q (θ 0 )D d (0, Σ ). 7

18 Proof of Lemma 0 Q (θ 0 )D = cos(α 0 n + β 0 n )ρ (X(n)) n= sin(α 0 n + β 0 n )ρ (X(n)) n= n[a 0 sin(α 0 n + β 0 n ) B 0 cos(α 0 n + β 0 n )ρ (X(n)) n [A 0 sin(α 0 n + β 0 n ) B 0 cos(α 0 n + β 0 n )ρ (X(n)) n= n= For investigation about [ Q (θ 0 )D let us concentrate on ρ (X(n)). We note that Eρ (X(n)) = 0 as ρ (x) is an odd function and X(n) has symmetric density f around zero. This gives E[ Q (θ 0 )D = 0. ow to calculate V ρ (X(n)) let us consider the function [ρ (x). then [ρ (x) = [ γx 4 4 4γx γx I + I 0<x x> γ γ + [ γx γx γx I + I x 0 x< γ γ = + [ [ γ 4 x 4 4γ 3 x 3 + 4γ x I 0<x γ + [ γx γx γx I, x 0 γ V [ρ (X(n)) = E[ρ (X(n)) = + R 0, where for some constant C, R 0 C γ 0 as. Using above calculated variance, the elements of [ Q (θ 0 )D satisfies the conditions of the Central Limit Theorem by Fuller(996). [ To find the asymptotic variance of cos(α 0 n + β 0 n )ρ (X(n)) we need to calculate for h = 0, ±, ±, lim h n= n= cos(α 0 n + β 0 n ) cos(α 0 (n + h) + β 0 (n + h) ). Using Lemma, and after some calculations it can be shown that for h = 0 lim h n= cos(α 0 n + β 0 n ) cos(α 0 (n + h) + β 0 (n + h) ) = 8

19 and it is 0 otherwise. Therefore, using the Central Limit Theorem of linear processes, see Fuller (996, page 3), the variance turns out to be. To find the variance [ of sin(α 0 n + β 0 n )ρ (X(n)) we need to calculate the above limits where n= both the cos terms are replaced by sin terms, and we will get similar result using Lemma. ow for all h, and for t = 0,,, we also get lim h t+ n= n t cos(α 0 n + β 0 n ) sin(α 0 (n + h) + β 0 (n + h) ) = 0 and the variance-covariances of the other terms can be obtained along the same line, using these limits. Finally we get [ Q (θ 0 )D d (0, Σ ). Proof of Lemma 5: Using multivariate Taylor series expansion we have Q ( θ) Q (θ 0 ) = ( θ θ 0 ) Q ( θ) (8) where θ is a point on line joining θ and θ 0. Since, Q ( θ) = 0, (8) can be written as Q (θ 0 )D = ( θ θ 0 )D [D Q ( θ)d (9) ote that [ D Q [ ( θ)d = D Q ( θ) Q [ (θ 0 ) D + D Q (θ 0 )D Using Lemma 9 and Lemma 8 we get in probability as i.e., θ θ 0 a.s. (9) gives, [ D Q [ ( θ)d lim D Q (θ 0 )D = f(0)σ Using Lemma 0 we obtain [ Q (θ 0 )D ( θ θ 0 )D = [ Q (θ 0 )D[D Q ( θ)d (0) d (0, Σ ). So, from (0), combining above two observations we will get the asymptotic distribution of ( θ θ 0 )D. Dividing by the expression becomes ( θ θ 0 )( D) = [ Q (θ 0 )D[D Q ( θ)d. () Then ( θ θ 0 )( D) 0 in probability, () 9

20 Theorem 3. If the Assumptions - are satisfied, then ( θ θ 0 )(D ) 0 in probability. Proof: ote that Σ = 4Σ. As lim D Q (θ 0 )D = f(0)σ and [ Q (θ 0 d )D (0, Σ ) then by (0), ( θ θ 0 )D d 4 (0, Σ 4f(0) ) = 4 (0, Σ). We note that combining Lemma 5 and Lemma 4 we get ( θ θ 0 )D d f(0) 4 (0, f(0) Σ). Using Lemma 4 we get ( θ θ)( D) 0 in probability. This along with () gives ( θ θ 0 )( D) 0 in probability. 4 umerical Results and Data Analysis 4. umerical Results: In this section we perform some simulation experiments to see how the LAD estimators behave for different sample sizes. We consider the following model parameters: A =.0,B =.0,α =.75,β =.05. X(n) s are assumed to be i.i.d. Gaussian random variables with mean 0 and variance σ. We have taken different sample sizes namely n = 5, 50, 75, 00 and σ = for our simulation experiments. We compute the average estimates (MEA), mean squared errors (MSE), variance (VAR) over 000 replications, and we also provide the asymptotic variances (ASYV) for comparison purposes. We further calculate the asymptotic confidence length (ACO) and coverage probability (CP). To calculate LAD estimators we use the methodology given in Section 3.. umerically the minimum value has been obtained by using the Downhill Simplex Algorithm, see for example Press et al. (996). Some of the points are quite clear from the simulation experiments. It is observed that as sample size increases the MSEs, variances and the biases decrease. It verifies the consistency properties of the LAD estimators. The asymptotic variances of the LAD estimators and the MSE s of the different estimators obtained over 000 0

21 Table : The results for LADs are reported, when n = 5, σ = PARA MEA MSE (.0433) ( ) ( ) ( ) VAR ( ) (.35586) ( ) ( 0.040) ASYV ( ) ( ) ( ) ( ) ACO ( ) ( ) ( ) ( ) CP ( ) ( ) ( ) ( ) Table : The results for LADs are reported, when n = 50, σ = PARA MEA MSE ( ) ( ) ( ) ( ) VAR ( ) ( ) ( ) ( ) ASYV ( ) ( ) ( ) ( ) ACO (.7344) (.4938) ( ) ( ) CP ( ) ( ) ( ) ( ) replications are quite close to each other particularly for large sample sizes. So the performances of the LAD estimators are quite satisfactory. 5 Conclusion In this paper we consider the least absolute deviation estimators for parameters of one dimensional chirp signal. It is observed that the LAD estimators are strongly consistent. Also we found the joint asymptotic normal distribution of the estimators. It is observed that LAD estimators are more efficient than LSE s in presence of additive heavy tailed errors. Acknowledgement Part of this work of the first author has been financially supported by the Center for Scientific and Industrial Research (CSIR), and part of this work of the second and

22 Table 3: The results for LADs are reported, when n = 75, σ = PARA MEA MSE ( ) ( 0.05) ( ) ( ) VAR ( 0.649) ( 0.05) ( ) ( ) ASYV ( ) ( ) ( ) ( ) ACO (.77607) ( ) ( ) ( ) CP ( ) ( ) ( ) ( ) Table 4: The results for LADs are reported, when n = 00, σ = PARA MEA MSE ( ) ( ) ( ) ( ) VAR ( ) ( ) ( ) ( ) ASYV ( ) ( ) ( ) ( ) ACO (.55703) (.5475) ( ) ( ) CP ( ) ( ) ( ) ( ) third authors have been supported by a grant from the Department of Science and Technology, Government of India. The authors would like to thank the referees for their constructive comments, which had helped to improve the earlier version of the manuscript. References [ Abatzoglou, T. (986), Fast maximum likelihood joint estimation of frequency and frequency rate, IEEE Transactions on Aerospace and Electronic Systems, vol., [ Djuric, P.M. and Kay, S.M. (990), Parameter estimation of chirp signals, IEEE Transactions on Acoustics, Speech and Signal Processing, vol. 38, 8-6. [3 Fuller, W. A. (996), Introduction to statistical Time Series, second Edition, John Wiley and Sons, ew York.

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