Nonlinear regression model with continuous time and weak dependent or long-range dependent stationary noise is considered. Strong consistency suffient conditions of M-estimates of regression parameters are obtained

Ivanov, A.V.

Orlovsky, I.V.

Наукова електронна бібліотека періодичних видань НАН України (Vernadsky National Library of Ukraine)

Theory of Stochastic ProcessesVol.13 (29), no.1-2, 2007, pp.86-97ALEXANDER V. IVANOV AND IGOR V. ORLOVSKYCONSISTENCY OF M-ESTIMATES IN GENERALNONLINEAR REGRESSION MODELSNonlinear regression model with continuous time and weak depen-dent or long-range dependent stationary noise is considered. Strongconsistency sufficient conditions of M -estimates of regression param-eters are obtained.1. IntroductionConsider a regression modelX(t) = g(y(t), θ) + ε(t), t ≥ 0, (1)where g(y, τ) is a non random function defined on Y ×Θc, Θc is the closurein Rq of an open set Θ, Y ⊂ Rm is a compact region of regression ex-periment design. Borel function y(t) : [0,∞) → Y is a design of regressionexperiment, θ = (θ1, ..., θq) ∈ Θc is an unknown parameter. Let ε(t), t ∈ R1be a random process satisfying the assumptionA1. ε(t), t ∈ R1 is a real valued mean-square continuous measurablestationary process with zero mean on a complete probability space (Ω,, P ).We do not assume function g(y, θ) to be a linear form of coordinates ofthe vector θ.Definition 1. M-estimate of unknown parameter θ obtained by theobservations X(t), t ∈ [0, T ), of the type (1), is said to be any randomvector θ̂T that minimizes in τ ∈ Θc the functional MT (τ) = 1TT∫0ρ(X(t) −g(y(t), τ))dt with continuous risk function ρ : R1 → R1.The consistency property of M-estimates for nonlinear regression modelwith independent identically distributed observation errors is consideredin [1]. Some facts on consistency of the least squares estimates and leastmoduli estimates can be found in [2].Sufficient conditions for strong consistency of M-estimates of an un-known parameter θ of the model (1) with random noise that satisfies weakor long-range dependence conditions are presented in this paper.2000 Mathematics Subject Classifications. Primary 62J02; Secondary 62J99Key words and phrases. Consistency, M-estimates, nonlinear regression model.86CONSISTENCY OF M-ESTIMATES 872. Assumptions and the main resultsLet us impose some restriction on the random process ε(t), t ∈ R1.A2. ε(t), t ∈ R1 is a strictly stationary process, such that for someδ > 0 μ2+δ = E|ε(0)|2+δ < ∞ and∞∫0(α(r))δ2+δ dr < ∞,whereα(r) = supA∈σ(−∞,s], B∈σ[s+r,∞)|P (AB) − P (A)P (B)|,σ(a, b] is σ-algebra generated by random variables (r.v.) {ε(t), t ∈ (a, b]}.Definition 2. If for symmetric r.v. ξ the probabilities P{|ξ − b| < x},x ∈ [0,∞) are nonincreasing functions of the variable b ∈ [0,∞), then wesay that ξ is a symmetric and unimodal r.v..A3. ε(0) is a symmetric and unimodal r.v. with the distribution func-tion (d.f.) F (x).Let B be a σ-algebra of Borel subsets of Y . For any A ∈ BμT (A) = T−1m{t ∈ [0, T ] : y(t) ∈ A},where m is Lebesgue measure on [0,∞).Let Δg(y, τ) = g(y, θ) − g(y, τ) and vθ(ε) = {τ ∈ Rq : ‖τ − θ‖ < ε}.B1. The measures μT are weakly converge, as T → ∞, to some measureμ: μT =⇒ μ and for any ε > 0 μ {y ∈ Y : Δg(y, τ) = 0} < 1 for eachτ /∈ vθ(ε).Example. Assume {yi}i≥1 ⊂ Y to be some sequence and y(t) = yi, t ∈[i − 1; i), i = 1, 2, ... . Introduce the measureμT =1T[T ]∑i=1δyi +{T}Tδy[T ]+1,where [T ] and {T} are integer and fractional parts of T . Then, if 1nn∑i=1δyi ⇒μ as n → ∞, then μT ⇒ μ as T → ∞.Requirement on the measure μ in the condition B1 can be written asfollows: for any ε > 0 μ{y ∈ Y : g(y, τ) = g(y, θ)} > 0 for each τ /∈ vθ(ε).Suppose that the measure μ is absolutely continuous with respect toLebesgue measure l on Y , furthermore l(Y ) > 0 and μ has the density f(y)separated from zero: infy∈Y f(y) ≥ f∗ > 0. Thenμ{y ∈ Y : g(y, τ) = g(y, θ)} =∫{y∈Y : g(y,τ)=g(y,θ)}f(y)dy ≥88 ALEXANDER V. IVANOV AND IGOR V. ORLOVSKY≥ f∗l{y ∈ Y : g(y, τ) = g(y, θ)} > 0,if l{y ∈ Y : g(y, τ) = g(y, θ)} > 0. But the last inequality is the propertyof the regression function to distinguish parameters.Definition 3. Function J(·) : R1 → R1 is called symmetric, if thereexists some point b0 ∈ R1(which is called the center of symmetry) andsome function ϕ(·) : [0,∞) → R1 such that J(b) = ϕ(|b − b0|). If ϕ is amonotonically nondecreasing function and ϕ(x) > ϕ(0) for x > 0, then J iscalled unimodal and the center of symmetry is called the mode.Impose some restriction on risk function. Let Eρ(ε(t)) = Eρ(ε(0)) < ∞.C1. ρ(x) is continuous unimodal, with mode in zero, function such thatρ(0) = 0.C2. There exists c > 0 such that |ρ(x1) − ρ(x2)| ≤ c|x1 − x2| for anyx1, x2 ∈ R1.Assume alsoA4.∞∫0[P{|ε(0)| < z} − P{|ε(0)− b| < z}] dρ(z) > 0, b > 0.Note that from C1 it follows that ρ(x) is monotonically nondecreasingfunction in the region x ≥ 0. It means that Lesbegue-Stilties integral in A4exists. Moreover, from A3 it follows that the difference in square bracketsA4 is nonnegative.Theorem 1. Suppose that assumptions A1-A4, B1, C1 and C2 arefulfilled. Then M-estimate θ̂T → θ a.s. as T → ∞.To state the second result of the paper we need to introduce additionalcondition on ε(t).Definition 4. Stationary process ε(t), t ∈ R1 Eε(t) = 0 is called aprocess with long-range dependence ifEε(0)ε(t) = B(t) =L(|t|)|t|α , 0 < α < 1, (2)where L(t) : [0,∞) → [0,∞) is a slowly varying function (at infinity).A5. Gaussian random process ε(t), t ∈ R1 is a process with long-rangedependence, B(0) = 1.Theorem 2. Suppose that assumptions A1, A4, A5, B1, C1 and C2are fulfilled. Then M-estimate θ̂T → θ a.s. as T → ∞.3. Auxiliary assertionsSetδT (τ) = QT (τ) − EQT (τ), ΔT (τ) = QT (τ) − QT (θ).Definition 5. An unknown parameter θ is said to be identifiable, if forany ε > 0 there exist the numbers T0 = T0(ε) and δ = δ(ε) > 0 such thatEΔT (τ) > δ when T > T0 and τ /∈ vθ(ε).CONSISTENCY OF M-ESTIMATES 89Lemma 1. Assume that θ is identifiable parameter andsupτ∈Θc|δT (τ)| −→T→∞0 a.s., (3)then θ̂T −→ θ a.s. as T → ∞.Proof. Let us denote by Ω1 the event of the probability 1, for which thecondition (3) is fulfilled. For elementary events ω ∈ Ω1 from the definitionof the estimate θ̂T we haveΔT (θ̂T ) ≤ 0. (4)Suppose that for some fixed ω ∈ Ω1 θ̂T −→θ as T → ∞. It means thatthere exists some number ε0 > 0 and the sequence of numbers Tn ↑ ∞ asn → ∞ such that for n > n(ε0)∥∥∥θ̂Tn − θ∥∥∥ ≥ ε0. As for these Tn (4) alsoholds, then infτ /∈vθ(ε0) ΔTn(τ) ≤ 0.Let Tn ≥ T0(ε0) and for n > n(ε0) supτ∈Θc |δTn(τ)| < δ04 , where δ0 =δ(ε0). Then for n > n(ε0)0 ≥ infτ /∈vθ(ε0)ΔTn(τ) = infτ /∈vθ(ε0)(δTn(τ) + EΔTn(τ)) − δTn(θ)≥ infτ /∈vθ(ε0)δTn(τ) + infτ /∈vθ(ε0)EΔTn(τ) − δTn(θ)≥ infτ∈ΘcδTn(τ) + infτ /∈vθ(ε0)EΔTn(τ) − δTn(θ) >δ02.We obtain contradiction. Hence, for ω ∈ Ω1 θ̂T −→ θ as T → ∞. Introduce function J(b) = Eρ(ε(t) − b) = Eρ(ε(0) − b), b ∈ R1.The next lemma states sufficient conditions of identifiability of parame-ter θ.Lemma 2. An unknown parameter θ is identifiable if(i) for any ε > 0 there exist T0 = T0(ε) and x = x(ε) > 0 such that forany T > T0 and any τ /∈ vθ(ε) μT {y ∈ Y : |Δg(y, τ)| > x} > x;(ii) J(b) is unimodal;(iii) J(b) > J(0) for any b = 0.Proof. It is easily seen that under the conditions (ii) and (iii) the modeis in b = 0. Furthermore,EΔT (τ) =1TT∫0[J(Δg(y(t), τ)) − J(0)] dt. (5)Fix some ε > 0 and consider numbers T0 and x taken from the condition (i)of the Lemma. From the condition (ii) it follows that the right hand side of90 ALEXANDER V. IVANOV AND IGOR V. ORLOVSKYthe relation (5) permits the lower bound1TT∫0[J(Δg(y(t), τ)) − J(0)] dt ≥ (J(x) − J(0)) 1TT∫0χ(x,∞) (|Δg(y(t), τ)|)dt= (J(x) − J(0))μT {y ∈ Y : |Δg(y, τ)| > x} ,where χA(x) is the indicator of the set A.From (i) and (ii) it follows that in the definition of the identifiability ofparameter one can set δ = x(J(x) − J(0)). Further we formulate some sufficient conditions on the validity of Lem-mas 1 and 2.Lemma 3. If the assumption B1 holds, then the condition (i) of theLemma 2 fulfiles.Proof. Let ε > 0 be an arbitrary number. It is necessary to show thatthere exists some numbers T0 and x > 0 such thatμT {y ∈ Y : |Δg(y, τ)| > x} > x, T > T0, τ /∈ vθ(ε). (6)Assume that (6) does not hold. Then there exist some sequences Tn ↑ ∞as n → ∞ and τn ∈ Θc\vθ(ε) such thatμTn{y ∈ Y : |Δg(y, τn)| > n−1} ≤ n−1, n ≥ 1. (7)As the set Θc\vθ(ε) is compact, there exists some point τ ∗ ∈ Θc\vθ(ε) andthe sequence nk, k ≥ 1 such that τnk → τ ∗ as k → ∞.Let δ > 0 be an arbitrary fixed number. Then there exists some numberkδ such that for k > kδ, uniformly in y ∈ Y ,|Δg(y, τnk) − Δg(y, τ ∗)| ≤δ2. (8)Thanks to (8), for k > kδ{|Δg(y, τ ∗)| > δ} ⊂ {|Δg(y, τ ∗) − Δg(y, τnk)| + |Δg(y, τnk)| > δ}⊂{|Δg(y, τnk)| >δ2}. (9)Taking into account the inequality (7) for nk >2δone hasμTnk{y ∈ Y : |Δg(y, τnk)| >δ2}≤ 1nk. (10)Then, from (9) and (10) it follows thatμTnk {y ∈ Y : |Δg(y, τ ∗)| > δ} ≤ n−1k , (11)CONSISTENCY OF M-ESTIMATES 91which is true for any k > k′δ = max(kδ, min{k : nk >2δ}).Denote by Yδ = {y ∈ Y : |Δg(y, τ ∗)| ≤ δ}. From (11) it follows thatμTnk (Yδ) > 1 − n−1k for all k > k′δ.As Yδ is a closed set, then thanks to weak convergence of μT to themeasure μ, we obtain (see, for example, [3], p. 21)limk→∞μTnk (Yδ) ≤ μ(Yδ), δ > 0.For δ ↓ 0, from the continuity of the measure μ it follows thatμ{y ∈ Y : Δg(y, τ) = 0} = 1. (12)But the relation (12) contradicts to the condition B1. Lemma 4. If the assumptions A3, A4 and C1 hold, then the conditions(ii) and (iii) of the Lemma 2 are fulfilled.Proof. Without loss of generality, assume that ρ(x), x ≥ 0 is strictlymonotonically increasing function. From the formula for the mean of thenonnegative r.v. (see, for example, [4], p. 190) one hasJ(b) − J(0) =∞∫0(P{ρ(ε(0)) < x} − P{ρ(ε(0) − b) < x}) dx =∞∫0(P{−ρ−1(x) < ε(0) < ρ−1(x)} − P {−ρ−1(x) < ε(0) − b < ρ−1(x)}) dx,where ρ−1(x) is the inverse of the function ρ(x), x ≥ 0.By the change of variable x = ρ(z), z ≥ 0 in the last integral,J(b) − J(0) =∞∫0(P {|ε(0)| < z} − P {|ε(0) − b| < z}) dρ(z) ==∞∫0(F (z) − F (z − b) − F (z + b) + F (z)) dρ(z), (13)where F (x) is the d.f. of the r.v. ε(0).The integral in the first equality of the relations (13) coincides with theexpression of A4, and the condition (iii) of Lemma 2 is fulfilled.From the symmetry of ρ and r.v. ε(0) it follows the symmetry of J(b).Denote by Δ2bF (z) = (F (z) − F (z − b)) − (F (z + b) − F (z)) , b, z ≥ 0.Then A4 can be rewritten in the form∞∫0Δ2bF (z)dρ(z) > 0, b > 0.92 ALEXANDER V. IVANOV AND IGOR V. ORLOVSKYFrom (13) it follows thatΔ2bF (z) = P{|ε(0)| < z} − P{|ε(0)− b| < z}.Consider for b2 > b1 the differenceJ(b2) − J(b1) =∞∫0(Δ2b2F (z) − Δ2b1F (z))dρ(z).It is easily seen thatΔ2b2F (z) − Δ2b1F (z) = P{|ε(0)− b1| < z} − P{|ε(0)− b2| < z} ≥ 0from the unimodality of the r.v. ε(0). It means that J(b2)− J(b1) ≥ 0, andthe condition (ii) of Lemma 2 is a corollary of A3 and C1. Assume that the d.f. F (x) is continuously differentiable and the densityof the distribution p(x) is an even strictly decreasing for x ≥ 0 function.Suppose that a continuous even function ρ(x) is such that ρ(0) = 0 andstrictly monotonically increasing for x ≥ 0. Then one can use Lemma 10.2of the book [3], p. 217-218, and for any b = 0J(b) − J(0) = Eρ(ε(0) − b) − Eρ(ε(0)) > 0,and the integral in A4 is strictly positive.Consider next sufficient conditions of the uniform convergence in (3) ofLemma 1.Lemma 5. Suppose the condition C2 fulfiles andδT (τ) −→T→∞0 a.s., τ ∈ Θc, (14)then (3) holds.Proof. From C2 it follows that for τ1, τ2 ∈ Θc|QT (τ1) − QT (τ2)| ≤ cTT∫0|g(y(t), τ1) − g(y(t), τ2)|dt.Similarly, from C2 for τ1, τ2 ∈ Θc one has|δT (τ1) − δT (τ2)| ≤ 2cTT∫0|g(y(t), τ1) − g(y(t), τ2)|dt.Hence, the family of functions {δT (τ) : ω ∈ Ω, T > 0} is an equicontin-uous on the set Θc. So for any δ > 0 there exists a finite number of pointsτ1, ..., τk ∈ Θc such thatsupτ∈Θc|δT (τ)| ≤ max1≤j≤k|δT (τj)| + δ, ω ∈ Ω, T > 0.From (14) it follows that max1≤j≤k |δT (τj)| −→ 0 a.s. as T → ∞, and,hence, supτ∈Θc |δT (τ)| −→ 0 a.s. as T → ∞. CONSISTENCY OF M-ESTIMATES 934. Proof of Theorem 1We shall prove that (14) holds under the assumptions of Theorem 1.Using the notationξ(t) = ρ(ε(t) − Δg(y(t), τ)) − Eρ(ε(t) − Δg(y(t), τ)), τ ∈ Θcone hasδT (τ) =1TT∫0ξ(t)dt, Eδ2T (τ) =1T 2T∫0T∫0Eξ(t)ξ(s)dtds ≤≤ 10T 2T∫0T∫0[Eρ2+δ(ε(t) − Δg(y(t), τ))] 12+δ ×× [Eρ2+δ(ε(s) − Δg(y(s), τ))] 12+δ α δ2+δ (|t − s|)dtds. (15)To obtain (15) the Davidov inequality has been used with p = q = 2+δ, r =1 + 2δ(see [5], and also Lemma 1.6.2 of the book [6]).As ρ(0) = 0, then from the condition C2 one obtainsEρ2+δ (ε(t) − Δg(y(t), τ)) ≤ c2+δE |ε(0) − Δg(y(t), τ)|2+δ .By obvious inequalities|a + b|κ ≤ 2κ−1 (|a|κ + |b|κ) , |a + b| 1κ ≤ |a| 1κ + |b| 1κ , κ = 2 + δ, (16)[Eρ2+δ (ε(t) − Δg(y(t), τ))] 12+δ ≤ 2 1+δ2+δ c (μ 12+δ2+δ + |Δg(y(t), τ)|) ,i.e.Eδ2T (τ) ≤ 2δ2+δ c220T 2T∫0T∫0αδ2+δ (|t − s|)[μ12+δ2+δ + |Δg(y(t), τ)|]××[μ12+δ2+δ + |Δg(y(s), τ)|]dtds ≤≤ 2 δ2+δ c2 20T 2T∫0T∫0αδ2+δ (|t − s|)[μ12+δ2+δ + |Δg(y(t), τ)|]2dtds.Using the first inequality of (16) with κ = 2,Eδ2T (τ) ≤ 2δ2+δ c240T 2T∫0T∫0αδ2+δ (|t − s|)[μ22+δ2+δ + |Δg(y(t), τ)|2]dtds.94 ALEXANDER V. IVANOV AND IGOR V. ORLOVSKYIt remains to estimate two integrals, namely:I1 =1T 2T∫0T∫0αδ2+δ (|t − s|)dtds ≤ 1T 2T∫0dsT∫−Tαδ2+δ (|t|)dt = O(T−1)as T → ∞, under assumption A2. On the other hand,I2 =1T 2T∫0T∫0αδ2+δ (|t − s|)|Δg(y(t), τ)|2dtds≤⎛⎝2 ∞∫0αδ2+δ (s)ds⎞⎠ 1T 2T∫0|Δg(y(t), τ)|2dt. (17)As g(y, τ) is continuous function on the compact Y ×Θc , the right handside of the inequality (17) is of the order O(T−1) as T → ∞.Thus, Eδ2T (τ) = O(T−1) as T → ∞, and δT (τ) −→ 0 in probability asT → ∞ .Note that for the sequence Tn = n2, n ≥ 1∞∑n=1Eδ2Tn(τ) < ∞, i.e.δTn(τ)−→n→∞ 0 a.s.If T ∈ [Tn, Tn+1], then|δT (τ)| ≤ supTn≤T≤Tn+1|δT (τ) − δTn(τ)| + |δTn(τ)| ,and the Theorem will be proved, if supTn≤T≤Tn+1|δT (τ) − δTn(τ)| −→n→∞ 0 a.s.ObviouslyδT (τ) − δTn(τ) =1TT∫0ξ(t)dt− 1TnTn∫0ξ(t)dt ==(1T− 1Tn) Tn∫0ξ(t)dt +1TT∫Tnξ(t)dt = I3 + I4.Furthermore, for T ∈ [Tn, Tn+1]|I3| ≤ Tn+1 − TnTn|δTn(τ)| −→n→∞0 a.s.;|I4| ≤ 1TnTn+1∫Tn|ξ(t)|dt ≤ 1TnTn+1∫Tnρ (ε(t) − Δg(y(t), τ)) dt+CONSISTENCY OF M-ESTIMATES 95+1TnTn+1∫TnEρ (ε(t) − Δg(y(t), τ)) dt = I5 + I6.As under the Lipshits condition C2ρ (ε(t) − Δg(y(t), τ)) ≤ c (|ε(t)| + |Δg(y(t), τ)|) ,thenI5 ≤ cTnTn+1∫Tn|ε(t)|dt + cTnTn+1∫Tn|Δg(y(t), τ)|dt = I7 + I8,I8 = c⎛⎝Tn+1Tn· 1Tn+1Tn+1∫0|Δg(y(t), τ)|dt− 1TnTn∫0|Δg(y(t), τ)|dt⎞⎠ .From the assumption B1 of the Theorem it follows1TnTn∫0|Δg(y(t), τ)|dt =∫Y|Δg(y, τ)|μTn(dy) −→n→∞∫Y|Δg(y, τ)|μ(dy),then I8 −→ 0 as n → ∞.On the other hand,I7 = c⎛⎝ 1TnTn+1∫Tn(|ε(t)| − E|ε(t)|) dt + E|ε(0)|Tn+1 − TnTn⎞⎠ −→n→∞0 a.s.by Davidov inequality.Similarly, it can be shown that I6 −→ 0 as n → ∞.Consequently, (14) is fulfilled. The validity of Theorem 1 follows nowfrom the Lemmas 1-5 proved above. 5. Proof of Theorem 2Similarly to proof of Theorem 1 we need to proof that (14) holds. Thenthe result of Theorem 2 will follow from the Lemmas 1-5.Consider a random processG(ε(t), t) = ρ(ε(t) − Δg(y(t), τ)). (18)From C2 and A5EG2(ε(t), t) ≤ c2E |ε(t) − Δg(y(t), τ)|2 = c2 (1 + |Δg(y(t), τ)|2) ≤ C < ∞(19)96 ALEXANDER V. IVANOV AND IGOR V. ORLOVSKYuniformly in t ≥ 0 and τ ∈ Θc. Therefore in Gilbert space L2(R1, ϕ(u)du),where ϕ(x) = 1√2πe−x22 is a standard Gaussian density, there exists an ex-pansion (see, for example, [6])G(u, t) =∞∑m=0Cm(t)m!Hm(u), Cm(t) =∫R1G(u, t)Hm(u)ϕ(u)du, m ≥ 0by Chebyshev-Hermite polynomialsHm(u) = (−1)meu22dmdume−u22 , m ≥ 0. (20)Note that C0(t) = Eρ(ε(0) − Δg(y(t), τ)) = J (Δg(y(t), τ)).Thanks to relationsEHm(ε(t))Hk(ε(s)) = δkmm!Bm(t − s), (21)where δkm is Kroneker delta we haveEξ(t)ξ(s) = cov (G(ε(t), t), G(ε(s), s)) =∞∑m=1Cm(t)Cm(s)m!Bm(t − s).Hence, taking into account that B(0) = 1, we obtainEδ2T (τ) =∞∑m=11m!1T 2T∫0T∫0Cm(t)Cm(s)Bm(t − s)dtds≤∞∑m=11m!1T 2T∫0T∫0C2m(t)Bm(t − s)dtds≤ 1T 2T∫0T∫0( ∞∑m=1C2m(t)m!)B(t − s)dtds.Note that, thanks to (19),∞∑m=1C2m(t)m!= EG2(ε(0), t) − (EG(ε(0), t))2 = DG(ε(0), t) ≤ C < ∞,andEδ2T (τ) ≤ C1T 2T∫0T∫0B(t − s)dtds.CONSISTENCY OF M-ESTIMATES 97On the other hand, as T → ∞,1T 2T∫0T∫0B(t−s)dtds =1∫01∫0B (T (t − s)) dtds = 1T α1∫01∫0L (T |t − s|)|t − s|α dtds∼⎛⎝ 1∫01∫0dtds|t − s|α⎞⎠ L(T )T α=2(1 − α)(2 − α)L(T )T αby the properties of the slowly varying function (see, for example [7],[8]).For the sequence Tn = n1α+ν , where ν > 0 is some number,∞∑n=1L(Tn)T αn<∞, and so δTn(τ) −→ 0 a.s., as n → ∞.Taking into account the proof of Theorem 1, it remains to show that1TnTn∫0(|ε(t)| − E|ε(t)|) dt −→n→∞0 a.s. (22)But the proof of (22) is similar to the previous reasoning for G(ε(t), t).Bibliography1. F. Liese and I. Vajda Consistency of M -estimates in general regressionmodels, Journal of Multivariate Analysis, 50, 1, (1994), 93-114.2. A. V. Ivanov Asymptotic Theory of Nonlinear Regression, Kluwer Aca-demic Publishers, Dordrecht, (1997).3. Ibrahimov I.A. and Hasminsky R.Z. Asymptotic theory of estimation, Nau-ka, Moskow, (1979). (in Russian)4. Feller V. Introduction to probability theory and its applications, Mir, Mos-kow, volume 2, (1967). (in Russian)5. Davidov U.A. About convergence of the distributions, generated by station-ary random processes, TVIP, Vol. 15, 3, (1970), 498-509.(in Russian)6. Ivanov A.V. and Leonenko N.N. Statistical analysis of random fields, Vys-cha Shkola, Kyiv,(1986). (in Russian)7. Seneta E. Regularly varying function, Nauka, Moskow, (1985). (in Russian)8. Ivanov A.V. and Orlovsky I.V. Lp-Estimates in Nonlinear Regression withLong-Range Dependence, Theory of Stochastic Processes, Vol. 7(23), . 3-4,(2002), 38-49.National technical university of Ukraine, ”KPI”. Peremogi avenue37, KievE-mail address: ivanov@paligora.kiev.uaNational technical university of Ukraine, ”KPI”. Peremogi avenue37, KievE-mail address: avalon@ln.ua

Consistency of M-estimates in general nonlinear regression models

http://dspace.nbuv.gov.ua/xmlui/bitstream/123456789/4479/1/2007_13_1-2_9.pdf

Consistency of M-estimates in general nonlinear regression models

Abstract

Similar works

Full text

Available Versions

Наукова електронна бібліотека періодичних видань НАН України (Vernadsky National Library of Ukraine)