Research Journal of Mathematics and Statistics 6(1): 1-5, 2014 ISSN: 2042-2024, e-ISSN: 2040-7505 © Maxwell Scientific Organization, 2014 Submitted: September 18, 2013 Accepted: November 23, 2013 Published: February 25, 2014 Estimate by the L2 Norm of a Parameter Poisson Intensity Discontinuous Demba Bocar Ba UFR SET-Université de Thiès BP 967, Senegal Abstract: The model considered has two types of discontinuities. The first parameter is a parameter of scale; the second is a parameter of translation. The aim of study is to show that the minimum distance estimator is consistent and asymptotically normal. The problem of estimation studied in this work is the same time singular and regular. It is singular because the intensity is a discontinuous function and we obtains the consistency with the discontinuity and the study of asymptotic normality is based on the regularity of the function Λ(ππ, π‘π‘). Keywords: Asymptotic properties, minimum distance estimator, parameter estimation, Poisson process, singular estimation INTRODUCTION We also studied the asymptotic behaviour of the maximum likelihood estimator and the Bayesian estimator (Ba et al., 2008). The present study is devoted to the dimensional case i.e., we suppose that the intensity function ππ(ππ, π‘π‘) depends to 2 dimensional parameter ππ = (πΎπΎ1 , πΎπΎ2 ). Our objective is to study the Minimum Distance Estimator (MDE) built from the norm L2. We show that the MDE is consistent and asymptotically normal. We studied the proprieties of the estimators using other norms and others models. Moreover in the case not hilbertien, it was shown in the studies of Kutoyants and Liese (1992) that we cannot have the asymptotic normality. In many areas of everyday life we use the inhomogeneous Poisson processes; there exists a large literature describing the different point processes and particularly Poisson with special attention to parameter estimation problems. The properties of the estimators (Maximum Likelihood (MLE), Bayesian (BE), Minimum Distance (MDE) are usually studied in the case of smooth w.r.t., parameter situation i.e., the intensity function has one or two derivative, w.r.t. unknown parameter. According to general theory all these estimators in regular (smooth) case are consistent and asymptotically normal see, Kutoyants (1998). Moreover, the first two estimators are asymptotically efficient in usual sense. In non regular case, when, say intensity function is discontinuous, the properties of estimators are quite different. Remember that the MDE is a particular case of the minimum contrast estimator if we consider the function ||X() − Λ()|| as a contrast see Le (1972) for details. We have several possibilities of the choice of the space H. In the case H = L2 (µ) the measure µ can also be chosen in different ways continuous, discrete, etc. Others definitions of the MDE can also be realized. Note that the asymptotic behavior of the estimator depends strongly of the chosen metric. We are mainly interested in the properties of the MDE in the case H = L2 (µ). Particularly we show that under regularity conditions the MDE are consistent, asymptotically normal. For another metric this estimator is also consistent but its limit distribution is not Gaussian (Kutoyants and Liese, 1992). We did a study on Regular properties of the MDE for Non Smooth Model of Poisson Process (Ba and Dabye, 2009). STATISTICAL MODEL We observe ππ trajectories ππππ = (ππππ (π‘π‘), π‘π‘ ∈ [0, ππ]) ππ = 1, . . . , ππ of the Poisson process with intensity function ππ(ππππ , π‘π‘) = ππ(ππ1 π‘π‘ + ππ2 ) + ππ. Here ππππ = (ππ1 , ππ2 ) in the unknown parameter, which we have to estimate, the function ππ and the constant ππ are known and positive: π‘π‘ πΈπΈππππ ππππ (π‘π‘) = Λ(ππππ , π‘π‘) = ∫0 ππ(ππππ , π π )ππππ with ππ(ππππ , π‘π‘) = ππ(ππ1 π‘π‘ + ππ2 ) + ππ Hypothesis A: • • • 1 The function ππ( ) and the constant ππ known and positive. The function ππ( ) is continuously differentiable on [ππ1 , ππ1∗ ) ∪ (ππ1∗ , ππ2∗ ) ∪ (ππ2∗ , π½π½1 ππ + π½π½2 )], we suppose that π½π½2 < πππΌπΌ1 + πΌπΌ2 . The function ππ( ) have two jumps at ππ1∗ , ππ2∗ ∈ (π½π½2 , πππΌπΌ1 + πΌπΌ2 ) with ππ1∗ < ππ2∗ and ππ(ππππ∗ +) − ππ(ππππ∗ −) = ππππ ≠ 0 ππ = 1,2. Res. J. Math. Stat., 6(1): 1-5, 2014 This leads to ππ1 = ππ ′1 et ππ2 = ππ ′ 2 i.e., ππππ = ππ ′ ππ which contradicts the fact that ππππ ≠ ππ ′ ππ . So the lemma is proved. We define this below the minimum distance estimator of ππππ and we are intereste to their properties when ππ βΆ ∞. Let πΏπΏ2 ([0, ππ]) the Hilbert space with the ππ Demonstration of the theorem: We put ππππ (π‘π‘) = 1 ππ ∑ ππ (π‘π‘) and ππππ (π‘π‘) = √ππ (ππππ (π‘π‘) − Λ(ππππ , π‘π‘))which ππ ππ =1 ππ allows to write for all πΏπΏ > 0: 1/2 ||β( )|| = οΏ½∫0 β2 (π‘π‘)οΏ½ , then we define the minimum distance estimator as solution of the equation: 1 1 οΏ½οΏ½ ∑ππππ=1 ππππ − Λ(ππππ∗ ,⋅)οΏ½οΏ½ = ππππππ οΏ½οΏ½ ∑ππππ=1 ππππ − Λ(ππ,⋅)οΏ½ | ππ (ππ) STUDY OF THE CONSISTENCY |ππ −ππππ |≤πΏπΏ (ππ) The following theorem ensures the consistency of the MDE. ≤ ππππππ οΏ½2||ππππ ( )|| > √ππππ(πΏπΏ, ππππ )οΏ½ where, we have used the properties of the norm: Theorem 1: If the Hypothesis π΄π΄ is satisfied, then the estimator of the minimum distance ππππ∗ is consistent: for all πΏπΏ > 0: ππππ (ππ ) {ππππ∗ − ππππ | > πΏπΏ} ≤ 4 ππ ππ ∫0 π¬π¬(ππππ ,π‘π‘)ππππ ππ ππ(πΏπΏ,ππππ )2 οΏ½|ππππ ( ) − Λ(ππππ )|οΏ½ + οΏ½|Λ(ππππ ,⋅) − Λ(ππ, )|οΏ½ ≥ ||ππππ ( ) − Λ(ππ,⋅)|| ||ππππ ( ) − Λ(ππππ , )|| − ||Λ(ππππ ,⋅)|| ≤ ||ππππ ( ) − Λ(ππ,⋅)|| βΆ0 Using the lemma ππ(πΏπΏ, ππππ ) > 0. so, applying the inequality of Chebychev it follows: with ππ(πΏπΏ, ππππ ) = inf|ππ−ππππ |>πΏπΏ ||Λ(ππ,⋅) − Λ(ππππ ,⋅)|| (ππ) For the demonstration of this theorem, we need the following lemme. ππππ {2||ππππ ( )||√ππ ππ(πΏπΏ, ππππ ) ≤ Because, Lemme: If the Hypothesis π΄π΄ is satisfied, then for all πΏπΏ > 0, the function ππ(πΏπΏ, ππππ ) > 0. ππ ππ 1 √ππ βΆ0 2 ∑ππ1 ( ππππ (π‘π‘) − Λ(ππππ , π‘π‘)οΏ½ ππππ = ∫0 ( Λ(ππππ , π‘π‘)ππππ The latter term being bounded, which demonstrated the theorem. Asymptotic normality: In this section we will to prove the Asymptotic Normality of MDE. We define the functions πΉπΉ(ππ, π‘π‘) et πΊπΊ(ππ, π‘π‘) for all π‘π‘ ∈ [0, ππ] as: Λ(ππ ′ ππ , π‘π‘) = Λ(ππππ , π‘π‘) since Λ(ππ, π‘π‘) is continuous function in π‘π‘. ππ ππ We deduce that Λ(ππ ′ ππ , π‘π‘) = Λ(ππππ , π‘π‘) for all ππππ ππππ π‘π‘ ∈ [0, ππ] i.e.: πΉπΉ(ππ, π‘π‘) = − ππ2) = ππ(ππ ′1 π‘π‘ + ππ ′ 2 ) + ππ = ππ(ππ ′1 π‘π‘ + ππ ′ 2 ) 1 ππ1 ππ οΏ½∫0 ππ (ππ1 π π + ππ2 )ππππ − π‘π‘ ππ(ππ1 π‘π‘ + et πΊπΊ(ππ, π‘π‘) = ππ(ππ1 π‘π‘ + ππ2 ) − ππ(ππ2 ) Let us notice that the functions π π (ππ, π‘π‘) et πΊπΊ(ππ, π‘π‘) are respectively the partial derives formal of the function Λ(ππ, π‘π‘). ππ Let us put π΄π΄1,1 (ππ) = ∫0 πΉπΉ(ππ, π‘π‘)2 ππππ: ππ1 π‘π‘π π 1 + ππ1 = ππ1 ππ1 π‘π‘π π 2 + ππ2 = ππ2 we have: π‘π‘π π οΏ½ 1 π‘π‘π π 2 ππ = ∫0 πΈπΈππππ οΏ½ so for all π‘π‘ ∈ [0, ππ]: ππ1 ππ 1 οΏ½ οΏ½ 1 οΏ½ = οΏ½ππ οΏ½ 1 ππ2 2 ππππ (πΏπΏ,ππππ )2 ∫0 πΈπΈππππ (√ππ(ππππ (π‘π‘) − Λ(ππππ , π‘π‘)))2 ππππ ππππππ ||Λ(ππ,⋅) − Λ(ππππ )|| = ||Λ(ππ ′ ππ ) − Λ(ππππ )|| = 0 with means that: 4πΈπΈππ ππ ||ππππ ( )||2 πΈπΈππππ ||ππππ ( )||2 Demonstration: To show ππ(πΏπΏ, ππππ ) > 0, proceed by contradiction i.e., we suppose that for a πΏπΏ > 0, we have ππ(πΏπΏ, ππππ ) = 0. In this case, it exist ππππ ≠ ππ ′ ππ such: ππ(ππ1 π‘π‘ + ππ2 ) + ππ ππ(ππ1 π‘π‘ + ππ2 ) (ππ) ππππππ οΏ½οΏ½|ππππ∗ − ππππ |οΏ½ > πΏπΏοΏ½ ≤ ππππππ inf οΏ½|ππππ ( )– Λ(ππ,⋅)|οΏ½ > |ππ−ππππ |≤πΏπΏ οΏ½ οΏ½ inf οΏ½|ππππ ( )– Λ(ππ, )|οΏ½ ππ ππ Λ2,2 (ππ) = ∫0 πΊπΊ(ππ, π‘π‘)2 ππππ 2 π΄π΄12 ππ = π΄π΄2,1 = ∫0 πΉπΉ (ππ, π‘π‘) πΊπΊ(ππ, π‘π‘)ππππ Res. J. Math. Stat., 6(1): 1-5, 2014 ππ 2 π»π» (ππ) = ππππ1 ππ ππ1 ππ π‘π‘ ∫0 οΏ½∫0 ππ (ππ1 π π + 1 οΏ½ ∑ππππ=1 ππππ (π‘π‘) − ππ and ππ ππ πΆπΆ11 (ππ) = ∫0 ∫0 πΉπΉ (ππ, π‘π‘)πΉπΉ(ππ, π π )Λ(ππ, π π ∧ π‘π‘) ππππππππ πΆπΆ2,2 (ππ) = πΆπΆ1,2 (ππ) = ππ ππ ∫0 ∫0 πΊπΊ (ππ, π‘π‘)πΊπΊ(ππ, π π )Λ(ππ, π π ∧ π‘π‘) ππππππππ ππ ππ (ππ, π‘π‘)πΊπΊ(ππ, π π )Λ(ππ, π π πΆπΆ2,1 (ππ) = ∫0 ∫0 πΉπΉ Let be carrees matrix of order 2: and ππππππππ ∧ π‘π‘) ππ π»π» (ππ) = ππππ2 ππ 2 ππ − ∫0 [ ππ(ππ1 π‘π‘ ππ1 Λ(π‘π‘βπππ‘π‘ππ,π‘π‘ πππ‘π‘ π΄π΄11 (ππ) π΄π΄1,2 (ππ) οΏ½ π΄π΄(ππ) = οΏ½ π΄π΄1,1 (ππ) π΄π΄2,1 (ππ) ππ 1 ππ ∗ ∗ ∗ π‘π‘ + ππ2,ππ ) − ππ(ππ2,ππ )οΏ½ ∫0 οΏ½ππ(ππ1,ππ 1 οΏ½ ∑ππππ=1 ππππ (π‘π‘) − Λ(ππππ∗ , π‘π‘)οΏ½ ππππ = 0 ππ − We can write this system as: Hypothesis B: Théorème 2: If the Hypothesis π΄π΄ and π΅π΅ are satisfied then the minimum distance estimator is asymptotically normal i.e.: −1 − ππππ ) βΉ ππ(0, ππ πΆπΆ(ππππ )ππ (ππππ )) thus, ππ − Λ(ππ, π‘π‘)οΏ½ ππππ ππ ππ ππ = √ππ ∫0 πΊπΊ(ππππ∗ , π‘π‘)[Λ(ππππ∗ , π‘π‘) − Λ(ππππ , π‘π‘) ππππ Then, ππ ∫0 [ Λ(ππ + β, π‘π‘) − Λ(ππ, π‘π‘) − (β, Λ(ππ, π‘π‘) −(β, ΛΜ (ππ, π‘π‘)2 ππππ = ππ(|β|2 ) Using the continuity of this functions and the consistency of the estimator, we can write the system as: π»π»ππ (ππ)ππ=ππ2 = 0 The calculation of the partial derives by report at ππ1 and ππ2 gives us: ππ ∫0 ππππ (π‘π‘)πΊπΊ(ππππ∗ , π‘π‘) ππππ ππ= Θ ππππ ∑ππππ=1( ππππ (π‘π‘) − Λ(ππππ , π‘π‘)) = √ππ ∫0 πΉπΉ(ππππ∗ , π‘π‘)[Λ(ππππ∗ , π‘π‘) − Λ(ππππ , π‘π‘) ππππ = ππππππ inf π»π»ππ (ππ) π»π»ππ (ππ)ππ =ππ1 = 0 1 √ππ ∫0 ππππ (π‘π‘)πΉπΉ(ππππ∗ , π‘π‘) ππππ 2 ∗ ∗ , ππ2,ππ ) and ππ = (ππ1 , ππ2 ) Let us note that ππππ∗ = (ππ1,ππ the function π»π»ππ (ππ) is derivable. So the MDE ππππ∗ minimize the function π»π»ππ (ππ) and is consistent in ππππ element of Θ. It with a probability numeric. This estimator is a solution of system: ππ 1 ππππ (π‘π‘) = Let us remind that the minimum distance estimator ππππ∗ is a solution of the equation: ππππ ππ We introduce the process us ππππ () define by: Demonstration: We define the function π»π»ππ (ππ) by: ππππ∗ 1 ∫0 πΊπΊ (ππππ∗ , π‘π‘) οΏ½ππ ∑ππππ=1 ππππ (π‘π‘) − Λ(ππππ , π‘π‘) + Λ(ππππ , π‘π‘) − Λ(ππππ∗,π‘π‘)πππ‘π‘=0 π΄π΄11 (ππ) π΄π΄2,2 (ππ) ≠ (π΄π΄1,2 (ππ))2 π»π»ππ (ππ) = ππ ∫0 πΉπΉ (ππππ∗ , π‘π‘) οΏ½ππ ∑ππππ=1 ππππ (π‘π‘) − Λ(ππππ , π‘π‘) + Λ(ππππ , π‘π‘) − Λ(ππππ∗,π‘π‘)πππ‘π‘=0 To prove the asymptotic normality, we suppose that: ∑ππππ=1 ππππ (π‘π‘) π‘π‘ ππ ππ οΏ½∫0 πΉπΉ (ππ, π‘π‘)πΊπΊ(ππ, π‘π‘)πππποΏ½ ππ 1 ∫0 οΏ½ππ ππ οΏ½ ∑ππππ=1 ππππ (π‘π‘) − Λ(ππππ∗ , π‘π‘)οΏ½ ππππ = 0 ππ ππ ∫0 πΉπΉ (ππ, π‘π‘)2 ππππ ∫0 πΊπΊ (ππ, π‘π‘)2 ππππ 2 −1 1 + ππ2 − ππ(ππ2 )] οΏ½ ∑ππππ=1 ππππ (π‘π‘) − ∗ ∗ ∗ ∗ π π + ππ2,ππ )ππππ − π‘π‘π‘π‘(ππ1,ππ π‘π‘ + ππ2,ππ )οΏ½ ∫0 οΏ½∫0 ππ (ππ1,ππ the determinant of the matrix π΄π΄(ππ) is: √ππ(ππππ∗ Λ(ππ, π‘π‘)οΏ½ ππππ thus using (3) and (4), we obtain: πΆπΆ (ππ) πΆπΆ12 (ππ) πΆπΆ(ππ) = οΏ½ 11 οΏ½ πΆπΆ21 (ππ) πΆπΆ2,2 (ππ) ππeπ‘π‘(π΄π΄(ππ)) = ππ2 )ππππ − π‘π‘π‘π‘(ππ1 π‘π‘ + ππ2 )οΏ½ × ππ 3 ∫0 ππππ (π‘π‘)πΉπΉ(ππππ , π‘π‘)ππππ ππ = ∫ πΉπΉ(ππππ , π‘π‘)(π’π’ππ∗ , ΛΜ (ππππ , π‘π‘)ππππ + ππ(1) 0 Res. J. Math. Stat., 6(1): 1-5, 2014 ππ πΆπΆβ,ππ (ππππ ) = πΈπΈππππ (ππ (β) ππ (ππ) ) ∫0 ππππ (π‘π‘)πΊπΊ(ππππ , π‘π‘)ππππ ππ = ∫ πΊπΊ(ππππ , π‘π‘)(π’π’ππ∗ , ΛΜ (ππππ , π‘π‘)) ππππ + ππ(1) 0 We obtain: ∗ ∗ where we put π’π’ππ∗ = (π’π’1ππ , π’π’∗ 2ππ)π‘π‘ , π’π’1,ππ = √ππ(ππ ∗ 1, ππ − ∗ ∗ ππ1 ) et π’π’2,ππ = (√ππ(ππ2,ππ − ππ2 ) we introduce the vector (1) (2) ππππ = (ππππ , ππππ )ππ with, (1) ππππ (2) ππππ (1) ππππ (2) ππππ and, ππ = ∫0 πΉπΉ (ππππ , π‘π‘) ππππ (π‘π‘) ππππ ππ ππ 1 = = ππ ∑ππππ=1( ππππ (π‘π‘) − Λ(ππππ , π‘π‘)οΏ½ πΉπΉ(ππππ , π‘π‘) ππππ we obtain for example: ππ (1) π π πππ π πππ‘π‘ 1 √ππ 1 √ππ as √ππ(ππππ∗ − ππππ ) = π΅π΅(ππππ ) ππππ + ππ(1) ππ ∑ππππ=1 ∫0 ( ππππ (π‘π‘) − Λ(ππππ , π‘π‘)πΊπΊ(ππππ , π‘π‘)ππππ we deduce: π’π’ππ∗ = √ππ(ππππ∗ − ππππ ) βΉ π©π©(0, π΅π΅(ππππ ) πΆπΆ(ππππ ) π΅π΅(ππππ )π‘π‘ ) i.e.: √ππ(ππππ∗ − ππππ ) βΉ π©π©(0, π΄π΄−1 (ππππ ) πΆπΆ(ππππ ) π΄π΄−1 (ππππ )π‘π‘ ) Let us note that the terms of the matrix are: ππ So the asymptotic normality is proved. ππ ππ (1) = ∫0 οΏ½∫π‘π‘ πΉπΉ(ππππ , π π ) πππποΏ½ ππππ(Λ(ππππ , π‘π‘)) ππ (2) = ππ ππ ∫0 οΏ½∫π‘π‘ πΊπΊ(ππππ , π π )πππποΏ½ We have the following relation: ππ ππππ βΉ π©π©(0, πΆπΆ(ππππ )) ∑ππππ=1 ππππ(2) with ππππ(1) √ππ ππ = ∫0 ( ππππ (π‘π‘) − Λ(ππππ , π‘π‘))πΊπΊ(ππππ , π‘π‘)ππππ (1) (2) πΈπΈππππ ππππ ππππ = πΆπΆβ,ππ (ππππ ) = ππ = ∫0 ∫0 πΉπΉ(ππππ , π‘π‘)πΊπΊ(ππππ , π π )Λ(ππππ , π‘π‘ ∧ Using the central theorem limit, we obtain: = ∫0 οΏ½ ) ∑ππππ=1( ππππ (π‘π‘) − Λ(ππππ , π‘π‘)οΏ½ πΉπΉ(ππππ , π‘π‘) ππππ = (2) πΈπΈππππ ππ1 ππ1 ∫0 πΊπΊ(ππππ , π‘π‘) ππππ (π‘π‘) ππππ 1 π‘π‘ πΈπΈππππ [ππ1 (π‘π‘) − Λ(ππππ , π‘π‘)][ππ1 (π π ) − Λ(ππππ , π π ) = Λ(ππππ , π‘π‘ ∧ π π ] ∑ππ ππ (1) with ππππ(1) √ππ ππ =1 ππ ππ ∫0 ( ππππ (π‘π‘) − Λ(ππππ , π‘π‘))πΉπΉ(ππππ , π‘π‘)ππππ ππ ππ = ∫0 ∫π π πΊπΊ(ππππ , π‘π‘)ππππ ππππ(π π ) On the other hand using: ∑ππππ=1 ∫0 ( ππππ (π‘π‘) − Λ(ππππ , π‘π‘)πΉπΉ(ππππ , π‘π‘) ππππ 1 ππ ππ = ∫0 ∫π π πΉπΉ(ππππ , π‘π‘) ππππππππ(π π ) (2) ππ1 ∫0 πΉπΉ (ππππ , π‘π‘)ππππ (π‘π‘)ππππ √ππ ππ ππ (2) We show that ππ πΈπΈππππ (ππππ∗ − ππππ )(ππππ∗ − ππππ )π‘π‘ = π΅π΅(ππππ )πΆπΆ(ππππ ) π΅π΅(ππππ )π‘π‘ + ππ(1) We have: 1 ππ For ππ1 we have: π’π’ππ∗ = π΅π΅(ππππ )ππππ + ππ(1) √ππ ππ = ∫0 ∫π π 10≤π π ≤π‘π‘ [ππππ1 (π π ) − ππ(ππππ , π‘π‘) ππππ] πΉπΉ(ππππ , π‘π‘) ππππ = ∫0 ∫0 πΉπΉ(ππππ , π‘π‘) ππππ[ππππ1 (π π ) − ππ(ππππ , π‘π‘) ππππ] From the hypothesis π΅π΅, the matrix of order 2 π΄π΄(ππππ ) is invertible thus we can write π΅π΅(ππππ ) = π΄π΄(ππππ )−1 , so: ππ ππ = ∫0 [ ππ1 (π‘π‘) − Λ(ππππ , π‘π‘)] πΉπΉ(ππππ , π‘π‘) ππππ π΄π΄ππ (ππππ ) π’π’ππ∗ = ππππ + ππ(1) = ∫0 οΏ½ 2 ππ ππ ππ (1) ππ1 or under matrix shape: ππ ππ πΆπΆ2,2 (ππππ ) = ∫0 οΏ½∫π‘π‘ πΊπΊ(ππππ , π π ) πππποΏ½ ππ(ππππ , π‘π‘) ππππ ∫0 οΏ½∫π‘π‘ πΉπΉ(ππππ , π π ) ππππ ∫0 πΊπΊ (ππππ , ππ) πππποΏ½ ππ(ππππ , π‘π‘) ππππ ∗ ∗ π΄π΄1,1 (ππππ )π’π’1,ππ ) + π΄π΄1,2 (ππππ )π’π’2,ππ + ππ(1) ∗ π΄π΄2,1 (ππππ )π’π’1,ππ π‘π‘) + π΄π΄2,2 (ππππ ) + ππ(1) = 2 ππ = ∫0 οΏ½∫π‘π‘ πΉπΉ(ππππ , π π ) πππποΏ½ ππ(ππππ , π‘π‘) ππππ πΆπΆ1,2 (ππππ ) = πΆπΆ2,1 (ππππ ) = = ∫0 πΊπΊ (ππππ , π‘π‘) ππππ (π‘π‘) ππππ we obtain = ππ πΆπΆ1,1 (ππππ ) REFERENCES πππποΏ½Λ(ππππ , π‘π‘)οΏ½ Ba, D.B. and A.S. 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