Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2013 Article ID 910828 9 pageshttpdxdoiorg1011552013910828

Research ArticleOutlier Detection in Adaptive Functional-CoefficientAutoregressive Models Based on Extreme Value Theory

Ping Chen1 Ling Dong2 Wanyi Chen3 and Jin-Guan Lin1

1 Department of Mathematics Southeast University Nanjing Jiangsu 210096 China2 School of Finance and Statistics East China Normal University Shanghai 200241 China3Department of Industrial Engineering and Operations Research Columbia University New York NY 10027 USA

Correspondence should be addressed to Jin-Guan Lin jglinseueducn

Received 26 January 2013 Accepted 12 March 2013

Academic Editor Ming Li

Copyright copy 2013 Ping Chen et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

This paper proposes several test statistics to detect additive or innovative outliers in adaptive functional-coefficient autoregressive(AFAR)models based on extreme value theory and likelihood ratio tests All the test statistics follow a tractable asymptotic Gumbeldistribution Also we propose an asymptotic critical value on a fixed significance level and obtain an asymptotic 119901-value for testingwhich is used to detect outliers in time series Simulation studies indicate that the extreme value method for detecting outliers inAFAR models is effective both for AO and IO for a lone outlier and multiple outliers and for separate outliers and outlier patchesFurthermore it is shown that our procedure can reduce possible effects of masking and swamping

1 Introduction

Outlier detection and analysis play important roles in practi-cal applications For instance outlier detection can be appliedto anomaly detection in computer networks financial timeseries and data series in geosciences as can be seen from [1Chap 1] and [2] Other examples include the study on lossof customers in the commercial field and the detection andtracking of financial crime as credit card fraud all of whichinvolve and exploit the useful information provided by thepresence of outliers On the other hand outliers in dynamicsystems or engineering time series [3] can have adverse effectson model identification and parameter estimation whereeliminating outliers is necessary in the statistical modelingof time series for the purpose of preprocessing data seefor example [4] Some studies have even shown that theemerging of outliers generates certain nonlinear time seriesSeveral procedures are available in the literature to deal withproblems related to outliers Chen [5] developed a methodfor detecting additive outliers in bilinear time series whichbelong to the family of fractal time series models [6] Cai etal [7] studied the functional-coefficient regression modelsof nonlinear time series Battaglia [8] discovered a way to

identify and estimate outliers in functional autoregressivetime series Battaglia and Orfei [9] addressed the issue ofoutlier detection and estimation in nonlinear time seriesChen et al [10] and Chen et al [11] discussed the detection ofoutlier patches change point and outliers in bilinear models

Extreme value theory and likelihood ratio tests have beenused in the detection of outliers and time series analysisFor instance Martin [12] conducted extreme value analysison the optimal level cross-prediction of linear Gaussianprocesses Zhu and Ling [13] performed likelihood ratiotests on the structural change from an AR(p) model toa threshold AR(p) model Furthermore based on extremevalue theory Chareka et al [14] proposed an alternative testmethod for the detection of additive outliers inGaussian timeseries On the other hand some scholars such as Fung etal [15] and Rıo [16] focused their studies on special casesof outliers detection It is commonly agreed that the key ofoutlier detection lies in determining whether the test statisticexceeds a critical value that is the threshold under a givensignificance level However explanations for the selectionof threshold in many literatures are ambiguous and thethreshold itself can hardly be controlled under a certainlevel of significance In this paper we propose an asymptotic

2 Mathematical Problems in Engineering

critical value on a fixed significance level which is used todetect additive and innovative outliers in adaptive functional-coefficient autoregressive (AFAR) models (see eg Fan andYao [17])

This paper is structured as follows In Section 2 weconsider AFAR models with additive or innovative outliersand their estimation In Section 3 several procedures are pro-posed to detect outliers in the AFAR models based on ex-treme value theory and likelihood ratio tests In Section 4we present some simulation studies and demonstrate theeffectiveness of the proposed method through an empiricalway Concluding remarks are summarized in Section 5

2 Outliers Models and Test Statistics

We now consider the AFAR model

119909119905=cos (119886

1119909119905minus2)

1198862+ 11988631199092119905minus2

119909119905minus1

+ (1198864+ 1198865exp (119886

61199092

119905minus2)) 119909119905minus2

+ 120576119905

(1)

which can be written as

119909119905=

2

sum

119895=1

119891119895(119909119905minus2) 119909119905minus119895

+ 120576119905 (2)

where 120576119905is a Gaussian white noise with mean zero and vari-

ance 1205902 and 1198911(119909119905minus2) = cos(119886

1119909119905minus2)(1198862+ 11988631199092

119905minus2) 1198912(119909119905minus2) =

1198864+ 1198865exp(11988661199092

119905minus2)

Suppose that 119909119905119899

119905=1is a zero-mean stationary process

followingmodel (2) description and there is an outlier at time119905 = 119902 whose influence magnitude is 120596

119902 then the observed

series 119910119905may be presented as follows

(1) Innovative outlier (IO) model

119910119905= 119909119905 1 le 119905 lt 119902

119910119905=

2

sum

119895=1

119891119895(119910119905minus2) 119910119905minus119895

+ 120578119905 119902 le 119905 le 119899

120578119905= 120576119905+ 120596

IO119902sdot 120575 (119905 minus 119902)

(3)

(2) Additive outlier (AO) model

119910119905= 119909119905+ 120596

AO119902

sdot 120575 (119905 minus 119902) 119905 = 1 119899 (4)

where 120575(119905) is the Kronecker symbol If 119905 = 0 then 120575(119905) = 1else 120575(119905) = 0

It is obvious that the detection for IO or AO may betransformed to test whether the null hypothesis119867

0 120596

IO119902= 0

or11986710158400 120596

AO119902

= 0 is true under a certain level of significanceThis can be solved by knowing the distributions of IO

119902and

AO119902

under the null hypothesis We estimate 120596119902by using

the maximum likelihood method The maximum likelihoodfunction of observations is the positive ratio of themaximumlikelihood function of residuals For a given 119902 assume thatthe 119891119895and corresponding parameters are known then we

can obtain the estimation of 120596119902by maximizing the likelihood

function For convenience suppose that there is no outlierin the two former observations Denote the residual ofobservations by 120578

119905 for any 119896 le 0 and let 120576

119896= 0 119909

119896=

0 and 119891119896(sdot) = 0 120576

1 1205762be obtained by letting 120576

119905= 119909119905minus

sum2

119895=1119891119895(119909119905minus2)119909119905minus119895

Under initial conditions mentioned abovethe conditional likelihood function of 120596

119902is given by

119897 (119910 | 120596) = 119897 (119909) = 119897 (120576) =

119899

prod

119905=3

(21205871205902)minus12

exp(minus1205762

119905

21205902)

= (21205871205902)minus(119899minus2)2

exp(minussum119899

119905=31205762

119905

21205902)

(5)

We can obtain the maximum likelihood estimation of 120596119902at

the minimum of sum119899119905=3

1205762

119905 It is an accurate estimation for the

linear model However it is just an approximation for thenonlinear model For 119905 gt 2 the residual of observations isgiven by 120578

119905= 119910119905minus sum2

119895=1119891119895(119910119905minus2)119910119905minus119895

We then discuss theestimations of 120596IO

119902and 120596AO

119902for model (2)

(1) Assume that there is an IO at 119905 = 119902 and its influencemagnitude is 120596IO

119902 From (3) it follows that 120578

119902= 120576119902+ 120596

IO119902

and120578119905= 120576119905(119905 = 119902) where sum119899

119905=31205762

119905= sum119905 = 119902

1205782

119905+ (120578119902minus 120596

IO119902)2 So the

maximum likelihood estimation of 120596IO119902

is given by

IO119902= 120578119902= 119910119902minus

2

sum

119895=1

119891119895(119910119902minus2) 119910119902minus119895 (6)

For known 119891119895and 1205902 we have IO

119902sim 119873(120596

IO119902 1205902) which is

similar to Battaglia [8] Under the null hypothesis1198670 120596

IO119902=

0 IO119902

sim 119873(0 1205902) we obtain likelihood radio test statistic

IO119902

= (IO119902120590) sim119873(0 1) by standardizing IO

119902 When 119891

119895and

1205902 are unknown they can be replaced by their consistent

estimations 2IO = (119899 minus 2)minus1sum119899

119905=3119905 = 1199022

119905and

119905= 119910119905minus

sum2

119895=1119895(119910119905minus2)119910119905minus119895

Similarly we also have that

IO119902=IO119902

IOsim 119873 (0 1) (7)

(2) Assume that there is an AO at 119905 = 119902 and its influencemagnitude is 120596AO

119902 From (4) it follows that

120578119902= 119910119902minus

2

sum

119895=1

119891119895(119910119902minus2) 119910119902minus119895

= 119909119902+ 120596

AO119902

minus

2

sum

119895=1

119891119895(119909119902minus2) 119909119902minus119895

= 120576119902+ 120596

AO119902

(8)

Mathematical Problems in Engineering 3

When 119905 lt 119902 120578119905

= 119910119905minus sum2

119895=1119891119895(119910119905minus2)119910119905minus119895

= 119909119905minus

sum2

119895=1119891119895(119909119905minus2)119909119905minus119895

= 120576119905 if 119905 gt 119902 that is 119896 gt 0 then 120578

119902+119896=

119910119902+119896

minus sum2

119895=1119891119895(119910119902+119896minus2

)119910119902+119896minus119895

if 119896 = 1 we have that

120578119902+1

= 119910119902+1

minus

2

sum

119895=1

119891119895(119910119902minus1) 119910119902+1minus119895

= 119909119902+1

minus

2

sum

119895=1

119891119895(119909119902minus1) (119909119902+1minus119895

+ 120596AO119902120575 (1 minus 119895))

= 120576119902+1

minus

2

sum

119895=1

119891119895(119909119902minus1) 120596

AO119902120575 (1 minus 119895)

= 120576119902+1

minus 1198911(119909119902minus1) 120596

AO119902

= 120576119902+1

minus 1198911(119910119902minus1) 120596

AO119902

(9)

and if 119896 = 2 we have that

120578119902+2

= 119910119902+2

minus

2

sum

119895=1

119891119895(119910119902) 119910119902+2minus119895

= 119909119902+2

minus

2

sum

119895=1

119891119895(119909119902) 119909119902+2minus119895

+

2

sum

119895=1

119891119895(119909119902) 119909119902+2minus119895

minus

2

sum

119895=1

119891119895(119910119902) 119910119902+2minus119895

= 120576119902+2

minus (1198911(119910119902) minus 1198911(119909119902)) 119910119902+1

+ 1198912(119909119902) (119910119902minus 120596

AO119902) minus 1198912(119910119902) 119910119902

= 120576119902+2

minus 1198911015840

1(119909119902+ 1205791120596AO119902) 120596

AO119902119910119902+1

minus (1198912(119910119902) minus 1198912(119909119902)) 119910119902minus 1198912(119909119902) 120596

AO119902

= 120576119902+2

minus 1198911015840

1(119909119902+ 1205791120596AO119902) 120596

AO119902119910119902+1

minus 1198911015840

2(119909119902+ 1205792120596AO119902) 120596

AO119902119910119902minus 1198912(119909119902) 120596

AO119902

= 120576119902+2

minus 120596AO119902

sdot [

[

1198912(119909119902) +

2

sum

119895=1

1198911015840

119895(119909lowast

119902119895) 119910119902+2minus119895

]

]

(10)

where 119909lowast119902119895

= 119909119902+ 120579119895120596AO119902

0 lt 120579119895lt 1 and 119895 = 1 2 If 119896 gt 2

then 120578119902+119896

= 120576119902+119896

from which it follows that

120578119902+119896

= 119888119896120596AO119902

+ 120576119902+119896

119896 = 1 2 (11)

where 1198881= minus1198911(119910119902minus1) and 119888

2= minus[119891

2(119909119902)+sum2

119895=11198911015840

119895(119909lowast

119902119895)119910119902+2minus119895

]so we have that

119899

sum

119905=3

1205762

119905=

119902minus1

sum

119905=3

1205762

119905+ 1205762

119902+

119902+2

sum

119905=119902+1

1205762

119905+

119899

sum

119905=119902+3

1205762

119905

=

119902minus1

sum

119905=3

1205782

119905+ (120578119902minus 120596

AO119902)2

+

2

sum

119896=1

(120578119902+119896

minus 119888119896120596AO119902)2

+

119899

sum

119905=119902+3

1205782

119905

=

119902minus1

sum

119905=3

1205782

119905+

119899

sum

119905=119902+3

1205782

119905+ (1 + 119888

2

1+ 1198882

2) 120596

AO119902

2

minus 2 (120578119902+ 1198881120578119902+1

+ 1198882120578119902+2) 120596

AO119902

+ (1205782

119902+ 1205782

119902+1+ 1205782

119902+2)

(12)

By minimizing the above expression we obtain

120596AO119902

=120578119902+ 1198881120578119902+1

+ 1198882120578119902+2

1 + 11988821+ 11988822

(13)

where 1198881may be obtained by estimation of minus119891

1(119910119902minus1) while

1198882is complicated and difficult to be confirmed Nevertheless

Battaglia [8] indicated that we could estimate 1198882by minus119891

2(119910119902)

which is convenient and effective If all 119891119895(sdot) and 120590

2 areknown then we have

AO119902

=120578119902minus 1198911(119910119902minus1) 120578119902+1

minus 1198912(119910119902) 120578119902+2

Δ119902

Δ119902= 1 + 119891

2

1(119910119902minus1) + 1198912

2(119910119902)

(14)

Similar to Battaglia [8] we obtain

AO119902

sim 119873(120596AO1199021205902

Δ119902

) (15)

Under the null hypothesis11986710158400 120596

AO119902

= 0 we obtain likelihood

radio test statistics AO119902

= (AO119902(120590radicΔ 119902)) sim 119873(0 1) by stan-

dardizing AO119902

In practice if119891119895(sdot) and1205902 are unknownwhich

can be replaced by their consistent estimations 2AO = (119899 minus

2)minus1[sum119899

119905=32

119905minus (

AO119902)2sdot sum2

119895=12

119895(119910119902+119895minus2

)] we likewise obtain

AO119902

=AO119902

AOradicΔ119902

sim 119873 (0 1) (16)

It is indicated by (6) and (13) that the best estimate of 120596is 120578119902at time 119902 for the case of IO the best estimate of 120596 is

the linear combination of errors 120578119902 120578119902+1

and 120578119902+2

at time 119902in the case of AO If the location 119902 of outlier is known thedistributions of likelihood ratio test statistics

IO119902

and AO119902

aregiven by (7) and (16) However in practice the location ofoutlier is unknown thus we should inspect the magnitudeof test statistic at every time point By replacing

IO119902

and

AO119902

with IO119905

and AO119905

the following theorem indicates that

the processes of IO119905 and

AO119905 are both Gaussian process

respectively

Theorem 1 (1) Under the hypothesis 1198670 120596

IO119905

= 0 that isthere is no IO at any time IO

119905 is a stationary zero-mean

Gaussian process with variance 1205902 and its autocovariancefunction is 119903IO(ℎ) = 120590

2sdot 119868 (ℎ = 0) where 119868

119860(119909) is the index

function If 119909 isin 119860 then 119868119860(119909) = 1 and else 119868

119860(119909) = 0

4 Mathematical Problems in Engineering

(2) Under the hypothesis 11986710158400 120596

AO119905

= 0 that is there isno AO at any time AO

119905 is a stationary zero-mean Gaussian

process with variance 1205902Δ119905 and its autocovariance function

is

119903AO

(119905 ℎ) = 1205902Δminus1

119905Δminus1

119905+ℎ[

[

2

sum

119895=ℎ+1

119891119895(119910119905+119895minus2

) 119891119895minusℎ

(119910119905+119895minus2

)

minus119891ℎ(119910119905+ℎminus2

) ]

]

119868 (1 le ℎ le 2)

(17)

Proof The Gaussian character is shown by (7) and (16) Wederive their autocovariance function as follows

(1) Under the null hypothesis1198670 120596

IO119905= 0 we have 120578

119905=

120576119905and

119903IO(ℎ) = cov (IO

119905

IO119905+ℎ) = cov (120578

119905 120578119905+ℎ)

= cov (120576119905 120576119905+ℎ) = 1205902sdot 119868 (ℎ = 0)

(18)

(2) Under the null hypothesis 11986710158400 120596

AO119905

= 0 we have120578119905= 120576119905and

119903AO

(119905 ℎ) = cov (AO119905

AO119905+ℎ)

= cov(Δminus1119905(120578119905minus

2

sum

119895=1

119891119895(119910119905+119895minus2

) 120578119905+119895)

Δminus1

119905+ℎ(120578119905+ℎ

minus

2

sum

119895=1

119891119895(119910119905+ℎ+119895minus2

) 120578119905+ℎ+119895

))

= Δminus1

119905Δminus1

119905+ℎ[

[

cov (120578119905 120578119905+ℎ)

minus

2

sum

119895=1

119891119895(119910119905+ℎ+119895minus2

) cov (120578119905 120578119905+ℎ+119895

)

minus

2

sum

119895=1

119891119895(119910119905+119895minus2

) cov (120578119905+119895 120578119905+ℎ)

+ cov(2

sum

119895=1

119891119895(119910119905+119895minus2

) 120578119905+119895

2

sum

119895=1

119891119895(119910119905+ℎ+119895minus2

) 120578119905+ℎ+119895

)]

]

= 1205902Δminus1

119905Δminus1

119905+ℎ[

[

2

sum

119895=ℎ+1

119891119895(119910119905+119895minus2

) 119891119895minusℎ

(119910119905+119895minus2

)

minus119891ℎ(119910119905+ℎminus2

) ]

]

119868 (1 le ℎ le 2)

(19)

We know that IO119905 is not only Gaussian but also a

stationary process byTheorem 1 Also it is serial independentunder the null hypothesis 119867

0 120596

IO119905

= 0 However despitethat AO

119905 is Gaussian it is not stationary due to the fact

that its autocovariance function 119903AO(119905 ℎ) is dependent on 119905Besides AO

119905 is truncated in finite steps which means that

its maximum correlation is within two steps that is AO119905

isindependent with AO

119905+ℎwhen ℎ gt 2 By standardizing IO

119905

and AO119905 we can obtain

IO119905 and

AO119905 It is obvious that

IO119905 is a standardized stationary Gaussian process under the

hypothesis1198670 120596

IO119905= 0 Also

AO119905 is a standardized Gaus-

sian process under the hypothesis11986710158400 120596

AO119905

= 0 but it is notstationary whereas its autocovariance function is truncatedin finite steps

3 Detection of Outliers Based on ExtremeValue Theory

As similar to Chareka et al [14] and Leadbetter and Rootzen[18] we obtain the following

Lemma 2 (see [18]) Assume that 119883119899 is a stationary zero-

mean Gaussian time series and its autocorrelation functionis 120588(ℎ) Let 119885

119899= max119883

1 1198832 119883

119899 and if the Berman

condition limℎrarrinfin

120588(ℎ) sdot log ℎ = 0 is satisfied then one hasthat 119875(((119885

119899minus 119888119899)119889119899) le 119909) rarr Λ(119909) = expminus119890minus119909 119899 rarr infin

where 119889119899= (2 log 119899)minus12 and 119888

119899= (1119889

119899) minus (119889

1198992)(log log 119899 +

log(4120587))

Lemma 3 (see [18]) Assume that 119883119899 is a stationary zero-

mean Gaussian time series and let 119860119899= max119899

119905=1|119883119905| and

1198852119899

= max2119899119905=1119883119905 and if 119883

119899satisfies the Berman condition

then 119860119899and 119885

2119899have the same distribution

Theorem 4 Let 120578IO = max119905=12119899

|IO119905| and 120578AO =

max119905=12119899

|AO119905| and under the hypothesis 119867

0 120596

IO119905

= 0

and11986710158400 120596

AO119905

= 0 one has that

119875(120578IO minus 11988821198991198892119899

le 119909) 997888rarr Λ (119909)

119875 (120578AO minus 11988821198991198892119899

le 119909) 997888rarr Λ (119909) 119899 997888rarr infin

(20)

where 1198892119899

= [2 log(2119899)]minus12 and 1198882119899

= (11198892119899) minus (119889

2119899

2)[log log(2119899) + log(4120587)]

Proof It follows from Theorem 1 that the autocorrelationfunction of

IO119905 is 120588IO(ℎ) = 119868 (ℎ = 0) under the hypothesis

1198670 120596

IO119905

= 0 so limℎrarrinfin

120588IO(ℎ) sdot log ℎ = 0 that is the Ber-

man condition is satisfied Again IO119905 is a stationary zero-

mean Gaussian time series Hence from Lemma 2 we havethat 119875(((max119899

119905=1

IO119905 minus 119888119899)119889119899) le 119909) rarr Λ(119909) Again from

Lemma 3 we know that the distribution of 120578IO =

max119899119905=1|

IO119905| is the same as that of max2119899

119905=1

IO119905 so we

Mathematical Problems in Engineering 5

have that 119875(((120578IO minus 1198882119899)1198892119899) le 119909) rarr Λ(119909) Similarly underthe hypothesis1198671015840

0 120596

AO119905

= 0 from Theorem 1 we know thatthe autocorrelation function of

AO119905

is

120588AO

(119905 ℎ) = Δminus12

119905sdot Δminus12

119905+ℎsdot [

[

2

sum

119895=ℎ+1

119891119895(119910119905+119895minus2

) 119891119895minusℎ

(119910119905+119895minus2

)

minus119891ℎ(119910119905+ℎminus2

) ]

]

119868 (1 le ℎ le 2)

(21)

where limℎrarrinfin

120588AO(ℎ) sdot log ℎ = 0 that is the Berman

condition is satisfied Therefore similar to the statementsabove it must hold that 119875(((120578AO minus 119888

2119899)1198892119899) le 119909) rarr

Λ(119909)

Lemma 5 (see [14]) Assume that 119883119899 is a stationary zero-

mean Gaussian time series with variance 1 and autocorrelationfunction 120588(ℎ) and let 119872

119899= max1198832

1 119883

2

119899 and if 120588(ℎ)

satisfies the Berman condition then one has 119875(((119872119899minus119890119899)2) le

119909) rarr Λ(119909) 119899 rarr infin where 119890119899= 2 log 119899 minus log(log 119899) minus log120587

Theorem 6 Let 120599IO = max119899119905=1(

IO119905)2 and 120599AO =

max119899119905=1(

AO119905)2 under the hypothesis 119867

0 120596

IO119905

= 0 and1198671015840

0 120596

AO119905

= 0 and if the corresponding Berman conditionsfor

IO119905 and

AO119905 are satisfied then one has

119875(120599IO minus 119890119899

2le 119909) 997888rarr Λ (119909)

119875 (120599AO minus 119890119899

2le 119909) 997888rarr Λ (119909) 119899 997888rarr infin

(22)

where 119890119899= 2 log 119899 minus log(log 119899) minus log120587

Proof We know that the Berman conditions for 120582IO119905 and

120582AO119905 are satisfied under the hypothesis 119867

0 120596

IO119905

= 0 and1198671015840

0 120596

AO119905

= 0 by the inferential process of Theorem 4 Thusthe conclusion is also obtained by Lemma 5

Let 120572 be the test significance level and 119910119899= (1 + (1 minus

120572)1119899)2Wedenote the 119905-distribution functionwith 119899degrees

of freedom by 119867119899(119905) Let 119909

119899= 119867minus1

119899minus2(119910119899) and 119890

119899= (119899(119899 minus

1))1199092

119899minus 2Λminus1(1 minus 120572) where Λminus1(1 minus 120572) is the 1 minus 120572 quantile of

the Gumbel distribution Similarly we have the following

Theorem 7 Under the conditions above one has that

119875(119872119899minus 119890119899

2le 119909) 997888rarr Λ (119909)

119875 (120599IO minus 119890119899

2le 119909) 997888rarr Λ (119909)

119875 (120599AO minus 119890119899

2le 119909) 997888rarr Λ (119909) 119899 997888rarr infin

(23)

At this point it is convenient to introduce two morepieces of notation 119860IO

119905= ((|

IO119905| minus 1198882119899)1198892119899) and 119860

AO119905

=

((|AO119905|minus 1198882119899)1198892119899) which we call absolute value test statistics

119878IO119905

= (((IO119905)2minus 119890119899)2) and 119878AO

119905= (((

AO119905)2minus 119890119899)2) which

we call square test statistics 119862IO119905

= (((IO119905)2minus 119890119899)2) and

119862AO119905

= (((AO119905)2minus 119890119899)2) which we call adjusted square test

statistic We likewise obtain

119875(119899max119905=1

119860IO119905 le 119909) 997888rarr Λ (119909)

119875 (119899max119905=1

119860AO119905 le 119909) 997888rarr Λ (119909)

119875 (119899max119905=1

119878IO119905 le 119909) 997888rarr Λ (119909) 119899 997888rarr infin

119875(119899max119905=1

119878AO119905 le 119909) 997888rarr Λ (119909)

119875 (119899max119905=1

119862IO119905 le 119909) 997888rarr Λ (119909)

119875 (119899max119905=1

119862AO119905 le 119909) 997888rarr Λ (119909) 119899 997888rarr infin

(24)

If there are several outliers the main idea is to first detectthemaximum outlier by using ourmethod and to obtain newtest statistics by deleting its effectThen we go on to detect thenext outlier and repeat the procedure till there is no outlierWhen two types of outliers appear the test is

1198670 120596

IO119905=0 120596

AO119905

=0 versus 1198671 120596

IO119905

= 0 or 120596AO119905

= 0

119905 = 1 119899

(25)

For details we provide the following steps using absolutevalue test statistics to detect the AO and IO in AFAR modelas follows (a) Let test statistics 119879

1= max119899

119905=1119860

IO119905 and 119879

2=

max119899119905=1119860

AO119905 (b) Calculate

IO119905

and AO119905

at every time pointand calculate themaximum 119905

1= 119860

IO1199021

= max119899119905=1119860

IO119905 and 119905

2=

119860AO1199022

= max119899119905=1119860

AO119905 of absolute value test statistics (c) Let

119862119881 = Λminus1(1 minus 120572) = minus log(minus log(1 minus 120572)) if 119905

1= max119905

1 1199052 gt

119862119881 then we believe that the observation is an IO at 119905 = 1199021

and if 1199052= max119905

1 1199052 gt 119862119881 then we believe that the

observation is an AO at 119905 = 1199022 else we believe there is no

IO or AO (d) Calculate the 119901-value 1199011= 1198751198670

(1198791ge 1199051) =

1 minusΛ(1199051) and 119901

2= 1198751198670

(1198792ge 1199052) = 1 minusΛ(119905

2) and if 119901

1le 120572 or

1199012le 120572 then reject the hypothesis119867

0 believing there is an IO

or an AO Furthermore we decide whether it is an IO or anAO by its minimal 119901-value (e) Delete the effect of detectedoutliers and detect the next outlier Repeat the above steps tillthere is no outlier

4 Simulation Studies

Example 8 In this simulation we consider the followingmodel where there is only one IO The sample sizes are

6 Mathematical Problems in Engineering

40

30

20

10

0

minus10

0 20 40 60 80 100(a)

80

60

40

20

0

minus20

0 20 40 60 80 100(b)

80

60

40

20

0

minus20

0 20 40 60 80 100(c)

Figure 1 Example 9 first detection

2520151050minus5

minus10

0 20 40 60 80 100(a)

50

40

30

20

10

0

minus10

0 20 40 60 80 100(b)

50

40

30

20

10

0

minus10

0 20 40 60 80 100(c)

Figure 2 Example 9 second detection

119899 = 50 100 500 1000 respectively and 120596IO1198992

= 4 We have

119910119905=cos (15119910

119905minus2)

1 + 1199102119905minus2

119910119905minus1

+ (minus exp(minus1199102

119905minus2

2))119910119905minus2

+ 120578119905

120578119905= 120576119905+ 120596

IO1198992

sdot 120575 (119905 minus119899

2) 120576

119905sim 119873 (0 05)

(26)

For 119899 = 50 All the three test statistics detect an outlierat 119905 = 25 The 119901-values for believing it is an IO are 119901(119860IO

) =

48414119890 minus 6 119901(119878IO) = 18326119890 minus 8 and 119901(119862IO) = 33739119890 minus

8 respectively The 119901-values for believing it is an AO are119901(119860

AO) = 31509119890 minus 4 119901(119878AO) = 47392119890 minus 5 and 119901(119862AO

) =

87247119890minus5 respectively Because the 119901-values for believing itis an IO are smaller than the119901-values for believing it is anAOwe believe it is an IO and its influence magnitude is 31994

Mathematical Problems in Engineering 7

10

5

0

minus5

minus10

0 20 40 60 80 100(a)

0 20 40 60 80 100

1086420minus2

minus4

(b)

0 20 40 60 80 100

1086420minus2

minus4

(c)

Figure 3 Example 9 fifth detection

420minus2

minus4

minus6

minus8

minus10

0 20 40 60 80 100(a)

0 20 40 60 80 100

3210minus1

minus2

minus3

minus4

(b)

0 20 40 60 80 100

3210minus1

minus2

minus3

minus4

(c)

Figure 4 Example 9 sixth detection

The results are similar for 119899 = 100 500 1000 respectively wethus omit the details here For different test statistics their 119901-values for believing it is an IO are summarized as follows inTable 1

Example 9 There are four IOs and one AO in the time seriesThese IOs appear at 119905 = 49 119905 = 50 119905 = 51 and 119905 = 52

sequentially Their sizes are 3 6 5 and 4 respectively One

AO appears at 119905 = 70 alone and its size is 7 119899 = 100 Themodel is as follows

119909119905=cos (15119909

119905minus2)

1 + 1199092119905minus2

119909119905minus1

+ (minus exp (minus051199092119905minus2)) 119909119905minus2

+ 120576119905 120576119905sim 119873 (0 06)

(27)

where 119910119905is obtained by (3) or (4)

8 Mathematical Problems in Engineering

Table 1

119901-values Absolute valuestatistic Square statistic Adjusted square

statistic119899 = 50 48414119890 minus 6 18326119890 minus 8 33739119890 minus 8

119899 = 100 10027119890 minus 6 16592119890 minus 9 21995119890 minus 9

119899 = 500 19065119890 minus 6 26461119890 minus 8 260394119890 minus 8

119899 = 1000 28048119890 minus 7 13358119890 minus 9 12590119890 minus 9

First we detect an AO at 119905 = 70 and its size is 68904The three test statistics and their corresponding critical valuesare shown in Figure 1 where Figure 1(a) corresponds tothe absolute value test statistics Figure 1(b) corresponds tosquare test statistics and Figure 1(c) corresponds to adjustedsquare test statistics Symbol ldquordquo denotes IO and ldquolowastrdquo denotesAO Parallel broken line denotes critical value which isuniform in the figures Deleting the effect of this AO and thencontinuing to detect other outliers in the series we observean IO at 119905 = 50 whose size is 60285 see details in Figure 2Deleting the effect of the aforementioned two outliers andthen continuing to detect other outliers in the series weobserve an IO at 119905 = 51 whose size is 48119 Deletingthe effect of the aforementioned three outliers and thencontinuing to detect other outliers in the series we observean IO at 119905 = 52 whose size is 40024 Deleting the effectof the aforementioned four outliers and then continuing todetect other outlier in the series we detect an IO at 119905 =

49 whose size is 29046 see Figure 3 Deleting the effectof the aforementioned five outliers and then continuing todetect other outliers in the series we have not detected anyoutlier see Figure 4 Considering the length of paper we omitsome figures here The result is consistent with the advanceenactment

5 Conclusions

The FAR model is mainly featured by the model-dependent-variable which in one way or another limits the scope ofits applications As a generalization of the class of modelsAFAR model clearly covers a larger range of objects thanthe FAR model which makes it possible to reduce modelingbiases [17] via choosing a proper model-dependent direc-tion This paper is concerned with detecting AO and IOin AFAR models using extreme value methods We derivethe asymptotic distribution of test statistics and provide acontrol for significance level which serves as an extensionand improvement of existing methods Based on severalsimulation studies we give conclusion remarks as follows(a)The extreme value method for detecting outliers in AFARmodels is tractable and effective not only for IO and AO butalso for separate outliers and outlier patches Furthermoreit is shown that our method can reduce possible effects ofmasking and swamping (b) When applying extreme valuetheory to detect outliers with a relatively small size samplesat hand the employment of square test statistics works betterthan that of adjusted square test statistics as well as absolutevalue test statistics While following the increments in thesamples size the detecting effect of adjusted square test

statistics also increases against that of square test statistics andabsolute value test statistics (eg see Table 1) (c) Selection ofmodel parameters and the magnitude of outliers have hugeinfluences on the effect of detection

Acknowledgments

The authors sincerely wish to thank the editor and the threereferees for their insightful suggestions which have led toimproving the early version of the paper The research issupported by the National Natural Science Foundation ofChina (11171065) the Natural Science Foundation of JiangsuProvince (BK2011058) and Research Fund for the DoctoralProgram of Higher Education of China (20120092110021)

References

[1] C C AggarwalOutlier Analysis Springer New York NY USA2013

[2] X JieM LiWZhao and S YChen ldquoBoundmaxima as a trafficfeature under DDOS flood attacksrdquo Mathematical Problems inEngineering vol 2012 Article ID 419319 20 pages 2012

[3] M Li andW Zhao ldquoOn bandlimitedness and lag-limitedness offractional Gaussian noiserdquo Physica A vol 392 no 9 pp 1955ndash1961 2013

[4] M Li Y-Q Chen J-Y Li and W Zhao ldquoHolder scales of sealevelrdquoMathematical Problems in Engineering Article ID 86370722 pages 2012

[5] C W S Chen ldquoDetection of additive outliers in bilinear timeseriesrdquo Computational Statistics amp Data Analysis vol 24 no 3pp 283ndash294 1997

[6] M Li ldquoFractal time seriesmdasha tutorial reviewrdquo MathematicalProblems in Engineering Article ID 157264 26 pages 2010

[7] Z Cai J Fan and Q Yao ldquoFunctional-coefficient regressionmodels for nonlinear time seriesrdquo Journal of the AmericanStatistical Association vol 95 no 451 pp 941ndash956 2000

[8] F Battaglia ldquoOutliers in functional autoregressive time seriesrdquoStatistics amp Probability Letters vol 72 no 4 pp 323ndash332 2005

[9] F Battaglia and L Orfei ldquoOutlier detection and estimation innonlinear time seriesrdquo Journal of Time Series Analysis vol 26no 1 pp 107ndash121 2005

[10] P Chen L Li Y Liu and J-G Lin ldquoDetection of outliers andpatches in bilinear time series modelsrdquoMathematical Problemsin Engineering Article ID 580583 10 pages 2010

[11] P Chen J Yang and L Y Li ldquoSynthetic Detection of ChangePoint andOutliers in Bilinear Time SeriesModelsrdquo InternationalJournal of Systems Science In press

[12] R A Martin ldquoExtreme value analysis of optimal level-crossingprediction for linear Gaussian processesrdquo Journal of Time SeriesAnalysis vol 33 no 4 pp 583ndash607 2012

[13] K Zhu and S Ling ldquoLikelihood ratio tests for the structuralchange of an AR(p)model to a threshold AR(p)modelrdquo Journalof Time Series Analysis vol 33 no 2 pp 223ndash232 2012

[14] P Chareka F Matarise and R Turner ldquoA test for additiveoutliers applicable to long-memory time seriesrdquo Journal ofEconomic Dynamics amp Control vol 30 no 4 pp 595ndash621 2006

[15] W-K Fung Z-Y Zhu B-C Wei and X He ldquoInfluencediagnostics and outlier tests for semiparametric mixedmodelsrdquoJournal of the Royal Statistical Society B vol 64 no 3 pp 565ndash579 2002

Mathematical Problems in Engineering 9

[16] A M Rıo Extreme value theory-based P values in time seriesoutlier detection [PhD thesis] University of Wisconsin Madi-son 2005

[17] J Fan and Q Yao Nonlinear time series Springer Series inStatistics Springer New York NY USA 2003

[18] M R Leadbetter andH Rootzen ldquoExtremal theory for stochas-tic processesrdquo The Annals of Probability vol 16 no 2 pp 431ndash478 1988

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Mathematical Problems in Engineering

critical value on a fixed significance level which is used todetect additive and innovative outliers in adaptive functional-coefficient autoregressive (AFAR) models (see eg Fan andYao [17])

This paper is structured as follows In Section 2 weconsider AFAR models with additive or innovative outliersand their estimation In Section 3 several procedures are pro-posed to detect outliers in the AFAR models based on ex-treme value theory and likelihood ratio tests In Section 4we present some simulation studies and demonstrate theeffectiveness of the proposed method through an empiricalway Concluding remarks are summarized in Section 5

2 Outliers Models and Test Statistics

We now consider the AFAR model

119909119905=cos (119886

1119909119905minus2)

1198862+ 11988631199092119905minus2

119909119905minus1

+ (1198864+ 1198865exp (119886

61199092

119905minus2)) 119909119905minus2

+ 120576119905

(1)

which can be written as

119909119905=

2

sum

119895=1

119891119895(119909119905minus2) 119909119905minus119895

+ 120576119905 (2)

where 120576119905is a Gaussian white noise with mean zero and vari-

ance 1205902 and 1198911(119909119905minus2) = cos(119886

1119909119905minus2)(1198862+ 11988631199092

119905minus2) 1198912(119909119905minus2) =

1198864+ 1198865exp(11988661199092

119905minus2)

Suppose that 119909119905119899

119905=1is a zero-mean stationary process

followingmodel (2) description and there is an outlier at time119905 = 119902 whose influence magnitude is 120596

119902 then the observed

series 119910119905may be presented as follows

(1) Innovative outlier (IO) model

119910119905= 119909119905 1 le 119905 lt 119902

119910119905=

2

sum

119895=1

119891119895(119910119905minus2) 119910119905minus119895

+ 120578119905 119902 le 119905 le 119899

120578119905= 120576119905+ 120596

IO119902sdot 120575 (119905 minus 119902)

(3)

(2) Additive outlier (AO) model

119910119905= 119909119905+ 120596

AO119902

sdot 120575 (119905 minus 119902) 119905 = 1 119899 (4)

where 120575(119905) is the Kronecker symbol If 119905 = 0 then 120575(119905) = 1else 120575(119905) = 0

It is obvious that the detection for IO or AO may betransformed to test whether the null hypothesis119867

0 120596

IO119902= 0

or11986710158400 120596

AO119902

= 0 is true under a certain level of significanceThis can be solved by knowing the distributions of IO

119902and

AO119902

under the null hypothesis We estimate 120596119902by using

the maximum likelihood method The maximum likelihoodfunction of observations is the positive ratio of themaximumlikelihood function of residuals For a given 119902 assume thatthe 119891119895and corresponding parameters are known then we

can obtain the estimation of 120596119902by maximizing the likelihood

function For convenience suppose that there is no outlierin the two former observations Denote the residual ofobservations by 120578

119905 for any 119896 le 0 and let 120576

119896= 0 119909

119896=

0 and 119891119896(sdot) = 0 120576

1 1205762be obtained by letting 120576

119905= 119909119905minus

sum2

119895=1119891119895(119909119905minus2)119909119905minus119895

Under initial conditions mentioned abovethe conditional likelihood function of 120596

119902is given by

119897 (119910 | 120596) = 119897 (119909) = 119897 (120576) =

119899

prod

119905=3

(21205871205902)minus12

exp(minus1205762

119905

21205902)

= (21205871205902)minus(119899minus2)2

exp(minussum119899

119905=31205762

119905

21205902)

(5)

We can obtain the maximum likelihood estimation of 120596119902at

the minimum of sum119899119905=3

1205762

119905 It is an accurate estimation for the

linear model However it is just an approximation for thenonlinear model For 119905 gt 2 the residual of observations isgiven by 120578

119905= 119910119905minus sum2

119895=1119891119895(119910119905minus2)119910119905minus119895

We then discuss theestimations of 120596IO

119902and 120596AO

119902for model (2)

(1) Assume that there is an IO at 119905 = 119902 and its influencemagnitude is 120596IO

119902 From (3) it follows that 120578

119902= 120576119902+ 120596

IO119902

and120578119905= 120576119905(119905 = 119902) where sum119899

119905=31205762

119905= sum119905 = 119902

1205782

119905+ (120578119902minus 120596

IO119902)2 So the

maximum likelihood estimation of 120596IO119902

is given by

IO119902= 120578119902= 119910119902minus

2

sum

119895=1

119891119895(119910119902minus2) 119910119902minus119895 (6)

For known 119891119895and 1205902 we have IO

119902sim 119873(120596

IO119902 1205902) which is

similar to Battaglia [8] Under the null hypothesis1198670 120596

IO119902=

0 IO119902

sim 119873(0 1205902) we obtain likelihood radio test statistic

IO119902

= (IO119902120590) sim119873(0 1) by standardizing IO

119902 When 119891

119895and

1205902 are unknown they can be replaced by their consistent

estimations 2IO = (119899 minus 2)minus1sum119899

119905=3119905 = 1199022

119905and

119905= 119910119905minus

sum2

119895=1119895(119910119905minus2)119910119905minus119895

Similarly we also have that

IO119902=IO119902

IOsim 119873 (0 1) (7)

(2) Assume that there is an AO at 119905 = 119902 and its influencemagnitude is 120596AO

119902 From (4) it follows that

120578119902= 119910119902minus

2

sum

119895=1

119891119895(119910119902minus2) 119910119902minus119895

= 119909119902+ 120596

AO119902

minus

2

sum

119895=1

119891119895(119909119902minus2) 119909119902minus119895

= 120576119902+ 120596

AO119902

(8)

Mathematical Problems in Engineering 3

When 119905 lt 119902 120578119905

= 119910119905minus sum2

119895=1119891119895(119910119905minus2)119910119905minus119895

= 119909119905minus

sum2

119895=1119891119895(119909119905minus2)119909119905minus119895

= 120576119905 if 119905 gt 119902 that is 119896 gt 0 then 120578

119902+119896=

119910119902+119896

minus sum2

119895=1119891119895(119910119902+119896minus2

)119910119902+119896minus119895

if 119896 = 1 we have that

120578119902+1

= 119910119902+1

minus

2

sum

119895=1

119891119895(119910119902minus1) 119910119902+1minus119895

= 119909119902+1

minus

2

sum

119895=1

119891119895(119909119902minus1) (119909119902+1minus119895

+ 120596AO119902120575 (1 minus 119895))

= 120576119902+1

minus

2

sum

119895=1

119891119895(119909119902minus1) 120596

AO119902120575 (1 minus 119895)

= 120576119902+1

minus 1198911(119909119902minus1) 120596

AO119902

= 120576119902+1

minus 1198911(119910119902minus1) 120596

AO119902

(9)

and if 119896 = 2 we have that

120578119902+2

= 119910119902+2

minus

2

sum

119895=1

119891119895(119910119902) 119910119902+2minus119895

= 119909119902+2

minus

2

sum

119895=1

119891119895(119909119902) 119909119902+2minus119895

+

2

sum

119895=1

119891119895(119909119902) 119909119902+2minus119895

minus

2

sum

119895=1

119891119895(119910119902) 119910119902+2minus119895

= 120576119902+2

minus (1198911(119910119902) minus 1198911(119909119902)) 119910119902+1

+ 1198912(119909119902) (119910119902minus 120596

AO119902) minus 1198912(119910119902) 119910119902

= 120576119902+2

minus 1198911015840

1(119909119902+ 1205791120596AO119902) 120596

AO119902119910119902+1

minus (1198912(119910119902) minus 1198912(119909119902)) 119910119902minus 1198912(119909119902) 120596

AO119902

= 120576119902+2

minus 1198911015840

1(119909119902+ 1205791120596AO119902) 120596

AO119902119910119902+1

minus 1198911015840

2(119909119902+ 1205792120596AO119902) 120596

AO119902119910119902minus 1198912(119909119902) 120596

AO119902

= 120576119902+2

minus 120596AO119902

sdot [

[

1198912(119909119902) +

2

sum

119895=1

1198911015840

119895(119909lowast

119902119895) 119910119902+2minus119895

]

]

(10)

where 119909lowast119902119895

= 119909119902+ 120579119895120596AO119902

0 lt 120579119895lt 1 and 119895 = 1 2 If 119896 gt 2

then 120578119902+119896

= 120576119902+119896

from which it follows that

120578119902+119896

= 119888119896120596AO119902

+ 120576119902+119896

119896 = 1 2 (11)

where 1198881= minus1198911(119910119902minus1) and 119888

2= minus[119891

2(119909119902)+sum2

119895=11198911015840

119895(119909lowast

119902119895)119910119902+2minus119895

]so we have that

119899

sum

119905=3

1205762

119905=

119902minus1

sum

119905=3

1205762

119905+ 1205762

119902+

119902+2

sum

119905=119902+1

1205762

119905+

119899

sum

119905=119902+3

1205762

119905

=

119902minus1

sum

119905=3

1205782

119905+ (120578119902minus 120596

AO119902)2

+

2

sum

119896=1

(120578119902+119896

minus 119888119896120596AO119902)2

+

119899

sum

119905=119902+3

1205782

119905

=

119902minus1

sum

119905=3

1205782

119905+

119899

sum

119905=119902+3

1205782

119905+ (1 + 119888

2

1+ 1198882

2) 120596

AO119902

2

minus 2 (120578119902+ 1198881120578119902+1

+ 1198882120578119902+2) 120596

AO119902

+ (1205782

119902+ 1205782

119902+1+ 1205782

119902+2)

(12)

By minimizing the above expression we obtain

120596AO119902

=120578119902+ 1198881120578119902+1

+ 1198882120578119902+2

1 + 11988821+ 11988822

(13)

where 1198881may be obtained by estimation of minus119891

1(119910119902minus1) while

1198882is complicated and difficult to be confirmed Nevertheless

Battaglia [8] indicated that we could estimate 1198882by minus119891

2(119910119902)

which is convenient and effective If all 119891119895(sdot) and 120590

2 areknown then we have

AO119902

=120578119902minus 1198911(119910119902minus1) 120578119902+1

minus 1198912(119910119902) 120578119902+2

Δ119902

Δ119902= 1 + 119891

2

1(119910119902minus1) + 1198912

2(119910119902)

(14)

Similar to Battaglia [8] we obtain

AO119902

sim 119873(120596AO1199021205902

Δ119902

) (15)

Under the null hypothesis11986710158400 120596

AO119902

= 0 we obtain likelihood

radio test statistics AO119902

= (AO119902(120590radicΔ 119902)) sim 119873(0 1) by stan-

dardizing AO119902

In practice if119891119895(sdot) and1205902 are unknownwhich

can be replaced by their consistent estimations 2AO = (119899 minus

2)minus1[sum119899

119905=32

119905minus (

AO119902)2sdot sum2

119895=12

119895(119910119902+119895minus2

)] we likewise obtain

AO119902

=AO119902

AOradicΔ119902

sim 119873 (0 1) (16)

It is indicated by (6) and (13) that the best estimate of 120596is 120578119902at time 119902 for the case of IO the best estimate of 120596 is

the linear combination of errors 120578119902 120578119902+1

and 120578119902+2

at time 119902in the case of AO If the location 119902 of outlier is known thedistributions of likelihood ratio test statistics

IO119902

and AO119902

aregiven by (7) and (16) However in practice the location ofoutlier is unknown thus we should inspect the magnitudeof test statistic at every time point By replacing

IO119902

and

AO119902

with IO119905

and AO119905

the following theorem indicates that

the processes of IO119905 and

AO119905 are both Gaussian process

respectively

Theorem 1 (1) Under the hypothesis 1198670 120596

IO119905

= 0 that isthere is no IO at any time IO

119905 is a stationary zero-mean

Gaussian process with variance 1205902 and its autocovariancefunction is 119903IO(ℎ) = 120590

2sdot 119868 (ℎ = 0) where 119868

119860(119909) is the index

function If 119909 isin 119860 then 119868119860(119909) = 1 and else 119868

119860(119909) = 0

4 Mathematical Problems in Engineering

(2) Under the hypothesis 11986710158400 120596

AO119905

= 0 that is there isno AO at any time AO

119905 is a stationary zero-mean Gaussian

process with variance 1205902Δ119905 and its autocovariance function

is

119903AO

(119905 ℎ) = 1205902Δminus1

119905Δminus1

119905+ℎ[

[

2

sum

119895=ℎ+1

119891119895(119910119905+119895minus2

) 119891119895minusℎ

(119910119905+119895minus2

)

minus119891ℎ(119910119905+ℎminus2

) ]

]

119868 (1 le ℎ le 2)

(17)

Proof The Gaussian character is shown by (7) and (16) Wederive their autocovariance function as follows

(1) Under the null hypothesis1198670 120596

IO119905= 0 we have 120578

119905=

120576119905and

119903IO(ℎ) = cov (IO

119905

IO119905+ℎ) = cov (120578

119905 120578119905+ℎ)

= cov (120576119905 120576119905+ℎ) = 1205902sdot 119868 (ℎ = 0)

(18)

(2) Under the null hypothesis 11986710158400 120596

AO119905

= 0 we have120578119905= 120576119905and

119903AO

(119905 ℎ) = cov (AO119905

AO119905+ℎ)

= cov(Δminus1119905(120578119905minus

2

sum

119895=1

119891119895(119910119905+119895minus2

) 120578119905+119895)

Δminus1

119905+ℎ(120578119905+ℎ

minus

2

sum

119895=1

119891119895(119910119905+ℎ+119895minus2

) 120578119905+ℎ+119895

))

= Δminus1

119905Δminus1

119905+ℎ[

[

cov (120578119905 120578119905+ℎ)

minus

2

sum

119895=1

119891119895(119910119905+ℎ+119895minus2

) cov (120578119905 120578119905+ℎ+119895

)

minus

2

sum

119895=1

119891119895(119910119905+119895minus2

) cov (120578119905+119895 120578119905+ℎ)

+ cov(2

sum

119895=1

119891119895(119910119905+119895minus2

) 120578119905+119895

2

sum

119895=1

119891119895(119910119905+ℎ+119895minus2

) 120578119905+ℎ+119895

)]

]

= 1205902Δminus1

119905Δminus1

119905+ℎ[

[

2

sum

119895=ℎ+1

119891119895(119910119905+119895minus2

) 119891119895minusℎ

(119910119905+119895minus2

)

minus119891ℎ(119910119905+ℎminus2

) ]

]

119868 (1 le ℎ le 2)

(19)

We know that IO119905 is not only Gaussian but also a

stationary process byTheorem 1 Also it is serial independentunder the null hypothesis 119867

0 120596

IO119905

= 0 However despitethat AO

119905 is Gaussian it is not stationary due to the fact

that its autocovariance function 119903AO(119905 ℎ) is dependent on 119905Besides AO

119905 is truncated in finite steps which means that

its maximum correlation is within two steps that is AO119905

isindependent with AO

119905+ℎwhen ℎ gt 2 By standardizing IO

119905

and AO119905 we can obtain

IO119905 and

AO119905 It is obvious that

IO119905 is a standardized stationary Gaussian process under the

hypothesis1198670 120596

IO119905= 0 Also

AO119905 is a standardized Gaus-

sian process under the hypothesis11986710158400 120596

AO119905

= 0 but it is notstationary whereas its autocovariance function is truncatedin finite steps

3 Detection of Outliers Based on ExtremeValue Theory

As similar to Chareka et al [14] and Leadbetter and Rootzen[18] we obtain the following

Lemma 2 (see [18]) Assume that 119883119899 is a stationary zero-

mean Gaussian time series and its autocorrelation functionis 120588(ℎ) Let 119885

119899= max119883

1 1198832 119883

119899 and if the Berman

condition limℎrarrinfin

120588(ℎ) sdot log ℎ = 0 is satisfied then one hasthat 119875(((119885

119899minus 119888119899)119889119899) le 119909) rarr Λ(119909) = expminus119890minus119909 119899 rarr infin

where 119889119899= (2 log 119899)minus12 and 119888

119899= (1119889

119899) minus (119889

1198992)(log log 119899 +

log(4120587))

Lemma 3 (see [18]) Assume that 119883119899 is a stationary zero-

mean Gaussian time series and let 119860119899= max119899

119905=1|119883119905| and

1198852119899

= max2119899119905=1119883119905 and if 119883

119899satisfies the Berman condition

then 119860119899and 119885

2119899have the same distribution

Theorem 4 Let 120578IO = max119905=12119899

|IO119905| and 120578AO =

max119905=12119899

|AO119905| and under the hypothesis 119867

0 120596

IO119905

= 0

and11986710158400 120596

AO119905

= 0 one has that

119875(120578IO minus 11988821198991198892119899

le 119909) 997888rarr Λ (119909)

119875 (120578AO minus 11988821198991198892119899

le 119909) 997888rarr Λ (119909) 119899 997888rarr infin

(20)

where 1198892119899

= [2 log(2119899)]minus12 and 1198882119899

= (11198892119899) minus (119889

2119899

2)[log log(2119899) + log(4120587)]

Proof It follows from Theorem 1 that the autocorrelationfunction of

IO119905 is 120588IO(ℎ) = 119868 (ℎ = 0) under the hypothesis

1198670 120596

IO119905

= 0 so limℎrarrinfin

120588IO(ℎ) sdot log ℎ = 0 that is the Ber-

man condition is satisfied Again IO119905 is a stationary zero-

mean Gaussian time series Hence from Lemma 2 we havethat 119875(((max119899

119905=1

IO119905 minus 119888119899)119889119899) le 119909) rarr Λ(119909) Again from

Lemma 3 we know that the distribution of 120578IO =

max119899119905=1|

IO119905| is the same as that of max2119899

119905=1

IO119905 so we

Mathematical Problems in Engineering 5

have that 119875(((120578IO minus 1198882119899)1198892119899) le 119909) rarr Λ(119909) Similarly underthe hypothesis1198671015840

0 120596

AO119905

= 0 from Theorem 1 we know thatthe autocorrelation function of

AO119905

is

120588AO

(119905 ℎ) = Δminus12

119905sdot Δminus12

119905+ℎsdot [

[

2

sum

119895=ℎ+1

119891119895(119910119905+119895minus2

) 119891119895minusℎ

(119910119905+119895minus2

)

minus119891ℎ(119910119905+ℎminus2

) ]

]

119868 (1 le ℎ le 2)

(21)

where limℎrarrinfin

120588AO(ℎ) sdot log ℎ = 0 that is the Berman

condition is satisfied Therefore similar to the statementsabove it must hold that 119875(((120578AO minus 119888

2119899)1198892119899) le 119909) rarr

Λ(119909)

Lemma 5 (see [14]) Assume that 119883119899 is a stationary zero-

mean Gaussian time series with variance 1 and autocorrelationfunction 120588(ℎ) and let 119872

119899= max1198832

1 119883

2

119899 and if 120588(ℎ)

satisfies the Berman condition then one has 119875(((119872119899minus119890119899)2) le

119909) rarr Λ(119909) 119899 rarr infin where 119890119899= 2 log 119899 minus log(log 119899) minus log120587

Theorem 6 Let 120599IO = max119899119905=1(

IO119905)2 and 120599AO =

max119899119905=1(

AO119905)2 under the hypothesis 119867

0 120596

IO119905

= 0 and1198671015840

0 120596

AO119905

= 0 and if the corresponding Berman conditionsfor

IO119905 and

AO119905 are satisfied then one has

119875(120599IO minus 119890119899

2le 119909) 997888rarr Λ (119909)

119875 (120599AO minus 119890119899

2le 119909) 997888rarr Λ (119909) 119899 997888rarr infin

(22)

where 119890119899= 2 log 119899 minus log(log 119899) minus log120587

Proof We know that the Berman conditions for 120582IO119905 and

120582AO119905 are satisfied under the hypothesis 119867

0 120596

IO119905

= 0 and1198671015840

0 120596

AO119905

= 0 by the inferential process of Theorem 4 Thusthe conclusion is also obtained by Lemma 5

Let 120572 be the test significance level and 119910119899= (1 + (1 minus

120572)1119899)2Wedenote the 119905-distribution functionwith 119899degrees

of freedom by 119867119899(119905) Let 119909

119899= 119867minus1

119899minus2(119910119899) and 119890

119899= (119899(119899 minus

1))1199092

119899minus 2Λminus1(1 minus 120572) where Λminus1(1 minus 120572) is the 1 minus 120572 quantile of

the Gumbel distribution Similarly we have the following

Theorem 7 Under the conditions above one has that

119875(119872119899minus 119890119899

2le 119909) 997888rarr Λ (119909)

119875 (120599IO minus 119890119899

2le 119909) 997888rarr Λ (119909)

119875 (120599AO minus 119890119899

2le 119909) 997888rarr Λ (119909) 119899 997888rarr infin

(23)

At this point it is convenient to introduce two morepieces of notation 119860IO

119905= ((|

IO119905| minus 1198882119899)1198892119899) and 119860

AO119905

=

((|AO119905|minus 1198882119899)1198892119899) which we call absolute value test statistics

119878IO119905

= (((IO119905)2minus 119890119899)2) and 119878AO

119905= (((

AO119905)2minus 119890119899)2) which

we call square test statistics 119862IO119905

= (((IO119905)2minus 119890119899)2) and

119862AO119905

= (((AO119905)2minus 119890119899)2) which we call adjusted square test

statistic We likewise obtain

119875(119899max119905=1

119860IO119905 le 119909) 997888rarr Λ (119909)

119875 (119899max119905=1

119860AO119905 le 119909) 997888rarr Λ (119909)

119875 (119899max119905=1

119878IO119905 le 119909) 997888rarr Λ (119909) 119899 997888rarr infin

119875(119899max119905=1

119878AO119905 le 119909) 997888rarr Λ (119909)

119875 (119899max119905=1

119862IO119905 le 119909) 997888rarr Λ (119909)

119875 (119899max119905=1

119862AO119905 le 119909) 997888rarr Λ (119909) 119899 997888rarr infin

(24)

If there are several outliers the main idea is to first detectthemaximum outlier by using ourmethod and to obtain newtest statistics by deleting its effectThen we go on to detect thenext outlier and repeat the procedure till there is no outlierWhen two types of outliers appear the test is

1198670 120596

IO119905=0 120596

AO119905

=0 versus 1198671 120596

IO119905

= 0 or 120596AO119905

= 0

119905 = 1 119899

(25)

For details we provide the following steps using absolutevalue test statistics to detect the AO and IO in AFAR modelas follows (a) Let test statistics 119879

1= max119899

119905=1119860

IO119905 and 119879

2=

max119899119905=1119860

AO119905 (b) Calculate

IO119905

and AO119905

at every time pointand calculate themaximum 119905

1= 119860

IO1199021

= max119899119905=1119860

IO119905 and 119905

2=

119860AO1199022

= max119899119905=1119860

AO119905 of absolute value test statistics (c) Let

119862119881 = Λminus1(1 minus 120572) = minus log(minus log(1 minus 120572)) if 119905

1= max119905

1 1199052 gt

119862119881 then we believe that the observation is an IO at 119905 = 1199021

and if 1199052= max119905

1 1199052 gt 119862119881 then we believe that the

observation is an AO at 119905 = 1199022 else we believe there is no

IO or AO (d) Calculate the 119901-value 1199011= 1198751198670

(1198791ge 1199051) =

1 minusΛ(1199051) and 119901

2= 1198751198670

(1198792ge 1199052) = 1 minusΛ(119905

2) and if 119901

1le 120572 or

1199012le 120572 then reject the hypothesis119867

0 believing there is an IO

or an AO Furthermore we decide whether it is an IO or anAO by its minimal 119901-value (e) Delete the effect of detectedoutliers and detect the next outlier Repeat the above steps tillthere is no outlier

4 Simulation Studies

Example 8 In this simulation we consider the followingmodel where there is only one IO The sample sizes are

6 Mathematical Problems in Engineering

40

30

20

10

0

minus10

0 20 40 60 80 100(a)

80

60

40

20

0

minus20

0 20 40 60 80 100(b)

80

60

40

20

0

minus20

0 20 40 60 80 100(c)

Figure 1 Example 9 first detection

2520151050minus5

minus10

0 20 40 60 80 100(a)

50

40

30

20

10

0

minus10

0 20 40 60 80 100(b)

50

40

30

20

10

0

minus10

0 20 40 60 80 100(c)

Figure 2 Example 9 second detection

119899 = 50 100 500 1000 respectively and 120596IO1198992

= 4 We have

119910119905=cos (15119910

119905minus2)

1 + 1199102119905minus2

119910119905minus1

+ (minus exp(minus1199102

119905minus2

2))119910119905minus2

+ 120578119905

120578119905= 120576119905+ 120596

IO1198992

sdot 120575 (119905 minus119899

2) 120576

119905sim 119873 (0 05)

(26)

For 119899 = 50 All the three test statistics detect an outlierat 119905 = 25 The 119901-values for believing it is an IO are 119901(119860IO

) =

48414119890 minus 6 119901(119878IO) = 18326119890 minus 8 and 119901(119862IO) = 33739119890 minus

8 respectively The 119901-values for believing it is an AO are119901(119860

AO) = 31509119890 minus 4 119901(119878AO) = 47392119890 minus 5 and 119901(119862AO

) =

87247119890minus5 respectively Because the 119901-values for believing itis an IO are smaller than the119901-values for believing it is anAOwe believe it is an IO and its influence magnitude is 31994

Mathematical Problems in Engineering 7

10

5

0

minus5

minus10

0 20 40 60 80 100(a)

0 20 40 60 80 100

1086420minus2

minus4

(b)

0 20 40 60 80 100

1086420minus2

minus4

(c)

Figure 3 Example 9 fifth detection

420minus2

minus4

minus6

minus8

minus10

0 20 40 60 80 100(a)

0 20 40 60 80 100

3210minus1

minus2

minus3

minus4

(b)

0 20 40 60 80 100

3210minus1

minus2

minus3

minus4

(c)

Figure 4 Example 9 sixth detection

The results are similar for 119899 = 100 500 1000 respectively wethus omit the details here For different test statistics their 119901-values for believing it is an IO are summarized as follows inTable 1

Example 9 There are four IOs and one AO in the time seriesThese IOs appear at 119905 = 49 119905 = 50 119905 = 51 and 119905 = 52

sequentially Their sizes are 3 6 5 and 4 respectively One

AO appears at 119905 = 70 alone and its size is 7 119899 = 100 Themodel is as follows

119909119905=cos (15119909

119905minus2)

1 + 1199092119905minus2

119909119905minus1

+ (minus exp (minus051199092119905minus2)) 119909119905minus2

+ 120576119905 120576119905sim 119873 (0 06)

(27)

where 119910119905is obtained by (3) or (4)

8 Mathematical Problems in Engineering

Table 1

119901-values Absolute valuestatistic Square statistic Adjusted square

statistic119899 = 50 48414119890 minus 6 18326119890 minus 8 33739119890 minus 8

119899 = 100 10027119890 minus 6 16592119890 minus 9 21995119890 minus 9

119899 = 500 19065119890 minus 6 26461119890 minus 8 260394119890 minus 8

119899 = 1000 28048119890 minus 7 13358119890 minus 9 12590119890 minus 9

First we detect an AO at 119905 = 70 and its size is 68904The three test statistics and their corresponding critical valuesare shown in Figure 1 where Figure 1(a) corresponds tothe absolute value test statistics Figure 1(b) corresponds tosquare test statistics and Figure 1(c) corresponds to adjustedsquare test statistics Symbol ldquordquo denotes IO and ldquolowastrdquo denotesAO Parallel broken line denotes critical value which isuniform in the figures Deleting the effect of this AO and thencontinuing to detect other outliers in the series we observean IO at 119905 = 50 whose size is 60285 see details in Figure 2Deleting the effect of the aforementioned two outliers andthen continuing to detect other outliers in the series weobserve an IO at 119905 = 51 whose size is 48119 Deletingthe effect of the aforementioned three outliers and thencontinuing to detect other outliers in the series we observean IO at 119905 = 52 whose size is 40024 Deleting the effectof the aforementioned four outliers and then continuing todetect other outlier in the series we detect an IO at 119905 =

49 whose size is 29046 see Figure 3 Deleting the effectof the aforementioned five outliers and then continuing todetect other outliers in the series we have not detected anyoutlier see Figure 4 Considering the length of paper we omitsome figures here The result is consistent with the advanceenactment

5 Conclusions

The FAR model is mainly featured by the model-dependent-variable which in one way or another limits the scope ofits applications As a generalization of the class of modelsAFAR model clearly covers a larger range of objects thanthe FAR model which makes it possible to reduce modelingbiases [17] via choosing a proper model-dependent direc-tion This paper is concerned with detecting AO and IOin AFAR models using extreme value methods We derivethe asymptotic distribution of test statistics and provide acontrol for significance level which serves as an extensionand improvement of existing methods Based on severalsimulation studies we give conclusion remarks as follows(a)The extreme value method for detecting outliers in AFARmodels is tractable and effective not only for IO and AO butalso for separate outliers and outlier patches Furthermoreit is shown that our method can reduce possible effects ofmasking and swamping (b) When applying extreme valuetheory to detect outliers with a relatively small size samplesat hand the employment of square test statistics works betterthan that of adjusted square test statistics as well as absolutevalue test statistics While following the increments in thesamples size the detecting effect of adjusted square test

statistics also increases against that of square test statistics andabsolute value test statistics (eg see Table 1) (c) Selection ofmodel parameters and the magnitude of outliers have hugeinfluences on the effect of detection

Acknowledgments

The authors sincerely wish to thank the editor and the threereferees for their insightful suggestions which have led toimproving the early version of the paper The research issupported by the National Natural Science Foundation ofChina (11171065) the Natural Science Foundation of JiangsuProvince (BK2011058) and Research Fund for the DoctoralProgram of Higher Education of China (20120092110021)

References

[1] C C AggarwalOutlier Analysis Springer New York NY USA2013

[2] X JieM LiWZhao and S YChen ldquoBoundmaxima as a trafficfeature under DDOS flood attacksrdquo Mathematical Problems inEngineering vol 2012 Article ID 419319 20 pages 2012

[3] M Li andW Zhao ldquoOn bandlimitedness and lag-limitedness offractional Gaussian noiserdquo Physica A vol 392 no 9 pp 1955ndash1961 2013

[4] M Li Y-Q Chen J-Y Li and W Zhao ldquoHolder scales of sealevelrdquoMathematical Problems in Engineering Article ID 86370722 pages 2012

[5] C W S Chen ldquoDetection of additive outliers in bilinear timeseriesrdquo Computational Statistics amp Data Analysis vol 24 no 3pp 283ndash294 1997

[6] M Li ldquoFractal time seriesmdasha tutorial reviewrdquo MathematicalProblems in Engineering Article ID 157264 26 pages 2010

[7] Z Cai J Fan and Q Yao ldquoFunctional-coefficient regressionmodels for nonlinear time seriesrdquo Journal of the AmericanStatistical Association vol 95 no 451 pp 941ndash956 2000

[8] F Battaglia ldquoOutliers in functional autoregressive time seriesrdquoStatistics amp Probability Letters vol 72 no 4 pp 323ndash332 2005

[9] F Battaglia and L Orfei ldquoOutlier detection and estimation innonlinear time seriesrdquo Journal of Time Series Analysis vol 26no 1 pp 107ndash121 2005

[10] P Chen L Li Y Liu and J-G Lin ldquoDetection of outliers andpatches in bilinear time series modelsrdquoMathematical Problemsin Engineering Article ID 580583 10 pages 2010

[11] P Chen J Yang and L Y Li ldquoSynthetic Detection of ChangePoint andOutliers in Bilinear Time SeriesModelsrdquo InternationalJournal of Systems Science In press

[12] R A Martin ldquoExtreme value analysis of optimal level-crossingprediction for linear Gaussian processesrdquo Journal of Time SeriesAnalysis vol 33 no 4 pp 583ndash607 2012

[13] K Zhu and S Ling ldquoLikelihood ratio tests for the structuralchange of an AR(p)model to a threshold AR(p)modelrdquo Journalof Time Series Analysis vol 33 no 2 pp 223ndash232 2012

[14] P Chareka F Matarise and R Turner ldquoA test for additiveoutliers applicable to long-memory time seriesrdquo Journal ofEconomic Dynamics amp Control vol 30 no 4 pp 595ndash621 2006

[15] W-K Fung Z-Y Zhu B-C Wei and X He ldquoInfluencediagnostics and outlier tests for semiparametric mixedmodelsrdquoJournal of the Royal Statistical Society B vol 64 no 3 pp 565ndash579 2002

Mathematical Problems in Engineering 9

[16] A M Rıo Extreme value theory-based P values in time seriesoutlier detection [PhD thesis] University of Wisconsin Madi-son 2005

[17] J Fan and Q Yao Nonlinear time series Springer Series inStatistics Springer New York NY USA 2003

[18] M R Leadbetter andH Rootzen ldquoExtremal theory for stochas-tic processesrdquo The Annals of Probability vol 16 no 2 pp 431ndash478 1988

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Mathematical Problems in Engineering

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 3

When 119905 lt 119902 120578119905

= 119910119905minus sum2

119895=1119891119895(119910119905minus2)119910119905minus119895

= 119909119905minus

sum2

119895=1119891119895(119909119905minus2)119909119905minus119895

= 120576119905 if 119905 gt 119902 that is 119896 gt 0 then 120578

119902+119896=

119910119902+119896

minus sum2

119895=1119891119895(119910119902+119896minus2

)119910119902+119896minus119895

if 119896 = 1 we have that

120578119902+1

= 119910119902+1

minus

2

sum

119895=1

119891119895(119910119902minus1) 119910119902+1minus119895

= 119909119902+1

minus

2

sum

119895=1

119891119895(119909119902minus1) (119909119902+1minus119895

+ 120596AO119902120575 (1 minus 119895))

= 120576119902+1

minus

2

sum

119895=1

119891119895(119909119902minus1) 120596

AO119902120575 (1 minus 119895)

= 120576119902+1

minus 1198911(119909119902minus1) 120596

AO119902

= 120576119902+1

minus 1198911(119910119902minus1) 120596

AO119902

(9)

and if 119896 = 2 we have that

120578119902+2

= 119910119902+2

minus

2

sum

119895=1

119891119895(119910119902) 119910119902+2minus119895

= 119909119902+2

minus

2

sum

119895=1

119891119895(119909119902) 119909119902+2minus119895

+

2

sum

119895=1

119891119895(119909119902) 119909119902+2minus119895

minus

2

sum

119895=1

119891119895(119910119902) 119910119902+2minus119895

= 120576119902+2

minus (1198911(119910119902) minus 1198911(119909119902)) 119910119902+1

+ 1198912(119909119902) (119910119902minus 120596

AO119902) minus 1198912(119910119902) 119910119902

= 120576119902+2

minus 1198911015840

1(119909119902+ 1205791120596AO119902) 120596

AO119902119910119902+1

minus (1198912(119910119902) minus 1198912(119909119902)) 119910119902minus 1198912(119909119902) 120596

AO119902

= 120576119902+2

minus 1198911015840

1(119909119902+ 1205791120596AO119902) 120596

AO119902119910119902+1

minus 1198911015840

2(119909119902+ 1205792120596AO119902) 120596

AO119902119910119902minus 1198912(119909119902) 120596

AO119902

= 120576119902+2

minus 120596AO119902

sdot [

[

1198912(119909119902) +

2

sum

119895=1

1198911015840

119895(119909lowast

119902119895) 119910119902+2minus119895

]

]

(10)

where 119909lowast119902119895

= 119909119902+ 120579119895120596AO119902

0 lt 120579119895lt 1 and 119895 = 1 2 If 119896 gt 2

then 120578119902+119896

= 120576119902+119896

from which it follows that

120578119902+119896

= 119888119896120596AO119902

+ 120576119902+119896

119896 = 1 2 (11)

where 1198881= minus1198911(119910119902minus1) and 119888

2= minus[119891

2(119909119902)+sum2

119895=11198911015840

119895(119909lowast

119902119895)119910119902+2minus119895

]so we have that

119899

sum

119905=3

1205762

119905=

119902minus1

sum

119905=3

1205762

119905+ 1205762

119902+

119902+2

sum

119905=119902+1

1205762

119905+

119899

sum

119905=119902+3

1205762

119905

=

119902minus1

sum

119905=3

1205782

119905+ (120578119902minus 120596

AO119902)2

+

2

sum

119896=1

(120578119902+119896

minus 119888119896120596AO119902)2

+

119899

sum

119905=119902+3

1205782

119905

=

119902minus1

sum

119905=3

1205782

119905+

119899

sum

119905=119902+3

1205782

119905+ (1 + 119888

2

1+ 1198882

2) 120596

AO119902

2

minus 2 (120578119902+ 1198881120578119902+1

+ 1198882120578119902+2) 120596

AO119902

+ (1205782

119902+ 1205782

119902+1+ 1205782

119902+2)

(12)

By minimizing the above expression we obtain

120596AO119902

=120578119902+ 1198881120578119902+1

+ 1198882120578119902+2

1 + 11988821+ 11988822

(13)

where 1198881may be obtained by estimation of minus119891

1(119910119902minus1) while

1198882is complicated and difficult to be confirmed Nevertheless

Battaglia [8] indicated that we could estimate 1198882by minus119891

2(119910119902)

which is convenient and effective If all 119891119895(sdot) and 120590

2 areknown then we have

AO119902

=120578119902minus 1198911(119910119902minus1) 120578119902+1

minus 1198912(119910119902) 120578119902+2

Δ119902

Δ119902= 1 + 119891

2

1(119910119902minus1) + 1198912

2(119910119902)

(14)

Similar to Battaglia [8] we obtain

AO119902

sim 119873(120596AO1199021205902

Δ119902

) (15)

Under the null hypothesis11986710158400 120596

AO119902

= 0 we obtain likelihood

radio test statistics AO119902

= (AO119902(120590radicΔ 119902)) sim 119873(0 1) by stan-

dardizing AO119902

In practice if119891119895(sdot) and1205902 are unknownwhich

can be replaced by their consistent estimations 2AO = (119899 minus

2)minus1[sum119899

119905=32

119905minus (

AO119902)2sdot sum2

119895=12

119895(119910119902+119895minus2

)] we likewise obtain

AO119902

=AO119902

AOradicΔ119902

sim 119873 (0 1) (16)

It is indicated by (6) and (13) that the best estimate of 120596is 120578119902at time 119902 for the case of IO the best estimate of 120596 is

the linear combination of errors 120578119902 120578119902+1

and 120578119902+2

at time 119902in the case of AO If the location 119902 of outlier is known thedistributions of likelihood ratio test statistics

IO119902

and AO119902

aregiven by (7) and (16) However in practice the location ofoutlier is unknown thus we should inspect the magnitudeof test statistic at every time point By replacing

IO119902

and

AO119902

with IO119905

and AO119905

the following theorem indicates that

the processes of IO119905 and

AO119905 are both Gaussian process

respectively

Theorem 1 (1) Under the hypothesis 1198670 120596

IO119905

= 0 that isthere is no IO at any time IO

119905 is a stationary zero-mean

Gaussian process with variance 1205902 and its autocovariancefunction is 119903IO(ℎ) = 120590

2sdot 119868 (ℎ = 0) where 119868

119860(119909) is the index

function If 119909 isin 119860 then 119868119860(119909) = 1 and else 119868

119860(119909) = 0

4 Mathematical Problems in Engineering

(2) Under the hypothesis 11986710158400 120596

AO119905

= 0 that is there isno AO at any time AO

119905 is a stationary zero-mean Gaussian

process with variance 1205902Δ119905 and its autocovariance function

is

119903AO

(119905 ℎ) = 1205902Δminus1

119905Δminus1

119905+ℎ[

[

2

sum

119895=ℎ+1

119891119895(119910119905+119895minus2

) 119891119895minusℎ

(119910119905+119895minus2

)

minus119891ℎ(119910119905+ℎminus2

) ]

]

119868 (1 le ℎ le 2)

(17)

Proof The Gaussian character is shown by (7) and (16) Wederive their autocovariance function as follows

(1) Under the null hypothesis1198670 120596

IO119905= 0 we have 120578

119905=

120576119905and

119903IO(ℎ) = cov (IO

119905

IO119905+ℎ) = cov (120578

119905 120578119905+ℎ)

= cov (120576119905 120576119905+ℎ) = 1205902sdot 119868 (ℎ = 0)

(18)

(2) Under the null hypothesis 11986710158400 120596

AO119905

= 0 we have120578119905= 120576119905and

119903AO

(119905 ℎ) = cov (AO119905

AO119905+ℎ)

= cov(Δminus1119905(120578119905minus

2

sum

119895=1

119891119895(119910119905+119895minus2

) 120578119905+119895)

Δminus1

119905+ℎ(120578119905+ℎ

minus

2

sum

119895=1

119891119895(119910119905+ℎ+119895minus2

) 120578119905+ℎ+119895

))

= Δminus1

119905Δminus1

119905+ℎ[

[

cov (120578119905 120578119905+ℎ)

minus

2

sum

119895=1

119891119895(119910119905+ℎ+119895minus2

) cov (120578119905 120578119905+ℎ+119895

)

minus

2

sum

119895=1

119891119895(119910119905+119895minus2

) cov (120578119905+119895 120578119905+ℎ)

+ cov(2

sum

119895=1

119891119895(119910119905+119895minus2

) 120578119905+119895

2

sum

119895=1

119891119895(119910119905+ℎ+119895minus2

) 120578119905+ℎ+119895

)]

]

= 1205902Δminus1

119905Δminus1

119905+ℎ[

[

2

sum

119895=ℎ+1

119891119895(119910119905+119895minus2

) 119891119895minusℎ

(119910119905+119895minus2

)

minus119891ℎ(119910119905+ℎminus2

) ]

]

119868 (1 le ℎ le 2)

(19)

We know that IO119905 is not only Gaussian but also a

stationary process byTheorem 1 Also it is serial independentunder the null hypothesis 119867

0 120596

IO119905

= 0 However despitethat AO

119905 is Gaussian it is not stationary due to the fact

that its autocovariance function 119903AO(119905 ℎ) is dependent on 119905Besides AO

119905 is truncated in finite steps which means that

its maximum correlation is within two steps that is AO119905

isindependent with AO

119905+ℎwhen ℎ gt 2 By standardizing IO

119905

and AO119905 we can obtain

IO119905 and

AO119905 It is obvious that

IO119905 is a standardized stationary Gaussian process under the

hypothesis1198670 120596

IO119905= 0 Also

AO119905 is a standardized Gaus-

sian process under the hypothesis11986710158400 120596

AO119905

= 0 but it is notstationary whereas its autocovariance function is truncatedin finite steps

3 Detection of Outliers Based on ExtremeValue Theory

As similar to Chareka et al [14] and Leadbetter and Rootzen[18] we obtain the following

Lemma 2 (see [18]) Assume that 119883119899 is a stationary zero-

mean Gaussian time series and its autocorrelation functionis 120588(ℎ) Let 119885

119899= max119883

1 1198832 119883

119899 and if the Berman

condition limℎrarrinfin

120588(ℎ) sdot log ℎ = 0 is satisfied then one hasthat 119875(((119885

119899minus 119888119899)119889119899) le 119909) rarr Λ(119909) = expminus119890minus119909 119899 rarr infin

where 119889119899= (2 log 119899)minus12 and 119888

119899= (1119889

119899) minus (119889

1198992)(log log 119899 +

log(4120587))

Lemma 3 (see [18]) Assume that 119883119899 is a stationary zero-

mean Gaussian time series and let 119860119899= max119899

119905=1|119883119905| and

1198852119899

= max2119899119905=1119883119905 and if 119883

119899satisfies the Berman condition

then 119860119899and 119885

2119899have the same distribution

Theorem 4 Let 120578IO = max119905=12119899

|IO119905| and 120578AO =

max119905=12119899

|AO119905| and under the hypothesis 119867

0 120596

IO119905

= 0

and11986710158400 120596

AO119905

= 0 one has that

119875(120578IO minus 11988821198991198892119899

le 119909) 997888rarr Λ (119909)

119875 (120578AO minus 11988821198991198892119899

le 119909) 997888rarr Λ (119909) 119899 997888rarr infin

(20)

where 1198892119899

= [2 log(2119899)]minus12 and 1198882119899

= (11198892119899) minus (119889

2119899

2)[log log(2119899) + log(4120587)]

Proof It follows from Theorem 1 that the autocorrelationfunction of

IO119905 is 120588IO(ℎ) = 119868 (ℎ = 0) under the hypothesis

1198670 120596

IO119905

= 0 so limℎrarrinfin

120588IO(ℎ) sdot log ℎ = 0 that is the Ber-

man condition is satisfied Again IO119905 is a stationary zero-

mean Gaussian time series Hence from Lemma 2 we havethat 119875(((max119899

119905=1

IO119905 minus 119888119899)119889119899) le 119909) rarr Λ(119909) Again from

Lemma 3 we know that the distribution of 120578IO =

max119899119905=1|

IO119905| is the same as that of max2119899

119905=1

IO119905 so we

Mathematical Problems in Engineering 5

have that 119875(((120578IO minus 1198882119899)1198892119899) le 119909) rarr Λ(119909) Similarly underthe hypothesis1198671015840

0 120596

AO119905

= 0 from Theorem 1 we know thatthe autocorrelation function of

AO119905

is

120588AO

(119905 ℎ) = Δminus12

119905sdot Δminus12

119905+ℎsdot [

[

2

sum

119895=ℎ+1

119891119895(119910119905+119895minus2

) 119891119895minusℎ

(119910119905+119895minus2

)

minus119891ℎ(119910119905+ℎminus2

) ]

]

119868 (1 le ℎ le 2)

(21)

where limℎrarrinfin

120588AO(ℎ) sdot log ℎ = 0 that is the Berman

condition is satisfied Therefore similar to the statementsabove it must hold that 119875(((120578AO minus 119888

2119899)1198892119899) le 119909) rarr

Λ(119909)

Lemma 5 (see [14]) Assume that 119883119899 is a stationary zero-

mean Gaussian time series with variance 1 and autocorrelationfunction 120588(ℎ) and let 119872

119899= max1198832

1 119883

2

119899 and if 120588(ℎ)

satisfies the Berman condition then one has 119875(((119872119899minus119890119899)2) le

119909) rarr Λ(119909) 119899 rarr infin where 119890119899= 2 log 119899 minus log(log 119899) minus log120587

Theorem 6 Let 120599IO = max119899119905=1(

IO119905)2 and 120599AO =

max119899119905=1(

AO119905)2 under the hypothesis 119867

0 120596

IO119905

= 0 and1198671015840

0 120596

AO119905

= 0 and if the corresponding Berman conditionsfor

IO119905 and

AO119905 are satisfied then one has

119875(120599IO minus 119890119899

2le 119909) 997888rarr Λ (119909)

119875 (120599AO minus 119890119899

2le 119909) 997888rarr Λ (119909) 119899 997888rarr infin

(22)

where 119890119899= 2 log 119899 minus log(log 119899) minus log120587

Proof We know that the Berman conditions for 120582IO119905 and

120582AO119905 are satisfied under the hypothesis 119867

0 120596

IO119905

= 0 and1198671015840

0 120596

AO119905

= 0 by the inferential process of Theorem 4 Thusthe conclusion is also obtained by Lemma 5

Let 120572 be the test significance level and 119910119899= (1 + (1 minus

120572)1119899)2Wedenote the 119905-distribution functionwith 119899degrees

of freedom by 119867119899(119905) Let 119909

119899= 119867minus1

119899minus2(119910119899) and 119890

119899= (119899(119899 minus

1))1199092

119899minus 2Λminus1(1 minus 120572) where Λminus1(1 minus 120572) is the 1 minus 120572 quantile of

the Gumbel distribution Similarly we have the following

Theorem 7 Under the conditions above one has that

119875(119872119899minus 119890119899

2le 119909) 997888rarr Λ (119909)

119875 (120599IO minus 119890119899

2le 119909) 997888rarr Λ (119909)

119875 (120599AO minus 119890119899

2le 119909) 997888rarr Λ (119909) 119899 997888rarr infin

(23)

At this point it is convenient to introduce two morepieces of notation 119860IO

119905= ((|

IO119905| minus 1198882119899)1198892119899) and 119860

AO119905

=

((|AO119905|minus 1198882119899)1198892119899) which we call absolute value test statistics

119878IO119905

= (((IO119905)2minus 119890119899)2) and 119878AO

119905= (((

AO119905)2minus 119890119899)2) which

we call square test statistics 119862IO119905

= (((IO119905)2minus 119890119899)2) and

119862AO119905

= (((AO119905)2minus 119890119899)2) which we call adjusted square test

statistic We likewise obtain

119875(119899max119905=1

119860IO119905 le 119909) 997888rarr Λ (119909)

119875 (119899max119905=1

119860AO119905 le 119909) 997888rarr Λ (119909)

119875 (119899max119905=1

119878IO119905 le 119909) 997888rarr Λ (119909) 119899 997888rarr infin

119875(119899max119905=1

119878AO119905 le 119909) 997888rarr Λ (119909)

119875 (119899max119905=1

119862IO119905 le 119909) 997888rarr Λ (119909)

119875 (119899max119905=1

119862AO119905 le 119909) 997888rarr Λ (119909) 119899 997888rarr infin

(24)

If there are several outliers the main idea is to first detectthemaximum outlier by using ourmethod and to obtain newtest statistics by deleting its effectThen we go on to detect thenext outlier and repeat the procedure till there is no outlierWhen two types of outliers appear the test is

1198670 120596

IO119905=0 120596

AO119905

=0 versus 1198671 120596

IO119905

= 0 or 120596AO119905

= 0

119905 = 1 119899

(25)

For details we provide the following steps using absolutevalue test statistics to detect the AO and IO in AFAR modelas follows (a) Let test statistics 119879

1= max119899

119905=1119860

IO119905 and 119879

2=

max119899119905=1119860

AO119905 (b) Calculate

IO119905

and AO119905

at every time pointand calculate themaximum 119905

1= 119860

IO1199021

= max119899119905=1119860

IO119905 and 119905

2=

119860AO1199022

= max119899119905=1119860

AO119905 of absolute value test statistics (c) Let

119862119881 = Λminus1(1 minus 120572) = minus log(minus log(1 minus 120572)) if 119905

1= max119905

1 1199052 gt

119862119881 then we believe that the observation is an IO at 119905 = 1199021

and if 1199052= max119905

1 1199052 gt 119862119881 then we believe that the

observation is an AO at 119905 = 1199022 else we believe there is no

IO or AO (d) Calculate the 119901-value 1199011= 1198751198670

(1198791ge 1199051) =

1 minusΛ(1199051) and 119901

2= 1198751198670

(1198792ge 1199052) = 1 minusΛ(119905

2) and if 119901

1le 120572 or

1199012le 120572 then reject the hypothesis119867

0 believing there is an IO

or an AO Furthermore we decide whether it is an IO or anAO by its minimal 119901-value (e) Delete the effect of detectedoutliers and detect the next outlier Repeat the above steps tillthere is no outlier

4 Simulation Studies

Example 8 In this simulation we consider the followingmodel where there is only one IO The sample sizes are

6 Mathematical Problems in Engineering

40

30

20

10

0

minus10

0 20 40 60 80 100(a)

80

60

40

20

0

minus20

0 20 40 60 80 100(b)

80

60

40

20

0

minus20

0 20 40 60 80 100(c)

Figure 1 Example 9 first detection

2520151050minus5

minus10

0 20 40 60 80 100(a)

50

40

30

20

10

0

minus10

0 20 40 60 80 100(b)

50

40

30

20

10

0

minus10

0 20 40 60 80 100(c)

Figure 2 Example 9 second detection

119899 = 50 100 500 1000 respectively and 120596IO1198992

= 4 We have

119910119905=cos (15119910

119905minus2)

1 + 1199102119905minus2

119910119905minus1

+ (minus exp(minus1199102

119905minus2

2))119910119905minus2

+ 120578119905

120578119905= 120576119905+ 120596

IO1198992

sdot 120575 (119905 minus119899

2) 120576

119905sim 119873 (0 05)

(26)

For 119899 = 50 All the three test statistics detect an outlierat 119905 = 25 The 119901-values for believing it is an IO are 119901(119860IO

) =

48414119890 minus 6 119901(119878IO) = 18326119890 minus 8 and 119901(119862IO) = 33739119890 minus

8 respectively The 119901-values for believing it is an AO are119901(119860

AO) = 31509119890 minus 4 119901(119878AO) = 47392119890 minus 5 and 119901(119862AO

) =

87247119890minus5 respectively Because the 119901-values for believing itis an IO are smaller than the119901-values for believing it is anAOwe believe it is an IO and its influence magnitude is 31994

Mathematical Problems in Engineering 7

10

5

0

minus5

minus10

0 20 40 60 80 100(a)

0 20 40 60 80 100

1086420minus2

minus4

(b)

0 20 40 60 80 100

1086420minus2

minus4

(c)

Figure 3 Example 9 fifth detection

420minus2

minus4

minus6

minus8

minus10

0 20 40 60 80 100(a)

0 20 40 60 80 100

3210minus1

minus2

minus3

minus4

(b)

0 20 40 60 80 100

3210minus1

minus2

minus3

minus4

(c)

Figure 4 Example 9 sixth detection

The results are similar for 119899 = 100 500 1000 respectively wethus omit the details here For different test statistics their 119901-values for believing it is an IO are summarized as follows inTable 1

Example 9 There are four IOs and one AO in the time seriesThese IOs appear at 119905 = 49 119905 = 50 119905 = 51 and 119905 = 52

sequentially Their sizes are 3 6 5 and 4 respectively One

AO appears at 119905 = 70 alone and its size is 7 119899 = 100 Themodel is as follows

119909119905=cos (15119909

119905minus2)

1 + 1199092119905minus2

119909119905minus1

+ (minus exp (minus051199092119905minus2)) 119909119905minus2

+ 120576119905 120576119905sim 119873 (0 06)

(27)

where 119910119905is obtained by (3) or (4)

8 Mathematical Problems in Engineering

Table 1

119901-values Absolute valuestatistic Square statistic Adjusted square

statistic119899 = 50 48414119890 minus 6 18326119890 minus 8 33739119890 minus 8

119899 = 100 10027119890 minus 6 16592119890 minus 9 21995119890 minus 9

119899 = 500 19065119890 minus 6 26461119890 minus 8 260394119890 minus 8

119899 = 1000 28048119890 minus 7 13358119890 minus 9 12590119890 minus 9

First we detect an AO at 119905 = 70 and its size is 68904The three test statistics and their corresponding critical valuesare shown in Figure 1 where Figure 1(a) corresponds tothe absolute value test statistics Figure 1(b) corresponds tosquare test statistics and Figure 1(c) corresponds to adjustedsquare test statistics Symbol ldquordquo denotes IO and ldquolowastrdquo denotesAO Parallel broken line denotes critical value which isuniform in the figures Deleting the effect of this AO and thencontinuing to detect other outliers in the series we observean IO at 119905 = 50 whose size is 60285 see details in Figure 2Deleting the effect of the aforementioned two outliers andthen continuing to detect other outliers in the series weobserve an IO at 119905 = 51 whose size is 48119 Deletingthe effect of the aforementioned three outliers and thencontinuing to detect other outliers in the series we observean IO at 119905 = 52 whose size is 40024 Deleting the effectof the aforementioned four outliers and then continuing todetect other outlier in the series we detect an IO at 119905 =

49 whose size is 29046 see Figure 3 Deleting the effectof the aforementioned five outliers and then continuing todetect other outliers in the series we have not detected anyoutlier see Figure 4 Considering the length of paper we omitsome figures here The result is consistent with the advanceenactment

5 Conclusions

The FAR model is mainly featured by the model-dependent-variable which in one way or another limits the scope ofits applications As a generalization of the class of modelsAFAR model clearly covers a larger range of objects thanthe FAR model which makes it possible to reduce modelingbiases [17] via choosing a proper model-dependent direc-tion This paper is concerned with detecting AO and IOin AFAR models using extreme value methods We derivethe asymptotic distribution of test statistics and provide acontrol for significance level which serves as an extensionand improvement of existing methods Based on severalsimulation studies we give conclusion remarks as follows(a)The extreme value method for detecting outliers in AFARmodels is tractable and effective not only for IO and AO butalso for separate outliers and outlier patches Furthermoreit is shown that our method can reduce possible effects ofmasking and swamping (b) When applying extreme valuetheory to detect outliers with a relatively small size samplesat hand the employment of square test statistics works betterthan that of adjusted square test statistics as well as absolutevalue test statistics While following the increments in thesamples size the detecting effect of adjusted square test

statistics also increases against that of square test statistics andabsolute value test statistics (eg see Table 1) (c) Selection ofmodel parameters and the magnitude of outliers have hugeinfluences on the effect of detection

Acknowledgments

The authors sincerely wish to thank the editor and the threereferees for their insightful suggestions which have led toimproving the early version of the paper The research issupported by the National Natural Science Foundation ofChina (11171065) the Natural Science Foundation of JiangsuProvince (BK2011058) and Research Fund for the DoctoralProgram of Higher Education of China (20120092110021)

References

[1] C C AggarwalOutlier Analysis Springer New York NY USA2013

[2] X JieM LiWZhao and S YChen ldquoBoundmaxima as a trafficfeature under DDOS flood attacksrdquo Mathematical Problems inEngineering vol 2012 Article ID 419319 20 pages 2012

[3] M Li andW Zhao ldquoOn bandlimitedness and lag-limitedness offractional Gaussian noiserdquo Physica A vol 392 no 9 pp 1955ndash1961 2013

[4] M Li Y-Q Chen J-Y Li and W Zhao ldquoHolder scales of sealevelrdquoMathematical Problems in Engineering Article ID 86370722 pages 2012

[5] C W S Chen ldquoDetection of additive outliers in bilinear timeseriesrdquo Computational Statistics amp Data Analysis vol 24 no 3pp 283ndash294 1997

[6] M Li ldquoFractal time seriesmdasha tutorial reviewrdquo MathematicalProblems in Engineering Article ID 157264 26 pages 2010

[7] Z Cai J Fan and Q Yao ldquoFunctional-coefficient regressionmodels for nonlinear time seriesrdquo Journal of the AmericanStatistical Association vol 95 no 451 pp 941ndash956 2000

[8] F Battaglia ldquoOutliers in functional autoregressive time seriesrdquoStatistics amp Probability Letters vol 72 no 4 pp 323ndash332 2005

[9] F Battaglia and L Orfei ldquoOutlier detection and estimation innonlinear time seriesrdquo Journal of Time Series Analysis vol 26no 1 pp 107ndash121 2005

[10] P Chen L Li Y Liu and J-G Lin ldquoDetection of outliers andpatches in bilinear time series modelsrdquoMathematical Problemsin Engineering Article ID 580583 10 pages 2010

[11] P Chen J Yang and L Y Li ldquoSynthetic Detection of ChangePoint andOutliers in Bilinear Time SeriesModelsrdquo InternationalJournal of Systems Science In press

[12] R A Martin ldquoExtreme value analysis of optimal level-crossingprediction for linear Gaussian processesrdquo Journal of Time SeriesAnalysis vol 33 no 4 pp 583ndash607 2012

[13] K Zhu and S Ling ldquoLikelihood ratio tests for the structuralchange of an AR(p)model to a threshold AR(p)modelrdquo Journalof Time Series Analysis vol 33 no 2 pp 223ndash232 2012

[14] P Chareka F Matarise and R Turner ldquoA test for additiveoutliers applicable to long-memory time seriesrdquo Journal ofEconomic Dynamics amp Control vol 30 no 4 pp 595ndash621 2006

[15] W-K Fung Z-Y Zhu B-C Wei and X He ldquoInfluencediagnostics and outlier tests for semiparametric mixedmodelsrdquoJournal of the Royal Statistical Society B vol 64 no 3 pp 565ndash579 2002

Mathematical Problems in Engineering 9

[16] A M Rıo Extreme value theory-based P values in time seriesoutlier detection [PhD thesis] University of Wisconsin Madi-son 2005

[17] J Fan and Q Yao Nonlinear time series Springer Series inStatistics Springer New York NY USA 2003

[18] M R Leadbetter andH Rootzen ldquoExtremal theory for stochas-tic processesrdquo The Annals of Probability vol 16 no 2 pp 431ndash478 1988

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Mathematical Problems in Engineering

(2) Under the hypothesis 11986710158400 120596

AO119905

= 0 that is there isno AO at any time AO

119905 is a stationary zero-mean Gaussian

process with variance 1205902Δ119905 and its autocovariance function

is

119903AO

(119905 ℎ) = 1205902Δminus1

119905Δminus1

119905+ℎ[

[

2

sum

119895=ℎ+1

119891119895(119910119905+119895minus2

) 119891119895minusℎ

(119910119905+119895minus2

)

minus119891ℎ(119910119905+ℎminus2

) ]

]

119868 (1 le ℎ le 2)

(17)

Proof The Gaussian character is shown by (7) and (16) Wederive their autocovariance function as follows

(1) Under the null hypothesis1198670 120596

IO119905= 0 we have 120578

119905=

120576119905and

119903IO(ℎ) = cov (IO

119905

IO119905+ℎ) = cov (120578

119905 120578119905+ℎ)

= cov (120576119905 120576119905+ℎ) = 1205902sdot 119868 (ℎ = 0)

(18)

(2) Under the null hypothesis 11986710158400 120596

AO119905

= 0 we have120578119905= 120576119905and

119903AO

(119905 ℎ) = cov (AO119905

AO119905+ℎ)

= cov(Δminus1119905(120578119905minus

2

sum

119895=1

119891119895(119910119905+119895minus2

) 120578119905+119895)

Δminus1

119905+ℎ(120578119905+ℎ

minus

2

sum

119895=1

119891119895(119910119905+ℎ+119895minus2

) 120578119905+ℎ+119895

))

= Δminus1

119905Δminus1

119905+ℎ[

[

cov (120578119905 120578119905+ℎ)

minus

2

sum

119895=1

119891119895(119910119905+ℎ+119895minus2

) cov (120578119905 120578119905+ℎ+119895

)

minus

2

sum

119895=1

119891119895(119910119905+119895minus2

) cov (120578119905+119895 120578119905+ℎ)

+ cov(2

sum

119895=1

119891119895(119910119905+119895minus2

) 120578119905+119895

2

sum

119895=1

119891119895(119910119905+ℎ+119895minus2

) 120578119905+ℎ+119895

)]

]

= 1205902Δminus1

119905Δminus1

119905+ℎ[

[

2

sum

119895=ℎ+1

119891119895(119910119905+119895minus2

) 119891119895minusℎ

(119910119905+119895minus2

)

minus119891ℎ(119910119905+ℎminus2

) ]

]

119868 (1 le ℎ le 2)

(19)

We know that IO119905 is not only Gaussian but also a

stationary process byTheorem 1 Also it is serial independentunder the null hypothesis 119867

0 120596

IO119905

= 0 However despitethat AO

119905 is Gaussian it is not stationary due to the fact

that its autocovariance function 119903AO(119905 ℎ) is dependent on 119905Besides AO

119905 is truncated in finite steps which means that

its maximum correlation is within two steps that is AO119905

isindependent with AO

119905+ℎwhen ℎ gt 2 By standardizing IO

119905

and AO119905 we can obtain

IO119905 and

AO119905 It is obvious that

IO119905 is a standardized stationary Gaussian process under the

hypothesis1198670 120596

IO119905= 0 Also

AO119905 is a standardized Gaus-

sian process under the hypothesis11986710158400 120596

AO119905

= 0 but it is notstationary whereas its autocovariance function is truncatedin finite steps

3 Detection of Outliers Based on ExtremeValue Theory

As similar to Chareka et al [14] and Leadbetter and Rootzen[18] we obtain the following

Lemma 2 (see [18]) Assume that 119883119899 is a stationary zero-

mean Gaussian time series and its autocorrelation functionis 120588(ℎ) Let 119885

119899= max119883

1 1198832 119883

119899 and if the Berman

condition limℎrarrinfin

120588(ℎ) sdot log ℎ = 0 is satisfied then one hasthat 119875(((119885

119899minus 119888119899)119889119899) le 119909) rarr Λ(119909) = expminus119890minus119909 119899 rarr infin

where 119889119899= (2 log 119899)minus12 and 119888

119899= (1119889

119899) minus (119889

1198992)(log log 119899 +

log(4120587))

Lemma 3 (see [18]) Assume that 119883119899 is a stationary zero-

mean Gaussian time series and let 119860119899= max119899

119905=1|119883119905| and

1198852119899

= max2119899119905=1119883119905 and if 119883

119899satisfies the Berman condition

then 119860119899and 119885

2119899have the same distribution

Theorem 4 Let 120578IO = max119905=12119899

|IO119905| and 120578AO =

max119905=12119899

|AO119905| and under the hypothesis 119867

0 120596

IO119905

= 0

and11986710158400 120596

AO119905

= 0 one has that

119875(120578IO minus 11988821198991198892119899

le 119909) 997888rarr Λ (119909)

119875 (120578AO minus 11988821198991198892119899

le 119909) 997888rarr Λ (119909) 119899 997888rarr infin

(20)

where 1198892119899

= [2 log(2119899)]minus12 and 1198882119899

= (11198892119899) minus (119889

2119899

2)[log log(2119899) + log(4120587)]

Proof It follows from Theorem 1 that the autocorrelationfunction of

IO119905 is 120588IO(ℎ) = 119868 (ℎ = 0) under the hypothesis

1198670 120596

IO119905

= 0 so limℎrarrinfin

120588IO(ℎ) sdot log ℎ = 0 that is the Ber-

man condition is satisfied Again IO119905 is a stationary zero-

mean Gaussian time series Hence from Lemma 2 we havethat 119875(((max119899

119905=1

IO119905 minus 119888119899)119889119899) le 119909) rarr Λ(119909) Again from

Lemma 3 we know that the distribution of 120578IO =

max119899119905=1|

IO119905| is the same as that of max2119899

119905=1

IO119905 so we

Mathematical Problems in Engineering 5

have that 119875(((120578IO minus 1198882119899)1198892119899) le 119909) rarr Λ(119909) Similarly underthe hypothesis1198671015840

0 120596

AO119905

= 0 from Theorem 1 we know thatthe autocorrelation function of

AO119905

is

120588AO

(119905 ℎ) = Δminus12

119905sdot Δminus12

119905+ℎsdot [

[

2

sum

119895=ℎ+1

119891119895(119910119905+119895minus2

) 119891119895minusℎ

(119910119905+119895minus2

)

minus119891ℎ(119910119905+ℎminus2

) ]

]

119868 (1 le ℎ le 2)

(21)

where limℎrarrinfin

120588AO(ℎ) sdot log ℎ = 0 that is the Berman

condition is satisfied Therefore similar to the statementsabove it must hold that 119875(((120578AO minus 119888

2119899)1198892119899) le 119909) rarr

Λ(119909)

Lemma 5 (see [14]) Assume that 119883119899 is a stationary zero-

mean Gaussian time series with variance 1 and autocorrelationfunction 120588(ℎ) and let 119872

119899= max1198832

1 119883

2

119899 and if 120588(ℎ)

satisfies the Berman condition then one has 119875(((119872119899minus119890119899)2) le

119909) rarr Λ(119909) 119899 rarr infin where 119890119899= 2 log 119899 minus log(log 119899) minus log120587

Theorem 6 Let 120599IO = max119899119905=1(

IO119905)2 and 120599AO =

max119899119905=1(

AO119905)2 under the hypothesis 119867

0 120596

IO119905

= 0 and1198671015840

0 120596

AO119905

= 0 and if the corresponding Berman conditionsfor

IO119905 and

AO119905 are satisfied then one has

119875(120599IO minus 119890119899

2le 119909) 997888rarr Λ (119909)

119875 (120599AO minus 119890119899

2le 119909) 997888rarr Λ (119909) 119899 997888rarr infin

(22)

where 119890119899= 2 log 119899 minus log(log 119899) minus log120587

Proof We know that the Berman conditions for 120582IO119905 and

120582AO119905 are satisfied under the hypothesis 119867

0 120596

IO119905

= 0 and1198671015840

0 120596

AO119905

= 0 by the inferential process of Theorem 4 Thusthe conclusion is also obtained by Lemma 5

Let 120572 be the test significance level and 119910119899= (1 + (1 minus

120572)1119899)2Wedenote the 119905-distribution functionwith 119899degrees

of freedom by 119867119899(119905) Let 119909

119899= 119867minus1

119899minus2(119910119899) and 119890

119899= (119899(119899 minus

1))1199092

119899minus 2Λminus1(1 minus 120572) where Λminus1(1 minus 120572) is the 1 minus 120572 quantile of

the Gumbel distribution Similarly we have the following

Theorem 7 Under the conditions above one has that

119875(119872119899minus 119890119899

2le 119909) 997888rarr Λ (119909)

119875 (120599IO minus 119890119899

2le 119909) 997888rarr Λ (119909)

119875 (120599AO minus 119890119899

2le 119909) 997888rarr Λ (119909) 119899 997888rarr infin

(23)

At this point it is convenient to introduce two morepieces of notation 119860IO

119905= ((|

IO119905| minus 1198882119899)1198892119899) and 119860

AO119905

=

((|AO119905|minus 1198882119899)1198892119899) which we call absolute value test statistics

119878IO119905

= (((IO119905)2minus 119890119899)2) and 119878AO

119905= (((

AO119905)2minus 119890119899)2) which

we call square test statistics 119862IO119905

= (((IO119905)2minus 119890119899)2) and

119862AO119905

= (((AO119905)2minus 119890119899)2) which we call adjusted square test

statistic We likewise obtain

119875(119899max119905=1

119860IO119905 le 119909) 997888rarr Λ (119909)

119875 (119899max119905=1

119860AO119905 le 119909) 997888rarr Λ (119909)

119875 (119899max119905=1

119878IO119905 le 119909) 997888rarr Λ (119909) 119899 997888rarr infin

119875(119899max119905=1

119878AO119905 le 119909) 997888rarr Λ (119909)

119875 (119899max119905=1

119862IO119905 le 119909) 997888rarr Λ (119909)

119875 (119899max119905=1

119862AO119905 le 119909) 997888rarr Λ (119909) 119899 997888rarr infin

(24)

If there are several outliers the main idea is to first detectthemaximum outlier by using ourmethod and to obtain newtest statistics by deleting its effectThen we go on to detect thenext outlier and repeat the procedure till there is no outlierWhen two types of outliers appear the test is

1198670 120596

IO119905=0 120596

AO119905

=0 versus 1198671 120596

IO119905

= 0 or 120596AO119905

= 0

119905 = 1 119899

(25)

For details we provide the following steps using absolutevalue test statistics to detect the AO and IO in AFAR modelas follows (a) Let test statistics 119879

1= max119899

119905=1119860

IO119905 and 119879

2=

max119899119905=1119860

AO119905 (b) Calculate

IO119905

and AO119905

at every time pointand calculate themaximum 119905

1= 119860

IO1199021

= max119899119905=1119860

IO119905 and 119905

2=

119860AO1199022

= max119899119905=1119860

AO119905 of absolute value test statistics (c) Let

119862119881 = Λminus1(1 minus 120572) = minus log(minus log(1 minus 120572)) if 119905

1= max119905

1 1199052 gt

119862119881 then we believe that the observation is an IO at 119905 = 1199021

and if 1199052= max119905

1 1199052 gt 119862119881 then we believe that the

observation is an AO at 119905 = 1199022 else we believe there is no

IO or AO (d) Calculate the 119901-value 1199011= 1198751198670

(1198791ge 1199051) =

1 minusΛ(1199051) and 119901

2= 1198751198670

(1198792ge 1199052) = 1 minusΛ(119905

2) and if 119901

1le 120572 or

1199012le 120572 then reject the hypothesis119867

0 believing there is an IO

or an AO Furthermore we decide whether it is an IO or anAO by its minimal 119901-value (e) Delete the effect of detectedoutliers and detect the next outlier Repeat the above steps tillthere is no outlier

4 Simulation Studies

Example 8 In this simulation we consider the followingmodel where there is only one IO The sample sizes are

6 Mathematical Problems in Engineering

40

30

20

10

0

minus10

0 20 40 60 80 100(a)

80

60

40

20

0

minus20

0 20 40 60 80 100(b)

80

60

40

20

0

minus20

0 20 40 60 80 100(c)

Figure 1 Example 9 first detection

2520151050minus5

minus10

0 20 40 60 80 100(a)

50

40

30

20

10

0

minus10

0 20 40 60 80 100(b)

50

40

30

20

10

0

minus10

0 20 40 60 80 100(c)

Figure 2 Example 9 second detection

119899 = 50 100 500 1000 respectively and 120596IO1198992

= 4 We have

119910119905=cos (15119910

119905minus2)

1 + 1199102119905minus2

119910119905minus1

+ (minus exp(minus1199102

119905minus2

2))119910119905minus2

+ 120578119905

120578119905= 120576119905+ 120596

IO1198992

sdot 120575 (119905 minus119899

2) 120576

119905sim 119873 (0 05)

(26)

For 119899 = 50 All the three test statistics detect an outlierat 119905 = 25 The 119901-values for believing it is an IO are 119901(119860IO

) =

48414119890 minus 6 119901(119878IO) = 18326119890 minus 8 and 119901(119862IO) = 33739119890 minus

8 respectively The 119901-values for believing it is an AO are119901(119860

AO) = 31509119890 minus 4 119901(119878AO) = 47392119890 minus 5 and 119901(119862AO

) =

87247119890minus5 respectively Because the 119901-values for believing itis an IO are smaller than the119901-values for believing it is anAOwe believe it is an IO and its influence magnitude is 31994

Mathematical Problems in Engineering 7

10

5

0

minus5

minus10

0 20 40 60 80 100(a)

0 20 40 60 80 100

1086420minus2

minus4

(b)

0 20 40 60 80 100

1086420minus2

minus4

(c)

Figure 3 Example 9 fifth detection

420minus2

minus4

minus6

minus8

minus10

0 20 40 60 80 100(a)

0 20 40 60 80 100

3210minus1

minus2

minus3

minus4

(b)

0 20 40 60 80 100

3210minus1

minus2

minus3

minus4

(c)

Figure 4 Example 9 sixth detection

The results are similar for 119899 = 100 500 1000 respectively wethus omit the details here For different test statistics their 119901-values for believing it is an IO are summarized as follows inTable 1

Example 9 There are four IOs and one AO in the time seriesThese IOs appear at 119905 = 49 119905 = 50 119905 = 51 and 119905 = 52

sequentially Their sizes are 3 6 5 and 4 respectively One

AO appears at 119905 = 70 alone and its size is 7 119899 = 100 Themodel is as follows

119909119905=cos (15119909

119905minus2)

1 + 1199092119905minus2

119909119905minus1

+ (minus exp (minus051199092119905minus2)) 119909119905minus2

+ 120576119905 120576119905sim 119873 (0 06)

(27)

where 119910119905is obtained by (3) or (4)

8 Mathematical Problems in Engineering

Table 1

119901-values Absolute valuestatistic Square statistic Adjusted square

statistic119899 = 50 48414119890 minus 6 18326119890 minus 8 33739119890 minus 8

119899 = 100 10027119890 minus 6 16592119890 minus 9 21995119890 minus 9

119899 = 500 19065119890 minus 6 26461119890 minus 8 260394119890 minus 8

119899 = 1000 28048119890 minus 7 13358119890 minus 9 12590119890 minus 9

First we detect an AO at 119905 = 70 and its size is 68904The three test statistics and their corresponding critical valuesare shown in Figure 1 where Figure 1(a) corresponds tothe absolute value test statistics Figure 1(b) corresponds tosquare test statistics and Figure 1(c) corresponds to adjustedsquare test statistics Symbol ldquordquo denotes IO and ldquolowastrdquo denotesAO Parallel broken line denotes critical value which isuniform in the figures Deleting the effect of this AO and thencontinuing to detect other outliers in the series we observean IO at 119905 = 50 whose size is 60285 see details in Figure 2Deleting the effect of the aforementioned two outliers andthen continuing to detect other outliers in the series weobserve an IO at 119905 = 51 whose size is 48119 Deletingthe effect of the aforementioned three outliers and thencontinuing to detect other outliers in the series we observean IO at 119905 = 52 whose size is 40024 Deleting the effectof the aforementioned four outliers and then continuing todetect other outlier in the series we detect an IO at 119905 =

49 whose size is 29046 see Figure 3 Deleting the effectof the aforementioned five outliers and then continuing todetect other outliers in the series we have not detected anyoutlier see Figure 4 Considering the length of paper we omitsome figures here The result is consistent with the advanceenactment

5 Conclusions

The FAR model is mainly featured by the model-dependent-variable which in one way or another limits the scope ofits applications As a generalization of the class of modelsAFAR model clearly covers a larger range of objects thanthe FAR model which makes it possible to reduce modelingbiases [17] via choosing a proper model-dependent direc-tion This paper is concerned with detecting AO and IOin AFAR models using extreme value methods We derivethe asymptotic distribution of test statistics and provide acontrol for significance level which serves as an extensionand improvement of existing methods Based on severalsimulation studies we give conclusion remarks as follows(a)The extreme value method for detecting outliers in AFARmodels is tractable and effective not only for IO and AO butalso for separate outliers and outlier patches Furthermoreit is shown that our method can reduce possible effects ofmasking and swamping (b) When applying extreme valuetheory to detect outliers with a relatively small size samplesat hand the employment of square test statistics works betterthan that of adjusted square test statistics as well as absolutevalue test statistics While following the increments in thesamples size the detecting effect of adjusted square test

statistics also increases against that of square test statistics andabsolute value test statistics (eg see Table 1) (c) Selection ofmodel parameters and the magnitude of outliers have hugeinfluences on the effect of detection

Acknowledgments

The authors sincerely wish to thank the editor and the threereferees for their insightful suggestions which have led toimproving the early version of the paper The research issupported by the National Natural Science Foundation ofChina (11171065) the Natural Science Foundation of JiangsuProvince (BK2011058) and Research Fund for the DoctoralProgram of Higher Education of China (20120092110021)

References

[1] C C AggarwalOutlier Analysis Springer New York NY USA2013

[2] X JieM LiWZhao and S YChen ldquoBoundmaxima as a trafficfeature under DDOS flood attacksrdquo Mathematical Problems inEngineering vol 2012 Article ID 419319 20 pages 2012

[3] M Li andW Zhao ldquoOn bandlimitedness and lag-limitedness offractional Gaussian noiserdquo Physica A vol 392 no 9 pp 1955ndash1961 2013

[4] M Li Y-Q Chen J-Y Li and W Zhao ldquoHolder scales of sealevelrdquoMathematical Problems in Engineering Article ID 86370722 pages 2012

[5] C W S Chen ldquoDetection of additive outliers in bilinear timeseriesrdquo Computational Statistics amp Data Analysis vol 24 no 3pp 283ndash294 1997

[6] M Li ldquoFractal time seriesmdasha tutorial reviewrdquo MathematicalProblems in Engineering Article ID 157264 26 pages 2010

[7] Z Cai J Fan and Q Yao ldquoFunctional-coefficient regressionmodels for nonlinear time seriesrdquo Journal of the AmericanStatistical Association vol 95 no 451 pp 941ndash956 2000

[8] F Battaglia ldquoOutliers in functional autoregressive time seriesrdquoStatistics amp Probability Letters vol 72 no 4 pp 323ndash332 2005

[9] F Battaglia and L Orfei ldquoOutlier detection and estimation innonlinear time seriesrdquo Journal of Time Series Analysis vol 26no 1 pp 107ndash121 2005

[10] P Chen L Li Y Liu and J-G Lin ldquoDetection of outliers andpatches in bilinear time series modelsrdquoMathematical Problemsin Engineering Article ID 580583 10 pages 2010

[11] P Chen J Yang and L Y Li ldquoSynthetic Detection of ChangePoint andOutliers in Bilinear Time SeriesModelsrdquo InternationalJournal of Systems Science In press

[12] R A Martin ldquoExtreme value analysis of optimal level-crossingprediction for linear Gaussian processesrdquo Journal of Time SeriesAnalysis vol 33 no 4 pp 583ndash607 2012

[13] K Zhu and S Ling ldquoLikelihood ratio tests for the structuralchange of an AR(p)model to a threshold AR(p)modelrdquo Journalof Time Series Analysis vol 33 no 2 pp 223ndash232 2012

[14] P Chareka F Matarise and R Turner ldquoA test for additiveoutliers applicable to long-memory time seriesrdquo Journal ofEconomic Dynamics amp Control vol 30 no 4 pp 595ndash621 2006

[15] W-K Fung Z-Y Zhu B-C Wei and X He ldquoInfluencediagnostics and outlier tests for semiparametric mixedmodelsrdquoJournal of the Royal Statistical Society B vol 64 no 3 pp 565ndash579 2002

Mathematical Problems in Engineering 9

[16] A M Rıo Extreme value theory-based P values in time seriesoutlier detection [PhD thesis] University of Wisconsin Madi-son 2005

[17] J Fan and Q Yao Nonlinear time series Springer Series inStatistics Springer New York NY USA 2003

[18] M R Leadbetter andH Rootzen ldquoExtremal theory for stochas-tic processesrdquo The Annals of Probability vol 16 no 2 pp 431ndash478 1988

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 5

have that 119875(((120578IO minus 1198882119899)1198892119899) le 119909) rarr Λ(119909) Similarly underthe hypothesis1198671015840

0 120596

AO119905

= 0 from Theorem 1 we know thatthe autocorrelation function of

AO119905

is

120588AO

(119905 ℎ) = Δminus12

119905sdot Δminus12

119905+ℎsdot [

[

2

sum

119895=ℎ+1

119891119895(119910119905+119895minus2

) 119891119895minusℎ

(119910119905+119895minus2

)

minus119891ℎ(119910119905+ℎminus2

) ]

]

119868 (1 le ℎ le 2)

(21)

where limℎrarrinfin

120588AO(ℎ) sdot log ℎ = 0 that is the Berman

condition is satisfied Therefore similar to the statementsabove it must hold that 119875(((120578AO minus 119888

2119899)1198892119899) le 119909) rarr

Λ(119909)

Lemma 5 (see [14]) Assume that 119883119899 is a stationary zero-

mean Gaussian time series with variance 1 and autocorrelationfunction 120588(ℎ) and let 119872

119899= max1198832

1 119883

2

119899 and if 120588(ℎ)

satisfies the Berman condition then one has 119875(((119872119899minus119890119899)2) le

119909) rarr Λ(119909) 119899 rarr infin where 119890119899= 2 log 119899 minus log(log 119899) minus log120587

Theorem 6 Let 120599IO = max119899119905=1(

IO119905)2 and 120599AO =

max119899119905=1(

AO119905)2 under the hypothesis 119867

0 120596

IO119905

= 0 and1198671015840

0 120596

AO119905

= 0 and if the corresponding Berman conditionsfor

IO119905 and

AO119905 are satisfied then one has

119875(120599IO minus 119890119899

2le 119909) 997888rarr Λ (119909)

119875 (120599AO minus 119890119899

2le 119909) 997888rarr Λ (119909) 119899 997888rarr infin

(22)

where 119890119899= 2 log 119899 minus log(log 119899) minus log120587

Proof We know that the Berman conditions for 120582IO119905 and

120582AO119905 are satisfied under the hypothesis 119867

0 120596

IO119905

= 0 and1198671015840

0 120596

AO119905

= 0 by the inferential process of Theorem 4 Thusthe conclusion is also obtained by Lemma 5

Let 120572 be the test significance level and 119910119899= (1 + (1 minus

120572)1119899)2Wedenote the 119905-distribution functionwith 119899degrees

of freedom by 119867119899(119905) Let 119909

119899= 119867minus1

119899minus2(119910119899) and 119890

119899= (119899(119899 minus

1))1199092

119899minus 2Λminus1(1 minus 120572) where Λminus1(1 minus 120572) is the 1 minus 120572 quantile of

the Gumbel distribution Similarly we have the following

Theorem 7 Under the conditions above one has that

119875(119872119899minus 119890119899

2le 119909) 997888rarr Λ (119909)

119875 (120599IO minus 119890119899

2le 119909) 997888rarr Λ (119909)

119875 (120599AO minus 119890119899

2le 119909) 997888rarr Λ (119909) 119899 997888rarr infin

(23)

At this point it is convenient to introduce two morepieces of notation 119860IO

119905= ((|

IO119905| minus 1198882119899)1198892119899) and 119860

AO119905

=

((|AO119905|minus 1198882119899)1198892119899) which we call absolute value test statistics

119878IO119905

= (((IO119905)2minus 119890119899)2) and 119878AO

119905= (((

AO119905)2minus 119890119899)2) which

we call square test statistics 119862IO119905

= (((IO119905)2minus 119890119899)2) and

119862AO119905

= (((AO119905)2minus 119890119899)2) which we call adjusted square test

statistic We likewise obtain

119875(119899max119905=1

119860IO119905 le 119909) 997888rarr Λ (119909)

119875 (119899max119905=1

119860AO119905 le 119909) 997888rarr Λ (119909)

119875 (119899max119905=1

119878IO119905 le 119909) 997888rarr Λ (119909) 119899 997888rarr infin

119875(119899max119905=1

119878AO119905 le 119909) 997888rarr Λ (119909)

119875 (119899max119905=1

119862IO119905 le 119909) 997888rarr Λ (119909)

119875 (119899max119905=1

119862AO119905 le 119909) 997888rarr Λ (119909) 119899 997888rarr infin

(24)

If there are several outliers the main idea is to first detectthemaximum outlier by using ourmethod and to obtain newtest statistics by deleting its effectThen we go on to detect thenext outlier and repeat the procedure till there is no outlierWhen two types of outliers appear the test is

1198670 120596

IO119905=0 120596

AO119905

=0 versus 1198671 120596

IO119905

= 0 or 120596AO119905

= 0

119905 = 1 119899

(25)

For details we provide the following steps using absolutevalue test statistics to detect the AO and IO in AFAR modelas follows (a) Let test statistics 119879

1= max119899

119905=1119860

IO119905 and 119879

2=

max119899119905=1119860

AO119905 (b) Calculate

IO119905

and AO119905

at every time pointand calculate themaximum 119905

1= 119860

IO1199021

= max119899119905=1119860

IO119905 and 119905

2=

119860AO1199022

= max119899119905=1119860

AO119905 of absolute value test statistics (c) Let

119862119881 = Λminus1(1 minus 120572) = minus log(minus log(1 minus 120572)) if 119905

1= max119905

1 1199052 gt

119862119881 then we believe that the observation is an IO at 119905 = 1199021

and if 1199052= max119905

1 1199052 gt 119862119881 then we believe that the

observation is an AO at 119905 = 1199022 else we believe there is no

IO or AO (d) Calculate the 119901-value 1199011= 1198751198670

(1198791ge 1199051) =

1 minusΛ(1199051) and 119901

2= 1198751198670

(1198792ge 1199052) = 1 minusΛ(119905

2) and if 119901

1le 120572 or

1199012le 120572 then reject the hypothesis119867

0 believing there is an IO

or an AO Furthermore we decide whether it is an IO or anAO by its minimal 119901-value (e) Delete the effect of detectedoutliers and detect the next outlier Repeat the above steps tillthere is no outlier

4 Simulation Studies

Example 8 In this simulation we consider the followingmodel where there is only one IO The sample sizes are

6 Mathematical Problems in Engineering

40

30

20

10

0

minus10

0 20 40 60 80 100(a)

80

60

40

20

0

minus20

0 20 40 60 80 100(b)

80

60

40

20

0

minus20

0 20 40 60 80 100(c)

Figure 1 Example 9 first detection

2520151050minus5

minus10

0 20 40 60 80 100(a)

50

40

30

20

10

0

minus10

0 20 40 60 80 100(b)

50

40

30

20

10

0

minus10

0 20 40 60 80 100(c)

Figure 2 Example 9 second detection

119899 = 50 100 500 1000 respectively and 120596IO1198992

= 4 We have

119910119905=cos (15119910

119905minus2)

1 + 1199102119905minus2

119910119905minus1

+ (minus exp(minus1199102

119905minus2

2))119910119905minus2

+ 120578119905

120578119905= 120576119905+ 120596

IO1198992

sdot 120575 (119905 minus119899

2) 120576

119905sim 119873 (0 05)

(26)

For 119899 = 50 All the three test statistics detect an outlierat 119905 = 25 The 119901-values for believing it is an IO are 119901(119860IO

) =

48414119890 minus 6 119901(119878IO) = 18326119890 minus 8 and 119901(119862IO) = 33739119890 minus

8 respectively The 119901-values for believing it is an AO are119901(119860

AO) = 31509119890 minus 4 119901(119878AO) = 47392119890 minus 5 and 119901(119862AO

) =

87247119890minus5 respectively Because the 119901-values for believing itis an IO are smaller than the119901-values for believing it is anAOwe believe it is an IO and its influence magnitude is 31994

Mathematical Problems in Engineering 7

10

5

0

minus5

minus10

0 20 40 60 80 100(a)

0 20 40 60 80 100

1086420minus2

minus4

(b)

0 20 40 60 80 100

1086420minus2

minus4

(c)

Figure 3 Example 9 fifth detection

420minus2

minus4

minus6

minus8

minus10

0 20 40 60 80 100(a)

0 20 40 60 80 100

3210minus1

minus2

minus3

minus4

(b)

0 20 40 60 80 100

3210minus1

minus2

minus3

minus4

(c)

Figure 4 Example 9 sixth detection

The results are similar for 119899 = 100 500 1000 respectively wethus omit the details here For different test statistics their 119901-values for believing it is an IO are summarized as follows inTable 1

Example 9 There are four IOs and one AO in the time seriesThese IOs appear at 119905 = 49 119905 = 50 119905 = 51 and 119905 = 52

sequentially Their sizes are 3 6 5 and 4 respectively One

AO appears at 119905 = 70 alone and its size is 7 119899 = 100 Themodel is as follows

119909119905=cos (15119909

119905minus2)

1 + 1199092119905minus2

119909119905minus1

+ (minus exp (minus051199092119905minus2)) 119909119905minus2

+ 120576119905 120576119905sim 119873 (0 06)

(27)

where 119910119905is obtained by (3) or (4)

8 Mathematical Problems in Engineering

Table 1

119901-values Absolute valuestatistic Square statistic Adjusted square

statistic119899 = 50 48414119890 minus 6 18326119890 minus 8 33739119890 minus 8

119899 = 100 10027119890 minus 6 16592119890 minus 9 21995119890 minus 9

119899 = 500 19065119890 minus 6 26461119890 minus 8 260394119890 minus 8

119899 = 1000 28048119890 minus 7 13358119890 minus 9 12590119890 minus 9

First we detect an AO at 119905 = 70 and its size is 68904The three test statistics and their corresponding critical valuesare shown in Figure 1 where Figure 1(a) corresponds tothe absolute value test statistics Figure 1(b) corresponds tosquare test statistics and Figure 1(c) corresponds to adjustedsquare test statistics Symbol ldquordquo denotes IO and ldquolowastrdquo denotesAO Parallel broken line denotes critical value which isuniform in the figures Deleting the effect of this AO and thencontinuing to detect other outliers in the series we observean IO at 119905 = 50 whose size is 60285 see details in Figure 2Deleting the effect of the aforementioned two outliers andthen continuing to detect other outliers in the series weobserve an IO at 119905 = 51 whose size is 48119 Deletingthe effect of the aforementioned three outliers and thencontinuing to detect other outliers in the series we observean IO at 119905 = 52 whose size is 40024 Deleting the effectof the aforementioned four outliers and then continuing todetect other outlier in the series we detect an IO at 119905 =

49 whose size is 29046 see Figure 3 Deleting the effectof the aforementioned five outliers and then continuing todetect other outliers in the series we have not detected anyoutlier see Figure 4 Considering the length of paper we omitsome figures here The result is consistent with the advanceenactment

5 Conclusions

The FAR model is mainly featured by the model-dependent-variable which in one way or another limits the scope ofits applications As a generalization of the class of modelsAFAR model clearly covers a larger range of objects thanthe FAR model which makes it possible to reduce modelingbiases [17] via choosing a proper model-dependent direc-tion This paper is concerned with detecting AO and IOin AFAR models using extreme value methods We derivethe asymptotic distribution of test statistics and provide acontrol for significance level which serves as an extensionand improvement of existing methods Based on severalsimulation studies we give conclusion remarks as follows(a)The extreme value method for detecting outliers in AFARmodels is tractable and effective not only for IO and AO butalso for separate outliers and outlier patches Furthermoreit is shown that our method can reduce possible effects ofmasking and swamping (b) When applying extreme valuetheory to detect outliers with a relatively small size samplesat hand the employment of square test statistics works betterthan that of adjusted square test statistics as well as absolutevalue test statistics While following the increments in thesamples size the detecting effect of adjusted square test

statistics also increases against that of square test statistics andabsolute value test statistics (eg see Table 1) (c) Selection ofmodel parameters and the magnitude of outliers have hugeinfluences on the effect of detection

Acknowledgments

The authors sincerely wish to thank the editor and the threereferees for their insightful suggestions which have led toimproving the early version of the paper The research issupported by the National Natural Science Foundation ofChina (11171065) the Natural Science Foundation of JiangsuProvince (BK2011058) and Research Fund for the DoctoralProgram of Higher Education of China (20120092110021)

References

[1] C C AggarwalOutlier Analysis Springer New York NY USA2013

[2] X JieM LiWZhao and S YChen ldquoBoundmaxima as a trafficfeature under DDOS flood attacksrdquo Mathematical Problems inEngineering vol 2012 Article ID 419319 20 pages 2012

[3] M Li andW Zhao ldquoOn bandlimitedness and lag-limitedness offractional Gaussian noiserdquo Physica A vol 392 no 9 pp 1955ndash1961 2013

[4] M Li Y-Q Chen J-Y Li and W Zhao ldquoHolder scales of sealevelrdquoMathematical Problems in Engineering Article ID 86370722 pages 2012

[5] C W S Chen ldquoDetection of additive outliers in bilinear timeseriesrdquo Computational Statistics amp Data Analysis vol 24 no 3pp 283ndash294 1997

[6] M Li ldquoFractal time seriesmdasha tutorial reviewrdquo MathematicalProblems in Engineering Article ID 157264 26 pages 2010

[7] Z Cai J Fan and Q Yao ldquoFunctional-coefficient regressionmodels for nonlinear time seriesrdquo Journal of the AmericanStatistical Association vol 95 no 451 pp 941ndash956 2000

[8] F Battaglia ldquoOutliers in functional autoregressive time seriesrdquoStatistics amp Probability Letters vol 72 no 4 pp 323ndash332 2005

[9] F Battaglia and L Orfei ldquoOutlier detection and estimation innonlinear time seriesrdquo Journal of Time Series Analysis vol 26no 1 pp 107ndash121 2005

[10] P Chen L Li Y Liu and J-G Lin ldquoDetection of outliers andpatches in bilinear time series modelsrdquoMathematical Problemsin Engineering Article ID 580583 10 pages 2010

[11] P Chen J Yang and L Y Li ldquoSynthetic Detection of ChangePoint andOutliers in Bilinear Time SeriesModelsrdquo InternationalJournal of Systems Science In press

[12] R A Martin ldquoExtreme value analysis of optimal level-crossingprediction for linear Gaussian processesrdquo Journal of Time SeriesAnalysis vol 33 no 4 pp 583ndash607 2012

[13] K Zhu and S Ling ldquoLikelihood ratio tests for the structuralchange of an AR(p)model to a threshold AR(p)modelrdquo Journalof Time Series Analysis vol 33 no 2 pp 223ndash232 2012

[14] P Chareka F Matarise and R Turner ldquoA test for additiveoutliers applicable to long-memory time seriesrdquo Journal ofEconomic Dynamics amp Control vol 30 no 4 pp 595ndash621 2006

[15] W-K Fung Z-Y Zhu B-C Wei and X He ldquoInfluencediagnostics and outlier tests for semiparametric mixedmodelsrdquoJournal of the Royal Statistical Society B vol 64 no 3 pp 565ndash579 2002

Mathematical Problems in Engineering 9

[16] A M Rıo Extreme value theory-based P values in time seriesoutlier detection [PhD thesis] University of Wisconsin Madi-son 2005

[17] J Fan and Q Yao Nonlinear time series Springer Series inStatistics Springer New York NY USA 2003

[18] M R Leadbetter andH Rootzen ldquoExtremal theory for stochas-tic processesrdquo The Annals of Probability vol 16 no 2 pp 431ndash478 1988

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Mathematical Problems in Engineering

40

30

20

10

0

minus10

0 20 40 60 80 100(a)

80

60

40

20

0

minus20

0 20 40 60 80 100(b)

80

60

40

20

0

minus20

0 20 40 60 80 100(c)

Figure 1 Example 9 first detection

2520151050minus5

minus10

0 20 40 60 80 100(a)

50

40

30

20

10

0

minus10

0 20 40 60 80 100(b)

50

40

30

20

10

0

minus10

0 20 40 60 80 100(c)

Figure 2 Example 9 second detection

119899 = 50 100 500 1000 respectively and 120596IO1198992

= 4 We have

119910119905=cos (15119910

119905minus2)

1 + 1199102119905minus2

119910119905minus1

+ (minus exp(minus1199102

119905minus2

2))119910119905minus2

+ 120578119905

120578119905= 120576119905+ 120596

IO1198992

sdot 120575 (119905 minus119899

2) 120576

119905sim 119873 (0 05)

(26)

For 119899 = 50 All the three test statistics detect an outlierat 119905 = 25 The 119901-values for believing it is an IO are 119901(119860IO

) =

48414119890 minus 6 119901(119878IO) = 18326119890 minus 8 and 119901(119862IO) = 33739119890 minus

8 respectively The 119901-values for believing it is an AO are119901(119860

AO) = 31509119890 minus 4 119901(119878AO) = 47392119890 minus 5 and 119901(119862AO

) =

87247119890minus5 respectively Because the 119901-values for believing itis an IO are smaller than the119901-values for believing it is anAOwe believe it is an IO and its influence magnitude is 31994

Mathematical Problems in Engineering 7

10

5

0

minus5

minus10

0 20 40 60 80 100(a)

0 20 40 60 80 100

1086420minus2

minus4

(b)

0 20 40 60 80 100

1086420minus2

minus4

(c)

Figure 3 Example 9 fifth detection

420minus2

minus4

minus6

minus8

minus10

0 20 40 60 80 100(a)

0 20 40 60 80 100

3210minus1

minus2

minus3

minus4

(b)

0 20 40 60 80 100

3210minus1

minus2

minus3

minus4

(c)

Figure 4 Example 9 sixth detection

The results are similar for 119899 = 100 500 1000 respectively wethus omit the details here For different test statistics their 119901-values for believing it is an IO are summarized as follows inTable 1

Example 9 There are four IOs and one AO in the time seriesThese IOs appear at 119905 = 49 119905 = 50 119905 = 51 and 119905 = 52

sequentially Their sizes are 3 6 5 and 4 respectively One

AO appears at 119905 = 70 alone and its size is 7 119899 = 100 Themodel is as follows

119909119905=cos (15119909

119905minus2)

1 + 1199092119905minus2

119909119905minus1

+ (minus exp (minus051199092119905minus2)) 119909119905minus2

+ 120576119905 120576119905sim 119873 (0 06)

(27)

where 119910119905is obtained by (3) or (4)

8 Mathematical Problems in Engineering

Table 1

119901-values Absolute valuestatistic Square statistic Adjusted square

statistic119899 = 50 48414119890 minus 6 18326119890 minus 8 33739119890 minus 8

119899 = 100 10027119890 minus 6 16592119890 minus 9 21995119890 minus 9

119899 = 500 19065119890 minus 6 26461119890 minus 8 260394119890 minus 8

119899 = 1000 28048119890 minus 7 13358119890 minus 9 12590119890 minus 9

First we detect an AO at 119905 = 70 and its size is 68904The three test statistics and their corresponding critical valuesare shown in Figure 1 where Figure 1(a) corresponds tothe absolute value test statistics Figure 1(b) corresponds tosquare test statistics and Figure 1(c) corresponds to adjustedsquare test statistics Symbol ldquordquo denotes IO and ldquolowastrdquo denotesAO Parallel broken line denotes critical value which isuniform in the figures Deleting the effect of this AO and thencontinuing to detect other outliers in the series we observean IO at 119905 = 50 whose size is 60285 see details in Figure 2Deleting the effect of the aforementioned two outliers andthen continuing to detect other outliers in the series weobserve an IO at 119905 = 51 whose size is 48119 Deletingthe effect of the aforementioned three outliers and thencontinuing to detect other outliers in the series we observean IO at 119905 = 52 whose size is 40024 Deleting the effectof the aforementioned four outliers and then continuing todetect other outlier in the series we detect an IO at 119905 =

49 whose size is 29046 see Figure 3 Deleting the effectof the aforementioned five outliers and then continuing todetect other outliers in the series we have not detected anyoutlier see Figure 4 Considering the length of paper we omitsome figures here The result is consistent with the advanceenactment

5 Conclusions

The FAR model is mainly featured by the model-dependent-variable which in one way or another limits the scope ofits applications As a generalization of the class of modelsAFAR model clearly covers a larger range of objects thanthe FAR model which makes it possible to reduce modelingbiases [17] via choosing a proper model-dependent direc-tion This paper is concerned with detecting AO and IOin AFAR models using extreme value methods We derivethe asymptotic distribution of test statistics and provide acontrol for significance level which serves as an extensionand improvement of existing methods Based on severalsimulation studies we give conclusion remarks as follows(a)The extreme value method for detecting outliers in AFARmodels is tractable and effective not only for IO and AO butalso for separate outliers and outlier patches Furthermoreit is shown that our method can reduce possible effects ofmasking and swamping (b) When applying extreme valuetheory to detect outliers with a relatively small size samplesat hand the employment of square test statistics works betterthan that of adjusted square test statistics as well as absolutevalue test statistics While following the increments in thesamples size the detecting effect of adjusted square test

statistics also increases against that of square test statistics andabsolute value test statistics (eg see Table 1) (c) Selection ofmodel parameters and the magnitude of outliers have hugeinfluences on the effect of detection

Acknowledgments

The authors sincerely wish to thank the editor and the threereferees for their insightful suggestions which have led toimproving the early version of the paper The research issupported by the National Natural Science Foundation ofChina (11171065) the Natural Science Foundation of JiangsuProvince (BK2011058) and Research Fund for the DoctoralProgram of Higher Education of China (20120092110021)

References

[1] C C AggarwalOutlier Analysis Springer New York NY USA2013

[2] X JieM LiWZhao and S YChen ldquoBoundmaxima as a trafficfeature under DDOS flood attacksrdquo Mathematical Problems inEngineering vol 2012 Article ID 419319 20 pages 2012

[3] M Li andW Zhao ldquoOn bandlimitedness and lag-limitedness offractional Gaussian noiserdquo Physica A vol 392 no 9 pp 1955ndash1961 2013

[4] M Li Y-Q Chen J-Y Li and W Zhao ldquoHolder scales of sealevelrdquoMathematical Problems in Engineering Article ID 86370722 pages 2012

[5] C W S Chen ldquoDetection of additive outliers in bilinear timeseriesrdquo Computational Statistics amp Data Analysis vol 24 no 3pp 283ndash294 1997

[6] M Li ldquoFractal time seriesmdasha tutorial reviewrdquo MathematicalProblems in Engineering Article ID 157264 26 pages 2010

[7] Z Cai J Fan and Q Yao ldquoFunctional-coefficient regressionmodels for nonlinear time seriesrdquo Journal of the AmericanStatistical Association vol 95 no 451 pp 941ndash956 2000

[8] F Battaglia ldquoOutliers in functional autoregressive time seriesrdquoStatistics amp Probability Letters vol 72 no 4 pp 323ndash332 2005

[9] F Battaglia and L Orfei ldquoOutlier detection and estimation innonlinear time seriesrdquo Journal of Time Series Analysis vol 26no 1 pp 107ndash121 2005

[10] P Chen L Li Y Liu and J-G Lin ldquoDetection of outliers andpatches in bilinear time series modelsrdquoMathematical Problemsin Engineering Article ID 580583 10 pages 2010

[11] P Chen J Yang and L Y Li ldquoSynthetic Detection of ChangePoint andOutliers in Bilinear Time SeriesModelsrdquo InternationalJournal of Systems Science In press

[12] R A Martin ldquoExtreme value analysis of optimal level-crossingprediction for linear Gaussian processesrdquo Journal of Time SeriesAnalysis vol 33 no 4 pp 583ndash607 2012

[13] K Zhu and S Ling ldquoLikelihood ratio tests for the structuralchange of an AR(p)model to a threshold AR(p)modelrdquo Journalof Time Series Analysis vol 33 no 2 pp 223ndash232 2012

[14] P Chareka F Matarise and R Turner ldquoA test for additiveoutliers applicable to long-memory time seriesrdquo Journal ofEconomic Dynamics amp Control vol 30 no 4 pp 595ndash621 2006

[15] W-K Fung Z-Y Zhu B-C Wei and X He ldquoInfluencediagnostics and outlier tests for semiparametric mixedmodelsrdquoJournal of the Royal Statistical Society B vol 64 no 3 pp 565ndash579 2002

Mathematical Problems in Engineering 9

[16] A M Rıo Extreme value theory-based P values in time seriesoutlier detection [PhD thesis] University of Wisconsin Madi-son 2005

[17] J Fan and Q Yao Nonlinear time series Springer Series inStatistics Springer New York NY USA 2003

[18] M R Leadbetter andH Rootzen ldquoExtremal theory for stochas-tic processesrdquo The Annals of Probability vol 16 no 2 pp 431ndash478 1988

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 7

10

5

0

minus5

minus10

0 20 40 60 80 100(a)

0 20 40 60 80 100

1086420minus2

minus4

(b)

0 20 40 60 80 100

1086420minus2

minus4

(c)

Figure 3 Example 9 fifth detection

420minus2

minus4

minus6

minus8

minus10

0 20 40 60 80 100(a)

0 20 40 60 80 100

3210minus1

minus2

minus3

minus4

(b)

0 20 40 60 80 100

3210minus1

minus2

minus3

minus4

(c)

Figure 4 Example 9 sixth detection

The results are similar for 119899 = 100 500 1000 respectively wethus omit the details here For different test statistics their 119901-values for believing it is an IO are summarized as follows inTable 1

Example 9 There are four IOs and one AO in the time seriesThese IOs appear at 119905 = 49 119905 = 50 119905 = 51 and 119905 = 52

sequentially Their sizes are 3 6 5 and 4 respectively One

AO appears at 119905 = 70 alone and its size is 7 119899 = 100 Themodel is as follows

119909119905=cos (15119909

119905minus2)

1 + 1199092119905minus2

119909119905minus1

+ (minus exp (minus051199092119905minus2)) 119909119905minus2

+ 120576119905 120576119905sim 119873 (0 06)

(27)

where 119910119905is obtained by (3) or (4)

8 Mathematical Problems in Engineering

Table 1

119901-values Absolute valuestatistic Square statistic Adjusted square

statistic119899 = 50 48414119890 minus 6 18326119890 minus 8 33739119890 minus 8

119899 = 100 10027119890 minus 6 16592119890 minus 9 21995119890 minus 9

119899 = 500 19065119890 minus 6 26461119890 minus 8 260394119890 minus 8

119899 = 1000 28048119890 minus 7 13358119890 minus 9 12590119890 minus 9

First we detect an AO at 119905 = 70 and its size is 68904The three test statistics and their corresponding critical valuesare shown in Figure 1 where Figure 1(a) corresponds tothe absolute value test statistics Figure 1(b) corresponds tosquare test statistics and Figure 1(c) corresponds to adjustedsquare test statistics Symbol ldquordquo denotes IO and ldquolowastrdquo denotesAO Parallel broken line denotes critical value which isuniform in the figures Deleting the effect of this AO and thencontinuing to detect other outliers in the series we observean IO at 119905 = 50 whose size is 60285 see details in Figure 2Deleting the effect of the aforementioned two outliers andthen continuing to detect other outliers in the series weobserve an IO at 119905 = 51 whose size is 48119 Deletingthe effect of the aforementioned three outliers and thencontinuing to detect other outliers in the series we observean IO at 119905 = 52 whose size is 40024 Deleting the effectof the aforementioned four outliers and then continuing todetect other outlier in the series we detect an IO at 119905 =

49 whose size is 29046 see Figure 3 Deleting the effectof the aforementioned five outliers and then continuing todetect other outliers in the series we have not detected anyoutlier see Figure 4 Considering the length of paper we omitsome figures here The result is consistent with the advanceenactment

5 Conclusions

The FAR model is mainly featured by the model-dependent-variable which in one way or another limits the scope ofits applications As a generalization of the class of modelsAFAR model clearly covers a larger range of objects thanthe FAR model which makes it possible to reduce modelingbiases [17] via choosing a proper model-dependent direc-tion This paper is concerned with detecting AO and IOin AFAR models using extreme value methods We derivethe asymptotic distribution of test statistics and provide acontrol for significance level which serves as an extensionand improvement of existing methods Based on severalsimulation studies we give conclusion remarks as follows(a)The extreme value method for detecting outliers in AFARmodels is tractable and effective not only for IO and AO butalso for separate outliers and outlier patches Furthermoreit is shown that our method can reduce possible effects ofmasking and swamping (b) When applying extreme valuetheory to detect outliers with a relatively small size samplesat hand the employment of square test statistics works betterthan that of adjusted square test statistics as well as absolutevalue test statistics While following the increments in thesamples size the detecting effect of adjusted square test

statistics also increases against that of square test statistics andabsolute value test statistics (eg see Table 1) (c) Selection ofmodel parameters and the magnitude of outliers have hugeinfluences on the effect of detection

Acknowledgments

The authors sincerely wish to thank the editor and the threereferees for their insightful suggestions which have led toimproving the early version of the paper The research issupported by the National Natural Science Foundation ofChina (11171065) the Natural Science Foundation of JiangsuProvince (BK2011058) and Research Fund for the DoctoralProgram of Higher Education of China (20120092110021)

References

[1] C C AggarwalOutlier Analysis Springer New York NY USA2013

[2] X JieM LiWZhao and S YChen ldquoBoundmaxima as a trafficfeature under DDOS flood attacksrdquo Mathematical Problems inEngineering vol 2012 Article ID 419319 20 pages 2012

[3] M Li andW Zhao ldquoOn bandlimitedness and lag-limitedness offractional Gaussian noiserdquo Physica A vol 392 no 9 pp 1955ndash1961 2013

[4] M Li Y-Q Chen J-Y Li and W Zhao ldquoHolder scales of sealevelrdquoMathematical Problems in Engineering Article ID 86370722 pages 2012

[5] C W S Chen ldquoDetection of additive outliers in bilinear timeseriesrdquo Computational Statistics amp Data Analysis vol 24 no 3pp 283ndash294 1997

[6] M Li ldquoFractal time seriesmdasha tutorial reviewrdquo MathematicalProblems in Engineering Article ID 157264 26 pages 2010

[7] Z Cai J Fan and Q Yao ldquoFunctional-coefficient regressionmodels for nonlinear time seriesrdquo Journal of the AmericanStatistical Association vol 95 no 451 pp 941ndash956 2000

[8] F Battaglia ldquoOutliers in functional autoregressive time seriesrdquoStatistics amp Probability Letters vol 72 no 4 pp 323ndash332 2005

[9] F Battaglia and L Orfei ldquoOutlier detection and estimation innonlinear time seriesrdquo Journal of Time Series Analysis vol 26no 1 pp 107ndash121 2005

[10] P Chen L Li Y Liu and J-G Lin ldquoDetection of outliers andpatches in bilinear time series modelsrdquoMathematical Problemsin Engineering Article ID 580583 10 pages 2010

[11] P Chen J Yang and L Y Li ldquoSynthetic Detection of ChangePoint andOutliers in Bilinear Time SeriesModelsrdquo InternationalJournal of Systems Science In press

[12] R A Martin ldquoExtreme value analysis of optimal level-crossingprediction for linear Gaussian processesrdquo Journal of Time SeriesAnalysis vol 33 no 4 pp 583ndash607 2012

[13] K Zhu and S Ling ldquoLikelihood ratio tests for the structuralchange of an AR(p)model to a threshold AR(p)modelrdquo Journalof Time Series Analysis vol 33 no 2 pp 223ndash232 2012

[14] P Chareka F Matarise and R Turner ldquoA test for additiveoutliers applicable to long-memory time seriesrdquo Journal ofEconomic Dynamics amp Control vol 30 no 4 pp 595ndash621 2006

[15] W-K Fung Z-Y Zhu B-C Wei and X He ldquoInfluencediagnostics and outlier tests for semiparametric mixedmodelsrdquoJournal of the Royal Statistical Society B vol 64 no 3 pp 565ndash579 2002

Mathematical Problems in Engineering 9

[16] A M Rıo Extreme value theory-based P values in time seriesoutlier detection [PhD thesis] University of Wisconsin Madi-son 2005

[17] J Fan and Q Yao Nonlinear time series Springer Series inStatistics Springer New York NY USA 2003

[18] M R Leadbetter andH Rootzen ldquoExtremal theory for stochas-tic processesrdquo The Annals of Probability vol 16 no 2 pp 431ndash478 1988

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Mathematical Problems in Engineering

Table 1

119901-values Absolute valuestatistic Square statistic Adjusted square

statistic119899 = 50 48414119890 minus 6 18326119890 minus 8 33739119890 minus 8

119899 = 100 10027119890 minus 6 16592119890 minus 9 21995119890 minus 9

119899 = 500 19065119890 minus 6 26461119890 minus 8 260394119890 minus 8

119899 = 1000 28048119890 minus 7 13358119890 minus 9 12590119890 minus 9

First we detect an AO at 119905 = 70 and its size is 68904The three test statistics and their corresponding critical valuesare shown in Figure 1 where Figure 1(a) corresponds tothe absolute value test statistics Figure 1(b) corresponds tosquare test statistics and Figure 1(c) corresponds to adjustedsquare test statistics Symbol ldquordquo denotes IO and ldquolowastrdquo denotesAO Parallel broken line denotes critical value which isuniform in the figures Deleting the effect of this AO and thencontinuing to detect other outliers in the series we observean IO at 119905 = 50 whose size is 60285 see details in Figure 2Deleting the effect of the aforementioned two outliers andthen continuing to detect other outliers in the series weobserve an IO at 119905 = 51 whose size is 48119 Deletingthe effect of the aforementioned three outliers and thencontinuing to detect other outliers in the series we observean IO at 119905 = 52 whose size is 40024 Deleting the effectof the aforementioned four outliers and then continuing todetect other outlier in the series we detect an IO at 119905 =

49 whose size is 29046 see Figure 3 Deleting the effectof the aforementioned five outliers and then continuing todetect other outliers in the series we have not detected anyoutlier see Figure 4 Considering the length of paper we omitsome figures here The result is consistent with the advanceenactment

5 Conclusions

The FAR model is mainly featured by the model-dependent-variable which in one way or another limits the scope ofits applications As a generalization of the class of modelsAFAR model clearly covers a larger range of objects thanthe FAR model which makes it possible to reduce modelingbiases [17] via choosing a proper model-dependent direc-tion This paper is concerned with detecting AO and IOin AFAR models using extreme value methods We derivethe asymptotic distribution of test statistics and provide acontrol for significance level which serves as an extensionand improvement of existing methods Based on severalsimulation studies we give conclusion remarks as follows(a)The extreme value method for detecting outliers in AFARmodels is tractable and effective not only for IO and AO butalso for separate outliers and outlier patches Furthermoreit is shown that our method can reduce possible effects ofmasking and swamping (b) When applying extreme valuetheory to detect outliers with a relatively small size samplesat hand the employment of square test statistics works betterthan that of adjusted square test statistics as well as absolutevalue test statistics While following the increments in thesamples size the detecting effect of adjusted square test

statistics also increases against that of square test statistics andabsolute value test statistics (eg see Table 1) (c) Selection ofmodel parameters and the magnitude of outliers have hugeinfluences on the effect of detection

Acknowledgments

The authors sincerely wish to thank the editor and the threereferees for their insightful suggestions which have led toimproving the early version of the paper The research issupported by the National Natural Science Foundation ofChina (11171065) the Natural Science Foundation of JiangsuProvince (BK2011058) and Research Fund for the DoctoralProgram of Higher Education of China (20120092110021)

References

[1] C C AggarwalOutlier Analysis Springer New York NY USA2013

[2] X JieM LiWZhao and S YChen ldquoBoundmaxima as a trafficfeature under DDOS flood attacksrdquo Mathematical Problems inEngineering vol 2012 Article ID 419319 20 pages 2012

[3] M Li andW Zhao ldquoOn bandlimitedness and lag-limitedness offractional Gaussian noiserdquo Physica A vol 392 no 9 pp 1955ndash1961 2013

[4] M Li Y-Q Chen J-Y Li and W Zhao ldquoHolder scales of sealevelrdquoMathematical Problems in Engineering Article ID 86370722 pages 2012

[5] C W S Chen ldquoDetection of additive outliers in bilinear timeseriesrdquo Computational Statistics amp Data Analysis vol 24 no 3pp 283ndash294 1997

[6] M Li ldquoFractal time seriesmdasha tutorial reviewrdquo MathematicalProblems in Engineering Article ID 157264 26 pages 2010

[7] Z Cai J Fan and Q Yao ldquoFunctional-coefficient regressionmodels for nonlinear time seriesrdquo Journal of the AmericanStatistical Association vol 95 no 451 pp 941ndash956 2000

[8] F Battaglia ldquoOutliers in functional autoregressive time seriesrdquoStatistics amp Probability Letters vol 72 no 4 pp 323ndash332 2005

[9] F Battaglia and L Orfei ldquoOutlier detection and estimation innonlinear time seriesrdquo Journal of Time Series Analysis vol 26no 1 pp 107ndash121 2005

[10] P Chen L Li Y Liu and J-G Lin ldquoDetection of outliers andpatches in bilinear time series modelsrdquoMathematical Problemsin Engineering Article ID 580583 10 pages 2010

[11] P Chen J Yang and L Y Li ldquoSynthetic Detection of ChangePoint andOutliers in Bilinear Time SeriesModelsrdquo InternationalJournal of Systems Science In press

[12] R A Martin ldquoExtreme value analysis of optimal level-crossingprediction for linear Gaussian processesrdquo Journal of Time SeriesAnalysis vol 33 no 4 pp 583ndash607 2012

[13] K Zhu and S Ling ldquoLikelihood ratio tests for the structuralchange of an AR(p)model to a threshold AR(p)modelrdquo Journalof Time Series Analysis vol 33 no 2 pp 223ndash232 2012

[14] P Chareka F Matarise and R Turner ldquoA test for additiveoutliers applicable to long-memory time seriesrdquo Journal ofEconomic Dynamics amp Control vol 30 no 4 pp 595ndash621 2006

[15] W-K Fung Z-Y Zhu B-C Wei and X He ldquoInfluencediagnostics and outlier tests for semiparametric mixedmodelsrdquoJournal of the Royal Statistical Society B vol 64 no 3 pp 565ndash579 2002

Mathematical Problems in Engineering 9

[16] A M Rıo Extreme value theory-based P values in time seriesoutlier detection [PhD thesis] University of Wisconsin Madi-son 2005

[17] J Fan and Q Yao Nonlinear time series Springer Series inStatistics Springer New York NY USA 2003

[18] M R Leadbetter andH Rootzen ldquoExtremal theory for stochas-tic processesrdquo The Annals of Probability vol 16 no 2 pp 431ndash478 1988

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 9

[16] A M Rıo Extreme value theory-based P values in time seriesoutlier detection [PhD thesis] University of Wisconsin Madi-son 2005

[17] J Fan and Q Yao Nonlinear time series Springer Series inStatistics Springer New York NY USA 2003

[18] M R Leadbetter andH Rootzen ldquoExtremal theory for stochas-tic processesrdquo The Annals of Probability vol 16 no 2 pp 431ndash478 1988

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of