EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 7, No. 3, 2014, 230-245 ISSN 1307-5543 www.jpam.com Total Last Squars Fttng th Thr-Paramtr Invrs Wbull Dnsty Dragan Juć, Darja Marovć Dpartmnt of Mathmatcs, J.J. Strossmayr Unvrsty of Osj, Trg Ljudvta Gaja 6, HR-31 000 Osj, Croata Abstract. Th focus of ths papr s on a nonlnar wghtd total last squars fttng problm for th thr-paramtr nvrs Wbull dnsty whch s frquntly mployd as a modl n rlablty and lftm studs. As a man rsult, a thorm on th xstnc of th total last squars stmator s obtand, as wll as ts gnralzaton n th l q norm (1 q ). 2010 Mathmatcs Subjct Classfcatons: 65D10, 62J02, 62G07, 62N05 Ky Words and Phrass: nvrs Wbull dnsty, total last squars, total last squars stmat, xstnc problm, data fttng 1. Introducton Th probablty dnsty functon of th random varabl T havng a thr-paramtr nvrs Wbull dstrbuton (IWD) wth locaton paramtrα 0, scal paramtrη>0and shap paramtrβ> 0 s gvn by f(t;α,β,η)= β η β+1 η t α ( η t α )β t>α 0 t α. (1) If α = 0, th rsultng dstrbuton s calld th two-paramtr nvrs Wbull dstrbuton. Ths modl was dvlopd by Erto[6]. Th IWD s vry flxbl and by an approprat choc of th shap paramtrβ th dnsty curv can assum a wd varty of shaps (s Fg. 1). Th dnsty functon s strctly ncrasng on(α, t m ] and strctly dcrasng on[t m, ), whr t m =α+η(1+1/β) 1/β. Ths mpls that th dnsty functon s unmodal wth th maxmum valu at t m. Ths s n contrast to th standard Wbull modl whr th shap s thr dcrasng (forβ 1) or unmodal (for β > 1). Whn β = 1, th IWD bcoms an nvrs xponntal dstrbuton; Corrspondng author. Emal addrsss: jucd@mathos.hr (D. Juć),darja@mathos.hr (D. Marovć) http://www.jpam.com 230 c 2014 EJPAM All rghts rsrvd.
D. Juć, D. Marovć/Eur. J. Pur Appl. Math, 7 (2014), 230-245 231 whnβ= 2, t s dntcal to th nvrs Raylgh dstrbuton; whnβ= 0.5, t approxmats th nvrs Gamma dstrbuton. That s th rason why th IWD s a frquntly usd modl n rlablty and lftm studs (s.g. Cohn and Whttn[5], Lawls[18], Murthy t al.[21], Nlson[22]). f(t) β=0.5 t Fgur 1: Plots of th nvrs Wbull dnsty for som valus ofβ and by assumngα=0and η=1.2 β=3 β=2 β=1 In practc, th unnown paramtrs α, β and η of th thr-paramtr nvrs Wbull dnsty (1) ar not nown n advanc and must b stmatd from a random sampl t 1,..., t n consstng of n obsrvatons of th thr-paramtr nvrs Wbull random varabl T. Thr s no unqu way to stmat th unnown paramtrs and many dffrnt mthods hav bn proposd n th ltratur (s.g. Abbas t al.[1], Lawlss[18], Marušć t al.[20], Murthy t al.[21], Nlson[22], Slvrman[26], Smth and Naylor[27, 28], Tapa and Thompson[29]). A vry popular mthod for paramtr stmaton s th last squars mthod. Th nonlnar wghtd ordnary last squars (OLS) fttng problm for th thr-paramtr nvrs Wbull dnsty s consdrd by Marušć t al. [20]. In ths papr w consdr th nonlnar wghtd total last squars (TLS) fttng problm for th thr-paramtr nvrs Wbull dnsty functon. Th structur of th papr s as follows. In Scton 2 w brfly dscrb th TLS mthod and prsnt our man rsult (Thorm 1) whch guarants th xstnc of th TLS stmator for th thr-paramtrc nvrs Wbull dnsty. Its gnralzaton n th l q norm (1 q ) s gvn n Thorm 2. All proofs ar gvn n Scton 3. 2. Th TLS Fttng Problm for th Thr-paramtr Invrs Wbull Dnsty Both th OLS and th TLS mthod rqur th ntal nonparamtrc dnsty stmats ˆf whch nd to b as good as possbl (s.g. Slvrman[26], Marušć t al.[20]). Suppos w ar gvn th ponts(t, y ), = 1,..., n, n>3, whr 0 t 1 t 2... t n ar obsrvatons of th nonngatv thr-paramtr nvrs Wbull random varabl T and y := ˆf(t ) ar th rspctv dnsty stmats.
D. Juć, D. Marovć/Eur. J. Pur Appl. Math, 7 (2014), 230-245 232 Th goal of th OLS mthod (s.g.[2, 3, 8, 11, 13, 14, 19, 25]) s to choos th unnown paramtrs of dnsty functon (1) such that th wghtd sum of squard dstancs btwn th modl and th data s as small as possbl. To b mor prcs, lt w > 0, = 1,..., n, b th data wghts whch dscrb th assumd rlatv accuracy of th data. Th unnown paramtrsα,β andηhav to b stmatd by mnmzng th functonal on th st S(α,β,η)= n w [f(t ;α,β,η) y ] 2 =1 := (α,β,η) 3 :α 0;β,η>0. A pont(α,β,η ) such that S(α,β,η )=nf (α,β,η) S(α,β,η) s calld th OLS stmator, f t xsts. As w hav alrady mntond, ths problm has bn solvd by Marušć t al.[20]. In th OLS approach th obsrvatons t of th ndpndnt varabl ar assumd to b xact and only th stmats y of th dnsty (dpndnt varabl) ar subjct to random rrors. Unfortunatly, ths assumpton dos not sm to b vry ralstc n practc, and many rrors (samplng rrors, human rrors, modlng rrors and nstrumnt rrors) prvnt us from nowng t xactly. In such stuaton, whn also th obsrvatons of th ndpndnt varabl contans rrors, t sms rasonabl to stmat th unnown paramtrs so that th wghtd sum of squars of all rrors s mnmzd. Ths approach, nown as th total last squars (TLS) mthod, s a natural gnralzaton of th OLS mthod (s.g. [7]). In th statstcs ltratur, th TLS approach s nown as rrors-n-varabls rgrsson or orthogonal dstanc rgrsson, and n numrcal analyss t was frst consdrd by Golub and Van Loan[9]. Th TLS mthod can b dscrbd as follows. Lt w, p > 0, = 1,..., n, b som wghts. If w assum that y contans unnown addtv rrorǫ and that t has unnown addtv rrorδ, thn th mathmatcal modl bcoms y = f(t +δ ;α,β,η)+ǫ, = 1,..., n. Th unnown paramtrsα,β andηof dnsty functon (1) hav to b stmatd by mnmzng th wghtd sum of squars of all rrors,.. by mnmzng th functonal (s.g. [4, 7, 10, 17, 24]) T(α,β,η,δ)= n w [f(t +δ ;α,β,η) y ] 2 + =1 n p δ 2 (2) on th st n. A pont(α,β,η ) n s calld th total last squars stmator (TLS stmator) of th unnown paramtrs(α, β, η) for th thr-paramtr nvrs Wbull dnsty, f thr xstsδ n such that =1 T(α,β,η,δ )= nf (α,β,η,δ) n T(α,β,η,δ). Numrcal mthods for solvng th nonlnar TLS problm ar dscrbd n Boggs t al. [4] and Schwtlc and Tllr[24]. As n th cas of th OLS approach, bfor th tratv
D. Juć, D. Marovć/Eur. J. Pur Appl. Math, 7 (2014), 230-245 233 mnmzaton of th sum of squars t s stll ncssary to as whthr th TLS stmator xsts. In th cas of nonlnar TLS problms t s stll xtrmly dffcult to answr ths quston (s.g.[3, 7, 12, 15 17]). Th dffrnc btwn th OLS and th TLS approach s llustratd n Fg. 2. Gomtrcally, f w = p for all = 1,..., n, mnmzaton of functonal T corrsponds to mnmzaton of th wghtd sum of squars of dstancs from data ponts to th modl curv. y (t,y ) y (t,y ) (t +δ,f(t +δ ;α,β,η)) t (a) OLS (b) TLS Fgur 2: Th dffrnc btwn th OLS and TLS approachs t Our man xstnc rsult for th TLS problm for th thr-paramtr nvrs Wbull dnsty s gvn n th nxt thorm. Thorm 1. Lt th ponts(t, y ), = 1,..., n, n>3, b gvn, such that 0 t 1 t 2... t n and y > 0, = 1,..., n. Furthrmor, lt w, p > 0, = 1,..., n, b som wghts. Thn thr xsts a pont(α,β,η,δ ) n such that.. th TLS stmator xsts. T(α,β,η,δ )= nf (α,β,η,δ) n T(α,β,η,δ), Th proof s gvn n Scton 3. Th followng total l q norm (q 1) gnralzaton of Thorm 1 holds tru. Thorm 2. Suppos 1 q. Lt th ponts and wghts b th sam as n Thorm 1. Dfn n n T q (α,β,η,δ) := w f(t +δ ;α,β,η) y q + p δ q. (3) Thn thr xsts a pont(α q,β q,η q,δ q ) n such that =1 =1 T q (α q,β q,η q,δ q )= nf T q(α,β,η,δ). (α,β,η,δ) n Th proof of ths thorm s omttd as t s smlar to that of Thorm 1. It suffcs to rplac th l 2 norm wth th l q norm. Thrby all parts of th proof rman th sam.
D. Juć, D. Marovć/Eur. J. Pur Appl. Math, 7 (2014), 230-245 234 3. Proof of Thorm 1 Bfor startng th proof of Thorm 1, w nd som prlmnary rsults. Lmma 1. Suppos w ar gvn data(w, t, y ), I :={1,..., n}, n>3, such that 0 t 1 t 2... t n and y > 0, I. Lt w, p > 0, I, b som wghts. Gvn any ral numbr q, 1 q, and any nonmpty subst I 0 of I, lt Σ I0 := w y q + p t τ 0 q, I\I 0 I 0 whr τ 0 argmn x n p t x q. =1 Thn thr xsts a pont n n at whch functonal T q dfnd by (3) attans a valu lss than Σ I0. Summaton I 0 s to b undrstood as follows: Th sum ovr thos ndcs nfor whch I 0. If thr ar no such ndcs, th sum s mpty; followng th usual convnton, w dfn t to b zro. Summaton I\I 0 has smlar manngs. It s asy to vrfy that t 1 mn I0 t τ 0 max I0 t t n. Not that for th cas whn q=2,τ 0 s a wll nown wghtd arthmtc man, and for th cas whn q=1,τ 0 s a wghtd mdan of th data (s.g. Sabo and Sctovs[23]). Proof. Sncτ 0 s an lmnt of th closd ntrval[t 1, t n ], thr xsts r {1,..., n} such that τ 0 (t r 1, t r ], whr t 0 = 0 by dfnton. Lt us frst choos ral y 0 such that 0 y 0 mn I and thn dfn functonsα,β,η :(0,1) by: b β(b) :=τ 0 y 0 b η(b) :=τ 0 b 1/β(b), y (4) α(b) :=τ 0 η(b)b 1/2β(b) =η(b) b 1/β(b) b 1/2β(b). Clarly, functonsβ andηar postv. Furthrmor, by usng th nqualty b 1/β(b) b 1/2β(b) > 0, whch holds for vry b (0,1), t s asy to show that functonαs also postv. Thus, w hav showd that(α(b),β(b),η(b)) for all b (0,1). Lt us now assocat wth ach ral b (0,1) a thr-paramtrc nvrs Wbull dnsty functon β(b) η(b) β(b) β(b) f(t;α(β),β(b),η(b))= t α(b) t α(b) η(b) t α(b) t>α(b) (5) 0 t α(b).
D. Juć, D. Marovć/Eur. J. Pur Appl. Math, 7 (2014), 230-245 235 Ths functon has maxmum at th pont whr α(b)+η(b)(1+1/β(b)) 1/β(b) =τ 0 ǫ(b), ǫ(b) :=η(b) b 1/2β(b) 1+ 1 1/β(b). β(b) It s strctly ncrasng on(α(b),τ 0 ǫ(b)] and strctly dcrasng on[τ 0 ǫ(b), ). Furthrmor, by a straghtforward calculaton, t can b vrfd that Now w ar gong to show that Frst, n vw of (8) and (9), w obtan f(τ 0 +α(b);α(b),β(b),η(b))= y 0, (6) lmβ(b)=, b 0 (7) lm 0, b 0 (8) lmα(b)=0. b 0 (9) lm f(t;α(b),β(b),η(b))=0, t τ 0. (10) b 0 η(b) lm = τ 0 b 0 t α(b) t. β(b) Ifτ 0 t, thn from (7) and (9) t follows radly that lm b 0 η(b) t α(b) = 1 and lm b 0 β(b) lm b 0 η(b) t α(b) β(b)= 0, and thrfor β(b) η(b) β(b) f(t;α(b),β(b),η(b))= lm η(b) β(b) t α(b) = 0. b 0 t α(b) t α(b) Ifτ 0 > t, thn thr xsts a suffcntly grat 0 such that η(b) 0 t α(b) for vry suffcntly small b>0. Now, by usng th nqualty x x (x 0) w obtan η(b) β(b) β(b) 0 β(b), t α(b) b 0, and thrfor, for any b 0 w hav 0f(t;α(b),β(b),η(b))= β(b) η(b) β(b) η(b) β(b) t α(b) t α(b) t α(b)
D. Juć, D. Marovć/Eur. J. Pur Appl. Math, 7 (2014), 230-245 236 Snc 1 η(b) β(b) (0 +1)β(b) η(b) t α(b). t α(b) t α(b) η(b) β(b) (0 +1)β(b) lm η(b) t α(b) = 0, b 0 t α(b) thn from th abov-mntond nqualty t follows that lm f(t;α(b),β(b),η(b))=0, t>τ 0. b 0 Thus, w provd th dsrd lmts (10). Not that f(τ 0 ;α(b),β(b),η(b))= β(b) η(b) β(b) η(b) β(b) τ 0 α(b) τ 0 α(b) τ 0 α(b) = β(b) b b = τ 0 y 0 b b τ 0 α(b) (τ 0 α(b)) b, from whr tang th lmt as b 0 t follows that lm f(τ 0;α(b),β(b),η(b))=. (11) b 0 Du to (9), (10) and (11), w may suppos that b s suffcntly small, so that 0α(b) t 1 (12) 0 f(t ;α(b),β(b),η(b)) y, f t τ 0 (13) f(τ 0 ;α(b),β(b),η(b))>max I y. (14) Lt us now show that for ach I 0 and for vry b (0,1) thr xsts a unqu numbr τ (b) such that (s Fgur 3) t τ (b)τ 0 ǫ(b)τ 0, f t τ 0 t τ (b) t +ǫ(b), f t =τ 0 (15) τ 0 τ (b) t, f t >τ 0 and f(τ (b);α(b),β(b),η(b))= y. (16) Frst, snc th functon t f(t ;α(b),β(b),η(b)) has maxmum at th pontτ 0 ǫ(b) and t s strctly ncrasng on(α(b),τ 0 ǫ(b)] and strctly dcrasng on[τ 0 ǫ(b), ), by usng (4), (6), (13) and (14) w obtan f(t ;α(b),β(b),η(b)) y f(τ 0 ǫ(b);α(b),β(b),η(b)), f t τ 0 f(t ;α(b),β(b),η(b)) y f(τ 0 ;α(b),β(b),η(b)), f t >τ 0 f(t +ǫ(b);α(b),β(b),η(b)) y f(t ;α(b),β(b),η(b)), f t =τ 0.
D. Juć, D. Marovć/Eur. J. Pur Appl. Math, 7 (2014), 230-245 237 Th xstnc of th dsrd numbrsτ (b), I 0, follows from th wll-nown Intrmdat Valu Thorm whch stats that a contnuous ral functon assums all ntrmdat valus on a closd ntrval, whl unqunss follows from monotoncty. f(t) (t, y ) (τ (b), y ) (τ j (b), y j ) (t j, y j ) τ 0 ε(b) τ 0 τ 0 +ε(b) Fgur 3:, j I 0, t τ 0, t j >τ 0 ; 0δ (b)=τ (b) t τ 0 ǫ(b) t τ 0 t ; τ 0 t j τ j (b) t j =δ j (b)0 t Sttng (16) bcoms δ (b) := τ (b) t, f I 0 0, f I\I 0, (17) f(t +δ (b);α(b),β(b),η(b))= y, I 0. (18) Not that only on of th followng two cass can occur: () I 0 =1, or () I 0 >1. Cas (): I 0 =1. In ths cas w havτ 0 = t r. It follows from (15) that 0δ r (b)ǫ(b). Wthout loss of gnralty, n addton to (12)-(14) w may suppos that b s suffcntly small, so that t r 1 +ǫ(b) t r 1+t r t r ǫ(b) 2 and f((t r 1 + t r )/2;α(b),β(b),η(b))mn y. I Du to ths two addtonal assumptons and th fact that th functon t f(t ;α(b),β(b),η(b)) s strctly ncrasng on(α(b), t r ǫ(b)] and strctly dcrasng on[t r ǫ(b), ), w dduc:
D. Juć, D. Marovć/Eur. J. Pur Appl. Math, 7 (2014), 230-245 238 If t t r, thn whras f t > t r, thn 0f(t ;α(b) δ r (b),β(b),η(b))= f(t +δ r (b);α(b),β(b),η(b)) f(t +ǫ(b);α(b),β(b),η(b)) f(t r 1 +ǫ(b);α(b),β(b),η(b)) f((t r 1 + t r )/2;α(b),β(b),η(b))mn I y y, (19) 0f(t ;α(b) δ r (b),β(b),η(b))= f(t +δ r (b);α(b),β(b),η(b)) f(t ;α(b),β(b),η(b)) y. (20) Thus, t follows from (18), (19) and (20) that, for vry b (0,1), T q (α(b) δ r (b),β(b),η(b),0)= n w f(t ;α(b) δ r (b),β(b),η(b)) y q =1 n w y q =Σ I0 Cas I 0 >1. Not that only on of th followng two subcass can occur: () τ 0 t for all I 0, or () τ 0 = t r for som r I 0. =1 r Subcas (): In ths subcas, t follows from (13), (15), (17) and (18) that, for vry b (0,1), T q (α(b),β(b),η(b),δ(b))= f(t ;α(b),β(b),η(b)) y q + I\I 0 w w y q + p t τ 0 q =Σ I0. I\I 0 I 0 I 0 p δ (b) q Subcas (): Assum thatτ 0 = t r for som r I 0. Lt ndx s I 0 b such that t s τ 0. Thn by (15), for vry b (0,1), and thrfor 0δ r (b)ǫ(b) and 0δ s (b) t r ǫ(b) t s p s δ s (b) q + p r δ r (b) q p s t r ǫ(b) t s q + p r ǫ q (b). It can b asly shown that th abov rght-hand sd s lss than p s t r t s q whnvr b s small nough. Thrfor, for vry small nough b w hav T q (α(b),β(b),η(b),δ(b))= f(t ;α(b),β(b),η(b)) y q I\I 0 w
D. Juć, D. Marovć/Eur. J. Pur Appl. Math, 7 (2014), 230-245 239 + p r δ r (b) q + p s δ s (b) q + p δ (b) q Ths complts th proof of th lmma. I 0 \{r,s} w y q + p t τ 0 q =Σ I0. I\I 0 I 0 Proof of Thorm 1. Proof. Snc functonal T s nonngatv, thr xsts T := nf (α,β,η,δ) n T(α,β,η,δ). To complt th proof t should b shown that thr xsts a pont(α,β,η,δ ) n such that T(α,β,η,δ )= T. Lt(α,β,,δ ) b a squnc n n, such that T = lm T(α,β,,δ )= lm = lm t +δ α w y 2 + t +δ >α w f(t +δ ;α 2+,β, ) y p (δ )2 I β η β+1 w t +δ α I β 2 y + p (δ )2. (21) I whr I={1,..., n}. Th summaton t +δ α (or t +δ >α ) s to b undrstood as follows: Th sum ovr thos ndcs nfor whch t +δ α (or t +δ >α ). If thr ar no such ponts t, th sum s mpty; followng th usual convnton, w dfn t to b zro. Thr s no loss of gnralty n assumng that all squncs(α ),(β ),( ),(δ 1 ),...,(δ n ) ar monoton. Ths s possbl bcaus th squnc(α,β,,δ 1,...,δ n ) has a subsqunc(α l,β l,η l,δ l 1,...,δl n ), such that all ts componnt squncs ar monoton; and snc lm T(α l,β l,η l,δ l )=lm T(α,β,,δ )= T. Snc ach monoton squnc of ral numbrs convrgs n th xtndd ral numbr systm, dfn α := lm α, β := lm β, η := lm, δ := lm δ =(δ 1,...,δ n ). Not that 0 α,β,η, bcaus(α,β, ). Also not thatδ for ach = 1,..., n. Indd, f δ = for som, thn t would follow from (21) that T =, whch s mpossbl. To complt th proof t s nough to show that(α,β,η ),.. that 0 α andβ,η (0, ). Th contnuty of th functonal T wll thn mply that T = lm T(α,β,,δ )= T(α,β,η,δ ).
D. Juć, D. Marovć/Eur. J. Pur Appl. Math, 7 (2014), 230-245 240 It rmans to show that(α,β,η ). Th proof wll b don n fv stps. In stp 1 w wll show thatα t n. In stp 2 w wll show thatβ 0. Th proof thatη wll b don n stp 3. In stp 4 w prov thatη 0. Fnally, n stp 5 w show thatβ. Stp 1. Ifα t n, from (21) t follows that T = n =1 w y 2 + I p δ 2. Snc accordng to Lmma 1 (for q=2and I 0 ={1}) thr xsts a pont n n at whch functonal T attans a valu smallr thanσ I0 and sncσ I0 n =1 w y 2 + I p δ 2, ths mans that n ths way (α t n ) functonal T cannot attan ts nfmum. Thus, w hav provd thatα t n. Bfor contnung th proof, lt us ntroduc som notaton and ma on rmar. Frst lt us dfn Iα, f I α I 0 := {1}, othrws whr I α :={ I : t +δ =α }. Lt us not that Lmma 1 wth q=2 mpls that whrτ I0 = I 0 p t I 0 p. T w y 2 + p (t τ I0 ) 2 =:Σ I0, (22) I\I 0 I 0 By tang an approprat subsqunc of(α,β,,δ ), f ncssary, w may assum that f t +δ α, thn t +δ α for vry. Smlarly, f t +δ >α, w may assum that t +δ >α for vry. Du to ths, now t s asy to show that from (21) t follows that T w y 2 + lm t +δ α + I t +δ >α w β η β+1 t +δ α β 2 y p δ 2. (23) Stp 2. Ifβ = 0, thn by usng th nqualty x x (x 0) w obtan 0 β η β+1 t +δ α whrfrom t follows radly that β η β+1 lm t +δ α β β t +δ α, f t +δ >α, β = 0, f t +δ >α. Now, from (23) t follows that T w y 2 + I\I 0 p δ 2 I 0
D. Juć, D. Marovć/Eur. J. Pur Appl. Math, 7 (2014), 230-245 241 = w y 2 + p (t α ) 2 I\I 0 I 0 w y 2 + p (t τ I0 ) 2 =Σ I0, (24) I\I 0 I 0 whch contradcts (22). Thrfor, n ths way (β = 0) functonal T cannot attan ts nfmum. Thus, w hav provd thatβ 0. Th last nqualty n (24) follows drctly from a wll-nown fact that th quadratc functon x I 0 p (t x) 2 attans ts mnmum I 0 p (t τ I0 ) 2 at pontτ I0. Stp 3. Lt us show thatη. W prov ths by contradcton. Suppos on th contrary thatη =. Wthout loss of gnralty, w may thn assum that f t +δ >α, thn η for all. Thn from th nqualty x x (x 0) t follows that f t +δ >α, thn Thus, f t +δ >α, thn 0 β η β+1 t +δ α β β η β t +δ α,. 1 η 2β t +δ α t +δ α β β η β = η β t +δ t +δ α t +δ α α η β t +δ α. (25) η Furthrmor, snc lm t +δ = andβ η 0, w hav lm β α = η and thrfor lm 2β η β t +δ t +δ α α = 0, so that from (25) t follows that β η β+1 lm t +δ α Puttng th abov lmts nto (23), w mmdatly obtan T w y 2 + p δ 2 Σ I0, I\I 0 I β = 0, f t +δ >α. whch contradcts (22). Hnc w provd thatη. So far w hav shown thatα t n,β 0 andη. By usng ths, n th nxt stp w wll show thatη 0. Stp 4. Lt us show thatη 0. To s ths, suppos on th contrary thatη = 0. Thn only on of th followng two cass can occur:
D. Juć, D. Marovć/Eur. J. Pur Appl. Math, 7 (2014), 230-245 242 () η = 0 andβ (0, ), or () η = 0 andβ =. Now, w ar gong to show that functonal T cannot attan ts nfmum n thr of ths two cass, whch wll prov thatη 0. Cas ():η = 0 andβ (0, ). In ths cas w would hav lm β η β+1 t +δ α β β η β = lm t +δ α t +δ α = 0, f t +δ >α and hnc from (23) t would follow that T w y 2 + t +δ α I p δ 2 Σ I0 whch contradcts assumpton (22). Cas ():η = 0 andβ =. Snc 0, thr xsts a ral numbr L> 1 and suffcntly grat 0 such that f t +δ >α and > 0, thn /(t +δ α )1/L. Wthout loss of gnralty, w may assum that 0 = 1. Thus, f t +δ >α, thn 0 β η β+1 t +δ α 1 β t +δ α L β Furthrmor, snc from (26) t follows that β β η β = η β t +δ t +δ α t +δ α α β. (26) β lm L β = 0 and lm β η β+1 lm t +δ α β = 1, β β = 0, f t +δ >α. Fnally, from (23) w obtan T t +δ w y 2 α + I p δ 2 Σ I0, whch contradcts assumpton (22). Ths mans that n ths cas functonal T cannot attan ts nfmum. Thus, w hav provd thatη 0.
REFERENCES 243 Stp 5. It rmans to show thatβ. W prov ths by contradcton. Suppos that β =. Argung as n cas () from stp 4, t can b shown that β η β lm t +δ α t +δ α β = 0, f 0 η t +δ 1. (27) α η If t +δ > 1, thn thr xsts a suffcntly grat α 0 such that η 0. Now, by usng th nqualty x x (x 0) w obtan β β η 0β t +δ α,, and thrfor β η β 0 t +δ α t +δ α 1 η (0+1)β t +δ α t +δ α β β. (28) η Snc lm β =, w hav that and thrfor from (28) t follows that ( lm 0+1)β t +δ α β η β lm t +δ α t +δ α β = 0 β = 0, f From (23), (27) and (29) w would obtan T t +δ w y 2 α + I p δ 2 contradcts (22). Thus, w hav provd thatβ and compltd th proof. η t +δ 1. (29) α > Σ I0, whch Rfrncs [1] B. Abbas, A.H.E. Jahrom, J. Arat, and M. Hossnouchac. Estmatng th paramtrs of Wbull dstrbuton usng smulatd annalng algorthm. Appld Mathmatcs and Computaton, 183:85-93, 2006. [2] D.M. Bats and D.G. Watts. Nonlnar rgrsson analyss and ts applcatons. Wly, Nw Yor, 1988.
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