A note on characterization related to distributional properties of random translation, contraction and dilation of generalized order statistics

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1 PobSa Foum, Volume 6, July 213, Pages ISSN PobSa Foum is an e-jounal. Fo eails please visi A noe on chaaceizaion elae o isibuional popeies of anom anslaion, conacion an ilaion of genealize oe saisics Imiyaz A. Shah Depamen of Saisics, Univesiy of Kashmi, Sinaga Jammu an Kashmi-19 6, Inia Absac. The Elang-uncae exponenial isibuion has been chaaceize hough anslaion of wo non-ajacen genealize oe saisics gos an hen he chaaceizing esuls ae obaine fo Paeo isibuion hough ilaion of genealize oe saisics gos an powe funcion isibuion hough conacion of non-ajacen ual genealize oe saisics gos. Fuhe, he esuls ae euce fo oe saisics an ajacen ual genealize oe saisics an genealize oe saisics. 1. Inoucion Kamps 1995 inouce he concep of genealize oe saisics gos as follows: Le X 1, X 2,, X n be a sequence of inepenen an ienically isibue ii anom vaiables v wih he absoluely coninuous isibuion funcion f F x an he pobabiliy ensiy funcion pf fx, x a, b. Le n N, n 2, k >, m = m 1, m 2,, m n 1 R n 1, M = n 1 j= m j, such ha γ = k + n + M > fo all {1, 2,, n 1}. If m 1 = m 2 = = m n 1 = m, hen X, n, m, k is calle he h m gos an is pf is given as whee f X,n,m,k x = Cn 1 [F x]γ 1! n 1 γ n = k + n m + 1, 1 n, C n 1 = i=1 γ n i, 1 n. [ ] 1 [F x] m+1 1 fx, a < x < b, 1 m + 1 Base on he genealize oe saisics gos, Bukscha e al. 23 inouce he concep of he ual genealize oe saisics gos whee he pf of he h m gos X, n, m, k is given as f X,n,m,kx = Cn 1 [F x]γ 1! n 1 [ ] 1 [F x] m+1 1 fx, a < x < b, m + 1 Keywos. Oe saisics, Genealize oe saisics, Dual genealize oe saisics, Ranom anslaion, Conacion, Dilaion, Chaaceizaion of isibuions, Elang-uncae exponenial, Powe funcion, Paeo isibuions Receive: 4 Augus 212; Revise: 3 June 213; Accepe: 11 July 213 aess: masoom.immy3@gmail.com Imiyaz A. Shah

2 Shah / PobSa Foum, Volume 6, July 213, Pages which is obaine jus by eplacing F x = 1 F x by F x. If suppo of he isibuion F x be ove a, b, hen by convenion, we will wie X, n, m, k = a an X, n, m, k = b. Ahsanullah 26 has chaaceize exponenial isibuion une anom ilaion fo ajacen gos. In his pape, isibuional popeies of he gos have been use o chaaceize Elang-uncae exponenial isibuion fo non-ajacen gos une anom anslaion, ilaion an conacion, hus genealizing he esuls of Ahsanullah 26. One may also efe o Anol e al. 28, Beune an Kamps 28, Nevzoov 21, Navao 28, Oncel e al. 25, Wesolowski an Ahsanullah 24 an Casaño- Maínez e al. 212 fo he elae esuls. I may be seen ha if Y is a measuable funcion of X wih he elaion Y = hx, hen Y, n, m, k = hx, n, m, k 2 if h is an inceasing funcion, an Y, n, m, k = hx, n, m, k 3 if h is a eceasing funcion, whee X, n, m, k is he h m gos an X, n, m, k is he h m gos. We will enoe i X Elang uncae expβα λ if X has an Elang-uncae exponenial isibuion wih he f F x = [1 e βα λx ], x <, β >, λ >, whee α λ = 1 e λ. ii X P aβα λ if X has a Paeo isibuion wih he f F x = 1 x βα λ, 1 < x <, β >, λ >. iii X powβα λ if X has a powe funcion isibuion wih he f F x = x βα λ, < x < 1, β >, λ >. I may fuhe be noe ha if log X Elang uncae expβα λ, hen X P aβα λ, 4 an if log X Elang uncae expβα λ, hen X powβα λ 5 I has been assume hee houghou ha he f is iffeeniable w... is agumen. 2. Chaaceizaion esuls Theoem 2.1. Le X, n, m, k be he h m gos fom a sample wih absoluely coninuous f F x an pf fx. Then fo 1 < n 2 n 1, we have ha Xn 1 n 2 + j, n 1 j, m, k = X, n 2, m, k + Xn 1 n 2 j, n 1, m, k, j =, 1, whee Xn 1 n 2 j, n 1, m, k is inepenen of X, n 2, m, k if an only if he anom vaiable v X 1 has Elang-uncae expβα λ isibuion an X = Y enoes ha X an Y have he same f.

3 Shah / PobSa Foum, Volume 6, July 213, Pages Poof. To pove he necessay pa, le he momen geneaing funcion mgf of Xn 1 n 2 +, n 1, m, k be M Xn1 n 2 +,n 1, hen,m,k Xn 1 n 2 +, n 1, m, k = X, n 2, m, k + Y M Xn1 n 2 +,n 1,m,k = MX,n2,m,k M Y. Since fo he Elang-uncae expβα λ isibuion we have ha γ n 2 M = Cn2 Γ 1 m+1 βα λ m+1 X,n2,m,k m 1 = Γ heefoe βα λ m+1 + γn2 m+1 M Y = M X n1 n 2 +,n 1,m,k M = Cn1 Γ X,n2 m + 1 n1 n2 n,m,k Γ γ 2 as C n1 n = 1 n 2+ 1 Cn1 1 Cn1 n ; γn1 n 1 n 2+ = γ n2 i=1 m+1 an γ n1 n 1 n 2 1 γ n 2 m+1 βα λ γ n2 i 1, βα λ m+1 + βα λ m+1 + n 1 n 2 + = γ n2 + m + 1. Bu his is he mgf of Xn 1 n 2, n 1, m, k, he n 1 n 2 h m gos fom a sample of size n 1 awn fom Elang-uncae expβα λ an hence he esul. Fo he poof of sufficiency pa, we have by he convoluion meho f Xn1 n 2+,n 1,m,ky = as γ n1 n 1 n 2 = γ n1 1 n. = f X,n2,m,kxf Y y x x βα λ C n1 n 1 n 2 1!m + 1 n1 n2 1 Diffeeniaing boh he sies of 6 w... y, we ge y f Xn 1 n 2+,n 1,m,ky = βα λn 1 n 2 1m + 1 C n1 n 1 n 2 1!m + 1 n1 n2 1 [e βα λy x ] γn 1 [1 e βα λy x m+1 ] n1 n2 1 f X,n2,m,kxx, 6 βα λ γ n1 n 1 n 2 C n1 n 1 n 2 1!m + 1 n1 n2 1 βα λ [e βα λy x ] γn 1 +m+1 [1 e βα λy x m+1 ] n1 n2 2 f X,n2,m,kx x βα λ [e βα λy x ] γn 1 Now since, [1 e βα λy x m+1 ] n1 n2 1 f X,n2,m,kx x βα λ C n1 n f Xn1 n 2,n 1,m,kx = n 1 n 2 1!m + 1 n1 n2 1 [e βα λx ] γ n 1 [1 e βα λx m+1 ] n1 n2 2 [1 e βα λx m+1 ]

4 implying ha Shah / PobSa Foum, Volume 6, July 213, Pages βα λ C n1 n = n 1 n 2 1!m + 1 n1 n2 1 [e βα λx ] γ n 1 [1 e βαλx m+1 ] n1 n2 2 βα λ C n1 n n 1 n 2 1!m + 1 n1 n2 1 [e βα λx ] γ n 1 n 1 n +m+1 2 [1 e βαλx m+1 ] n1 n2 2 βα λ n 1 n 2 1m + 1C n1 n 1 n 2 1!m + 1 n1 n2 1 [e βαλx ] γ n 1 n 1 n +m+1 2 [1 e βαλx m+1 ] n1 n2 2 = Cn1 C n1 1 n 1 n 2 2 an afe noing ha C n1 n = γn1 1 C n1 1 This leas o o f Xn,n 1 1,m,kx m + 1n 1 n 2 1f Xn1 n 2,n 1,m,kx n an γn 1 1 n 2 2 m+1 +n 1 n 2 j+ = γn1 j m+1 ; γn2 +j y f Xn 1 n 2+,n 1,m,ky = βα λ γ n1 1 [f Xn1 n 2+ 1,n 1 1,m,ky f Xn1 n 2+,n 1,m,ky], = γn1 n 1 n 2++j. f Xn1 n 2+,n 1,m,ky = βα λ γ n1 1 [F Xn1 n 2+ 1,n 1 1,m,ky F Xn1 n 2+,n 1,m,ky]. 7 Now Kamps, 1995 F Xn1 n 2+ 1,n 1 1,m,ky F Xn1 n 2+,n 1,m,ky = Theefoe, in view of 1, 7 an 8, we have ha fy F y = βα λ implying ha F y = e βα λy an hence he poof. C n1 1 n 1 n 2+ 2 n 1 n 2 + 1!m + 1 n1 n2+ 1 [ F y] γn 1 + [1 F y m+1 ] n1 n Coollay 2.2. Le X, n, m, k be he h m gos fom a sample wih absoluely coninuous f F x an pf fx. Then fo 1 < n 2 n 1, we have ha Xn 1 n 2 + j, n 1 j, m, k = X, n 2, m, kxn 1 n 2 j, n 1, m, k, j =, 1, 9 whee X j, n 1, m, k is inepenen of X, n 2, m, k if an only if X 1 has Paβα λ isibuion. Poof. Hee he pouc X, n 2, m, kx j, n 1, m, k in 9 is calle anom ilaion of X, n 2, m, k Beune an Kamps, 28. Noe ha log Xn 1 n 2 +, n 1, m, k = log X, n 2, m, k + log Xn 1 n 2, n 1 m, k Xn 1 n 2 +, n 1, m, k = X, n 2, m, kxn 1 n 2, n 1, m, k an he poof follows in view of 2 an 4.

5 Shah / PobSa Foum, Volume 6, July 213, Pages Remak 2.3. A j = an βα λ = α, we have X + 1, n 1, m, k = X, n 2, m, kx1, n 1, m, k as obaine by Beune an Kamps 28. Remak 2.4. Le X :n be he h oe saisic fom a sample wih absoluely coninuous f F x an pf fx. Then fo 1 < n 2 n 1, we have ha X n1 n 2+ j:n 1 j = X :n2. X n1 n 2 j:n 1, j =, 1, whee X n1 n 2 j:n 1 is inepenen of X :n2 if an only if he anom vaiable v X 1 has Paβα λ isibuion. A j = an βα λ = α, we have X +1:n1 = X:n1 1X 1:n1, as given in Casaño-Maínez e al This is of he fom X s:n1 = X:n2 V, again an unsolve poblem Anol e al., 28. Coollay 2.5. Le X, n, m, k be he h m gos fom a sample wih absoluely coninuous f F x, an pf fx. Then fo 1 < n 2 n 1, we have ha X n 1 n 2 + j, n 1 j, m, k = X, n 2, m, kx n 1 n 2 j, n 1, m, k, j =, 1, 1 whee X n 1 n 2 j, n 1, m, k an is inepenen of X, n 2, m, k if an only if he anom vaiable v X 1 has powβα λ isibuion. Poof. Hee he pouc X, n 2, m, kx n 1 n 2 j, n 1, m, k in 1 is calle anom conacion of X, n 2, m, k Beune an Kamps, 28. I may be noe ha log Xn 1 n 2 +, n 1, m, k = log X, n 2, m, k log Xn 1 n 2, n 1, m, k X n 1 n 2 +, n 1, m, k = X, n 2, m, kx n 1 n 2, n 1, m, k an he esul follows wih an appeal o 3 an 5. Remak 2.6. Le X :n be he h oe saisic fom a sample wih absoluely coninuous f F x an pf fx. Then fo 1 < n 2 n 1, we have ha X :n1 j = X :n2. X n2+1:n 1 j, j =, 1, whee X n2+1:n 1 j is inepenen of X :n2, if an only if he anom vaiable v X 1 has powβα λ isibuion. This is of he fom X :n1 = X:n2 W, which a = 1 an βα λ = α, euces o X 1:n1 = X1:n2 W. This is iscusse by Anol e al. 28. A j = an βα λ = α, we have X :n1 = X:n1 1X n1:n 1, whee X n1:n powαn 1 as given by Wesolowski an Ahsanullah 24 an Casaño-Maínez e al. Theoem 2.7. Le X, n, m, k be he h m gos fom a sample wih absoluely coninuous f F x an pf fx. Then fo 1 < n 2 n 1, Xn 1 n 2 + j, n 1 j, m, k = Xn 1 n 2, n 1, m, k + X, n 2 j, m, k, j =, 1, whee X, n 2 j, m, k is inepenen of Xn 1 n 2, n 1, m, k if an only if he anom vaiable v X 1 has Elang-uncae expβα λ isibuion.

6 Shah / PobSa Foum, Volume 6, July 213, Pages Poof. To pove of he necessay pa follows fom Theoem 2.1. To pove he sufficiency pa, we have o, f Xn1 n 2+,n 1,m,ky = = βα λ C n2 1 1!m f Xn1 n 2,n 1,m,kxf Y y x x Diffeeniaing boh he sies of 11 w... y, we ge y f Xn 1 n 2+,n 1,m,ky = βα λ 1m + 1 C n2 1 1!m [e βα λy x ] γn 2 [1 e αy x m+1 ] 1 f Xn1 n 2,n 1,m,kxx. 11 βα λ [e βαλy x ] γn 2 +m+1 [1 e βα λy x m+1 ] 2 f Xn1 n 2,n 1,m,kx x βα λ γ n2 C n2 1 1!m βα λ [e βαλy x ] γn 2 [1 e βα λy x m+1 ] 1 f Xn1 n 2,n 1,m,kx x = βα λ γ n2 [f Xn1 n 2+ 1,n 1,m,ky f Xn1 n 2+,n 1,m,ky] f Xn1 n 2+,n 1,m,ky = βα λ γ n2 [F Xn1 n 2+ 1,n 1,m,ky F Xn1 n 2+,n 1,m,ky]. 12 Now Kamps, 1995 F Xn1 n 2+ 1,n 1,m,ky F Xn1 n 2+,n 1,m,ky = Theefoe, in view of 1, 12 an 14, we have fy F y = βα λ implying ha F y = e βα λy an he Theoem is pove. C n1 n 1 n 2+ 2 n 1 n 2 + 1!m + 1 n1 n2+ 1 [ F y] γn 1 + [1 F y m+1 ] n1 n Coollay 2.8. Le X, n, m, k be he h m gos fom a sample wih absoluely coninuous f F x an pf fx. Then fo 1 < n 2 n 1, Xn 1 n 2 + j, n 1, m, k = Xn 1 n 2, n 1, m, kx, n 2 j, m, k, j =, 1, whee X, n 2 j, m, k is inepenen of X, n 1, m, k if an only if X 1 has Paβα λ isibuion. Poof. Consie log Xn 1 n 2 +, n 1, m, k = log Xn 1 n 2, n 1, m, k + log X, n 2, m, k Xn 1 n 2 +, n 1, m, k = Xn 1 n 2, n 1, m, kx, n 2, m, k an he poof follows in view of 2 an 4.

7 Shah / PobSa Foum, Volume 6, July 213, Pages Remak 2.9. Le X :n be he h oe saisic fom a sample wih absoluely coninuous he f F x an pf fx. Then fo 1 < n 2 n 1, we have ha X n1 n 2+ j:n 1 = Xn1 n 2:n 1 X :n2 j, j =, 1, whee X :n2 j is inepenen of X n1 n 2:n 1 if an only if he anom vaiable v X 1 has Paβα λ isibuion. A j = an βα λ = α, we have ha X n1 n 2+:n 1 = Xn1 n 2:n 1 X :n2, ha is X s:n1 = X:n1 X s :n1 as obaine by Casaño-Maínez e al Coollay 2.1. Le X, n, m, k be he h m gos fom a sample wih absoluely coninuous f F x an pf fx. Then fo 1 < n 2 n 1, X n 1 n 2 + j, n 1, m, k = X n 1 n 2, n 1, m, kx, n 2 j, m, k, j =, 1, 14 whee X, n 2 j, m, k an is inepenen of X n 1 n 2, n 1, m, k if an only if X 1 powβα λ. Poof. This can be shown by consieing log Xn 1 n 2 +, n 1, m, k = log Xn 1 n 2, n 1, m, k log X, n 2, m, k X n 1 n 2 +, n 2, m, k = X n 1 n 2, n 1, m, kx, n 2, m, k an he esul follows wih an appeal o 3 an 5. Remak Le X :n be he h oe saisic fom a sample wih absoluely coninuous he f F x an pf fx. Then fo 1 < n 2 n 1, we have ha X n2 j+1:n 1 = Xn2+1:n 1. X n2 j+1:n 2 j, j =, 1, whee X n2 j+1:n 2 j is inepenen of X n2+1:n 1 if an only if he anom vaiable v X 1 has powβα λ isibuion. A = 1 an βα λ = α, his euces X n1 j:n 1 = Xn2+1:n 1 X n1 j:n 1 j o X :n1 = X+1:n1 X :, whee X : powα as obaine by Navao 28 an Casaño-Maínez e al Acknowlegemens The auho is hankful o he Refeee an he Eio fo hei fuiful suggesions an commens. Refeences Ahsanullah, M. 26 The genealize oe saisics fom exponenial isibuion, Pak. J. Sais. 222, Anol, B.C., Casillo, E., Saabia, J.M. 28 Some chaaaceizaions involving unifom an powes of unifom anom vaiables, Sais. 426, Beune, E., Kamps, U. 28 Ranom conacion an anom ilaion of genealize oe saisics, Commun. Sais. Theo. Meh. 37, Bukscha, M., Came, E., Kamps, U. 23 Dual genealize oe saisics, Meon LXII, Casaño-Maínez, A., López-Blázquez, F., Salamanca-Miño, B. 21 Ranom anslaions, conacions an ilaions of oe saisics an ecos, Sais DOI: 1.18/ Kamps, U A Concep of Genealize Oe Saisics, B.G. Teubne, Suga. Nevzoov, V.B. 28 Recos: Mahemaical Theoy, Tanslaion of Mahemaical Monogaphs, 194, Povience: Ameican Mahemaical Sociey. Navao, J. 28 Chaaceizaions by powe conacion of oe saisics, Commun. Sais. Theo. Meh. 37, Oncel, S.Y., Ahsanullah, M., Aliev F.A., Aygun, F. 25 Swiching eco an oe saisics via anom conacion, Sais. Pobab. Le. 73, Wesolowski. J., Ahsanullah, M. 24 Swiching oe saisics hough anom powe conacions, Aus. N.Z. J. Sais. 46,