Appendix. Proof of relation (3) for α 0.05.

Transcription

1 Appendi. Proof of relation 3 for α.5. For the argument, we will need the following reult that follow from Lemma 1 Bakirov 1989 and it proof. Lemma 1 Let g,, 1 be a continuouly differentiable function uch that g >, g for all, 1, A.1 and let f,, 1 be a continuou function uch that f for ome [, 1, A.2 fd >. A.3 Then fgd >. For completene, we provide the proof of Lemma 1 below. Proof of Lemma 1. It uffice to how that i eay to ee that from condition A.2 and A.3 it follow that ɛ fgd > for all ɛ, 1. It ɛ [, 1. Thi, together with A.1, implie, uing integration by part, that ɛ fd > for all fgd = gɛ fd + g ɛ ɛ ɛ fd d > A.4 for all ɛ, 1. The proof i complete. Proof of relation 3 for α.5. Denote = A in Bakirov 1989, we have q cv2 cv 2 +q1. P t > cv = 1 π q q/2 d k k + 1 q k +,

2 where k = σ 2 k and, uing multiplication of all σ k, k = 1,..., q, by a contant, k k + 1 =. A.5 Let z = ma 1 k q k. Fiing all k, ecept i = z and j = y and conidering j = y a a function of z, one obtain that the derivative of the probability P t > with repect to z at the point = 1,..., q i given by ee Bakirov 1989 where M = y z 2π Lz, y, P, 1 U, d, A.6 Lz, y, v = A + B z + y + 2A B z + 2 y + 2 v, A.7 U, = q/21, A.8 q P, k + P, = k k + 1 k +, A.9 A = yz 1, B = y + z + 2. A.1 A in Bakirov 1989, from condition A.5 it follow that z + y + P,, and, therefore, the following etimate hold: 2πM 1 y z U, z + y + A + B 1 2 A B. A.11

3 Define the function f, y, z, = g, y, z, = According to A.11, thu, α 1+ 2A + A + B B 2 1 U, α [ z + y + ep 2πM y z ln 21 + z f, y, z, g, y, z, d [ ep. ln 21 + z the above-defined function f and g differ from the function f = A.12 α 1 A + B 2A B/ and g = U, α /z + y + in Bakirov 1989 by the additional factor e ln/21+z and e ln/21+z, repectively. Let u how, uing Lemma 1 that f, y, z, g, y, z, d >. Clearly, condition A.2 i atified for f if 2. We have ln g, y, z, 1 1 z z ince that + = q 1 + α k k + 1 α + 1 k q k k +1 k P, 1 y + A z + 1 y z z k z + k + 1 k + P, and k =. Now note k + k + i a conve function of, o that it i uniformly larger than it tangent at the point = 1, i.e for all [, 1. A k Therefore, i a conve function of, o that it i uniformly larger than it k + tangent at the point = 1, i.e. k k + k k k for all [, 1. k + 1 2

4 Alo, k z + 1 k + 1 k 2 k + 1 Uing the lat inequality, we have ln g, y, z, =, o that k k z 2 1 α z. 1 1 α 1 y z z z. A.15 From A.15 it follow that condition A.1 are atified for g if We have where f, y, z, y f, z, = 4 2α z. A.16 = f, z, g, z, d, α z 1 + 2z , Since [ g, z, = ep ln 21 + z. f, z, d = z 1 + 2z Γ.5Γ1 α + Γ1.5 α Γ.5Γ2 α [ z 1 + 2z Γ2.5 α 1 1 Γ.5Γ2 α Γ2.5 α 1.5 α = 1 α, A.17 the function f, z, and g, z, atify condition of Lemma 1 and, thu, f, y, z, / y > for > 5 4α αz 2 2α + 3 2αz. A.18

5 Obviouly, A.18 i atified if > 5 4α 2 2α. A.19 From the above we conclude that, under A.19, inequality A.3 hold if f,, z, d = f 1, z, g, z, d >, A.2 where f 1, z, = α +z+12+z 1. We have that the function f 1 and g atify the condition of Lemma 1 if that i, for f 1, z, d >, A.21 2 z zγ2 α Γ2.5 α + zγ1 α Γ1.5 α > A.22 or, equivalently, 2 z z1 α + z1.5 α >. A.23 Condition A.23 i atified if > Clearly, A.24 implie A α + 5 4αz 1 2α + 2 2αz. A.24 Firt, let = 5 + 5/ and let α = 4 /2 + 1/21 + z = 3/4 5/4 + 1/21 + z /21 + z. Condition A.16 i trivially atified. It i eay to check that inequality A.22 and, thu, A.19, are atified for z > z = z > z. = 3 + 5/ We conclude that A.13 hold for and Let u now how that A.13 i alo atified for and z z.

6 We firt make the following obervation. Let z and z 1 be uch that z z z 1 and let. Then f 1, z, g, z, d = g, z, f 1, z, g, z 1, g, z 1, d. Since ln g,z, g,z 1, = z z 1 1 A.25 from Lemma 1 it follow that it uffice to how that f 1, z, g, z 1, d > to be able to conclude that f 1, z, g, z, d >. In addition, f 1, z, z = α and, therefore, α d = 1 1Γ.5Γ2 α Γ.5Γ1 α = Γ2.5 α Γ1.5 α Γ.5Γ1 α 1 α α > Γ1.5 α if A.19 hold. Conequently, under condition A.19, by Lemma 1, f, z, g, z 1, d z and it uffice to how f, z, g, z 1, d > to conclude that, for all z [z, z 1, inequality A.2 and, therefore, A.13 hold. Finally, note that f, z, = α 1 z z 2 o that it, under A.19, it uffice to how f, z, g, z 1, d > A.26 to conclude that for all z [z, z 1 and, inequality A.13 i atified.

7 It i eay to check that condition A.16 and A.19 are atified for all and z if α = 5/2 /2 = 5 5/ From the above dicuion we conclude that, in order to how that A.13 hold for and z z, it uffice to check that, for a ufficiently mall and all 1 m z/ + 1, Im, = f, m 1, g, m, d >. Uing numerical computation, one can check that, for =.1 and all 1 m z/ + 1, Im, > I,.76 >. Thi implie that A.13 i atified for and z z, and, therefore, for all z. From the above argument it follow that it remain to conider the cae where i = /k, i = 1,..., k and i =, i = k + 1,..., n for ome 1 k n. That i, a in Bakirov 1989, we need to compare the quantitie Φ k = Φ k = P T k > where z k =. We have that k k 1 = 1 k π k/21 d zk + k1 1, < k n, A.27 where Φ k = 1 π f 1, z k, g, z k, d, f 1, z, = f 1, z, = f 1, z, g 1, z,, 1α z + z z z + ln 1 +, 1 [ g 1, z, = ep ln 21 + z, We have g 1, z, = [ k/22+α z + ep k+1/2 ln z [ln g 1, z k, = k 4 + 2α 2 k + 1 2z k + 2z k z k + 1 =

8 α 5/2 α 5/2 Uing again A.14, we obtain that [ln g 1, z k, + k + 1 [1 2 1 z k + 2z k + 1 = + k + 1z k 2z k + 2z k z k + 1. α 5/2 2z k z k + 1 = α 5/2 + k + 1 [ z k z k z k + 1 z k k + 1 [ 2 k + 1 kz k + 1 Evidently, the lat epreion i nonnegative for 5 2α. A in Bakirov 1989, we have that 1 2z k + 1. f Γ.5Γ2 α 1, z, hz, Γ2.5 α [ where hz = 21 α z + z z ln1 + z z 2 /2. It i not difficult to check that, for 1 = 3.7 and α = 5/2 1 /2 =.65, for h z = α α 1 + z z 21 α 41 α Since the function g 1, z, i nondecreaing, from Lemma 1 we conclude that f 1, z, > for all z 32 and 1. Let now z and z 1 be uch that z z z 1 and. Uing A.25, we conclude, a before that it uffice to how that f 1, z, g 1, z 1, d > for all z z z 1 and to be able to conclude that

9 for all z z z 1 and. In addition, ince f 1, z, g 1, z, d > f 1, z, we conclude that A.28 i atified if = 1α z > A.28 Iz = for all z z z 1. Finally, we note that it uffice to how that f 1, z, g 1, z 1, d > I z = to be able to conclude that 2 f1, z, z 2 g 1, z 1, d = 1α z + g 1, z 1, d > A.29 Iz > Iz + I z z z. Since, a i eay to ee, I = I =, from the above dicuion we conclude that, in order to how that A.28 hold for 1 and z z, it uffice to check that, for a ufficiently mall and all 1 m z/ + 1, Im, = g 1, m, m 1 + d >. Uing numerical computation, one can check that, for =.1 and all 1 m z/ + 1, Im, > I,.12 >. 1 and z z, and, therefore, for all z. Thi implie that A.13 i atified for Uing Lemma 1 we conclude that A.27 i increaing in < k n for all 1. Thi alo implie that 1 q 1 13 P T q1 > q > P T 14 >

10 for all q 15. Since, a we dicued before, A.13 i atified for all z if, where = 5 + 5/2 < 1 = 3.7, in view of A.12, thi complete the proof of the theorem for all q 14. Let u now how that A.13 i atified for 5 q 13 and q = q52α, q+1 where α = αq = 2+5q 4+5q 2. 4q Note that condition A.5 implie that ma k/ k + 1 /q, and, therefore, ince 1 k q / + 1 i increaing in, z = ma z k /q = z min. Conequently, 1 k q condition A.16 and A.19 are atified if and q. For α = 2 q/2 + 1/21 + z, condition A.16 and A.22 are atified for q and z > 2 q+6 q6 2 q4 q+3 and z > zq. = zq. Thi implie that A.13 hold for q In view of the above argument, for completion of the proof, it remain to check that, for 5 q 13, A.26 i atified with z = z q = q/q q, z1 = zq, α = αq and = q. Numerical calculation how that, indeed, for 5 q 13, with the above-defined α, z, z 1 and, Iq = f, z q, q g, zq, qd I13 >. A.3 We have that 5 = , 6 = , 7 = , 8 = , 9 = , 1 = , 11 = , 12 = and 13 = Numerical calculation how that qq 1 P T q1 > P T 5 > q q for 5 q 13. Conequently, the theorem hold for all q 5. The proof i complete

11 eference Bakirov, N. 1989: The Etrema of the Ditribution Function of Student atio for Obervation of Unequal Accuracy are Found, Journal of Soviet Mathematic, 44,