By Alexander V. Ivanov (auth.)

Let us imagine that an remark Xi is a random variable (r.v.) with values in 1 1 (1R1 , eight ) and distribution Pi (1R1 is the genuine line, and eight is the cr-algebra of its Borel subsets). allow us to additionally suppose that the unknown distribution Pi belongs to a 1 definite parametric family members {Pi() , () E e}. We name the triple £i = {1R1 , eight , Pi(), () E e} a statistical test generated by way of the statement Xi. n we will say statistical scan £n = {lRn, eight , P; ,() E e} is the made of the statistical experiments £i, i = 1, ... ,n if PO' = P () X ... X P () (IRn 1 n n is the n-dimensional Euclidean area, and eight is the cr-algebra of its Borel subsets). during this demeanour the test £n is generated by way of n autonomous observations X = (X1, ... ,Xn). during this booklet we learn the statistical experiments £n generated through observations of the shape j = 1, ... ,n. (0.1) Xj = g(j, (}) + cj, c c In (0.1) g(j, (}) is a non-random functionality outlined on e , the place e is the closure in IRq of the open set e ~ IRq, and C j are autonomous r. v .-s with universal distribution functionality (dJ.) P no longer reckoning on ().

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**Sample text**

17). 24) holds. Let 4JL2 ~ 1. 25) CHAPTER 1. 17) will hold. 18) holds for r > r1 miner : rs/(s+q) > 1 + r- 1 ). 25). • EXAMPLE 4: Let g(j, (J) = (J1 cos (J2 j, e = (0, A) x (h,7r T = [a, b] x [

0, A j ~ 1, < 00, b < A, h> 0,

h, s = 3. )1/2 = ( ~ ! J cos J 2+4 sin(J2 ' and uniformly with respect to (J E T. On the other hand, d2n «(J) = (J1 (Lj2 sin2 (J2j y/2 (J1 (n3 6 + n 2 _ ~ + n 2 sin(2n + 1)(J2 + ~ 4 12 4 sin (J2 cos2n(J2 4 sin2 (J2 _! j6 (1 + O(n-1)) n uniformly with (J E T.

21) 3; and its variance + g1)2, 9 ~ 0. Evidently Dlej + gl is a continuous function of g. Let us show that Dlej Since EICj + gl ~ 9-+00 + gl = 9 1 ~ 1'2 9+ -9 P (dx) then Die; + gl + g' -49 / g JL2· + 2 100 xP (dx), g+ (1- (f>(dX»),) + P(dx) [00 xP(dx) -4 ([00 XP(dx))2 -g 19+ < 4 [00 x 2p(dx) ~ 0. 22) g-+oo 19+ 3. 22) is quite clear. 24) . 24) is a consequence of the Chebyshev inequality (x(T) = 1). Let us consider the case s = 1. Let us show that the triangular array of r. -s = 1, ... 6 for a compact T.

22), we obtain p; { v(u) ~ sup UE( vC(r")\V(XTn))nU:; (0) rl < '~:::>(~r) + (n [r"/xTnl L (r I} 2 + 1)Sr- 28 x S(r) r=l [r" /xTnl X L r- 28 (r + l)s+qx- q(r). , if 8 2 > 8 + q. 13) will then be established if the constant x is chosen appropriately. Clearly the function r +1 ----;:2 x(r) decreases monotonically for r to take 1 = rs/(s+q) > O. 17). 24) holds. Let 4JL2 ~ 1. 25) CHAPTER 1. 17) will hold. 18) holds for r > r1 miner : rs/(s+q) > 1 + r- 1 ). 25). • EXAMPLE 4: Let g(j, (J) = (J1 cos (J2 j, e = (0, A) x (h,7r T = [a, b] x [

0, A j ~ 1, < 00, b < A, h> 0,

h, s = 3.