Ji J

In (28) and elsewhere, we use the convention that Y1 $(■) := 0. It is noteworthy (refer to Sukhatme [Suk37], see, e.g., Malmquist [Mal50] and pp. 20-21

in David [Dav81]) that, for each n > 1, {win ■ 1 < i < n} is a sequence of independent and exponentially distributed random variables. For convenience, we denote below by w = i = 1,...,n, a standard exponential random variable, fulfilling P(w > y) = e-y for y > 0.

Let g() be a measurable function on R+ = [0, x>). Below, we will assume that g £ G, where G = L2 (R+ ,e-udu) denotes the Banach space, with respect to the norm || • 112 of all such functions for which

Yi , n(g) = Y {- l0S Wj, n - Y + (wj, n - + ^ {g(Yj,n) - Vg]

where, for i = 0,...,n, i ii,n { - log Wj,n - Y + (Wj,n - l) |, (33)

We note for further use that the following inequalities hold for all gi,g2 £ G such that vgi = vg2. We have,

E(\Yn,n(gi) - Yn,n(g2)\2) = e(\Zn,n(gi ) - Zn,n (g2)\2) n 2

În,n = E { — log ^i,n — Y + ("i,n — , i=1 n

= E {— log (n(Ui,n — Ui-1,n)) — y} + Op(logn). (36)

Proof. We will make use of the following inequalities (see, e.g., 8.368, p.947 in Gradshteyn and Ryzhik [GR65]). We have, for each n > 1,

k=i which readily yield the following rough inequality. For each 1 < i < n,

Next, we write, via (27) and Taylor's formula, for 1 < i < n,

Ui,n - Ui-i,n = exp ( - Yi-i,n) - exp(-Yiin) = (Yin - Yi-i,n) exp ( - {Y^n - Pi,n(Yin - Yi-i,n)}), (40)

where pin fulfills 0 < pin < 1. By combining (28) with (40), we obtain readily that, as n ^ to, n n

E{ - l0g (n(Ui,n - Ui-i,n ))} = ^E ^ (Yi,n - Yi-i,n) - n log n i=i i=i n n

+ E { Yin - Pi,n(Yi,n - Yi-i,n)} = ^E { log Wi,n + log { n + Y }} i=i i=i n i .. n i .. n

Observe that n n

0 < ^^ |Pi,n(Yi,n - Yi-i,n)} < ^^ jYi,n - Yi-i,nj = Yn,n •

Tests of Fit based on Products of Spacings 127 Now, it is easily checked that, for each choice of c > 1, as n ^ tt,

P(Yn,n > clog n) < E P(y1 > clog n) = ne-clog n = ni-c ^ 0,

so that Ynn = OP(logn) as n ^ tt. Moreover, by (2), we have, as n ^ tt,

—i°g{ n \ <y\i+ 1 i ln-i +1 < ^ In n-i +1J

Was this article helpful?

0 0

Post a comment