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Stein's method, Malliavin calculus and infinite-dimensional Gaussian

Stein's method, Malliavin calculus and infinite-dimensional Gaussian

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By using an appropriate version of the so-called Jeulin Lemma (see [36, Lemma 1, p. 44]), one<br />

can prove that<br />

Z 1 Z 1<br />

W (t1 ; :::; t d ) 2<br />

<br />

dt 1 dt d = 1, a.s.-P.<br />

t 1 t d<br />

0<br />

0<br />

For " 2 (0; 1), we can now de…ne the r<strong>and</strong>om variable<br />

Z 1 Z 1<br />

W<br />

B" d (t1 ; :::; t d ) 2<br />

= <br />

dt 1 dt 2 :<br />

t 1 t d<br />

"<br />

"<br />

A st<strong>and</strong>ard computation yields E B" d <br />

= (log 1=") d , <strong>and</strong> also<br />

<br />

Var B"<br />

d (4 log 1=") d , as " ! 0:<br />

By setting<br />

eB " d , Bd " (log 1=") d<br />

,<br />

(4 log 1=") d 2<br />

one can therefore state the following generalization of Problem I, as stated at the end of Section<br />

2.1.<br />

Problem II. Prove that, as " ! 0, e B d "<br />

Law<br />

! N N (0; 1).<br />

Remark. See [21] for applications of quadratic functionals of Brownian sheets to tests of<br />

independence.<br />

11.1.2 Interlude: optimal rates for second chaos sequences<br />

In order to give an exhaustive answer to Problem II, we state (without proof) a result concerning<br />

sequences in the second Wiener chaos of a given isonormal <strong>Gaussian</strong> process X =<br />

fX (h) : h 2 Hg. It gives a simple criterion (based on cumulants) allowing to determine whether,<br />

for a sequence in the second chaos, the rate of convergence implied by (10.7) is optimal. Note that<br />

the forthcoming formula (11.1) is just a rewriting of (10.7), which we added for the convenience<br />

of the reader. We also use the notation<br />

(z) = P [N z] , where N N (0; 1) :<br />

Proposition 11.1 (See [58]) Let F n = I 2 (f n ), n 1, be such that f n 2 H 2 , <strong>and</strong> write<br />

(n)<br />

p = p (F n ), p 1. Assume that (n)<br />

2 = E(Fn) 2 ! 1 as n ! 1. Then, as n ! 1,<br />

Law (n)<br />

F n ! N N (0; 1) if <strong>and</strong> only if 4 ! 0. In this case, we have moreover<br />

s<br />

(n)<br />

4<br />

d Kol (F n ; N) <br />

6 + ((n) 2 1) 2 : (11.1)<br />

If, in addition, we have, as n ! 1,<br />

(n)<br />

2 1<br />

! 0; (11.2)<br />

(n)<br />

4<br />

6<br />

+ ( (n)<br />

2 1) 2<br />

56

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