Simple version of fourth moment theorem of Nualart and Peccati

Question

I'm trying to understand this theorem, which gives several conditions for a sequence of random variables in the $n$-th Weiner chaos to converge to a normal law. I am finding even the statements of the result difficult, though, and am trying to find the simplest possible form of this result.

I dislike the abstractness that most presentations work with, and am only interested in the Brownian motion case anyway. I prefer to look at everything in terms of iterated stochastic integrals. So, my understanding of the theorem is the following.

We have a sequence $X_j$ of r.v.'s of the form

$$ X_j = \int_0^T \int_0^{t_1} \int_0^{t_2} \ldots \int_0^{t_n} f_j(t_1, t_2, \ldots, t_n)dB_{t_n} dB_{t_{n-1}} \ldots dB_{t_1}, $$

where the $f_j$'s are deterministic functions, so that the $X_j$'s are all in the $n$-th Weiner chaos, for $n \geq 2$. Suppose also that $E[X_j^2] \to 1$ as $j \to \infty$. Then the following are equivalent.

1. $E[X_j^4] \to 3$ as $j \to \infty$.
2. $X_j$ converges to a $N(0,1)$ r.v. as $j \to \infty$.
3. $\lim_{j \to \infty}||f_j^{\otimes p}||^2 = 0$, for all $p=1, 2, \ldots, n-1$, where $f_j^{\otimes p}$ denotes a contraction in the Hilbert space.

My problem is the following. It seems that, in practice, to show convergence to a normal law the easiest thing to verify is condition 3, rather than condition 1. However, I don't understand what condition 3 really says, because I don't understand this contraction thing. It is generally defined abstractly but with all sorts of tensor notation, and it's more than I really need. Is it possible to give a simple statement for what 3 is saying? For instance, can $f_j^{\otimes p}$ just be expressed as a simple integral in the situation given, and if so what is the integral?

Thanks, Greg

Answer 1

Write $\tilde f$ for the symmetrisation of $f$, namely $$ \tilde f(t) = \frac{1}{n!} \sum_\sigma f(\sigma t)\;, $$ where $(\sigma t)_i = t_{\sigma(i)}$ for any permutation $\sigma$ of the $n$ arguments. Then what you call $f^{\otimes p}$ (but has nothing to do with the $p$-fold tensor product of $f$ with itself) is simply given by $$ f^{\otimes p}(s_1,\ldots,s_{2n-2p})= \int \tilde f(t_1,\ldots,t_p,s_1,\ldots,s_{n-p})\tilde f(t_1,\ldots,t_p,s_{n-p+1},\ldots,s_{2n-2p})\,dt_1\cdots dt_p\;. $$ The norm appearing in 3 is then the one of $L^2(\mathbb{R}^{2n-2p})$.