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pr.likelihood – Pedestrian proof of Gaussian chaos for order-two poynomial? Answer

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pr.likelihood – Pedestrian proof of Gaussian chaos for order-two poynomial?

Let $ell geqslant 1$. Let us deem $(g_n)_{n in mathbb{N}}$ identically distributed idependent actual gaussian variables and actual quantity $(a_{n_1,dots n_{ell}})_{(n_1, dots, n_{ell}s)inmathbb{N}^{ell}}$. We deem the random variable
$$G(omega)=sum_{n_1, dots, n_{ell}}a_{n_1, dots, n_{ell}} g_{n_1}(omega) cdots g_{n_{ell}}(omega)$$
the place we write $omega in Omega$ the likelihood area at hand.

It is named Gaussian polynomial chaos (or Wiener chaos)
$$|G(omega)|_{L^p_{Omega}} lesssim p^{frac{ell}{2}} |G(omega)|_{L^2_{Omega}}.$$

The benchmark proof of this outcomes makes use of hypercontractivity of the Ornstein-Uhlenbeck semigroup.

My query is the next: within the illustration $ell =2$ and $p=2k$ an plane integer,

Is it workable to show “directly by hand” that
$$mathbb{E}[|G(omega)|^{2k}] leqslant C ok^{2k}mathbb{E}[|G(omega)|^2]^ok.$$
If sure, is there a reference bespeak/article to search out it? Or is it workable to terminate my tried proof?

This doesn’t appear inconceivable. At least, when one expands the left-hand aspect, we’ve got
$$mathbb{E}[|G(omega)|^{2k}] = sum_{substack{n_1,dots ,n_k m_1, dots, m_k}}a_{n_1,m_1} dots a_{n_{2k},m_{2k}}mathbb{E}[g_{n_1}g_{m_1}cdots g_{n_{2k}}g_{m_{2k}}].$$

From there, one can employ the truth that $mathbb{E}[g^{2m+1}]=0$ for any $m$ to deduce that within the above, the expectation is non-vanishing provided that the indices ${n_i, m_i}$ emerge an plane variety of time. So specifically, all $n_i, m_i$ are paired.

The privilege hand aspect reads
$$ok^{2k}mathbb{E}[|G(omega)|^2]^ok = ok^{2k}sum_{substack{n_1,dots ,n_k m_1, dots, m_k}}a_{n_1,m_1} dots a_{n_{2k},m_{2k}}prod_{i=1}^kmathbb{E}[g_{n_i}g_{m_i}g_{n_{k+i}}g_{m_{k+i}}].$$

Also, comment that within the LHS, when all of the $g_i$ are equal, then we’ve got $mathbb{E}[g^{4k}] lesssim ok^{2k}$, which is the rectify magnitude which we search for.

However I’m not capable of show that $LHS leqslant RHS$. But I stay satisfied that there ought to breathe someplace to search out such a proof however I didn’t find it within the litterature.

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