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Let $X_ i$ be copies of the standard real Gaussian random variable $X$. Let $b$ be the expectation of $\log |X|$. Assume that the correlations $EX_ iX_ j$ are bounded by $\delta_ {|i-j|}$ in absolute value where $\delta_ k$ is a fast decreasing sequence (in the application I have in mind, $\delta_ k=\exp\{-ck^2\}$ but it should be a huge overkill).

Is it true that there exists a constant $C$ depending on the sequence $\delta_ k$ only such that for all $t_ i>0$, we have $E\prod_ i|X_ i|^{t_ i}\le exp\left( b\sum_ i t_ i+C\sum_ i t_ i^2 \right)$?

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3 Answers 3

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As a side note, it seems that we get the opposite inequality for free. If the X_{i} are independent, and we are looking at it for 1 to n, we get

$E[\Pi_{i=1}^{n} \vert X_{i} \vert] = E[\Pi_{i} exp(log(|X_{i}|))] = E[exp(\Sigma_{i} log(|X_{i}|)] \geq exp(E[\Sigma_{i} log(|X_{i}|)]) = exp(nb)$.

Also, you might notice that this doesn't depend at all on the independence of the $X_{i}$... or on the exponent being 1, since we are only taking the expectation of a sum, never a product.

I realize this doesn't answer your original question at all, so I was curious as to where the hypothesis came from. In particular, could you post a proof in that direction when the $X_{i}$ are independnt? Where does C come from?

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  • $\begingroup$ Yeah, the arithmetic-geometric mean inequality you mentioned is clear. As to the independent case, it boils down to showing the inequality for just one random variable (because the expectation of the product is the product of expectations then). But then it reduces to the observation that any $C^2$ function $F$ defined on $(-a,+\infty)$ for some $a>0$ that is $1$ at $0$ and grows not faster than $e^{Ct^2}$ at $+\infty$ satisfies $F(t)\le e^{tF'(0)+Ct^2}$ for $t>0$ with some $C>0$. $\endgroup$
    – fedja
    Commented Oct 27, 2009 at 11:45
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That was extremely difficult to parse. Until LaTeX support isn't yet enabled, please try to more simply! (e.g. don't use \left and \right)

Consider the independent case with t constant. I was hoping C = 0 would work, and the non-zero C only arises because of the dependence. This isn't the case, however. By Jensen's inequality:

EΠ|Xi|t = (E|X|t)n = (Eet log|X|)n ≥ etn Elog|X| = etbn.

Thus you need that C > 0 if you want that upper bound. As you pointed out in reply to unknown's comment, any C will work. My conjecture is that in the general case with the fast correlation decay, the result should hold for any constant C > 0.

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Here's my rewording of your question. Think of Y below as log|X|.

"Let φ(t) = EetY be the moment-generating function of Y. Suppose that for any C > 0,

φ(t) ≤ ebt + Ct².

If Yi are identical copies of Y with fast correlation decay EYiYj ≤ e-a|i-j|, then

Eet∑Yi ≤ ebnt + Cnt² for all C > 0,

where the sum is from 1 to n."

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  • $\begingroup$ It should be "for SOME $C>0$", not for any! You never get arbitrarily small values of $C$ unless your random variable is a constant (the problem is not with large $t_ i$ but with small ones, of course). I'm not sure if it is enough to have $Y_ i$ just weakly correlated unless they are all Gaussians (when weak correlations mean almost independence) but OK, if you prove that this stronger conjecture, I'll be even more happy. Also, you made all the powers the same. It is quite likely that it doesn't really matter for the proof but be aware that the original question was more general. $\endgroup$
    – fedja
    Commented Oct 28, 2009 at 2:09

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