Expectation of the product of almost independent Gaussians - MathOverflow

Expectation of the product of almost independent Gaussians

2009-10-26T15:46:00Z

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 $E\prod_ i|X_ i|^{t_ i}\le exp\left( b\sum_ i t_ i+C\sum_ i t_ i^2 \right)$?

Answer by unknown (google) for Expectation of the product of almost independent Gaussians

2009-10-27T05:47:23Z

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?

Answer by Tom LaGatta for Expectation of the product of almost independent Gaussians

2009-10-27T22:05:31Z

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|X_i|^t = (E|X|^t)ⁿ = (Ee^{t log|X|})ⁿ ≥ e^{tn Elog|X|} = e^tbn.

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.

Answer by Tom LaGatta for Expectation of the product of almost independent Gaussians

2009-10-27T22:38:55Z

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

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

φ(t) ≤ e^{bt + Ct²}.

If Y_i are identical copies of Y with fast correlation decay EY_iY_j ≤ e^-a|i-j|, then

Ee^t∑Y_i ≤ e^{bnt + Cnt²} for all C > 0,

where the sum is from 1 to n."