Expectation of the product of almost independent Gaussians - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T13:26:25Z http://mathoverflow.net/feeds/question/2628 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/2628/expectation-of-the-product-of-almost-independent-gaussians Expectation of the product of almost independent Gaussians fedja 2009-10-26T15:46:00Z 2009-11-20T17:33:04Z <p>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).</p> <p>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 $E\prod_ i|X_ i|^{t_ i}\le exp\left( b\sum_ i t_ i+C\sum_ i t_ i^2 \right)$?</p> http://mathoverflow.net/questions/2628/expectation-of-the-product-of-almost-independent-gaussians/2780#2780 Answer by unknown (google) for Expectation of the product of almost independent Gaussians unknown (google) 2009-10-27T05:47:23Z 2009-10-27T05:47:23Z <p>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</p> <p>$E[\Pi_{i=1}^{n} \vert X_{i} \vert] = E[\Pi_{i} exp(log(|X_{i}|))] = E[exp(\Sigma_{i} log(|X_{i}|)] \geq exp(E[\Sigma_{i} log(|X_{i}|)]) = exp(nb)$.</p> <p>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.</p> <p>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?</p> http://mathoverflow.net/questions/2628/expectation-of-the-product-of-almost-independent-gaussians/2908#2908 Answer by Tom LaGatta for Expectation of the product of almost independent Gaussians Tom LaGatta 2009-10-27T22:05:31Z 2009-10-27T22:05:31Z <p>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)</p> <p>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:</p> <p>E&Pi;|X<sub>i</sub>|<sup>t</sup> = (E|X|<sup>t</sup>)<sup>n</sup> = (Ee<sup>t log|X|</sup>)<sup>n</sup> &ge; e<sup>tn Elog|X|</sup> = e<sup>tbn</sup>.</p> <p>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.</p> http://mathoverflow.net/questions/2628/expectation-of-the-product-of-almost-independent-gaussians/2921#2921 Answer by Tom LaGatta for Expectation of the product of almost independent Gaussians Tom LaGatta 2009-10-27T22:38:55Z 2009-10-27T22:38:55Z <p>Here's my rewording of your question. Think of Y below as log|X|.</p> <p>"Let &phi;(t) = Ee<sup>tY</sup> be the moment-generating function of Y. Suppose that for any C > 0, </p> <p>&phi;(t) &le; e<sup>bt + Ct²</sup>. </p> <p>If Y<sub>i</sub> are identical copies of Y with fast correlation decay EY<sub>i</sub>Y<sub>j</sub> &le; e<sup>-a|i-j|</sup>, then </p> <p>Ee<sup>t&sum;Y<sub>i</sub></sup> &le; e<sup>bnt + Cnt²</sup> for all C > 0, </p> <p>where the sum is from 1 to n."</p>