The typical size of a random element in a Banach space - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T04:04:28Z http://mathoverflow.net/feeds/question/39792 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/39792/the-typical-size-of-a-random-element-in-a-banach-space The typical size of a random element in a Banach space Tom LaGatta 2010-09-23T20:45:18Z 2010-09-24T17:15:46Z <p>Let $X$ be a separable Banach space, and let $\mathbb P$ be a Radon probability measure on $X$ with zero mean and covariance operator $K : X^* \to X$. Let $x$ be an $X$-valued random variable with distribution $\mathbb P$. </p> <p>I would like a simple upper bound on the size of $\mathbb E \|x\|^2$ in terms of the operator norm $\|K\|$. In his textbook <em>The Concentration of Measure Phenomenon</em>, Ledoux proves that $$\mathbb E\|x\|^2 \le 4 \|K\|$$ as a consequence of a concentration inequality for a simpler example (where the vectors are sums of vectors with i.i.d. random coefficients $\eta_i$ such that <code>$|\eta_i| \le 1$</code> a.s.). Because of the boundedness assumption, the argument doesn't quite work in this setting, though I'm certain I could generalize it if I needed to. </p> <p>Nonetheless, I'm certain that the estimate I'm looking for is buried somewhere in the literature on concentration of measure (and in fact is probably due to Talagrand). Could you please point me in the right direction?</p> <p><b>Edit:</b> The inequality as I previously wrote it is incorrect. The correct inequality should be $$\operatorname{Var}(\|x\|) \le 4\|K\|,$$ implying $$\mathbb E\|x\|^2 \le 4\|K\| + \left( \mathbb E\|x\| \right)^2.$$ That is, the size of typical random element could be quite large (i.e. $\mathbb E\|x\| \gg 1$, as in Mark Meckes's example in the comments), but the deviation is only of the order $\|K\|^{1/2}$.</p> http://mathoverflow.net/questions/39792/the-typical-size-of-a-random-element-in-a-banach-space/39873#39873 Answer by Mark Meckes for The typical size of a random element in a Banach space Mark Meckes 2010-09-24T17:15:46Z 2010-09-24T17:15:46Z <p>The inequality can't be true without additional assumptions. To see this, let $X = \ell_2^n$ and let $x$ have a spherically symmetric distribution and let $R = \Vert x \Vert$. Then $R$ is an essentially arbitrary nonnegative random variable; indeed we could start by picking $R$ and defining $x=R\Theta$, where $\Theta$ is a uniform random vector in $S^{n-1}$ independent of $R$. </p> <p>Now $\mathrm{Var}(\Vert x \Vert) = \mathrm{Var} (R)$, and $K = \frac{1}{n} \mathbb{E}(R^2) I_n$, so $\Vert K \Vert = \frac{1}{n} \mathbb{E} (R^2)$. Thus for any fixed distribution of $R$, the inequality fails for sufficiently large $n$.</p>