What is known about the Gaussian measure of the unit ball in a Hilbert Space? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T23:04:13Z http://mathoverflow.net/feeds/question/36403 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/36403/what-is-known-about-the-gaussian-measure-of-the-unit-ball-in-a-hilbert-space What is known about the Gaussian measure of the unit ball in a Hilbert Space? RadonNikodym 2010-08-22T21:52:14Z 2010-08-25T06:39:39Z <p>Let $X$ be an infinite dimensional separable Hilbert Space with norm $||\cdot||$ and let $\mu$ be a Gaussian measure on $X$ such that $\mu(X) = 1$. What do we know about $\mu(B(0,1))$, where $B(0,1)$ is the unit ball w.r.t the norm?</p> <p>This seems to me like a fundamental question but I cannot seem to find anything. Any information/references would be most appreciated.</p> <p>EDIT: A related question which is of interest to me: Do there exist asymptotically tight bounds to $\int_{||u||> K}||u||^2 \mu(du)$?</p> http://mathoverflow.net/questions/36403/what-is-known-about-the-gaussian-measure-of-the-unit-ball-in-a-hilbert-space/36407#36407 Answer by Richard Borcherds for What is known about the Gaussian measure of the unit ball in a Hilbert Space? Richard Borcherds 2010-08-22T22:08:17Z 2010-08-22T22:08:17Z <p>There is no Gaussian measure on an infinite dimensional Hilbert space, or rather the Gaussian measure is identically zero. (Proof: If the Gaussian measure of a ball of radius r on a 1-dimensional Hilbert space is c&lt;1, then that of a ball in n dimensional is less than c<sup>n</sup>, so in infinite dimensions any ball has measure 0, so the measure of the whole space is 0.) You can put a non-zero Gaussian measure on a larger space (see <a href="http://en.wikipedia.org/wiki/Rigged_Hilbert_space" rel="nofollow">http://en.wikipedia.org/wiki/Rigged_Hilbert_space</a>) and the unit ball of Hilbert space is a subset of this, but has measure 0 by the above argument.</p> http://mathoverflow.net/questions/36403/what-is-known-about-the-gaussian-measure-of-the-unit-ball-in-a-hilbert-space/36417#36417 Answer by Peter Shor for What is known about the Gaussian measure of the unit ball in a Hilbert Space? Peter Shor 2010-08-22T23:22:00Z 2010-08-22T23:53:32Z <p>You can't talk about "the" Gaussian measure on an infinite-dimensional Hilbert space, for the same reason that you can't talk about a uniform probability distribution over all integers. It doesn't exist; see Richard's answer. However, there are a lot of non-uniform Gaussian measures on infinite dimensional Hilbert spaces. </p> <p>Consider the measure on $\mathbb{R}^\infty$ where the $j$th coordinate is a Gaussian with mean 0 and variance $\sigma_j^2$, where $\sum_{j=1}^{\infty} \sigma_j^2 &lt; \infty$ (and different coordinates are independent). This is almost surely bounded in the $\ell_2$ metric, and any projection onto a finite-dimensional space has a Gaussian distribution. The squared length of a vector drawn from this measure is a sum of squares of Gaussians, and so follows some kind of generalized $\chi$-square distribution. If I knew more about generalized $\chi$-square distributions, I might be able to tell you what the measure of the unit ball was.</p> <p>This kind of Gaussian distribution is very important in quantum optics. In fact, in quantum optics, a thermal state is Gaussian, so "the" Gaussian measure actually makes some sense.</p> http://mathoverflow.net/questions/36403/what-is-known-about-the-gaussian-measure-of-the-unit-ball-in-a-hilbert-space/36427#36427 Answer by Andreas Thom for What is known about the Gaussian measure of the unit ball in a Hilbert Space? Andreas Thom 2010-08-23T05:41:59Z 2010-08-23T06:12:19Z <p>In the book <a href="http://www.ams.org/mathscinet/search/publications.html?pg1=MR&amp;s1=MR2415834" rel="nofollow">Kazhdan’s Property (T)</a> (Appendix A7) by Bekka, de la Harpe and Valette the symmetric Fock space on a Hilbert space is $H$ studied as the analogue of a space of measurable functions on a Hilbert space $H$. This is called the <em>Gaussian construction</em> and quite important if one wants to pass from unitary representations of a group $G$ to actions of $G$ on a probability measure space. This is probably not quite what you want, but serves as a suitable replacement of the the Gaussian measure (on a finite-dimensional Hilbert space) for many purposes.</p> <p>In case $H$ is finite-dimensional, it precisely corresponds to the study of the Gaussian measure on $H$. Here, the correspondence is clear: If $G$ acts by unitary operators on $H$, then it preserves the Gaussian measure $\mu$ on $H$ and hence, there is an associated action on the probability space $(H,\mu)$.</p> http://mathoverflow.net/questions/36403/what-is-known-about-the-gaussian-measure-of-the-unit-ball-in-a-hilbert-space/36634#36634 Answer by Attar Reda for What is known about the Gaussian measure of the unit ball in a Hilbert Space? Attar Reda 2010-08-25T06:39:39Z 2010-08-25T06:39:39Z <p>I think you can see the articles entitled "concentration of measure phenomenon ". The idea is as follows: Let $(X,d,\mu)$ be a metric measure space, such as $\mu(X)=1$. Let $$\alpha(\epsilon) = \sup {\mu(X \backslash A_\epsilon) \ | \ \mu(A) = 1/2 }$$ where $$A_\epsilon = { x \ | \ d(x, A) &lt; \epsilon }$$ is the $\epsilon$-extension of a set $A$. The function $\alpha(.)$ is called the concentration rate of the space $E$. The following equivalent definition has many applications:$$\alpha(\epsilon) = \sup { \mu( { F >= M + \epsilon }) },$$ where the supremum is over all $1$-Lipschitz functions $F: X \to \mathbb{R}.$ For example the median (or Levy mean) $M = \mathop{Med}(F)$ is defined by the inequalities $$\mu ( F \geq M ) >= 1/2, \ \mu ( F &lt;= M ) \geq 1/2.$$ More precisely, the space $X$ exhibits a concentration phenomenon if $\alpha(\epsilon)$ decays very fast as $\epsilon$ grows. More formally, a family of metric measure spaces $(X_n,d_n,\mu_n)$ is called a Levy family if the corresponding concentration rates $\alpha(\epsilon)$ satisfy $$\forall \epsilon > 0 \ \ \alpha_n(\epsilon) \to 0,$$ and a normal Levy family if $$\forall \epsilon \to 0 \ \ \alpha_n(\epsilon) = O(\exp(-C n \epsilon^2))$$ for $C$ some positive constant. the last inequality is obtained bay applying the "Hoeffding inequality" and in the case of Hilbert space with concentration in small balls we do : $$\forall (x_1,x_2)\in X^2, \ \ d(x_1,x_2)=\|x_1,x_2\|&lt; r.$$</p>