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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?

This seems to me like a fundamental question but I cannot seem to find anything. Any information/references would be most appreciated.

EDIT: A related question which is of interest to me: Do there exist asymptotically tight bounds to $\int_{||u||> K}||u||^2 \mu(du)$?

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I note that the OP only says that $\mu$ is supposed to be a Gaussian measure on $X$, which is a perfectly well-defined notion; but the answers so fare are written as if the OP asked about the Gaussian measure, which is not. Of course, the answer to this will depend on the mean and covariance of $\mu$. – Mark Meckes Aug 22 '10 at 23:48
Look at the title of this question. I address both "the" Gaussian measure and "a" Gaussian measure in my answer. – Peter Shor Aug 23 '10 at 0:00
Good point, I didn't pay close attention to the title. – Mark Meckes Aug 23 '10 at 0:04
up vote 9 down vote accepted

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.

Consider the measure on $\mathbb{R}^\infty$ where the $j$th coordinate is a Gaussian with mean 0 and variance $\sigma_j^2$, where $\sum_{j=1}^{\infty} \sigma_j^2 < \infty$ (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.

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.

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Ok, your definition of "a" Gaussian measure matches mine. Sorry if I'm asking a dumb question, but what does a "non-uniform" Gaussian mean exactly? Thanks for the pointer on quantum optics, I am very interested in learning of applications of probability on infinite dimensional spaces. – RadonNikodym Aug 23 '10 at 0:32
As regards the generalized chi-square distribution idea. This is exactly what I chasing, however even on finite dimensional spaces the generalized chi-squared distribution has a quite complicated pdf, and things become hairy quite fast. – RadonNikodym Aug 23 '10 at 0:40
By "non-uniform" I meant that the variance isn't the same in all dimensions. I don't think this is standard terminology. I also think you're stuck with the generalized $\chi$-squared distribution, unfortunately. – Peter Shor Aug 23 '10 at 0:46
Aha! This is much clearer! Yes, of course, such a "uniform" Gaussian could not exist. In fact, you would need the variances in all dimensions to have a finite sum, which is equivalent to saying that the Covariance operator of the Gaussian measure is trace class. So, given a positive symmetric trace class operator, there is a Gaussian measure that has that covariance operator, however I think what you lose is unitary invariance, e.x if I permute the basis elements the corresponding Gaussian will be different. Thank you very much! – RadonNikodym Aug 23 '10 at 1:03
Sazonov's theorem</a> is a somewhat fancier way of saying what Peter Shor wrote in his answer; see – Richard Borcherds Aug 23 '10 at 1:13

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<1, then that of a ball in n dimensional is less than cn, 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 and the unit ball of Hilbert space is a subset of this, but has measure 0 by the above argument.

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Ok, there is something I am missing. Many books define a Gaussian measure on a Hilbert space $H$ to be a measure such that the push forward measure of every linear functional on $H$ is Gaussian on $\mathbb{R}$. One can then define the mean and covariance of the measure from the mean and variance of these real-valued distributions. So this isn't a Gaussian measure in some sense? – RadonNikodym Aug 22 '10 at 22:54
And yet, your reasoning is perfect... :( Are there non equivalent definitions of Gaussian measures on Hilbert Spaces. – RadonNikodym Aug 22 '10 at 23:00
See my answer for the definition of Gaussian measures on Hilbert spaces: one equivalent definition is that any projection onto a finite-dimensional space must be a Gaussian (with some covariance matrix $\Sigma$). – Peter Shor Aug 22 '10 at 23:28
Just to confirm: my answer is for the case of THE Gaussian measure, while Peter Shor's answer is for the case of A Gaussian measure. – Richard Borcherds Aug 23 '10 at 1:16

In the book Kazhdan’s Property (T) (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 Gaussian construction 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.

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 $(H,\mu)$.

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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 $$A_\epsilon = { x \ | \ d(x, A) < \epsilon }$$ 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 $$\mu ( F \geq M ) >= 1/2, \ \mu ( F <= M ) \geq 1/2.$$ 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 : $$\forall (x_1,x_2)\in X^2, \ \ d(x_1,x_2)=\|x_1,x_2\|< r.$$

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For more detail, you can see the book of Michel Ledoux - "The Concentration of Measure Phenomenon. – Attar Reda Aug 25 '10 at 6:47
Or, the site :… – Attar Reda Aug 25 '10 at 6:50

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