Comments on: Probability paradox

By: S.C. Kavassalis

S.C. Kavassalis — Fri, 30 Oct 2009 04:22:08 +0000

My comment is well behind the making of this post, but all I have to say is that this is an excellent discussion.

By: Gogi Pantsulaia

Gogi Pantsulaia — Sun, 17 May 2009 05:24:15 +0000

Here is a certain construction of a Gaussian probability measure $\gamma$ on infinite-dimensional Hilbert space $l^2$:
Let $\mu$ be a standard Gaussian measure of $R^{\infty}$.
Let $(A_k)_{k \in N}$ be such a sequence of positive numbers that
$0<\mu(\prod_{k \in N}[-A_k,+A_k])<1$.
We define a mapping $S:R^{\infty}\to R^{\infty}$ as follows
$S((x_k)_{k \in N})=(\frac{x_k}}{2^kA_k})_{k \in N}$.
Then $S(\prod_{k \in N}[-A_k,+A_k])\subset l_2$.
We set $\gamma(X)=\mu(S^{-1}(X))$ for every $X \in B(l_2)$, where $B(l_2)$ denotes a Borel $\sigma$-algebra of subsets of $l_2$.

Remark. Note that in your “Probability paradox” you try to construct a Gaussian measure $\gamma$ on $l_2$ by the identity mapping $f:R^{\infty} \to R^{\infty}$ (i.e., $f((x_k)_{k \in N}= (x_k)_{k \in N}$ for $(x_k)_{k \in N} \in R^{\infty}$) as follows:

$\gamma(X)=\mu(f^{-1}(X))$ for every $X \in B(l_2)$.

But your measure is identically zero on $l_2$ because
$\gamma(l_2)=\mu(f^{-1}(l_2))=\mu(l_2)=0$.

By: Gogi Pantsulaia

Gogi Pantsulaia — Sat, 16 May 2009 20:45:02 +0000

I want give a little comment on Mark Meckes’s
question:
“Also, do you find the nonexistence of such a measure more surprising than the nonexistence of “Lebesgue measure” on an infinite Banach space?”

In this context, I want inform you that there exist some translation-invariant Borel measures (which are not $\sigma$-finite) on the separable Banach space U with basis such that they get finite non-zero values on some compact subsets
and infinite values on every non-degenarate balls.
If $\mu$ is such measure then $\mu(X)=0$ implies that $X$ is “shy”. They are called as “generators of shy sets” on $U$.

It interesting also that in some non-separable Banach spaces(for example, on l^{\infty}-the vector space of all bounded real-valued sequences) there exists a translation-invariant (non-$\sigma$-finite) measures which are defined on the minimal $\sigma$-algebra generated by all balls and which get a numerical value 1 on the unite closed ball.

In Solovay model(ZF+DC+”every subset of $R$ is lebesgue measurable”) we have more strict result:
There exists a Borel measure on l^{\infty} with above mentioned properties.

By: Karthik Kalyanaraman

Karthik Kalyanaraman — Fri, 01 May 2009 07:08:02 +0000

To answer the question about real separable Banach spaces: the answer is yes there is notion of a Gaussian measure. The idea is to 1) exploit the fact that a real sep Banach space can be shown to be an “abstract Wiener space” i.e. contains a Hilbert space H dense in it and “measurable” wrrt to the Banach norm. (Notion of measurable norm needs to be made precuse) and 2) to then use a suitable generalization of the construction of the Wiener measure over C[0,1]: i.e. defining a Gaussian measure as a measure that is Gaussian in the normal sense when looking at cylinder sense over finite dimesional orthogonal projections of H … one then shows this has a unique sigma-additive extension..

A good reference on all this is Kuo’s Gaussian Measures in Banach Spaces (!) in the Lecture Notes in mathematics series by Springer

By: Noah Bray-Ali

Noah Bray-Ali — Wed, 10 Sep 2008 18:32:13 +0000

Quantum field theory is certainly hard, but I sincerely doubt that the considerations noted in the blog entry and in Prof. Tao’s reply are to the point. Consider the one-dimensional harmonic oscillator, perhaps the simplest quantum mechanical system there is. Despite it’s simplicity, its Hilbert space is infinite-dimensional, so all the considerations in the blog entry attach. Whatever the root of the problem with quantum field theory, it cannot be the non-existence of a Gaussian measure, for the very good reason that the measure does not exist for much simpler systems, such as the one-dimensional harmonic oscillator.

Moreover, the result, sketched in the blog entry, that the Gaussian measure of the whole infinite-dimensional Hilbert space vanishes, has a natural analog for the harmonic oscillator. Paths with finite action have zero measure. Sydney Coleman gives a nice argument for this result in an appendix to his lecture “The Uses of Instantons” (reprinted in Aspects of Symmetry by Cambridge University Press).

For what its worth, my own perspective on what makes quantum field theory hard is motivated by the work of Oded Schramm and collaborators that describe and define the self-similiar curves arising in quantum field theories of two-dimensional percolation and other critical phenomena (Please see Prof. Tao’s blog entry for Sept 3 for a description of this work and a note regarding the recent tragedy concerning Oded). There is good numerical evidence to suggest that self-similarity arises in higher-dimensions as well. Thus, making sense of quantum field theory in general probably entails describing and defining random, self-similar surfaces in two and higher dimensions, a topic of independent interest, and by no means trivial.

By: health

health — Fri, 15 Aug 2008 14:42:17 +0000

This is somthing I have been looking for a long time. Thanks!!!

By: Porridge

Porridge — Thu, 19 Jun 2008 09:08:49 +0000

Somehow i missed the point. Probably lost in translation Anyway … nice blog to visit.

cheers, Porridge.

By: B

Sun, 03 Feb 2008 00:53:06 +0000

hrm; I don’t know how to include TeX. The quantity I wanted to type is

\lim_{n \arrow \infty} \int_{R^n} f(

By: B

Sun, 03 Feb 2008 00:50:48 +0000

Do you know of anything that ‘looks like’ a Gaussian measure on Banach spaces in general (or especially seperable Banach spaces)? I ask because it is useful to have the following quantity defined on seperable Hilbert spaces: Let x_1, x_2, … be a complete orthnormal sequence, and then let, for a measurable function f and vector y:

\[\lim_{n \arrow \infty} \int_{R^n} f(\) e^{-\pi(a_1 ^2 + \cdots a_n ^2)} da_1 da_2 \cdots da_n\].

The resulting quantity won’t depend on the orthonormalization chosen, and can be used to prove all sorts of interesting results (uniform boundedness in Hilbert spaces, for instance). The question (maybe too vague of one) is whether you can replicate this idea on Banach spaces.

I have to apologize in advance if there is a standard construction yielding a ‘pseudo-measure’ like this; I know almost nothing about the subject.

By: mcguigan

mcguigan — Thu, 10 Jan 2008 17:32:53 +0000

If one replaces space by a finite lattice to define a measure there is difficulty with fermions. The continuum limit introduces extra modes which are unphysical. This is called fermion doubling. Elaborate methods are introduced in lattice field theories to address this.