MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Why are Gaussian valued random variables so important in the Geometry of Banach spaces? I am reading the monograph by Pisier - "Probabilistic Methods in the Geometry of Banach Spaces" and in the very first chapter - "Dvoretzky's theorem by Gaussian Methods" there are definitions using B valued gaussian random variables X (where B is the Banach space under consideration).

Intuitively, what is the reason that would make one look toward gaussian variables - as opposed to Bernoulli rv (which I guess are also used in several definitions).

share|cite|improve this question
up vote 3 down vote accepted

Gaussian measures is more or less the only ``naturally defined'' class of measures on infinite dimensional Banach spaces. There are no translation invariant (or even quasi-invariant) measures, so that one can not define measures by their densities with respect to a canonical one (like what one does by using the Lebesgue measure in the finite dimensional case).

As for the difference between Gaussian and Bernoulli random variables - this is a confusion based on a certain ambiguity of the probabilistic language. In fact, the qualifier "Bernoulli" can only be applied to a family of random variables (and is synonymous to i.i.d. - independent identically distributed), but not to a single random variable. If one talks about measures instead of random variables, then the difference becomes obvious - a Gaussian measure is a measure on a linear space, whereas a Bernoulli measure is a measure on a product space.

share|cite|improve this answer
Isn't Bernoulli usually IID 0-1 valued (in which case, perhaps the OP means Steinhaus, i.e. IID plus-minus 1) – Yemon Choi Nov 26 '12 at 10:36
Not necessarily - the usage varies. Sometimes people go as far as to distinguish between a "Bernoulli process" and a "Bernoulli scheme" (the values being 0,1 in the first case and arbitrary in the second case). Also sometimes one calls "Bernoulli" a random variable whose distribution is concentrated on 0,1. But I am pretty sure this is not what was meant by the OP. – R W Nov 26 '12 at 14:15
Thanks. That definitely helps! – Nirman Nov 26 '12 at 16:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.