Wikipedia:Reference desk/Archives/Mathematics/2012 March 20
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March 20
editName for a vector space with a zero quadratic form
editWhat does one call a vector space with a quadratic form that is identically zero? For example, a Grassmann algebra is an algebra over such a vector space. It is evidently not called a zero vector space, as this appears to be a vector space of zero dimensions. It is however technically an example of a zero algebra. Perhaps a null vector space? — Quondum☏✎ 19:17, 20 March 2012 (UTC)
- "Null vector space" seems like it would be easily confused with "null space". If I were to suggest a name, it would probably be "trivial quadratic form", but I can't cite any references for that. Rschwieb (talk) 00:27, 21 March 2012 (UTC)
- ... although mathematics seems to abound with such potential confusions (but I agree this one is severe). If one were seeking a term for the quadratic form, there are perhaps several that would do, including "zero quadratic form". I see from Quadratic form#Further definitions that the term "totally singular [vector] space" and from Isotropic quadratic form that "totally isotropic [vector] space" might apply. I was just trying to work out how to uniquely decompose a vector space into such a space and one in which the signature has no zeros, and realized I was having difficulty putting it into words. — Quondum☏✎ 05:49, 21 March 2012 (UTC)
- Ah sorry, I misunderstood the task. Did you succeed in your decomposition problem? Rschwieb (talk) 13:06, 21 March 2012 (UTC)
- Not too surprising, as I had not stated the task. The reference I found and you might like (the paper Construction of spinors via Witt decomposition and primitive idempotents, Rafał Abłamowicz 1995 in Clifford algebras and spinor structures: a special volume dedicated to the memory of Albert Crumeyrolle, 1919-1992, Rafał Abłamowicz, Pertti Lounesto) seems to tackle the problem head-on and calls it Witt decomposition (and it is a decomposition into an orthogonal direct sum of three rather than two spaces, but I must still wrap my head around that). Thanks very much for the reference [1] that you provided, which I've been reading all day. A very clear and comprehensive treatment of bilinear forms it is. — Quondum☏✎ 17:21, 21 March 2012 (UTC)
- I saw you write the Ablamowcz reference earlier, and thought it sounded like a good one. Witt decomposition is something I don't know about and would be glad to peruse. Rschwieb (talk) 18:26, 21 March 2012 (UTC)
- Not too surprising, as I had not stated the task. The reference I found and you might like (the paper Construction of spinors via Witt decomposition and primitive idempotents, Rafał Abłamowicz 1995 in Clifford algebras and spinor structures: a special volume dedicated to the memory of Albert Crumeyrolle, 1919-1992, Rafał Abłamowicz, Pertti Lounesto) seems to tackle the problem head-on and calls it Witt decomposition (and it is a decomposition into an orthogonal direct sum of three rather than two spaces, but I must still wrap my head around that). Thanks very much for the reference [1] that you provided, which I've been reading all day. A very clear and comprehensive treatment of bilinear forms it is. — Quondum☏✎ 17:21, 21 March 2012 (UTC)
- Ah sorry, I misunderstood the task. Did you succeed in your decomposition problem? Rschwieb (talk) 13:06, 21 March 2012 (UTC)
- ... although mathematics seems to abound with such potential confusions (but I agree this one is severe). If one were seeking a term for the quadratic form, there are perhaps several that would do, including "zero quadratic form". I see from Quadratic form#Further definitions that the term "totally singular [vector] space" and from Isotropic quadratic form that "totally isotropic [vector] space" might apply. I was just trying to work out how to uniquely decompose a vector space into such a space and one in which the signature has no zeros, and realized I was having difficulty putting it into words. — Quondum☏✎ 05:49, 21 March 2012 (UTC)
Modelling with Beta-Distributions
editHello Wikipedians,
I am working on a statistical modelling problem. I need to model a process which generates probabilities. My first idea was to use a general beta distribution. This has several advantages:
- The distribution is controlled by just two parameters which are easy to understand.
- I can vary mean and variance of the distribution.
- The beta distribution leads to formulae which are computationally acceptable.
As little is known about the real-life process which corresponds to that distribution, I am worried about introducing implicit assumptions by chosing the beta distribution.
Therefore I have the following questions for you:
- Are there classes of probability distributions on [0,1] which cannot be approximated by beta distributions? What do they look like?
- Are there other distributions with similar properties which might be worth to investigate?
Thanks a lot -- 109.193.1.228 (talk) 21:30, 20 March 2012 (UTC)
- In the list of probability distributions it says Kumaraswamy distribution is as versatile but simpler to use. Don't know if that helps. There's a generalisation of generalized beta distributions with example in this paper: www.federalreserve.gov/pubs/feds/1998/199818/199818pap.pdf And maybe the journal of simulation can help, here's an article about beta, johnson and Bézier distribution families: www.palgrave-journals.com/jos/journal/v4/n2/full/jos200931a.html But I don't know anything about the subject, so not sure if any of this is at all relevant..(no hyperlinks since wikipedia gives me hard to read captcha's when I try to post them.) 84.197.178.75 (talk) 22:50, 20 March 2012 (UTC)
If you really don't know anything at all about your distribution except that it is confined to the interval [0,1], I don't think there is any better choice than a spline function. Looie496 (talk) 00:02, 21 March 2012 (UTC)
- The beta distribution is conjugate to the binomial distribution, which means that it is very very useful for analyzing Bernoulli trials. The beta is very good at capturing priors. Only very contrived PDFs are badly described by the beta, in my experience. Robinh (talk) 02:16, 21 March 2012 (UTC)
- Binomial distributions are limiting cases for big populations of hypergeometric distributions
- Here k is the number of successes in a sample of n items taken from a population of N items including K successes. The fraction P=K/N is a probability parameter. The conditional distribution of K, knowing k, n and N, is
- These distributions have beta distributions as limiting cases for big populations. Bo Jacoby (talk) 11:38, 21 March 2012 (UTC).
- Binomial distributions are limiting cases for big populations of hypergeometric distributions