# Wikipedia:Reference desk/Archives/Mathematics/2009 April 17

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# April 17

## Is there an infinite number of different types of probability distribution?

Probability distributions seem to be discovered by individuals and often named after them, such as the Gaussian. See List of probability distributions. But the essence of the probability distribution is described by a mathematical formula. Are there therefore as many different probability distributions as there are maths formulas? I assume there is an infinate number of maths formulas. In other words, are the named probability distributions just a tiny subset of all possible probability distributions? 78.146.249.32 (talk) 11:07, 17 April 2009 (UTC)

Yes. Often the named probability distributions are the most important, although this does not imply that the other probability distributions are not. Generally, statisticians name distributions if they have significant applications in or outside of mathematics. The Gaussian distribution is important because it is natural (even a set of test scores tends to follow the Gaussian distribution). However, they are many other probability distributions that are named (as clicking the link you provided shows). Essentially, a probability distribution is much more that a formulae. For a start, it is formally a function with (generally) arbitrary domain (which is in the realm of measure theory) and range [0,1], satisfying certain properties. In basic mathematics, probability distributions are usually defined on a collection of subsets of the set of all real numbers. Note however, that in more advanced mathematics probability distributions are assumed to be defined on any arbitrary sigma algebra (clicking on the link is not necessary). The point to note is that probability distributions are not just formulae (because for a start they are defined on a collection of sets). In fact, they are actually measures. Hope this answers your question. I feel obliged, however, to notify you that looking up probability distribution on the internet, especially Wikipedia, may not give what you expect. In essence, probability theory is an advanced field of mathematics, much more than a formulae. --PST 12:38, 17 April 2009 (UTC)

## Functional form of the von-Neumann Morgenstern utility function - any help much appreciated!!!

I appreciate that this question is more micro economic related than pure maths but there is no social science section on the reference desk.

For my studies i need to use the von-Neumann Morgenstern utility function but i am not aware of its function form (i.e. a cobb douglas is Y=K^aL^b for example). I am not trying to get the answer to save me the work of doing so i am just trying to do background research.

Thank you very much for your time —Preceding unsigned comment added by 86.163.232.191 (talk) 16:33, 17 April 2009 (UTC)

For any Bernoulli utility function ${\displaystyle u(\cdot )}$, and any possible (prizes in correlation with) a random variable ${\displaystyle x\in \mathbb {R} }$, then for any lottery prospect with accumulated distribution ${\displaystyle F(\cdot )}$, the VN-M utility function is
${\displaystyle U\left(F\right)=\int u\left(x\right)dF\left(x\right)}$
...(for generality the Riemann-Stieltjes integral is used in this formulation).
In simpler terms, the Von-Neumann Morgenstern form has the following features:
• It is a utility over lotteries of prizes.
• Given the utility of each possible prize (e. g. by means of a Bernoulli function u), the VN-M utility U is built weighting each prize utility probabilistically, and adding up over the set of possible prizes.
For instance, in discrete terms, characterize the prizes ${\displaystyle x_{1},x_{2}\ldots x_{n}}$ and probabilities ${\displaystyle p_{1},p_{2}\ldots p_{n}}$ with ${\displaystyle \sum _{i}^{n}p_{i}=1}$. Then, by means of a Bernoulli function u(), you get the prizes' utility as ${\displaystyle u\left(x_{1}\right)}$, ${\displaystyle u\left(x_{2}\right)}$, etc. Then the VN-M corresponding utility function is
${\displaystyle U=p_{1}\cdot u\left(x_{1}\right)+p_{2}\cdot u\left(x_{2}\right)+\ldots p_{n}\cdot u\left(x_{n}\right)=\sum _{i}^{n}p_{i}\cdot u\left(x_{i}\right)}$. Pallida  Mors 18:23, 19 April 2009 (UTC)

## Class number 1

Hello, I need to find a list of quadratic imaginary fields with class number 1. And, I have found two lists that are slightly different. Is it just me not understanding something or is it that something is wrong in one of the sources?

The Wiki [Class_number_problem] says

The complete list of imaginary quadratic fields with class number one is ${\displaystyle \mathbf {Q} ({\sqrt {d}})}$ with d one of ${\displaystyle -3,-4,-7,-8,-11,-19,-43,-67,-163}$.

The Class Number page on MathWorld says this same thing. So, I assume this is the right answer. But, Introductory Algebraic Number Theory by Alaca and Williams has a list of class numbers for ${\displaystyle \mathbb {Q} ({\sqrt {k}})}$ for ${\displaystyle -195\leq k<0}$, k square free and it lists, in addition, -1 and -2. The rest agree (except that this does not list square-free k). Is this a definition thing or what?

Thanks StatisticsMan (talk) 19:08, 17 April 2009 (UTC)

${\displaystyle \mathbb {Q} ({\sqrt {-1}})=\mathbb {Q} ({\sqrt {-4}})}$ and ${\displaystyle \mathbb {Q} ({\sqrt {-2}})=\mathbb {Q} ({\sqrt {-8}})}$, so our article was correct if rather silly. The problem was that our article (like Mathworld) was listing discriminants of imaginary quadratic fields, and someone copied the same list into a different context. For k squarefree, the discriminant of ${\displaystyle \mathbb {Q} ({\sqrt {k}})}$ is either k or 4k, according as k is or is not congruent to 1 mod 4.Algebraist 19:26, 17 April 2009 (UTC)
(ec) They're really the same lists. ${\displaystyle \mathbb {Q} ({\sqrt {-1}})}$ is the same field as ${\displaystyle \mathbb {Q} ({\sqrt {-4}})}$, and similarly for -2 and -8. It is customary to list quadratic fields according to their discriminants, hence -4 and -8 as opposed to the squarefree -1 and -2. Bikasuishin (talk) 19:28, 17 April 2009 (UTC)