Wikipedia:Reference desk/Archives/Mathematics/2010 September 17

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September 17

Probability distribution with kurtosis parameter

Is there a probability distribution which is similar to the normal distribution but with the parameters being mean, standard deviation and kurtosis? Yaris678 (talk) 13:24, 17 September 2010 (UTC)

Pearson type VII distribution might fit the bill. --Salix (talk): 13:46, 17 September 2010 (UTC)

Spot on. Thanks! Yaris678 (talk) 18:46, 17 September 2010 (UTC)

Countability

If there exists a bijection between a set X and the natural numbers, making the set X countable, can the same bijection be used to show that any subset of X is countable? Thanks. asyndeton talk 17:35, 17 September 2010 (UTC)

I assume by countable you mean countably infinite. And my naive answer is no, if u change the domain or range of a function (e.g. by considering a subset of either), then it is no longer the same function. Zunaid 17:43, 17 September 2010 (UTC)

I think the obvious approach would be to use to the bijection to define functions satisfying the conditions of the Schroeder-Bernstein theorem. AndrewWTaylor (talk) 17:54, 17 September 2010 (UTC)

The restriction of the original function will define a bijection of the subset and a subset S of natural numbers. Compose it with an increasing enumeration of S to get a bijection with all natural numbers (or initial segment, if the subset is finite).—Emil J. 18:03, 17 September 2010 (UTC)

Let X = {x₁, x₂, x₃, ....} and suppose A is a subset of X. Then let

$a_{1}=x_{i}{\text{ where }}i=\min\{j:x_{j}\in A\}\,$

and let

$a_{n+1}=x_{i}{\text{ where }}i=\min\{j:x_{j}\in A{\text{ and }}x_{j}\not \in \{x_{1},\dots ,x_{n}\}\}.\,$

Keep going as long as A has not been exhausted. Then

$A=\{a_{1},a_{2},a_{3},\ldots \}.\,$

Therefore A is either finite or countably infinite. You don't need the Cantor–Bernstein theorem or the like. Michael Hardy (talk) 21:05, 17 September 2010 (UTC)

It's not even that complicated. The usual definition of a countable set X is that there exists a bijection between X and some subset of the natural numbers. So yes, Asyndeton, the same bijection can be used directly to show that any subset A of X is countable—just restrict the domain of the bijection to A and the codomain to the image of A. —Bkell (talk) 02:58, 18 September 2010 (UTC)

But if you define "countable" in that way, then you're still faced with the problem of proving that all countably infinite sets have the same cardinality. My way of doing it takes care of that. Michael Hardy (talk) 22:10, 18 September 2010 (UTC)

That's a separate problem, isn't it? We don't need to prove that in order to establish the claim in the original post, if we assume the usual definition of countable. —Bkell (talk) 22:10, 19 September 2010 (UTC)

Thanks for all your answers, especially yours Bkell; unfortunately I'm at a level that demands answers less hi-tech than Schroeder-Bernstein and the like. Restricting the domain and codomain seemed 'obvious' but being 'obvious' does not imply being correct and I wouldn't have thought to specifically mention restriction in my answer. Thanks again. asyndeton talk 12:04, 18 September 2010 (UTC)

A proof that in euclidean space pi is a constant.

I understand of course that in euclidean space (geometry) pi, the ratio between the circumference of a circle and its diameter is a constant. Further, I have been told that in some noneuclidean geometries that the value of pi varies with the size of the circle. I am looking for the proof, which I'm sure exists, that in Euclidean geometry (space) pi is a constant. Your help will be most appreciated. 71.112.10.31 (talk) 18:02, 17 September 2010 (UTC)

If I was going to construct my own proof, I would start with the fact that Euclidean spaces are vector spaces. Yaris678 (talk) 18:52, 17 September 2010 (UTC)

First try to show that for any two circles in the Euclidean plane, there's a mapping from the space to itself that multiplies all distances by the same constant that takes one of them to the other. (Actually, saying "same" and "constant" is redundant.) Michael Hardy (talk) 21:08, 17 September 2010 (UTC)

No, saying "same" and "constant" is not redundant; different distances could (in principle) be multiplied by different constants. They aren't, of course, which is the point. --COVIZAPIBETEFOKY (talk) 22:07, 17 September 2010 (UTC)

I think this is true for any normed space, including euclidean space. Define pi as the limit of an infinite sequence, using the Euclidean norm (or any norm - other norms will give a different constant) and assuming you have a circle of radius r. You will find the rs cancel. Yaris678 (talk) 06:16, 18 September 2010 (UTC)

Request for statistics article review

A new editor has written an article Multivariate kernel density estimation and requested feedback. The editor seems to have basic Wikipedia style under control; it needs the review of someone familiar with the field. (I'll also crosspost a couple specific editors.)--SPhilbrickT 21:12, 17 September 2010 (UTC)

Looks like a very nice article. The main improvement I can think of is that the exposition could use a few more wikilinks. 67.119.14.196 (talk) 05:42, 18 September 2010 (UTC)

Before I get to the main body of the article, I have an issue with the first sentence "is one of the most popular techniques for density estimation". I can only think of three techniques: histograms, parametric methods and kernal methods. Of these, I would suggest that kernal methods are the least popular, partly perhaps because they are the least well-known. I would probably say simply say "is a technique for density estimation". Yaris678 (talk) 06:30, 18 September 2010 (UTC) I have copied these posts to Wikipedia:Requests_for_feedback/2010_September_17#Multivariate_kernel_density_estimation. I suggest any further comments should go there. Yaris678 (talk) 06:52, 18 September 2010 (UTC)