Mathematics desk
< September 5	<< Aug \| September \| Oct >>	September 7 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

September 6

Evaluating a probability estimator

Hi. I have random variables X and Y, where X is nominal and Y takes values of 0 and 1. I have n observations of the form $(X_{i},Y_{i})\,\!$ , where several observations may share the same X. I denote $p=\mathrm {Pr} (Y=1|X)\,\!$ (so p is a random variable which becomes determined when given X), and ${\hat {p}}=f(X)$ is an estimator for p. I want to know how good this estimator is - specifically, I want to approximately know $\mathbb {E} ((p-{\hat {p}})^{2}).$ Is there a standard way of doing this?

I think I have reduced the problem to that of estimating $\mathbb {E} (p^{2})$ , in case this is any simpler. Thanks. -- Meni Rosenfeld (talk) 16:42, 6 September 2009 (UTC)[reply]

Actually I had an idea for this, but originally I thought it would be computationally intractable. After looking into it I see it's not that bad. The idea is to assume that p follows the beta distribution, find the MLE of the distribution parameters given the data, and calculate the mean and variance for those parameters. Does this sound reasonable? Is there a better way? -- Meni Rosenfeld (talk) 18:06, 6 September 2009 (UTC)[reply]

You might be interested in the following article: http://psychology.wikia.com/wiki/Inferential_statistics . Bo Jacoby (talk) 20:08, 6 September 2009 (UTC).[reply]

Ok, thanks. -- Meni Rosenfeld (talk) 18:39, 8 September 2009 (UTC)[reply]

Automorphisms of Z[x]

What are all the ring automorphisms of $\mathbb {Z} [x]$ ? I believe Z has only one: the identity. What is the general procedure of approaching the problem of determining Aut(R)? Thanks.--Shahab (talk) 22:01, 6 September 2009 (UTC)[reply]

Yes, the (unital) ring automorphism is determined by where you send x. x needs to be in the image, so x has to go to something of degree 1. I suspect it can go to ±x+n for any n in Z, and to no other place. If so, this is the infinite dihedral group, represented as the matrices { [ ±1, n ; 0, 1 ] : n in Z } under matrix multiplication. In general finding the automorphism group of fundamental, innocuous rings can be the source of long-standing open problems, such as (the strong) Nagata conjecture and the description of the Cremona group. I wrote a short thing on this a while back (the enwiki article is just a sentence). JackSchmidt (talk) 22:47, 6 September 2009 (UTC)[reply]

Well, it's clear the leading coefficient of the image of x must be ±1. If it is ax + n, then a poly in the image of the automorphism can be written as aQ(x) + constant, so it can't be surjective unless a is a unit.John Z (talk) 00:30, 7 September 2009 (UTC)[reply]