Testwiki:Reference desk/Archives/Mathematics/2009 September 6

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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 $p=\mathrm{P}\mathrm{r}(Y=1|X)$ (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 $𝔼((p-\hat{p}{)}^2).$ Is there a standard way of doing this?

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

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)

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

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

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)

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)

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)