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From |
Roger Newson <roger.newson@kcl.ac.uk> |

To |
statalist@hsphsun2.harvard.edu |

Subject |
Re: st: parametric vs. nonparametric estimators |

Date |
Wed, 16 Jun 2004 19:23:56 +0100 |

At 18:52 16/06/2004, Thomas wrote:

Probably not a great deal. A difference of half a percent in fitted value could easily be due to imperfect numerical precision during an iterative fit, which your "parametric" estimate uses and your "nonparametric" model doesn't.Dear Nick and Rich, maybe I miss something, but my problem is as follows: suppose I have a data set with 100 objects and two binary variables, X (sex=male (coded 1) or female (coded 0)) and Y (disease=absent (coded 0) or present (coded 1)) for example. My goal is to estimate the probability P(Y=1|X=1). Suppose 50 of the 100 persons are male and of this 10 have a disease, then my so called "nonparametric" estimate of P(Y=1|X=1) is 10/50=0.200. By nonparametric I mean, that no assumption about the distribution of Y is made. By using logistic regression, I assume that Y can be related to a latent variable Y* which has a logistic distribution. Now, for example, the logistic regression estimate of P(Y=1|X=1) is 0.201. What does the difference between 0.200 and 0.201 tell me?

Roger

--

Roger Newson

Lecturer in Medical Statistics

Department of Public Health Sciences

King's College London

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Tel: 020 7848 6648 International +44 20 7848 6648

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Email: roger.newson@kcl.ac.uk

Website: http://www.kcl-phs.org.uk/rogernewson

Opinions expressed are those of the author, not the institution.

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**References**:**RE: st: parametric vs. nonparametric estimators***From:*"Nick Cox" <n.j.cox@durham.ac.uk>

**AW: st: parametric vs. nonparametric estimators***From:*Thomas Mählmann <maehlmann@wiso.uni-koeln.de>

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