نتایج جستجو برای: nonlinear negative binomial regression

تعداد نتایج: 1038042  

Journal: :Journal of the Royal Statistical Society: Series C (Applied Statistics) 2009

2016
Shuai Fu

There is a growing interest in establishing the relationship between the count data y and numerous covariates x through a generalized linear model (GLM), such as explaining the road crash counts from the geometry and environmental factors. This paper proposes a hierarchical Bayesian method to deal with the negative binomial GLM. The Negative Binomial distribution is preferred for modeling nonne...

Journal: :The Stata Journal: Promoting communications on statistics and Stata 2014

Journal: :International Journal of Scientific Research in Science, Engineering and Technology 2020

2018
Muhammad Zubair Muhammad H. Tahir Gauss M. Cordeiro Ayman Alzaatreh Edwin M. M. Ortega

*Correspondence: [email protected] 3Departamento de Estatística, Universidade Federal de Pernambuco, PE 50740-540, Recife, Brazil Full list of author information is available at the end of the article Abstract We propose and study a new compounded model to extend the half-Cauchy and power-Cauchy distributions, which offers more flexibility in modeling lifetime data. The proposed model is ...

2013
Malina Zulkifli Noriszura Ismail Ahmad Mahir Razali

This paper proposes the application of the two well known models of negative binomial regression, namely the NB-1 and the NB-2, and the functional form of negative binomial regression, namely the NB-P, for the analysis of vehicle theft crime. The advantage of using the NB-P is that it parametrically nests both the NB-1 and the NB-2, and allows statistical tests of the NB-1 and the NB-2 models a...

2012
Mingyuan Zhou Lingbo Li David Dunson Lawrence Carin

Proof: The first and second derivatives of f(z) = − ln(1 − pz) are f ′(z) = p 1−pz and f ′′(z) = p (1−pz)2 = (f ′(z))2, respectively, and f j(0) = 0 for j ≥ 1, therefore d dzm f j(z) ∣∣ z=0 = 0 for j > m and thus (2) is true for j > m. When n = j = 1, we have d n dzn f j(z) ∣∣ z=0 = f ′ k(0) = p, ∑n j′=1 F (n, j ′) j! (j−j′!) [f ′(z)]nf j−j ′ (z) ∣∣∣ z=0 = p, and F (n, j)j!pn = p, therefore, (2...

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