estimate.regression: Estimate Regression Model Parameters

In LiYao-sfu/EDFtest: Goodness of Fit Based on Empirical Distribution Function

Description

estimate.normal.regression fits a standard linear model by OLS (ordinary least square); estimate.laplace.regression fits a standard linear model by LAD (least absolute deviations); the scale parameter is estimated by maximum likelihood. estimate.logistic.regression fits a standard linear model by maximum likelihood when the errors are modeled as having a logistic distribution. estimate.extremevalue.regression fits a standard linear model by maximum likelihood when the errors are modeled as having an extreme value distribution with standard cdf given by F(x) = exp(-exp(x)) . estimate.weibull.regression fits a Weibull model by maximum likelihood when the log of scale parameter is linearly predicted. The procedure is equivalent to taking logs and doing extreme value regression as above. estimate.gamma.regression estimate the parameters in a Gamma regression model which fits link(E(y))= x beta. "link" must be one of "log","identity", "inverse". A Gamma glm gets initial estimates which are improved by Maximum Likelihood. estimate.exp.regression estimate the parameters in an exponential regression model which fits link(E(y))= x beta. "link" must be one of "log","identity", "inverse". A Gamma glm is used for fitting.

Usage

estimate.normal.regression(y, x, fit, fit.intercept = TRUE)

estimate.gamma.regression(y, x, fit, fit.intercept = TRUE, link = "log")

estimate.exp.regression(y, x, fit, fit.intercept = TRUE, link = "log")

estimate.logistic.regression(y, x, fit, fit.intercept = TRUE, detail = FALSE)

estimate.laplace.regression(y, x, fit.intercept = TRUE)

estimate.weibull.regression(y, x, fit.intercept = TRUE)

estimate.extremevalue.regression(y, x, fit.intercept = TRUE)

Arguments

y	A univariate response.
x	A design matrix which is not expected to have a column of 1s.
fit.intercept	Logical; if TRUE, an intercept term is added.

Value

Estimated dispersion or sd or scale or inverse dispersion and coefficients of a linear regression.

Examples

n = 500
p = 3
beta = c(1,2,3)
x = rnorm(n*(p-1))
x = c(rep(1,n),x)
x = matrix(x,n,p)

# OLS regression
mean = x%*%(beta)
#apply the link to get the mean
#generate data with that mean,

sd = 2
y =rnorm(n,mean=mean,sd=sd)
estimate.normal.regression(y=y,x=x)

# LAD regression
library(L1pack)
y =L1pack::rlaplace(n=n, location=mean, scale=sd)
estimate.laplace.regression(y=y,x=x,FALSE)

# Gamma regression
alpha = 3
scale=exp(x %*% beta) / alpha
y =rgamma(n=n,shape=alpha,scale=scale)
estimate.gamma.regression(y=y,x=x,intercept=FALSE)

# Exponential regression
y =rexp(n=n,rate=1/exp(mean))
estimate.exp.regression(y=y,x=x)

# Weibull regression
y = rweibull(n=n,shape=1,scale=exp(mean))
estimate.weibull.regression(y=y,x=x)

# Extreme Value regression
y = log(rweibull(n=n,shape=1,scale=exp(mean)))
estimate.extremevalue.regression(y=y,x=x)

# Logistic regression
y = rlogis(n=n,location=mean,scale=2)
estimate.logistic.regression(y=y,x=x)

LiYao-sfu/EDFtest documentation built on Dec. 18, 2021, 4:35 a.m.