R: Equivalence between lm and glm
On May 31, 2013, at 17:10 , Stefano Sofia wrote:
I find difficult to understand why in lm(log(Y) ~ X) Y is assumed lognormal. I know that if Y ~ N then Z=exp(Y) ~ LN, and that if Y ~ LN then Z=log(Y) ~ N. In lm(log(Y) ~ X) I assume Y ~ N(mu, sigma^2), and then exp(Y) would be distributed by a LN, not l og(Y). Where is my mistake?
It is log(Y) that is assumed N(mu, sigma^2), and exp(log(Y)) is LN.
Moreover, in glm(Y ~ X, family=gaussian(link=log)) the regression is log(mu) = beta0 + beta1*X. In lm(log(Y) ~ X) the regression is exp(mu+(1/2)*sigma^2) = beta0 + beta1*X. Correct?
Probably not. (What is mu? If it is E(log(Y)), then it should just be just mu=beta0+beta1*X)
Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd.mes at cbs.dk Priv: PDalgd at gmail.com