Re: [R] question about the difference of AIC()

2021-05-12 Thread Søren Højsgaard
In the first model, I believe you estimate two parameters: the mean and the variance: > fm <- lm(y ~ 1) > 2*2 - 2 * logLik(fm) 'log Lik.' 40.49275 (df=2) > AIC(fm) [1] 40.49275 A zero mean model: fm0 <- lm(y ~ -1) > 2*1 - 2 * logLik(fm0) 'log Lik.' 39.00611 (df=1) > AIC(fm0) [1] 39.00611 Regard

Re: [R] question about the difference of AIC()

2021-05-12 Thread Jinsong Zhao
On 2021/5/12 19:49, Jinsong Zhao wrote: Hi there, I learned that AIC = 2 * npar - 2 * log(logLik(model)), where k is the number of estimated parameters in the model. k should be npar in the above sentence. Sorry for the mistake. For examle: > set.seed(123) > y <- rnorm(15) > fm <- lm(y

[R] question about the difference of AIC()

2021-05-12 Thread Jinsong Zhao
Hi there, I learned that AIC = 2 * npar - 2 * log(logLik(model)), where k is the number of estimated parameters in the model. For examle: > set.seed(123) > y <- rnorm(15) > fm <- lm(y ~ 1) In this example, npar should be 1, so, AIC is: > 2*1 - 2 * logLik(fm) 'log Lik.' 38.49275 (df=2) However