# Recalculate log-likelihood from a simple R lm model

I'm simply trying to recalculate with dnorm() the log-likelihood provided by the logLik function from a lm model (in R).

It works (almost perfectly) for high number of data (eg n=1000) :

> n <- 1000
> x <- 1:n
> set.seed(1)
> y <- 10 + 2*x + rnorm(n, 0, 2)
> mod <- glm(y ~ x, family = gaussian)
> logLik(mod)
'log Lik.' -2145.562 (df=3)
> sigma <- sqrt(summary(mod)$dispersion) > sum(log(dnorm(x = y, mean = predict(mod), sd = sigma)))  -2145.563 > sum(log(dnorm(x = resid(mod), mean = 0, sd = sigma)))  -2145.563  but for small datasets there are clear differences : > n <- 5 > x <- 1:n > set.seed(1) > y <- 10 + 2*x + rnorm(n, 0, 2) > > mod <- glm(y ~ x, family = gaussian) > logLik(mod) 'log Lik.' -8.915768 (df=3) > sigma <- sqrt(summary(mod)$dispersion)
> sum(log(dnorm(x = y, mean = predict(mod), sd = sigma)))
 -9.192832
> sum(log(dnorm(x = resid(mod), mean = 0, sd = sigma)))
 -9.192832


Because of small dataset effect I thought it could be due to the differences in residual variance estimates between lm and glm but using lm provides the same result as glm :

> modlm <- lm(y ~ x)
> logLik(modlm)
'log Lik.' -8.915768 (df=3)
>
> sigma <- summary(modlm)$sigma > sum(log(dnorm(x = y, mean = predict(modlm), sd = sigma)))  -9.192832 > sum(log(dnorm(x = resid(modlm), mean = 0, sd = sigma)))  -9.192832  Where am I wrong ? • With lm(), you are using$\sqrt{\hat\sigma}$instead of$\hat\sigma$. – Stéphane Laurent Oct 18 '13 at 22:40 • Thanks Stéphane for the correction but it still doesn't seem to work – Gilles Oct 19 '13 at 9:31 • try looking at the source code: stats:::logLik.glm – assumednormal Oct 19 '13 at 9:45 • I did this but this function just reverse the aic slot from the glm object to find back the log-likelihood. And I don't see anything about aic in the glm function... – Gilles Oct 19 '13 at 10:20 • I suspect this has something to do with LogLik and AIC (which are tied together at the hip) assuming that three parameters are being estimated (the slope, intercept, and dispersion/residual standard error) whereas the dispersion/residual standard error is calculated assuming two parameters are estimated (slope and intercept). – Tom Oct 19 '13 at 12:04 ## 1 Answer The logLik() function provides the evaluation of the log-likelihood by substituting the ML estimates of the parameters for the values of the unknown parameters. Now, the maximum likelihood estimates of the regression parameters (the$\beta_j$'s in$X{\boldsymbol \beta}$) coincide with the least-squares estimates, but the ML estimate of$\sigma$is$\sqrt{\frac{\sum \hat\epsilon_i^2}{n}}$, whereas you are using$\hat\sigma = \sqrt{\frac{\sum \hat\epsilon_i^2}{n-2}}$, that is the square root of the unbiased estimate of$\sigma^2$. > n <- 5 > x <- 1:n > set.seed(1) > y <- 10 + 2*x + rnorm(n, 0, 2) > modlm <- lm(y ~ x) > sigma <- summary(modlm)$sigma
>
>  # value of the likelihood with the "classical" sigma hat
>  sum(log(dnorm(x = y, mean = predict(modlm), sd = sigma)))
 -9.192832
>
>  # value of the likelihood with the ML sigma hat
>  sigma.ML <- sigma*sqrt((n-dim(model.matrix(modlm)))/n)
>  sum(log(dnorm(x = y, mean = predict(modlm), sd = sigma.ML)))
 -8.915768
>  logLik(modlm)
'log Lik.' -8.915768 (df=3)

• By the way you have to similarly be careful with the REML/ML option for lme/lmer models. – Stéphane Laurent Oct 19 '13 at 18:51
• (+1) Is it n-1 or is it indeed n-2 in the denominator of $\hat\sigma$ ? – Patrick Coulombe Oct 19 '13 at 20:07
• @PatrickCoulombe No : intercept + slope – Stéphane Laurent Oct 19 '13 at 20:28
• Ok, perfectly clear now. Thanks a lot ! But what do you mean with REML/ML (something to do with my last post on GuR I guess) ? Please explain (there maybe). I want to learn ! – Gilles Oct 19 '13 at 21:26
• The REML estimates of the variance components in a mixed models are like the "corrected for bias" ML estimates. I have not seen your post on GuR yet :) – Stéphane Laurent Oct 19 '13 at 22:17