Question

I am using the book “Generalized Linear Models and Extension” by Hardin and Hilbe (second edition, 2007) at the moment. The authors suggest that instead of OLS models, “the log link is generally used for response data that take only positive values on the continuous scale”. Of course they also suggest residual plots to check whether a “normal” linear model using an identity link can still be used.

I am trying to replicate in R what they do in the book in STATA. Indeed, I have no problems in STATA with the log link. However, when calling the same model using R’s glm-function, but specifying family=gaussian(link="log") I am asked to provide starting values. When I set them all equal to zero, I always get the message that the algorithm did not converge. Picking other values the message is sometimes the same, but more often I get:

Error in glm.fit(x = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,  :
     NA/NaN/Inf in 'x'

As I said, in STATA I can run these models without setting starting values and without errors. I tried many different models, and different datasets, but the problem is always the same (unless I only include one single independent variable). Could anyone tell me why this is the case, or what I do wrong, or why the suggested models from the book might not be appropriate? I’d appreciate any help, thanks!

Edit: As an example which reproduces the error consider the dataset which can be downloaded here. With this dataset loaded, I run the following model:

mod <- glm(betaplasma ~ age + vituse, family=gaussian(link="log"), data=data2, start=c(0,0,0))

This produces the the warning message that the algorithm did not converge.

Edit2: I was asked to also provide the STATA output for that model. Here it is:

. glm betaplasma age vituse, link(log)

Iteration 0:   log likelihood = -2162.1385  
Iteration 1:   log likelihood = -2096.4765  
Iteration 2:   log likelihood = -2076.2465  
Iteration 3:   log likelihood = -2076.2244  
Iteration 4:   log likelihood = -2076.2244  

Generalized linear models                          No. of obs      =       315
Optimization     : ML                              Residual df     =       312
                                                   Scale parameter =  31384.51
Deviance         =  9791967.359                    (1/df) Deviance =  31384.51
Pearson          =  9791967.359                    (1/df) Pearson  =  31384.51

Variance function: V(u) = 1                        [Gaussian]
Link function    : g(u) = ln(u)                    [Log]

                                                   AIC             =  13.20142
Log likelihood   = -2076.224437                    BIC             =   9790173

------------------------------------------------------------------------------
             |                 OIM
  betaplasma |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
         age |   .0056809   .0032737     1.74   0.083    -.0007354    .0120972
      vituse |   -.273027   .0650773    -4.20   0.000    -.4005762   -.1454779
       _cons |   5.467577   .2131874    25.65   0.000     5.049738    5.885417
------------------------------------------------------------------------------

Answer 1

As I said in my comment, it’s probably true that Stata has more robust (in the numerical, not the statistical sense) GLM fitting than R. That said, fitting this particular dataset doesn’t seem too hard.

Read data:

data2 <- read.table("http://lib.stat.cmu.edu/datasets/Plasma_Retinol",
         skip=30,nrows=315)
dnames <- c("age","sex","smokstat","quetelet","vituse","calories","fat","fiber",
           "alcohol","cholesterol","betadiet","retdiet","betaplasma","retplasma")
names(data2) <- dnames

Plot the data:

par(mfrow=c(1,2),las=1,bty="l")
with(data2,plot(betaplasma~age))
with(data2,boxplot(betaplasma~vituse))

It’s fairly easy to get these to fit by setting the starting value of the intercept parameter to something reasonable (i.e. something close to the mean of the data on the log scale: either of these works

mod <- glm(betaplasma ~ age + vituse, family=gaussian(link="log"), data=data2,
           start=c(10,0,0))
mod <- glm(betaplasma ~ age + vituse, family=gaussian(link="log"), data=data2,
           start=c(log(mean(data2$betaplasma)),0,0))

The latter case is probably a reasonable default strategy for starting log-link fits. The results (slightly abbreviated) match Stata’s very closely:

summary(mod)
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  5.467575   0.218360  25.039  < 2e-16 ***
## age          0.005681   0.003377   1.682   0.0935 .  
## vituse      -0.273027   0.065552  -4.165 4.03e-05 ***
## ---
## Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 
## 
## (Dispersion parameter for gaussian family taken to be 31385.26)
## 
##     Null deviance: 10515638  on 314  degrees of freedom
## Residual deviance:  9791967  on 312  degrees of freedom
## AIC: 4160.4
## 
## Number of Fisher Scoring iterations: 9

confint(mod)
##                     2.5 %      97.5 %
## (Intercept)  5.0364648709  5.87600710
## age         -0.0007913795  0.01211007
## vituse      -0.4075213916 -0.14995759

(R is using t rather than Z statistics for the p-values and (?) confidence intervals)

However, there are a few reasons I might not fit this model to these data. In particular, the assumption of constant variance (associated with the Gaussian model) is not very reasonable — these data seem better suited for a lognormal model (or equivalently, for just log-transforming and analyzing with a standard Gaussian model).

Plotting on a log(1+x) scale (there is a zero entry in the data):

with(data2,plot(log(1+betaplasma)~age))
with(data2,boxplot(log(1+betaplasma)~vituse))

Plotting with ggplot (this fits separate lines for each value of vituse rather than fitting an additive model)

library(ggplot)
theme_set(theme_bw())
(g1 <- qplot(age,1+betaplasma,colour=factor(vituse),data=data2)+
    geom_smooth(method="lm")+
    scale_y_log10())

View without ‘outlier’:

g1 %+% subset(data2,betaplasma>0)

Two other points: (1) it’s a bit odd that there’s a response with a value of 0 in this data set — not impossible, but odd; (2) it looks like vituse should be treated as a factor rather than as numeric (“1=Yes, fairly often, 2=Yes, not often, 3=No”) — possibly ordinal.