Analysing rank-ordered data using mlogit

Question

I am trying to analyse rank-ordered data using a logit model as described here (more specifically the model in 5.1). Right now I am not interested in the effect of any covariates but only want to asses the item specific differences by looking at the corresponding log odds. If I understand correctly I can achieve this by only including a constant term as the individual specific variable and removing any other intercepts. I am using the mlogit package in R and from what I have gathered from the vignette (especially the example for rank-ordered logit models in 2.8) I tried to do the following:

library(mlogit)
data("Game", package = "mlogit")
G <- mlogit.data(Game, shape="wide", choice="ch", varying=1:12, ranked=TRUE)
summary(mlogit(ch ~ 1 | -1 | -1,G))

This gets me the following error message:

Error in `rownames<-`(`*tmp*`, value = c("1.GameBoy", "1.GameCube", "1.PC",  :
attempt to set 'rownames' on an object with no dimensions

It is pretty obvious that I have a misconception about the way the mlogit formula has to be constructed so I tried to play around a bit and the only thing I got which looked like what I wanted was the following:

    Call:
mlogit(formula = ch ~ alt | -1 | -1, data = G, method = "nr", 
    print.level = 0)

Frequencies of alternatives:
    GameBoy    GameCube          PC PlayStation  PSPortable        Xbox 
    0.13846     0.13407     0.17363     0.18462     0.17363     0.19560 

nr method
4 iterations, 0h:0m:0s 
g'(-H)^-1g = 0.000575 
successive function values within tolerance limits 

Coefficients :
                        Estimate Std. Error t-value  Pr(>|t|)    
GameCube:(intercept)    0.058678   0.182572  0.3214  0.747911    
PC:(intercept)          1.275749   0.194292  6.5661 5.164e-11 ***
PlayStation:(intercept) 1.273903   0.191226  6.6618 2.705e-11 ***
PSPortable:(intercept)  0.622355   0.181645  3.4262  0.000612 ***
Xbox:(intercept)        1.401230   0.189758  7.3843 1.532e-13 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Log-Likelihood: -546.82

However, this is not very satisfactory since to me it seems strange to include the alternative as an individual specific covariate. Can anyone with experience of how to fit such a model in R using the mlogit package help?

P.S.: I was unsure whether this question is better suited for CV or StackOverflow but since the underlying question is a statistical one I chose to put it here.

Answer 1

My experience is still rather limited with mlogit package, but if I read Croissant vignette correctly (see the beginning of sec. 1.2 Model description, page 7), the alt variable in your model is specified as alternative specific with a generic coefficient and NOT as an individual specific covariate---those variables are placed between the pipes.

Answer 2

I used the model format you mentioned at the end and was able to reproduce the LRS listed for example 8.1 in the J Marden text (Analyzing and Modeling Rank Data):

> exData = t(matrix(c(c(1,2,4,3,5), c(2,1,4,3,5), c(2,3,5,4,1), rep(c(2,1,5,3,4), 3)), nrow=5))
> exData
     [,1] [,2] [,3] [,4] [,5]
[1,]    1    2    4    3    5
[2,]    2    1    4    3    5
[3,]    2    3    5    4    1
[4,]    2    1    5    3    4
[5,]    2    1    5    3    4
[6,]    2    1    5    3    4
> exTbl = as.data.table(exData)[, id := 1:.N][, melt(.SD, id.vars='id')][order(id, variable)][, setnames(.SD, 'value', 'ch')][, setnames(.SD, 'variable', 'alt')]
> exTbl %>% head
   id alt ch
1:  1  V1  1
2:  1  V2  2
3:  1  V3  4
4:  1  V4  3
5:  1  V5  5
6:  2  V1  2
> exLogit = mlogit.data(exTbl, shape='long', ranked=T, choice='ch', alt.var='alt', id.var='id')
> exLogit %>% head
     id alt    ch
1.V1  1  V1  TRUE
1.V2  1  V2 FALSE
1.V3  1  V3 FALSE
1.V4  1  V4 FALSE
1.V5  1  V5 FALSE
2.V2  1  V2  TRUE
> m1 = summary(mlogit(ch ~ alt | -1 | -1, exLogit, reflevel='V5'))
> m1

Call:
mlogit(formula = ch ~ alt | -1 | -1, data = exLogit, reflevel = "V5", 
    method = "nr", print.level = 0)

Frequencies of alternatives:
    V5     V1     V2     V3     V4 
0.1667 0.2500 0.2500 0.0833 0.2500 

nr method
6 iterations, 0h:0m:0s 
g(-H)^-1g = 4.08E-06 
successive function values within tolerance limits 

Coefficients :
               Estimate Std. Error t-value Pr(>|t|)   
V1:(intercept)    4.182      1.465    2.86   0.0043 **
V2:(intercept)    4.689      1.514    3.10   0.0020 **
V3:(intercept)   -0.726      0.867   -0.84   0.4024   
V4:(intercept)    2.061      1.133    1.82   0.0690 . 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Log-Likelihood: -14.4
> llNull = -sum(log(1:5)) * 6
> llNull
[1] -28.72
> m1$logLik
'log Lik.' -14.4 (df=4)
> LRS = 2 * (m1$logLik - llNull)
> LRS
'log Lik.' 28.64 (df=4)
> 
[0] 0:NvimR*    

I lifted the method to compute the log likelihood for the null model from another stats package, so it may/may not be correct. Although using it does make the result line up with the example 8.1 in the text, so it's likely correct.

I was surpised that the log likelihood value for the null model is only a function of the number of alternatives and number of subjects, and not influenced by the data observed. The data observed seems to only influence the log likelihood for the alternative model. If someone has an intuitive explanation as to why this is the case, please feel free to add that to this post.