R hmftest multinomial logit model " system is computationally singular"

Question

I have a multinomial logit model with two individual specific variables (first and age). I would like to conduct the hmftest to check if the IIA holds.

My dataset looks like this:

head(df)
  mode choice first age
1 both      1     0  24
2  pre      1     1  23
3 both      1     2  53
4 post      1     3  43
5   no      1     1  55
6 both      1     2  63

I adjusted it for the mlogit to:

   mode choice first age    idx
1   TRUE      1     0  24 1:both
2  FALSE      1     0  24   1:no
3  FALSE      1     0  24 1:post
4  FALSE      1     0  24  1:pre
5  FALSE      1     1  23 2:both
6  FALSE      1     1  23   2:no
7  FALSE      1     1  23 2:post
8   TRUE      1     1  23  2:pre
9   TRUE      1     2  53 3:both
10 FALSE      1     2  53   3:no

~~~ indexes ~~~~
   id1  id2
1    1 both
2    1   no
3    1 post
4    1  pre
5    2 both
6    2   no
7    2 post
8    2  pre
9    3 both
10   3   no
indexes:  1, 2

My original (full) model runs as follows:

full <- mlogit(mode ~ 0 | first + age, data = df_mlogit, reflevel = "no")

leading to the following result:

Call:
mlogit(formula = mode ~ 0 | first + age, data = df_mlogit, reflevel = "no", 
    method = "nr")

Frequencies of alternatives:choice
  no both post  pre 
 0.2  0.4  0.2  0.2 

nr method
18 iterations, 0h:0m:0s 
g'(-H)^-1g = 8.11E-07 
gradient close to zero 

Coefficients :
                    Estimate  Std. Error z-value Pr(>|z|)
(Intercept):both  2.0077e+01  1.0441e+04  0.0019   0.9985
(Intercept):post -4.1283e-01  1.4771e+04  0.0000   1.0000
(Intercept):pre   5.3346e-01  1.4690e+04  0.0000   1.0000
first1:both      -4.0237e+01  1.1059e+04 -0.0036   0.9971
first1:post      -8.9168e-01  1.4771e+04 -0.0001   1.0000
first1:pre       -6.6805e-01  1.4690e+04  0.0000   1.0000
first2:both      -1.9674e+01  1.0441e+04 -0.0019   0.9985
first2:post      -1.8975e+01  1.5683e+04 -0.0012   0.9990
first2:pre       -1.8889e+01  1.5601e+04 -0.0012   0.9990
first3:both      -2.1185e+01  1.1896e+04 -0.0018   0.9986
first3:post       1.9200e+01  1.5315e+04  0.0013   0.9990
first3:pre        1.9218e+01  1.5237e+04  0.0013   0.9990
age:both          2.1898e-02  2.9396e-02  0.7449   0.4563
age:post          9.3377e-03  2.3157e-02  0.4032   0.6868
age:pre          -1.2338e-02  2.2812e-02 -0.5408   0.5886

Log-Likelihood: -61.044
McFadden R^2:  0.54178 
Likelihood ratio test : chisq = 144.35 (p.value = < 2.22e-16)

To test for IIA, I exclude one alternative from the model (here "pre") and run the model as follows:

part <- mlogit(mode ~ 0 | first + age, data = df_mlogit, reflevel = "no", 
               alt.subset = c("no", "post", "both"))

leading to

Call:
mlogit(formula = mode ~ 0 | first + age, data = df_mlogit, alt.subset = c("no", 
    "post", "both"), reflevel = "no", method = "nr")

Frequencies of alternatives:choice
  no both post 
0.25 0.50 0.25 

nr method
18 iterations, 0h:0m:0s 
g'(-H)^-1g = 6.88E-07 
gradient close to zero 

Coefficients :
                    Estimate  Std. Error z-value Pr(>|z|)
(Intercept):both  1.9136e+01  6.5223e+03  0.0029   0.9977
(Intercept):post -9.2040e-01  9.2734e+03 -0.0001   0.9999
first1:both      -3.9410e+01  7.5835e+03 -0.0052   0.9959
first1:post      -9.3119e-01  9.2734e+03 -0.0001   0.9999
first2:both      -1.8733e+01  6.5223e+03 -0.0029   0.9977
first2:post      -1.8094e+01  9.8569e+03 -0.0018   0.9985
first3:both      -2.0191e+01  1.1049e+04 -0.0018   0.9985
first3:post       2.0119e+01  1.1188e+04  0.0018   0.9986
age:both          2.1898e-02  2.9396e-02  0.7449   0.4563
age:post          1.9879e-02  2.7872e-02  0.7132   0.4757

Log-Likelihood: -27.325
McFadden R^2:  0.67149 
Likelihood ratio test : chisq = 111.71 (p.value = < 2.22e-16)

However when I want to codnuct the hmftest then the following error occurs:

> hmftest(full, part)
Error in solve.default(diff.var) : 
  system is computationally singular: reciprocal condition number = 4.34252e-21

Does anyone have an idea where the problem might be?

Are there any predictors that have no variation when stratified by the dependent variable? — Bill O'Brien, Jul 21 '21 at 14:59
Both independent variables only vary by indiviuum, so for each choice the age will be the same f.ex., is that what you mean? — Anna Massfeller, Jul 21 '21 at 15:10

score 0 · Answer 1 · answered Jan 20 '22 at 15:11

0

I believe the issue here could be that the hmftest checks if the probability ratio of two alternatives depends only on the characteristics of these alternatives. Since there are only individual-level variables here, the test won't work in this case.

answered Jan 20 '22 at 15:11

Maria

1

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R hmftest multinomial logit model " system is computationally singular"

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