Exercise 3: Mixed logit model

Kenneth Train and Yves Croissant

2019-07-22

A sample of residential electricity customers were asked a series of choice experiments. In each experiment, four hypothetical electricity suppliers were described. The person was asked which of the four suppliers he/she would choose. As many as 12 experiments were presented to each person. Some people stopped before answering all 12. There are 361 people in the sample, and a total of 4308 experiments. In the experiments, the characteristics of each supplier were stated. The price of the supplier was either :

The length of contract that the supplier offered was also stated, in years (such as 1 year or 5 years.) During this contract period, the supplier guaranteed the prices and the buyer would have to pay a penalty if he/she switched to another supplier. The supplier could offer no contract in which case either side could stop the agreement at any time. This is recorded as a contract length of 0.

Some suppliers were also described as being a local company or a “well-known” company. If the supplier was not local or well-known, then nothing was said about them in this regard .

  1. Run a mixed logit model without intercepts and a normal distribution for the 6 parameters of the model, using 100 draws, halton sequences and taking into account the panel data structure.
library("mlogit")
data("Electricity", package = "mlogit")
Electr <- mlogit.data(Electricity, id = "id", choice = "choice", 
                      varying = 3:26, shape = "wide", sep = "")
Elec.mxl <- mlogit(choice ~ pf + cl + loc + wk + tod + seas | 0, Electr, 
              rpar=c(pf = 'n', cl = 'n', loc = 'n', wk = 'n', 
                     tod = 'n', seas = 'n'), 
              R = 100, halton = NA, panel = TRUE)
summary(Elec.mxl)
## 
## Call:
## mlogit(formula = choice ~ pf + cl + loc + wk + tod + seas | 0, 
##     data = Electr, start = strt, rpar = c(pf = "n", cl = "n", 
##         loc = "n", wk = "n", tod = "n", seas = "n"), R = 100, 
##     halton = NA, panel = TRUE)
## 
## Frequencies of alternatives:
##       1       2       3       4 
## 0.22702 0.26393 0.23816 0.27089 
## 
## bfgs method
## 1 iterations, 0h:0m:3s 
## g'(-H)^-1g = 1.45E-07 
## gradient close to zero 
## 
## Coefficients :
##          Estimate Std. Error z-value  Pr(>|z|)    
## pf      -0.973386   0.034324 -28.359 < 2.2e-16 ***
## cl      -0.205557   0.013323 -15.428 < 2.2e-16 ***
## loc      2.075724   0.080430  25.808 < 2.2e-16 ***
## wk       1.475645   0.065168  22.644 < 2.2e-16 ***
## tod     -9.052539   0.287218 -31.518 < 2.2e-16 ***
## seas    -9.103759   0.289043 -31.496 < 2.2e-16 ***
## sd.pf    0.219943   0.010840  20.291 < 2.2e-16 ***
## sd.cl    0.378303   0.018489  20.461 < 2.2e-16 ***
## sd.loc   1.482974   0.081305  18.240 < 2.2e-16 ***
## sd.wk    1.000057   0.074182  13.481 < 2.2e-16 ***
## sd.tod   2.289477   0.110731  20.676 < 2.2e-16 ***
## sd.seas  1.180869   0.109007  10.833 < 2.2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Log-Likelihood: -3952.5
## 
## random coefficients
##      Min.     1st Qu.     Median       Mean     3rd Qu. Max.
## pf   -Inf  -1.1217350 -0.9733857 -0.9733857 -0.82503650  Inf
## cl   -Inf  -0.4607187 -0.2055571 -0.2055571  0.04960449  Inf
## loc  -Inf   1.0754736  2.0757241  2.0757241  3.07597467  Inf
## wk   -Inf   0.8011174  1.4756454  1.4756454  2.15017342  Inf
## tod  -Inf -10.5967678 -9.0525388 -9.0525388 -7.50830989  Inf
## seas -Inf  -9.9002427 -9.1037589 -9.1037589 -8.30727517  Inf
    1. Using the estimated mean coefficients, determine the amount that a customer with average coefficients for price and length is willing to pay for an extra year of contract length.
coef(Elec.mxl)['cl'] / coef(Elec.mxl)['pf']
##        cl 
## 0.2111774

The mean coefficient of length is -0.20. The consumer with this average coefficient dislikes having a longer contract. So this person is willing to pay to reduce the length of the contract. The mean price coefficient is -0.97. A customer with these coefficients is willing to pay 0.20/0.97=0.21, or one-fifth a cent per kWh extra to have a contract that is one year shorter.

  1. Determine the share of the population who are estimated to dislike long term contracts (ie have a negative coefficient for the length.) \begin{answer}[5]
pnorm(- coef(Elec.mxl)['cl'] / coef(Elec.mxl)['sd.cl'])
##        cl 
## 0.7065611

The coefficient of length is normally distributed with mean -0.20 and standard deviation 0.35. The share of people with coefficients below zero is the cumulative probability of a standardized normal deviate evaluated at 0.20/0.35=0.57. Looking 0.57 up in a table of the standard normal distribution, we find that the share below 0.57 is 0.72. About seventy percent of the population are estimated to dislike long-term contracts.

  1. The price coefficient is assumed to be normally distributed in these runs. This assumption means that some people are assumed to have positive price coefficients, since the normal distribution has support on both sides of zero. Using your estimates from exercise 1, determine the share of customers with positive price coefficients. As you can see, this is pretty small share and can probably be ignored. However, in some situations, a normal distribution for the price coefficient will give a fairly large share with the wrong sign. Revise the program to make the price coefficient fixed rather than random. A fixed price coefficient also makes it easier to calculate the distribution of willingness to pay (wtp) for each non-price attribute. If the price coefficient is fixed, the distribtion of wtp for an attribute has the same distribution as the attribute’s coefficient, simply scaled by the price coefficient. However, when the price coefficient is random, the distribution of wtp is the ratio of two distributions, which is harder to work with.
pnorm(- coef(Elec.mxl)['pf'] / coef(Elec.mxl)['sd.pf'])
##        pf 
## 0.9999952

The price coefficient is distributed normal with mean -0.97 and standard deviation 0.20. The cumulative standard normal distribution evaluated at 0.97/0.20=4.85 is more than 0.999, which means that more than 99.9% of the population are estimated to have negative price coefficients. Essentially no one is estimated to have a positive price coefficient.

Elec.mxl2 <- mlogit(choice ~ pf + cl + loc + wk + tod + seas | 0, Electr, 
                   rpar = c(cl = 'n', loc = 'n', wk = 'n', 
                            tod = 'n', seas = 'n'), 
                   R = 100, halton = NA,  panel = TRUE)
summary(Elec.mxl2)
## 
## Call:
## mlogit(formula = choice ~ pf + cl + loc + wk + tod + seas | 0, 
##     data = Electr, start = strt, rpar = c(cl = "n", loc = "n", 
##         wk = "n", tod = "n", seas = "n"), R = 100, halton = NA, 
##     panel = TRUE)
## 
## Frequencies of alternatives:
##       1       2       3       4 
## 0.22702 0.26393 0.23816 0.27089 
## 
## bfgs method
## 1 iterations, 0h:0m:2s 
## g'(-H)^-1g = 4.6E-08 
## gradient close to zero 
## 
## Coefficients :
##          Estimate Std. Error z-value  Pr(>|z|)    
## pf      -0.879902   0.032759 -26.860 < 2.2e-16 ***
## cl      -0.217059   0.013673 -15.875 < 2.2e-16 ***
## loc      2.092298   0.081067  25.809 < 2.2e-16 ***
## wk       1.490902   0.065230  22.856 < 2.2e-16 ***
## tod     -8.581835   0.282912 -30.334 < 2.2e-16 ***
## seas    -8.583281   0.280347 -30.617 < 2.2e-16 ***
## sd.cl    0.373477   0.018018  20.728 < 2.2e-16 ***
## sd.loc   1.558857   0.087696  17.776 < 2.2e-16 ***
## sd.wk    1.050810   0.078023  13.468 < 2.2e-16 ***
## sd.tod   2.694660   0.120798  22.307 < 2.2e-16 ***
## sd.seas  1.950728   0.104766  18.620 < 2.2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Log-Likelihood: -3961.7
## 
## random coefficients
##      Min.     1st Qu.     Median       Mean     3rd Qu. Max.
## cl   -Inf  -0.4689658 -0.2170594 -0.2170594  0.03484691  Inf
## loc  -Inf   1.0408650  2.0922983  2.0922983  3.14373157  Inf
## wk   -Inf   0.7821415  1.4909018  1.4909018  2.19966212  Inf
## tod  -Inf -10.3993553 -8.5818346 -8.5818346 -6.76431383  Inf
## seas -Inf  -9.8990266 -8.5832805 -8.5832805 -7.26753448  Inf
  1. You think that everyone must like using a known company rather than an unknown one, and yet the normal distribution implies that some people dislike using a known company. Revise the program to give the coefficient of a uniform distribution between zero and b,where b is estimated (do this with the price coefficient fixed).
Elec.mxl3 <- update(Elec.mxl, rpar = c(cl = 'n', loc = 'n', wk = 'u', 
                                       tod = 'n', seas = 'n'))

The price coefficient is uniformly distributed with parameters 1.541 and 1.585.

summary(Elec.mxl3)
## 
## Call:
## mlogit(formula = choice ~ pf + cl + loc + wk + tod + seas | 0, 
##     data = Electr, start = strt, rpar = c(cl = "n", loc = "n", 
##         wk = "u", tod = "n", seas = "n"), R = 100, halton = NA, 
##     panel = TRUE)
## 
## Frequencies of alternatives:
##       1       2       3       4 
## 0.22702 0.26393 0.23816 0.27089 
## 
## bfgs method
## 1 iterations, 0h:0m:2s 
## g'(-H)^-1g = 1.08E-07 
## gradient close to zero 
## 
## Coefficients :
##          Estimate Std. Error z-value  Pr(>|z|)    
## pf      -0.882229   0.032818 -26.883 < 2.2e-16 ***
## cl      -0.217128   0.013776 -15.761 < 2.2e-16 ***
## loc      2.099323   0.081056  25.900 < 2.2e-16 ***
## wk       1.509425   0.065496  23.046 < 2.2e-16 ***
## tod     -8.606979   0.282983 -30.415 < 2.2e-16 ***
## seas    -8.602396   0.280671 -30.649 < 2.2e-16 ***
## sd.cl    0.381070   0.018150  20.996 < 2.2e-16 ***
## sd.loc   1.593852   0.087802  18.153 < 2.2e-16 ***
## sd.wk    1.786373   0.125764  14.204 < 2.2e-16 ***
## sd.tod   2.719073   0.119356  22.781 < 2.2e-16 ***
## sd.seas  1.945381   0.103686  18.762 < 2.2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Log-Likelihood: -3956.7
## 
## random coefficients
##            Min.     1st Qu.     Median       Mean    3rd Qu.
## cl         -Inf  -0.4741561 -0.2171285 -0.2171285  0.0398992
## loc        -Inf   1.0242863  2.0993231  2.0993231  3.1743600
## wk   -0.2769485   0.6162382  1.5094248  1.5094248  2.4026115
## tod        -Inf -10.4409656 -8.6069790 -8.6069790 -6.7729924
## seas       -Inf  -9.9145353 -8.6023958 -8.6023958 -7.2902563
##          Max.
## cl        Inf
## loc       Inf
## wk   3.295798
## tod       Inf
## seas      Inf
rpar(Elec.mxl3, 'wk')
## uniform distribution with parameters 1.509 (center) and 1.786 (span)
summary(rpar(Elec.mxl3, 'wk'))
##       Min.    1st Qu.     Median       Mean    3rd Qu. 
## -0.2769485  0.6162382  1.5094248  1.5094248  2.4026115 
##       Max. 
##  3.2957982
plot(rpar(Elec.mxl3, 'wk'))

The upper bound is 3.13. The estimated price coefficient is -0.88 and so the willingness to pay for a known provided ranges uniformly from -0.05 to 3.55 cents per kWh.

  1. Rerun the model with a fixed coefficient for price and lognormal distributions for the coefficients of TOD and seasonal (since their coefficient should be negative for all people.) To do this, you need to reverse the sign of the TOD and seasonal variables, since the lognormal is always positive and you want the these coefficients to be always negative.

A lognormal is specified as \(\exp(b+se)\) where \(e\) is a standard normal deviate. The parameters of the lognormal are \(b\) and \(s\). The mean of the lognormal is \(\exp(b+0.5s^2)\) and the standard deviation is the mean times \(\sqrt{(\exp(s^2))-1}\).

Electr <- mlogit.data(Electricity, id = "id", choice = "choice", 
                      varying = 3:26, shape = "wide", sep = "",
                      opposite = c('tod', 'seas'))
Elec.mxl4 <- mlogit(choice ~ pf + cl + loc + wk + tod + seas | 0, Electr, 
              rpar = c(cl = 'n', loc = 'n', wk = 'u', tod = 'ln', seas = 'ln'), 
              R = 100, halton = NA, panel = TRUE)
summary(Elec.mxl4)
## 
## Call:
## mlogit(formula = choice ~ pf + cl + loc + wk + tod + seas | 0, 
##     data = Electr, start = strt, rpar = c(cl = "n", loc = "n", 
##         wk = "u", tod = "ln", seas = "ln"), R = 100, halton = NA, 
##     panel = TRUE)
## 
## Frequencies of alternatives:
##       1       2       3       4 
## 0.22702 0.26393 0.23816 0.27089 
## 
## bfgs method
## 1 iterations, 0h:0m:2s 
## g'(-H)^-1g = 6.24E-08 
## gradient close to zero 
## 
## Coefficients :
##          Estimate Std. Error z-value  Pr(>|z|)    
## pf      -0.868985   0.032350 -26.862 < 2.2e-16 ***
## cl      -0.211334   0.013569 -15.575 < 2.2e-16 ***
## loc      2.023876   0.080102  25.266 < 2.2e-16 ***
## wk       1.479118   0.064957  22.771 < 2.2e-16 ***
## tod      2.112378   0.033769  62.554 < 2.2e-16 ***
## seas     2.124205   0.033342  63.709 < 2.2e-16 ***
## sd.cl    0.373120   0.017710  21.068 < 2.2e-16 ***
## sd.loc   1.548511   0.086400  17.922 < 2.2e-16 ***
## sd.wk    1.521790   0.119993  12.682 < 2.2e-16 ***
## sd.tod   0.367077   0.019997  18.357 < 2.2e-16 ***
## sd.seas  0.275352   0.016875  16.317 < 2.2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Log-Likelihood: -3976.5
## 
## random coefficients
##            Min.    1st Qu.     Median       Mean     3rd Qu.
## cl         -Inf -0.4629994 -0.2113338 -0.2113338  0.04033179
## loc        -Inf  0.9794208  2.0238757  2.0238757  3.06833059
## wk   -0.0426715  0.7182234  1.4791184  1.4791184  2.24001328
## tod   0.0000000  6.4545718  8.2678801  8.8441019 10.59060830
## seas  0.0000000  6.9482054  8.3662477  8.6894950 10.07369478
##          Max.
## cl        Inf
## loc       Inf
## wk   3.000908
## tod       Inf
## seas      Inf
plot(rpar(Elec.mxl4, 'seas'))

  1. Rerun the same model as previously, but allowing now the correlation between random parameters. Compute the correlation matrix of the random parameters. Test the hypothesis that the random parameters are uncorrelated.
Elec.mxl5 <- update(Elec.mxl4, correlation = TRUE)
#summary(Elec.mxl5)
#cor.mlogit(Elec.mxl5)
#lrtest(Elec.mxl5, Elec.mxl4)
#waldtest(Elec.mxl5, correlation = FALSE)
#scoretest(Elec.mxl4, correlation = TRUE)

The three tests clearly reject the hypothesis that the random parameters are uncorrelated.