Author: Kenneth Train and Yves Croissant
Date: 2025-07-12

# Exercise 2: Nested logit model

#### Kenneth Train and Yves Croissant

#### 2025-07-12

The data set `HC` from `mlogit` contains data in `R` format on the choice of heating and central cooling system for 250 single-family, newly built houses in California.

The alternatives are:

- Gas central heat with cooling `gcc`,
- Electric central resistence heat with cooling `ecc`,
- Electric room resistence heat with cooling `erc`,
- Electric heat pump, which provides cooling also `hpc`,
- Gas central heat without cooling `gc`,
- Electric central resistence heat without cooling `ec`,
- Electric room resistence heat without cooling `er`.

Heat pumps necessarily provide both heating and cooling such that heat pump without cooling is not an alternative.

The variables are:

- `depvar` gives the name of the chosen alternative,
- `ich.alt` are the installation cost for the heating portion of the system,
- `icca` is the installation cost for cooling
- `och.alt` are the operating cost for the heating portion of the system
- `occa` is the operating cost for cooling
- `income` is the annual income of the household

Note that the full installation cost of alternative `gcc` is `ich.gcc+icca`, and similarly for the operating cost and for the other alternatives with cooling.

1. Run a nested logit model on the data for two nests and one log-sum coefficient that applies to both nests. Note that the model is specified to have the cooling alternatives (`gcc},`ecc}, `erc},`hpc}) in one nest and the non-cooling alternatives (`gc},`ec}, `er}) in another nest.

    library("mlogit")
    data("HC", package = "mlogit")
    HC <- dfidx(HC, varying = c(2:8, 10:16), choice = "depvar")
    cooling.modes <- idx(HC, 2) %in% c('gcc', 'ecc', 'erc', 'hpc')
    room.modes <- idx(HC, 2) %in% c('erc', 'er')
    # installation / operating costs for cooling are constants,
    # only relevant for mixed systems
    HC$icca[! cooling.modes] <- 0
    HC$occa[! cooling.modes] <- 0
    # create income variables for two sets cooling and rooms
    HC$inc.cooling <- HC$inc.room <- 0
    HC$inc.cooling[cooling.modes] <- HC$income[cooling.modes]
    HC$inc.room[room.modes] <- HC$income[room.modes]
    # create an intercet for cooling modes
    HC$int.cooling <- as.numeric(cooling.modes)
    # estimate the model with only one nest elasticity
    nl <- mlogit(depvar ~ ich + och +icca + occa + inc.room + inc.cooling + int.cooling | 0, HC,
                 nests = list(cooling = c('gcc','ecc','erc','hpc'),
                 other = c('gc', 'ec', 'er')), un.nest.el = TRUE)
    summary(nl)

    ##
    ## Call:
    ## mlogit(formula = depvar ~ ich + och + icca + occa + inc.room +
    ##     inc.cooling + int.cooling | 0, data = HC, nests = list(cooling = c("gcc",
    ##     "ecc", "erc", "hpc"), other = c("gc", "ec", "er")), un.nest.el = TRUE)
    ##
    ## Frequencies of alternatives:choice
    ##    ec   ecc    er   erc    gc   gcc   hpc
    ## 0.004 0.016 0.032 0.004 0.096 0.744 0.104
    ##
    ## bfgs method
    ## 11 iterations, 0h:0m:0s
    ## g'(-H)^-1g = 7.26E-06
    ## successive function values within tolerance limits
    ##
    ## Coefficients :
    ##              Estimate Std. Error z-value  Pr(>|z|)
    ## ich         -0.554878   0.144205 -3.8478 0.0001192 ***
    ## och         -0.857886   0.255313 -3.3601 0.0007791 ***
    ## icca        -0.225079   0.144423 -1.5585 0.1191212
    ## occa        -1.089458   1.219821 -0.8931 0.3717882
    ## inc.room    -0.378971   0.099631 -3.8038 0.0001425 ***
    ## inc.cooling  0.249575   0.059213  4.2149 2.499e-05 ***
    ## int.cooling -6.000415   5.562423 -1.0787 0.2807030
    ## iv           0.585922   0.179708  3.2604 0.0011125 **
    ## ---
    ## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
    ##
    ## Log-Likelihood: -178.12

1. The estimated log-sum coefficient is \(0.59\). What does this estimate tell you about the degree of correlation in unobserved factors over alternatives within each nest?

> The correlation is approximately \(1-0.59=0.41\). It’s a moderate correlation.

2. Test the hypothesis that the log-sum coefficient is 1.0 (the value that it takes for a standard logit model.) Can the hypothesis that the true model is standard logit be rejected?

> We can use a t-test of the hypothesis that the log-sum coefficient equal to 1. The t-statistic is :

     (coef(nl)['iv'] - 1) / sqrt(vcov(nl)['iv', 'iv'])

    ##        iv
    ## -2.304171

> The critical value of t for 95% confidence is 1.96. So we can reject the hypothesis at 95% confidence.

> We can also use a likelihood ratio test because the multinomial logit is a special case of the nested model.

    # First estimate the multinomial logit model
    ml <- update(nl, nests = NULL)
    lrtest(nl, ml)

    ## Likelihood ratio test
    ##
    ## Model 1: depvar ~ ich + och + icca + occa + inc.room + inc.cooling + int.cooling |
    ##     0
    ## Model 2: depvar ~ ich + och + icca + occa + inc.room + inc.cooling + int.cooling |
    ##     0
    ##   #Df  LogLik Df  Chisq Pr(>Chisq)
    ## 1   8 -178.12
    ## 2   7 -180.29 -1 4.3234    0.03759 *
    ## ---
    ## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

> Note that the hypothesis is rejected at 95% confidence, but not at 99% confidence.

2. Re-estimate the model with the room alternatives in one nest and the central alternatives in another nest. (Note that a heat pump is a central system.)

    nl2 <- update(nl, nests = list(central = c('ec', 'ecc', 'gc', 'gcc', 'hpc'),
                        room = c('er', 'erc')))
    summary(nl2)

    ##
    ## Call:
    ## mlogit(formula = depvar ~ ich + och + icca + occa + inc.room +
    ##     inc.cooling + int.cooling | 0, data = HC, nests = list(central = c("ec",
    ##     "ecc", "gc", "gcc", "hpc"), room = c("er", "erc")), un.nest.el = TRUE)
    ##
    ## Frequencies of alternatives:choice
    ##    ec   ecc    er   erc    gc   gcc   hpc
    ## 0.004 0.016 0.032 0.004 0.096 0.744 0.104
    ##
    ## bfgs method
    ## 10 iterations, 0h:0m:0s
    ## g'(-H)^-1g = 5.87E-07
    ## gradient close to zero
    ##
    ## Coefficients :
    ##              Estimate Std. Error z-value Pr(>|z|)
    ## ich          -1.13818    0.54216 -2.0993  0.03579 *
    ## och          -1.82532    0.93228 -1.9579  0.05024 .
    ## icca         -0.33746    0.26934 -1.2529  0.21024
    ## occa         -2.06328    1.89726 -1.0875  0.27681
    ## inc.room     -0.75722    0.34292 -2.2081  0.02723 *
    ## inc.cooling   0.41689    0.20742  2.0099  0.04444 *
    ## int.cooling -13.82487    7.94031 -1.7411  0.08167 .
    ## iv            1.36201    0.65393  2.0828  0.03727 *
    ## ---
    ## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
    ##
    ## Log-Likelihood: -180.02

1. What does the estimate imply about the substitution patterns across alternatives? Do you think the estimate is plausible?

> The log-sum coefficient is over 1. This implies that there is more substitution across nests than within nests. I don’t think this is very reasonable, but people can differ on their concepts of what’s reasonable.

2. Is the log-sum coefficient significantly different from 1?

\begin{answer}[5] The t-statistic is :

     (coef(nl2)['iv'] - 1) / sqrt(vcov(nl2)['iv', 'iv'])

    ##        iv
    ## 0.5535849

    lrtest(nl2, ml)

    ## Likelihood ratio test
    ##
    ## Model 1: depvar ~ ich + och + icca + occa + inc.room + inc.cooling + int.cooling |
    ##     0
    ## Model 2: depvar ~ ich + och + icca + occa + inc.room + inc.cooling + int.cooling |
    ##     0
    ##   #Df  LogLik Df  Chisq Pr(>Chisq)
    ## 1   8 -180.02
    ## 2   7 -180.29 -1 0.5268      0.468

> We cannot reject the hypothesis at standard confidence levels.

3. How does the value of the log-likelihood function compare for this model relative to the model in exercise 1, where the cooling alternatives are in one nest and the heating alternatives in the other nest.

    logLik(nl)

    ## 'log Lik.' -178.1247 (df=8)

    logLik(nl2)

    ## 'log Lik.' -180.0231 (df=8)

> The \(\ln L\) is worse (more negative.) All in all, this seems like a less appropriate nesting structure.

3. Rerun the model that has the cooling alternatives in one nest and the non-cooling alternatives in the other nest (like for exercise 1), with a separate log-sum coefficient for each nest.

    nl3 <- update(nl, un.nest.el = FALSE)

1. Which nest is estimated to have the higher correlation in unobserved factors? Can you think of a real-world reason for this nest to have a higher correlation?

> The correlation in the cooling nest is around 1-0.60 = 0.4 and that for the non-cooling nest is around 1-0.45 = 0.55. So the correlation is higher in the non-cooling nest. Perhaps more variation in comfort when there is no cooling. This variation in comfort is the same for all the non-cooling alternatives.

2. Are the two log-sum coefficients significantly different from each other? That is, can you reject the hypothesis that the model in exercise 1 is the true model?

> We can use a likelihood ratio tests with models `nl` and `nl3`.

    lrtest(nl, nl3)

    ## Likelihood ratio test
    ##
    ## Model 1: depvar ~ ich + och + icca + occa + inc.room + inc.cooling + int.cooling |
    ##     0
    ## Model 2: depvar ~ ich + och + icca + occa + inc.room + inc.cooling + int.cooling |
    ##     0
    ##   #Df  LogLik Df  Chisq Pr(>Chisq)
    ## 1   8 -178.12
    ## 2   9 -178.04  1 0.1758      0.675

> The restricted model is the one from exercise 1 that has one log-sum coefficient. The unrestricted model is the one we just estimated. The test statistics is 0.6299. The critical value of chi-squared with 1 degree of freedom is 3.8 at the 95% confidence level. We therefore cannot reject the hypothesis that the two nests have the same log-sum coefficient.

4. Rewrite the code to allow three nests. For simplicity, estimate only one log-sum coefficient which is applied to all three nests. Estimate a model with alternatives `gcc`, `ecc` and `erc` in a nest, `hpc` in a nest alone, and alternatives `gc`, `ec` and `er` in a nest. Does this model seem better or worse than the model in exercise 1, which puts alternative `hpc` in the same nest as alternatives `gcc`, `ecc` and `erc`?

    nl4 <- update(nl, nests=list(n1 = c('gcc', 'ecc', 'erc'), n2 = c('hpc'),
                        n3 = c('gc', 'ec', 'er')))
    summary(nl4)

    ##
    ## Call:
    ## mlogit(formula = depvar ~ ich + och + icca + occa + inc.room +
    ##     inc.cooling + int.cooling | 0, data = HC, nests = list(n1 = c("gcc",
    ##     "ecc", "erc"), n2 = c("hpc"), n3 = c("gc", "ec", "er")),
    ##     un.nest.el = TRUE)
    ##
    ## Frequencies of alternatives:choice
    ##    ec   ecc    er   erc    gc   gcc   hpc
    ## 0.004 0.016 0.032 0.004 0.096 0.744 0.104
    ##
    ## bfgs method
    ## 8 iterations, 0h:0m:0s
    ## g'(-H)^-1g = 3.71E-08
    ## gradient close to zero
    ##
    ## Coefficients :
    ##               Estimate Std. Error z-value  Pr(>|z|)
    ## ich          -0.838394   0.100546 -8.3384 < 2.2e-16 ***
    ## och          -1.331598   0.252069 -5.2827 1.273e-07 ***
    ## icca         -0.256131   0.145564 -1.7596   0.07848 .
    ## occa         -1.405656   1.207281 -1.1643   0.24430
    ## inc.room     -0.571352   0.077950 -7.3297 2.307e-13 ***
    ## inc.cooling   0.311355   0.056357  5.5247 3.301e-08 ***
    ## int.cooling -10.413384   5.612445 -1.8554   0.06354 .
    ## iv            0.956544   0.180722  5.2929 1.204e-07 ***
    ## ---
    ## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
    ##
    ## Log-Likelihood: -180.26

> The \(\ln L\) for this model is \(-180.26\), which is lower (more negative) than for the model with two nests, which got \(-178.12\).