r - 进行 Tobit 回归时的奇异性错误

Question

我正在尝试估计一个标准的 tobit 模型，该模型被审查为零。

变量是

因变量 : 幸福

自变量 :

城市(芝加哥，纽约)，

性别(男，女)，

就业(0=失业，1=就业)，

工作类型(失业，蓝色，白色)，

假期(失业，每周1天，每周2天)

“Worktype”和“Holiday”变量与“Employment”变量相互作用。

我正在使用 censReg tobit回归包。

censReg(Happiness ~ City + Gender + Employment:Worktype + Employment:Holiday)

但是 summary()返回以下错误。

Error in printCoefmat(coef(x, logSigma = logSigma), digits = digits) : 
  'x' must be coefficient matrix/data frame

为了找出原因，我运行了 OLS 回归。

有一些 NA 值，我认为这是因为模型设计和变量设置(某些变量似乎有奇点。而具有 'Employment' = 0 的人具有值 'Worktype' = Unemployed 、 'Holidays' = Unemployed 。这可能是原因？ )

lm(Happiness ~ City + Gender + Employment:Worktype + Employment:Holiday)


Coefficients: (2 not defined because of singularities)
                               Estimate Std. Error t value Pr(>|t|)  
(Intercept)                      41.750      9.697   4.305   0.0499 *
CityNew York                    -44.500     11.197  -3.974   0.0579 .
Gender1                           2.750     14.812   0.186   0.8698  
Employment:WorktypeUnemployed        NA         NA      NA       NA  
Employment:WorktypeBluecolor     35.000     17.704   1.977   0.1867  
Employment:WorktypeWhitecolor   102.750     14.812   6.937   0.0202 *
Employment:Holiday1 day a week  -70.000     22.394  -3.126   0.0889 .
Employment:Holiday2 day a week       NA         NA      NA       NA 

我怎样才能忽略 NA 值并无错误地运行 tobit 回归？

下面是可重现的代码。

Happiness <- c(0, 80, 39, 0, 69, 90, 100, 30)

 City <- as.factor(c("New York", "Chicago", "Chicago", "New York", "Chicago", 
"Chicago", "New York", "New York"))
 Gender <- as.factor(c(0, 1, 0, 1, 1, 1, 0, 1)) # 0 = man, 1 = woman.
 Employment <- c(0,1, 0, 0, 1 ,1 , 1 , 1) # 0 = unemployed, 1 = employed.
 Worktype <- as.factor(c(0, 2, 0, 0, 1, 1, 2,2))
 levels(Worktype) <- c("Unemployed", "Bluecolor", "Whitecolor")
 Holiday <- as.factor(c(0, 1, 0, 0, 2, 2, 2, 1))
 levels(Holiday) <- c("Unemployed", "1 day a week", "2 day a week")

 data <- data.frame(Happiness, City, Gender, Employment, Worktype, Holiday)
 reg <- lm(Happiness ~ City + Gender + Employment:Worktype +      
           Employment:Holiday)
 summary(reg)

 install.packages("censReg")
 library(censReg)
 tobitreg <- censReg(Happiness ~ City + Gender + Employment:Worktype +      
                     Employment:Holiday)
 summary(tobitreg)

Answer 1

如果您逐步调试对 censReg 的调用，则会达到以下 maxLik 优化:

result <- maxLik(censRegLogLikCross, start = start, 
      yVec = yVec, xMat = xMat, left = left, right = right, 
      obsBelow = obsBelow, obsBetween = obsBetween, obsAbove = obsAbove, 
      ...)

初始条件向量 start这是使用 OLS 回归确定的包含 NA正如您已经发现的那样，对于两个 coefs:

就业:工作类型失业

就业:每周放假2天

这会导致 maxLik返回NULL，并带有错误消息:

Return code 100: Initial value out of range.

summary函数得到这个 NULL这解释了您得到的最终错误消息。
要覆盖它，您可以设置 start范围 :

tobitreg <- censReg(formula = Happiness ~ City + Gender + Employment:Worktype +      
                      Employment:Holiday, start = rep(0,9) )
summary(tobitreg)

Call:
censReg(formula = Happiness ~ City + Gender + Employment:Worktype + 
    Employment:Holiday, start = rep(0, 9))

Observations:
         Total  Left-censored     Uncensored Right-censored 
             8              2              6              0 

Coefficients:
                               Estimate Std. error t value Pr(> t)
(Intercept)                      38.666        Inf       0       1
CityNew York                    -50.669        Inf       0       1
Gender1                        -360.633        Inf       0       1
Employment:WorktypeUnemployed     0.000        Inf       0       1
Employment:WorktypeBluecolor    345.674        Inf       0       1
Employment:WorktypeWhitecolor    56.210        Inf       0       1
Employment:Holiday1 day a week  346.091        Inf       0       1
Employment:Holiday2 day a week   55.793        Inf       0       1
logSigma                          1.794        Inf       0       1

Newton-Raphson maximisation, 141 iterations
Return code 1: gradient close to zero
Log-likelihood: -19.35431 on 9 Df

即使错误消息消失了，结果也不可靠:

错误 = 信息

梯度接近 0 :没有最优值，解决方案是超平面

回归中的 NA 系数表明这些系数与其他系数线性相关，因此您需要删除其中的一些以获得唯一解。
正如您所怀疑的，这是因为您只有 Employement = 0当 worktype = Unemployed ，因此模型无法估计 Employment:WorktypeUnemployed 的系数.与 Employment:Holiday 相同的问题系数。
所以我担心您正在评估的回归模型没有单一的最佳解决方案。
如果你摆脱了链接的变量，这有效:

tobitreg <- censReg(formula = Happiness ~ City + Gender + Employment )
summary(tobitreg)
Call:
censReg(formula = Happiness ~ City + Gender + Employment)

Observations:
         Total  Left-censored     Uncensored Right-censored 
             8              2              6              0 

Coefficients:
             Estimate Std. error t value  Pr(> t)    
(Intercept)   38.6141     5.7188   6.752 1.46e-11 ***
CityNew York -50.1813     6.4885  -7.734 1.04e-14 ***
Gender1      -70.3859     8.2943  -8.486  < 2e-16 ***
Employment   111.5672    10.0927  11.054  < 2e-16 ***
logSigma       1.7930     0.2837   6.320 2.61e-10 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Newton-Raphson maximisation, 8 iterations
Return code 1: gradient close to zero
Log-likelihood: -19.36113 on 5 Df