我在R中估计了一个带有mhurdle包的障碍模型，我在解释结果时遇到了一些困难。我想计算边际效应，但我不知道如何计算mhurdle估计后的边际效应。 例如，下面的代码
library("mhurdle")
data("Comics", package = "mhurdle")
Comics$incu <- with(Comics, income / cu)
Comics$incum <- with(Comics, incu / mean(incu))
m111dii <- mhurdle(comics ~ gender + educ | log(incum) + I(log(incum)^2) + I(log(incum)^3) + size | age, data = Comics, corr = "dii", dist = "n", method = 'bfgs')
我们得到以下结果，如包的作者所示
summary(m111dii)
Call:
mhurdle(formula = comics ~ gender + educ | log(incum) + I(log(incum)^2) +
I(log(incum)^3) + size | age, data = Comics, dist = "n",
corr = "dii", method = "bfgs")
Frequency of 0: 0.78287
BFGS maximisation method
117 iterations, 0h:0m:31s
g'(-H)^-1g = 4.48E-05
Coefficients :
Estimate Std. Error t-value Pr(>|t|)
h1.(Intercept) -1.755587 0.229749 -7.6413 2.154e-14 ***
h1.genderfemale -0.547607 0.130028 -4.2115 2.537e-05 ***
h1.educ 0.188445 0.014849 12.6904 < 2.2e-16 ***
h2.(Intercept) -40.533234 3.485117 -11.6304 < 2.2e-16 ***
h2.log(incum) 14.633088 2.897881 5.0496 4.428e-07 ***
h2.I(log(incum)^2) -3.777712 2.648899 -1.4261 0.153827
h2.I(log(incum)^3) -4.457532 1.701937 -2.6191 0.008816 **
h2.size 5.253440 0.952565 5.5150 3.487e-08 ***
h3.(Intercept) 11.318180 2.048510 5.5251 3.293e-08 ***
h3.age -0.149555 0.026846 -5.5709 2.534e-08 ***
sd 55.212046 1.343068 41.1089 < 2.2e-16 ***
corr12 -0.283790 0.048753 -5.8210 5.850e-09 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Log-Likelihood: -7322.1 on 12 Df
R^2 :
Coefficient of determination : 0.0048516
Likelihood ratio index : 0.032983
我的问题是如何解释结果以及如何计算边际效应。