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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。例如在下面的代码中

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')

如包作者所示，我们得到以下结果

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。

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