使用R的加权调查数据中logit的边际效应

时间:2016-07-21 00:02:03

标签: r survey marginal-effects

我试图估计一个logit模型的边际效应,其中我有几个二分解释变量。

让我们说通过

估算的模型
logit<- svyglm ( if_member ~ if_female + dummy_agegroup_2 + dummy_agegroup_3 + dummy_education_2 + dummy_education_3 + dummy_education_4, family = quasibinomial(link = "logit"), design = survey_design)

我知道调查包中的 marginpred 功能,但我不是很熟悉它。我在模型中只有二分类变量,所以我想知道如何通过这个函数来估计边际效应,特别是我不确定 predictat (一个数据框给出的值模型中的变量用于预测。)

1 个答案:

答案 0 :(得分:1)

您是在寻找边际效果还是边缘预测

顾名思义,marginpred()函数返回预测predictat的参数是一个包含控制变量和模型中变量的数据框。我要强调一点:控制变量应该被排除在模型之外。

library("survey")

odds2prob <- function(x) x / (x + 1)
prob2odds <- function(x) x / (1 - x)
expit <- function(x) odds2prob(exp(x))
logit <- function(x) log(prob2odds(x))

set.seed(1)

survey_data <- data.frame(
  if_female = rbinom(n = 100, size = 1, prob = 0.5), 
  agegroup = factor(sample(x = 1:3, size = 100, replace = TRUE)), 
  education = NA_integer_,
  if_member = NA_integer_)
survey_data["agegroup"] <- relevel(survey_data$agegroup, ref = 3)
# Different probabilities between female and male persons
survey_data[survey_data$if_female == 0, "education"] <- sample(
  x = 1:4, 
  size = sum(survey_data$if_female == 0), 
  replace = TRUE, 
  prob = c(0.1, 0.1, 0.5, 0.3))
survey_data[survey_data$if_female == 1, "education"] <-sample(
  x = 1:4, 
  size = sum(survey_data$if_female == 1), 
  replace = TRUE, 
  prob = c(0.1, 0.1, 0.3, 0.5))
survey_data["if_member"] <- rbinom(n = 100, size = 1, prob = 
                                     expit((survey_data$education - 3)/2))
survey_data["education"] <- factor(survey_data$education)
survey_data["education"] <- relevel(survey_data$education, ref = 3)
survey_design <- svydesign(ids = ~ 1, data = survey_data)

logit <- svyglm(if_member ~ if_female + agegroup + education, 
                family = quasibinomial(link = "logit"), 
                design = survey_design)
exp(cbind(`odds ratio` = coef(logit), confint(logit)))
newdf <- data.frame(if_female = 0:1, education = c(3, 3), agegroup =  = c(3, 3))
# Fails
mp <- marginpred(model = logit, adjustfor = ~ agegroup + education, 
                 predictat = newdf, se = TRUE, type = "response")
logit2 <- svyglm(if_member ~ if_female, 
                family = quasibinomial(link = "logit"), 
                design = survey_design)
mp <- marginpred(model = logit2, adjustfor = ~ agegroup + education, 
                 predictat = newdf, se = TRUE, type = "response")
# Probability for male and for female persons controlling for agegroup and education
cbind(prob = mp, confint(mp))

我用survey包评估边际效果

# Probability difference between female and male persons
# when agegroup and education are set to 3
svycontrast(full_model, quote(
  (exp(`(Intercept)` + if_female) / (exp(`(Intercept)` + if_female) + 1)) - 
  (exp(`(Intercept)`) / (exp(`(Intercept)`) + 1))))
# Can't use custom functions like expit :_(

可能有更聪明的方法,但我希望它有所帮助。

请注意,marginpred()预测的概率之间的差异与svycontrast()估算的差异不同。 marginpred()预测的概率似乎不受改变控制变量值的影响(例如, education = c(4, 4)代替education = c(3, 3)),但svycontrast()的估算值会受到回归模型隐含的影响。