Newey-West标准误差与Mean Groups/Fama-MacBeth估计

coc*_*mas 20 regression r standard-error

我试图让Newey-West标准错误与包中的pmg()(Mean Groups/Fama-MacBeth估算器)输出一起使用plm.

按照这里的例子:

require(foreign)
require(plm)
require(lmtest)
test <- read.dta("http://www.kellogg.northwestern.edu/faculty/petersen/htm/papers/se/test_data.dta")

fpmg <- pmg(y~x, test, index=c("firmid", "year")) # Time index in second position, unlike the example
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我可以coeftest直接使用,以获得Fama-MacBeth标准错误:

# Regular “Fama-MacBeth” standard errors
coeftest(fpmg)

# t test of coefficients:
#   
#             Estimate Std. Error t value Pr(>|t|)    
# (Intercept) 0.032470   0.071671   0.453   0.6505    
# x           0.969212   0.034782  27.866   <2e-16 ***
# ---
# Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
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但是,尝试使用Newey-West估算器失败了:

# Newey-West standard-errors
coeftest(fpmg, vcov = NeweyWest(fpmg, lag=3))

# Error in UseMethod("estfun") : 
#   no applicable method for 'estfun' applied to an object of class "c('pmg', 'panelmodel')"
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这似乎是plm包装中的一个缺点.你知道一种让这项工作的方法吗?我应该estfunpmg对象编写自己的代码吗?从头开始编码Newey-West估算器?或者我应该plm完全绕过包裹?

Kon*_*nos 2

目前,这对于包来说是不可能的plm

但是,您可以自己创建它们。

假设您有:

fpmg <- pmg(y~x, test, index = c('year', 'firmid'))
fpmg.coefficients <- fpmg$coefficients
# (Intercept)            x 
#  0.03127797   1.03558610 

coeftest(fpmg)
#             Estimate Std. Error t value Pr(>|t|)    
# (Intercept) 0.031278   0.023356  1.3392   0.1806    
# x           1.035586   0.033342 31.0599   <2e-16 ***
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然后您可以简单地自己创建估计器,如下所示:

the.years <- unique(test$year)
a.formula <- y ~ x


first.step <-  lapply(the.years, function(a.year) {
                temp.data <- test[test$year == a.year, ]
                an.lm <- lm(a.formula, data = temp.data)
                the.coefficients <- an.lm$coef
                the.results <- as.data.frame(cbind(a.year, t(the.coefficients)))
                the.results
                }) 

first.step.df <- do.call('rbind', first.step)

second.step.coefficients <- apply(first.step.df[, -1], 2, mean)
second.step.coefficients
# (Intercept)           x 
#  0.03127797  1.03558610 

identical(fpmg.coefficients, second.step.coefficients)
# [1] TRUE
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检查它们是否相同,以防万一。最后,您可以获得 Newey-West (1987),其平均值具有一个滞后调整的 t 统计量:

library(sandwich)
second.step.NW.sigma.sq <- apply(first.step.df[, -1], 2, 
                             function(x) sqrt(NeweyWest(lm(x ~ 1), 
                               lag = 1, prewhite = FALSE)['(Intercept)',       
                                 '(Intercept)']))
second.step.NW.sigma.sq
#  (Intercept)            x 
#   0.02438398   0.02859447
t.statistics.NW.lag.1 <- second.step.coefficients / second.step.NW.sigma.sq

t.statistics.NW.lag.1
# (Intercept)           x 
#    1.282726   36.216301
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更新

在我的回答中,我只包含了 t 统计量的“手动”计算,因为它的计算速度更快。coeftest()更通用的解决方案是使用包的函数计算 Newey-West 校正 t 统计量及其 p 值lmtest

coeftest(lm(first.step.df$'(Intercept)' ~ 1), vcov = NeweyWest(lm(first.step.df$'(Intercept)' ~ 1), lag = 1, prewhite = FALSE))
# t test of coefficients:
#             Estimate Std. Error t value Pr(>|t|)
# (Intercept) 0.031278   0.024384  1.2827   0.2316
coeftest(lm(first.step.df$x ~ 1), vcov = NeweyWest(lm(first.step.df$x ~ 1), lag = 1, prewhite = FALSE))
# t test of coefficients:
#             Estimate Std. Error t value  Pr(>|t|)    
# (Intercept) 1.035586   0.028594  36.216 4.619e-11 ***
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