R Probit回归边际效应 [英] R probit regression marginal effects

问题描述

我正在使用R复制研究并获得与 作者报道.但是，在某一时刻，我计算出的边际效应似乎很小.如果您能看一下我的推理和下面的代码，看看我是否在某一点或另一点上被误解，将不胜感激.

I am using R to replicate a study and obtain mostly the same results the author reported. At one point, however, I calculate marginal effects that seem to be unrealistically small. I would greatly appreciate if you could have a look at my reasoning and the code below and see if I am mistaken at one point or another.

我的样本包含24535个观测值，因变量"x028bin"是一个 二进制变量取值为0和1，此外还有10 解释变量.这些自变量中有9个具有数字水平，自变量"f025grouped"是一个由不同宗教派别组成的因素.

My sample contains 24535 observations, the dependent variable "x028bin" is a binary variable taking on the values 0 and 1, and there are furthermore 10 explaining variables. Nine of those independent variables have numeric levels, the independent variable "f025grouped" is a factor consisting of different religious denominations.

我想进行概率回归，其中包括针对宗教教派的假人，然后计算边际效应.为了做到这一点，我首先消除了缺失的值，并使用因变量和自变量之间的交叉表来验证没有小单元格或0单元格.然后，我运行了运行良好的概率模型，并且也获得了合理的结果:

I would like to run a probit regression including dummies for religious denomination and then compute marginal effects. In order to do so, I first eliminate missing values and use cross-tabs between the dependent and independent variables to verify that there are no small or 0 cells. Then I run the probit model which works fine and I also obtain reasonable results:

probit4AKIE <- glm(x028bin ~ x003 + x003squ + x025secv2 + x025terv2 + x007bin + x04chief + x011rec + a009bin + x045mod + c001bin + f025grouped, family=binomial(link="probit"), data=wvshm5red2delna, na.action=na.pass)

summary(probit4AKIE)

但是，当根据概率系数和比例因子以所有变量计算均值的边际效应时，我获得的边际效应太小(例如2.6042e-78). 代码如下:

However, when calculating marginal effects with all variables at their means from the probit coefficients and a scale factor, the marginal effects I obtain are much too small (e.g. 2.6042e-78). The code looks like this:

ttt <- cbind(wvshm5red2delna$x003,
wvshm5red2delna$x003squ,
wvshm5red2delna$x025secv2,
wvshm5red2delna$x025terv2,
wvshm5red2delna$x007bin,
wvshm5red2delna$x04chief,
wvshm5red2delna$x011rec,
wvshm5red2delna$a009bin,
wvshm5red2delna$x045mod,
wvshm5red2delna$c001bin,
wvshm5red2delna$f025grouped,
wvshm5red2delna$f025grouped,
wvshm5red2delna$f025grouped,
wvshm5red2delna$f025grouped,
wvshm5red2delna$f025grouped,
wvshm5red2delna$f025grouped,
wvshm5red2delna$f025grouped,
wvshm5red2delna$f025grouped,
wvshm5red2delna$f025grouped) #I put variable "f025grouped" 9 times because this variable consists of 9 levels

ttt <- as.data.frame(ttt)

xbar <- as.matrix(mean(cbind(1,ttt[1:19]))) #1:19 position of variables in dataframe ttt

betaprobit4AKIE <- probit4AKIE$coefficients

zxbar <- t(xbar) %*% betaprobit4AKIE

scalefactor <- dnorm(zxbar)

marginprobit4AKIE <- scalefactor * betaprobit4AKIE[2:20] #2:20 are the positions of variables in the output of the probit model 'probit4AKIE' (variables need to be in the same ordering as in data.frame ttt), the constant in the model occupies the first position

marginprobit4AKIE #in this step I obtain values that are much too small

很抱歉，我无法为您提供有效的示例，因为我的数据集是 太大了.任何评论将不胜感激.非常感谢.

I apologize that I can not provide you with a working example as my dataset is much too large. Any comment would be greatly appreciated. Thanks a lot.

最好

Tobias

推荐答案

@Gavin是正确的，最好在姊妹站点询问.

@Gavin is right and it's better to ask at the sister site.

无论如何，这是我解释概率系数的窍门.

In any case, here's my trick to interpret probit coefficients.

概率回归系数与logit系数相同，最大比例为(1.6).因此，如果概率模型的拟合度为Pr(y=1) = fi(.5 - .3*x)，则等效于逻辑模型Pr(y=1) = invlogit(1.6(.5 - .3*x)).

The probit regression coefficients are the same as the logit coefficients, up to a scale (1.6). So, if the fit of a probit model is Pr(y=1) = fi(.5 - .3*x), this is equivalent to the logistic model Pr(y=1) = invlogit(1.6(.5 - .3*x)).

然后我使用软件包arm的invlogit函数来制作图形.另一种可能性是将所有系数(包括截距)乘以1.6，然后应用"4除法"(请参见Gelman和Hill的书)，即将新系数除以4，您会发现预测差异的上限，对应于x的单位差异.

And I use this to make a graphic, using the function invlogit of package arm. Another possibility is just to multiply all coefficients (including the intercept) by 1.6, and then applying the 'divide by 4 rule' (see the book by Gelman and Hill), i.e, divide the new coefficients by 4, and you will find out an upper bound of the predictive difference corresponding to a unit difference in x.

这是一个例子.

x1 = rbinom(100,1,.5)
x2 = rbinom(100,1,.3)
x3 = rbinom(100,1,.9)
ystar = -.5  + x1 + x2 - x3 + rnorm(100)
y = ifelse(ystar>0,1,0)
probit = glm(y~x1 + x2 + x3, family=binomial(link='probit'))
xbar <- as.matrix(mean(cbind(1,ttt[1:3])))

# now the graphic, i.e., the marginal effect of x1, x2 and x3
library(arm)
curve(invlogit(1.6*(probit$coef[1] + probit$coef[2]*x + probit$coef[3]*xbar[3] + probit$coef[4]*xbar[4]))) #x1
curve(invlogit(1.6*(probit$coef[1] + probit$coef[2]*xbar[2] + probit$coef[3]*x + probit$coef[4]*xbar[4]))) #x2
curve(invlogit(1.6*(probit$coef[1] + probit$coef[2]*xbar[2] + probit$coef[3]*xbar[3] + probit$coef[4]*x))) #x3

这篇关于R Probit回归边际效应的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！