使用"bife"软件包的固定效果logit模型的拟合优度 [英] Goodness-of-fit for fixed effect logit model using 'bife' package
问题描述
我正在使用'bife'软件包在R中运行固定效果logit模型.但是，鉴于下面的结果，我无法计算任何拟合优度来测量模型的整体拟合.如果能在有限的信息下知道如何衡量拟合优度，我将不胜感激.我更喜欢卡方检验，但仍然找不到实现此方法的方法.
---------------------------------------------------------------
Fixed effects logit model
with analytical bias-correction
Estimated model:
Y ~ X1 +X2 + X3 + X4 + X5 | Z
Log-Likelihood= -9153.165
n= 20383, number of events= 5104
Demeaning converged after 6 iteration(s)
Offset converged after 3 iteration(s)
Corrected structural parameter(s):
Estimate Std. error t-value Pr(> t)
X1 -8.67E-02 2.80E-03 -31.001 < 2e-16 ***
X2 1.79E+00 8.49E-02 21.084 < 2e-16 ***
X3 -1.14E-01 1.91E-02 -5.982 2.24E-09 ***
X4 -2.41E-04 2.37E-05 -10.171 < 2e-16 ***
X5 1.24E-01 3.33E-03 37.37 < 2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
AIC= 18730.33 , BIC= 20409.89
Average individual fixed effects= 1.6716
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让DGP成为
n <- 1000
x <- rnorm(n)
id <- rep(1:2, each = n / 2)
y <- 1 * (rnorm(n) > 0)
，因此我们将处于零假设下.就像在?bife中所说的那样，当没有偏差校正时，除了速度以外，其他都与glm相同.因此，让我们从glm开始.
modGLM <- glm(y ~ 1 + x + factor(id), family = binomial())
modGLM0 <- glm(y ~ 1, family = binomial())
执行LR测试的一种方法是
library(lmtest)
lrtest(modGLM0, modGLM)
# Likelihood ratio test
#
# Model 1: y ~ 1
# Model 2: y ~ 1 + x + factor(id)
# #Df LogLik Df Chisq Pr(>Chisq)
# 1 1 -692.70
# 2 3 -692.29 2 0.8063 0.6682
但是我们也可以手动完成
1 - pchisq(c((-2 * logLik(modGLM0)) - (-2 * logLik(modGLM))),
modGLM0$df.residual - modGLM$df.residual)
# [1] 0.6682207
现在让我们继续bife.
library(bife)
modBife <- bife(y ~ x | id)
modBife0 <- bife(y ~ 1 | id)
此处modBife是完整规格，而modBife0仅具有固定效果.为了方便起见，
logLik.bife <- function(object, ...) object$logl_info$loglik
用于对数似然提取.然后我们可以将modBife0与modBife进行比较，如
1 - pchisq((-2 * logLik(modBife0)) - (-2 * logLik(modBife)), length(modBife$par$beta))
# [1] 1
而modGLM0和modBife可以通过运行进行比较
1 - pchisq(c((-2 * logLik(modGLM0)) - (-2 * logLik(modBife))),
length(modBife$par$beta) + length(unique(id)) - 1)
# [1] 0.6682207
它提供与以前相同的结果，即使使用bife我们默认情况下也具有偏差校正.
最后，作为奖励，我们可以模拟数据，并按预期进行测试.下面的1000次迭代表明，这两个测试(因为两个测试相同)确实确实拒绝了空值.
I am using the 'bife' package to run the fixed effect logit model in R. However, I cannot compute any goodness-of-fit to measure the model's overall fit given the result I have below. I would appreciate if I can know how to measure the goodness-of-fit given this limited information. I prefer chi-square test but still cannot find a way to implement this either.
---------------------------------------------------------------
Fixed effects logit model
with analytical bias-correction
Estimated model:
Y ~ X1 +X2 + X3 + X4 + X5 | Z
Log-Likelihood= -9153.165
n= 20383, number of events= 5104
Demeaning converged after 6 iteration(s)
Offset converged after 3 iteration(s)
Corrected structural parameter(s):
Estimate Std. error t-value Pr(> t)
X1 -8.67E-02 2.80E-03 -31.001 < 2e-16 ***
X2 1.79E+00 8.49E-02 21.084 < 2e-16 ***
X3 -1.14E-01 1.91E-02 -5.982 2.24E-09 ***
X4 -2.41E-04 2.37E-05 -10.171 < 2e-16 ***
X5 1.24E-01 3.33E-03 37.37 < 2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
AIC= 18730.33 , BIC= 20409.89
Average individual fixed effects= 1.6716
---------------------------------------------------------------
Let the DGP be
n <- 1000
x <- rnorm(n)
id <- rep(1:2, each = n / 2)
y <- 1 * (rnorm(n) > 0)
so that we will be under the null hypothesis. As it says in ?bife, when there is no bias-correction, everything is the same as with glm, except for the speed. So let's start with glm.
modGLM <- glm(y ~ 1 + x + factor(id), family = binomial())
modGLM0 <- glm(y ~ 1, family = binomial())
One way to perform the LR test is with
library(lmtest)
lrtest(modGLM0, modGLM)
# Likelihood ratio test
#
# Model 1: y ~ 1
# Model 2: y ~ 1 + x + factor(id)
# #Df LogLik Df Chisq Pr(>Chisq)
# 1 1 -692.70
# 2 3 -692.29 2 0.8063 0.6682
But we may also do it manually,
1 - pchisq(c((-2 * logLik(modGLM0)) - (-2 * logLik(modGLM))),
modGLM0$df.residual - modGLM$df.residual)
# [1] 0.6682207
Now let's proceed with bife.
library(bife)
modBife <- bife(y ~ x | id)
modBife0 <- bife(y ~ 1 | id)
Here modBife is the full specification and modBife0 is only with fixed effects. For convenience, let
logLik.bife <- function(object, ...) object$logl_info$loglik
for loglikelihood extraction. Then we may compare modBife0 with modBife as in
1 - pchisq((-2 * logLik(modBife0)) - (-2 * logLik(modBife)), length(modBife$par$beta))
# [1] 1
while modGLM0 and modBife can be compared by running
1 - pchisq(c((-2 * logLik(modGLM0)) - (-2 * logLik(modBife))),
length(modBife$par$beta) + length(unique(id)) - 1)
# [1] 0.6682207
which gives the same result as before, even though with bife we, by default, have bias correction.
Lastly, as a bonus, we may simulate data and see it the test works as it's supposed to. 1000 iterations below show that both test (since two tests are the same) indeed reject as often as they are supposed to under the null.
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