我想将 lmer（真的是 glmer）的输出与玩具二项式示例相匹配。我已经阅读了这些小插曲，并相信我了解发生了什么。

但显然我没有。卡住之后，我根据随机效应确定了“真相”，并单独估计了固定效应。我在下面包含此代码。要查看它是否合法，您可以注释掉+ Z %*% b.k它，它将与常规 glm 的结果相匹配。我希望借用一些脑力来弄清楚为什么在包含随机效应时我无法匹配 lmer 的输出。

# Setup - hard coding simple data set 
df <- data.frame(x1 = rep(c(1:5), 3), subject = sort(rep(c(1:3), 5)))
df$subject <- factor(df$subject)

# True coefficient values  
beta <- matrix(c(-3.3, 1), ncol = 1) # Intercept and slope, respectively 
u <- matrix(c(-.5, .6, .9), ncol = 1) # random effects for the 3 subjects 

# Design matrices Z (random effects) and X (fixed effects)
Z <- model.matrix(~ 0 + factor(subject), data = df)
X <- model.matrix(~ 1 + x1, data = df)

# Response  
df$y <- c(1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 1)
    y <- df$y

### Goal: match estimates from the following lmer output! 
library(lme4)
my.lmer <- lmer( y ~ x1 + (1 | subject), data = df, family = binomial)
summary(my.lmer)
ranef(my.lmer)

### Matching effort STARTS HERE 

beta.k <- matrix(c(-3, 1.5), ncol = 1) # Initial values (close to truth)
b.k <- matrix(c(1.82478, -1.53618, -.5139356), ncol = 1) # lmer's random effects

# Iterative Gauss-Newton algorithm
for (iter in 1:6) {
  lin.pred <- as.numeric(X %*% beta.k +  Z %*% b.k)
  mu.k <- plogis(lin.pred)
  variances <- mu.k * (1 - mu.k)
  W.k <- diag(1/variances)

  y.star <- W.k^(.5) %*% (y - mu.k)
  X.star <- W.k^(.5) %*% (variances * X)
  delta.k <- solve(t(X.star) %*% X.star) %*% t(X.star) %*% y.star

  # Gauss-Newton Update 
  beta.k <- beta.k + delta.k
  cat(iter, "Fixed Effects: ", beta.k, "\n")
}