混合模型中的参数数量

机器算法验证 r 混合模式 aic lme4-nlme
2022-03-22 09:58:42

如何判断混合模型中的随机效应将估计多少个参数？这是 nlme 中 lme 函数的示例：

library(nlme)
fm1 <- lme(fixed = distance ~ age, random = ~ 1 | Subject, data = Orthodont, method = "ML", control=lmeControl(opt = "optim")) 
fm2 <- lme(fixed = distance ~ age + Sex, random = ~ 1 | Subject, data = Orthodont, method = "ML", control=lmeControl(opt = "optim")) 
fm3 <- lme(fixed = distance ~ age, random = ~ age | Subject, data = Orthodont, method = "ML", control=lmeControl(opt = "optim")) 
fm4 <- lme(fixed = distance ~ age + Sex, random = ~ age | Subject, data = Orthodont, method = "ML", control=lmeControl(opt = "optim")) 
fm5 <- lme(fixed = distance ~ age + Sex, random = ~ age + Sex | Subject, data = Orthodont, method = "ML", control=lmeControl(opt = "optim"))

library(AICcmodavg)
aictab(list(fm1,fm2,fm3,fm4,fm5), c("fm1","fm2","fm3","fm4","fm5"))

有人可以帮我理解为什么 K 是 fm1-fm5 的意思吗？

     K   AICc Delta_AICc AICcWt Cum.Wt      LL
fm2  5 445.44       0.00   0.72   0.72 -217.43
fm4  7 447.96       2.51   0.21   0.93 -216.42
fm1  4 451.78       6.33   0.03   0.96 -221.69
fm3  6 452.04       6.60   0.03   0.99 -219.61
fm5 10 453.44       7.99   0.01   1.00 -215.58

特别是为什么fm5比fm4多了3个参数？

1个回答
  • 调频1

fixed = distance ~ age, random = ~ 1 | Subject

K = 4：截距系数，和age; 随机截距的方差和误差项。

  • FM2

fixed = distance ~ age + Sex, random = ~ 1 | Subject

K = 5：截距、age和的系数Sex；随机截距的方差和误差项。

  • 调频3

fixed = distance ~ age, random = ~ age | Subject

K = 6：截距系数，和age; 随机截距、age和误差项的方差；截距的随机效应和 之间的协方差age。

  • 调频4

fixed = distance ~ age + Sex, random = ~ age | Subject

K = 7：截距、age和的系数Sex；随机截距、age和误差项的方差；截距的随机效应和 之间的协方差age。

  • 调频5

fixed = distance ~ age + Sex, random = ~ age + Sex | Subject

K = 10：截距、age和的系数Sex；随机截距、age、Sex和误差项的方差；age截距和、截距和Sex、和age的随机效应之间的协方差Sex。

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