R - 计算组合分布的MLE

Question

我有一个具有1000个值的给定数据集，它是两个正态分布N（y1,1）和N（y2,1）的组合。密度如下所示：

我想计算数据集中N（y1,1）和N（y2,1）的部分，两者意味着y1和y2。这是我目前的做法：

z <- #Dataset as vector with 1000 entries#
lik <- function(mu1, mu2, part) -sum(part*dnorm(z, mu1, 1, log=TRUE) + (1-part)*dnorm(z, mu2, 1, log=TRUE))
mle <- mle(lik, start=list(mu1=-7, mu2=5, part=0.33))

但是这给了我以下错误消息：

Error in solve.default(oout$hessian) : 
    Lapack routine dgesv: system is exactly singular: U[1,1] = 0

Answer 1

我重新定义了使用log()代替参数log = TRUE的可能性。

奇怪的是，尽管有警告，但以下情况仍有效。请注意，它们是警告，而不是错误。

library(stats4)

set.seed(7850)    # Make the results reproducible
z <- sample(c(rnorm(333, -7, 1), rnorm(667, 5, 1)))

plot(density(z))

lik2 <- function(mu1, mu2, part) -sum(log(part*dnorm(z, mu1, 1) + (1-part)*dnorm(z, mu2, 1)))
mle2 <- mle(lik2, start = list(mu1 = -6, mu2 = 6, part = 1/2))
#Warning messages:
#1: In log(part * dnorm(z, mu1, 1) + (1 - part) * dnorm(z, mu2, 1)) :
#  NaNs produced
#2: In log(part * dnorm(z, mu1, 1) + (1 - part) * dnorm(z, mu2, 1)) :
#  NaNs produced
#3: In log(part * dnorm(z, mu1, 1) + (1 - part) * dnorm(z, mu2, 1)) :
#  NaNs produced
#4: In log(part * dnorm(z, mu1, 1) + (1 - part) * dnorm(z, mu2, 1)) :
#  NaNs produced

mle2
#
#Call:
#mle(minuslogl = lik2, start = list(mu1 = -6, mu2 = 6, part = 1/2))
#
#Coefficients:
#       mu1        mu2       part 
#-7.1091780  4.9377339  0.3330038