我想用gamma函数代替指数函数来计算R中基于95%似然的置信区间

Question

我想替换指数对数似然

explik <- function(x) {

  sum( dexp(data, 1/x, log=TRUE))
}

具有伽马负对数似然

gammalike <- function(x) {
  
  -sum(dgamma(data, shape=x, scale=2, log=TRUE))
}

在以下函数和脚本中

likeqn <- function(mu) {
  
  muhat <- mean(data)
  maxlik <- explik(muhat)
  explik(mu) - maxlik + 0.5 * 3.841
}

source("explik.R")
source("likeqn.R")


# Input data
data <- c(63, 130, 88, 120, 330, 188, 270, 222, 189, 116)

# Find the MLE of an exponential distribution
muhat <- mean(data)

# Solve likelihood equation numerically to find likelihood based CI
vlikeqn <- Vectorize(likeqn)
CIlower <- uniroot(vlikeqn, c(10,muhat) )$root
CIupper <- uniroot(vlikeqn, c(muhat,500) )$root
cbind(CIlower, muhat, CIupper)

# Plot log-likelihood function
x <- seq(80, 350)
vexplik <- Vectorize(explik)
plot(x, vexplik(x), type="l", xlab=expression(mu), ylab="log-likelihood")

我曾尝试替换log-lik函数，但结果似乎并不合理。

Answer 1

具有速率λ的分布博览会( lambda )是Gamma(1，1/lambda) (形状-比例)。

gammalike <- function(x) {
  sum(dgamma(data, shape=1, scale=x, log=TRUE))
}