使用优化使r中的可能性最大化时出错 [英] Error in using optim to maximise the likelihood in r

问题描述

因此，我具有以下功能:

So, I have these functions:

funk1 <- function(a,x,l,r) {
x^2*exp(-(l*(1-exp(-r*a))/r))}

funk2 <- function(x,l,r) {
sapply(x, function (s) {
integrate(funk1, lower = 0, upper = s, x=s, l=l, r=r)$value })}  

用来解释y中的数据

z <- data.frame(ts = 1:100,
            y = funk2(1:100, l = 1, r = 1) + rpois(100, 1:100))

我希望使用优化来最大化可能性，所以我定义了可能性函数:

I wish to use optim to maximise the likelihood, so I defined a likelihood function:

LL_funk <- function(l,r) { 
n=nrow(z)
R = sum((funk2(ts,l,r) - y)^2)
logl = -((n/2)*log(R))
return(-logl)
} 

我尝试使用优化来适应

fit <- optim(par=c(0.5,0.5), fn= LL_funk, method="Nelder-Mead")

但是我得到一个错误:

 Error in integrate(funk1, lower = 0, upper = s, x = s, l = l, r = r) : 
 a limit is missing 

我不确定为什么吗?我可以运行将funk2(x，l，r)拟合为y的nls

I am not sure why? I could run nls fitting funk2(x,l,r) to y

nls(y ~ funk2(ts,l,r), data = z, start = list(l = 0.5, r = 0.5))

这意味着funk2正在运行.我想这是我设计的LL功能的问题，我无法弄清楚！！请帮忙！

That means funk2 is working. I guess its the problem with LL function that I have designed, which I cant figure out!! Please Help!

推荐答案

是的！您的功能存在两个问题.这对我有用:

Yup! There were two problems with your function. This worked for me:

LL_funk <- function(params) { 
  n=nrow(z)
  l = params[1]
  r = params[2]
  R = sum((funk2(z$ts,l,r) - z$y)^2)
  logl = -((n/2)*log(R))
  return(-logl)
}

以前的问题:

• LL_funk仅接受1个参数，这是参数的向量.
• 在分配R的LHS中，ts和y实际上没有引用数据集中的列.

• LL_funk only takes 1 argument, which is the vector of parameters.
• In LHS of the assignment of R, ts and y were not actually referring to columns in your dataset.

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