r - R 中 optim() 的优化(L-BFGS-B 需要 'fn' 的有限值)

Question

我在 R 中使用 optim() 来解决涉及积分的可能性时遇到了一些问题。我收到一条错误消息，上面写着“optim(par = c(0.1, 0.1), LLL, method = "L-BFGS-B", lower = c(0, : L-BFGS-B needs finite values of 'fn '"。下面是我的代码:

s1=c(1384,1,1219,1597,2106,145,87,1535,290,1752,265,588,1188,160,745,237,479,39,99,56,1503,158,916,651,1064,166,635,19,553,51,79,155,85,1196,142,108,325  
     ,135,28,422,1032,1018,128,787,1704,307,854,6,896,902)


LLL=function (par) {

  integrand1 <- function(x){ (x-s1[i]+1)*dgamma(x, shape=par[1], rate=par[2]) }
  integrand2 <- function(x){ (-x+s1[i]+1)*dgamma(x, shape=par[1],rate=par[2]) }



  likelihood = vector() 

  for(i in 1:length(s1)) {likelihood[i] = 
    log( integrate(integrand1,lower=s1[i]-1,upper=s1[i])$value+ integrate(integrand2,lower=s1[i],upper=s1[i]+1)$value )  
  }

  like= -sum(likelihood)
  return(like)

}




optim(par=c(0.1,0.1),LLL,method="L-BFGS-B", lower=c(0,0))

感谢您的帮助。

最好的，

YM

Answer 1

在您提供的参数的下限处评估的目标函数是无穷大。

LLL(c(0,0))
# [1] Inf

这就是 L-BFGS-B 失败的原因。尝试不同的下限，例如 c(0.001,0.001)，您将得到一个解决方案。

optim(par=c(0.1,0.1),LLL,method="L-BFGS-B", lower=c(0.001,0.001))

$par
[1] 0.6865841 0.0010000

$value
[1] 369.5532

$counts
function gradient 
      14       14 

$convergence
[1] 0

$message
[1] "CONVERGENCE: REL_REDUCTION_OF_F <= FACTR*EPSMCH"

要获得参数的 95% 置信区间，试试这个:

res <- optim(par=c(0.1,0.1),LLL,method="L-BFGS-B", lower=c(0.005,0.005), hessian=TRUE)
n <- length(s1)
res$par # solution
# [1] 1.900928 0.005000
res$par - 1.96*sqrt(diag(solve(res$hessian)))/n # lower limit for 95% confint
# [1] 1.888152372 0.004963286
res$par + 1.96*sqrt(diag(solve(res$hessian)))/n # upper limit for 95% confint
# [1] 1.913703040 0.005036714

引用这篇文章:http://www.ms.uky.edu/~mai/sta321/MLEexample.pdf