R中的MLE错误:'vmmin'中的初始值不是有限的 [英] MLE error in R: initial value in 'vmmin' is not finite

问题描述

假设我有2个data.frame对象:

df1 <- data.frame(x = 1:100)
df1$y <- 20 + 0.3 * df1$x + rnorm(100)
df2 <- data.frame(x = 1:200000)
df2$y <- 20 + 0.3 * df2$x + rnorm(200000)

我想做MLE.使用df1，一切正常:

I want to do MLE. With df1 everything is ok:

LL1 <- function(a, b, mu, sigma) {
    R = dnorm(df1$y - a- b * df1$x, mu, sigma) 
    -sum(log(R))
}
library(stats4)
mle1 <- mle(LL1, start = list(a = 20, b = 0.3,  sigma=0.5),
        fixed = list(mu = 0))

> mle1
Call:
mle(minuslogl = LL1, start = list(a = 20, b = 0.3, sigma = 0.5), 
fixed = list(mu = 0))

Coefficients:
      a           b          mu       sigma 
23.89704180  0.07408898  0.00000000  3.91681382 

但是如果我要对df2执行相同的任务，则会收到错误消息:

But if I would do the same task with df2 I would receive an error:

LL2 <- function(a, b, mu, sigma) {
    R = dnorm(df2$y - a- b * df2$x, mu, sigma) 
    -sum(log(R))
}
mle2 <- mle(LL2, start = list(a = 20, b = 0.3,  sigma=0.5),
              fixed = list(mu = 0))
Error in optim(start, f, method = method, hessian = TRUE, ...) : 
  initial value in 'vmmin' is not finite

我该如何克服?

推荐答案

R的值在某些时候变为零；导致函数的非限定值最小化并返回错误.

The value of R becomes zero at some point; it leads to a non-finite value of the function to be minimized and returns an error.

使用参数log=TRUE可以更好地解决此问题，请参见下面的功能LL3.以下给出了一些警告，但返回了结果，其参数估计值接近真实参数.

Using the argument log=TRUE handles better this issue, see function LL3 below. The following gives some warnings but a result is returned, with parameter estimates close to the true parameters.

require(stats4)
set.seed(123)
e <- rnorm(200000)
x <- 1:200000
df3 <- data.frame(x)
df3$y <- 20 + 0.3 * df3$x + e
LL3 <- function(a, b, mu, sigma) {
  -sum(dnorm(df3$y - a- b * df3$x, mu, sigma, log=TRUE))
}
mle3 <- mle(LL3, start = list(a = 20, b = 0.3,  sigma=0.5),
  fixed = list(mu = 0))
Warning messages:
1: In dnorm(df3$y - a - b * df3$x, mu, sigma, log = TRUE) : NaNs produced
2: In dnorm(df3$y - a - b * df3$x, mu, sigma, log = TRUE) : NaNs produced
3: In dnorm(df3$y - a - b * df3$x, mu, sigma, log = TRUE) : NaNs produced
4: In dnorm(df3$y - a - b * df3$x, mu, sigma, log = TRUE) : NaNs produced
5: In dnorm(df3$y - a - b * df3$x, mu, sigma, log = TRUE) : NaNs produced
6: In dnorm(df3$y - a - b * df3$x, mu, sigma, log = TRUE) : NaNs produced
7: In dnorm(df3$y - a - b * df3$x, mu, sigma, log = TRUE) : NaNs produced
8: In dnorm(df3$y - a - b * df3$x, mu, sigma, log = TRUE) : NaNs produced

> mle3
Call:
mle(minuslogl = LL3, start = list(a = 20, b = 0.3, sigma = 0.5), 
    fixed = list(mu = 0))

Coefficients:
        a         b        mu     sigma 
19.999166  0.300000  0.000000  1.001803 

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