`optimize()`: Maximum likelihood estimation of rate of an exponential distribution

Question

I'm really struggling with understanding MLE calculations in R.

If I have a random sample of size 6 from the exp(λ) distribution results in observations:

x <- c(1.636, 0.374, 0.534, 3.015, 0.932, 0.179)

I calculated out the MLE as follows

mean(x)

and got 1.111667 (I'm not 100% certain I did this part right).

But when I try to code numeric calculation using RI either get errors or an answer that doesn't match.

lik <- function(lam) prod(dexp(x))   # likelihood function
nlik <- function(lam) -lik(lam)      # negative-likelihood function
optimize(nlik, x)

gives me

#$minimum
#[1] 3.014928
#
#$objective
#[1] -0.001268399

Originally I had

lik <-function(lam) prod(dexp(x, lambda=lam))   # likelihood function
nlik <- function(lam) -lik(lam)      # negative-likelihood function
optim(par=1, nlik)   # minimize nlik with starting parameter value=1

But I kept getting

#Error in dexp(x, lambda = lam) : 
#  unused argument (lambda = lam)
#In addition: Warning message:
#In optim(par = 1, nlik) :
#  one-dimensional optimization by Nelder-Mead is unreliable:
#use "Brent" or optimize() directly

Answer 1

So here is your observation vector

x <- c(1.636, 0.374, 0.534, 3.015, 0.932, 0.179)

I'm not sure why you minimize negative likelihood directly; often we work with negative log likelihood.

nllik <- function (lambda, obs) -sum(dexp(obs, lambda, log = TRUE))

When using optimize , set a lower and upper bound:

optimize(nllik, lower = 0, upper = 10, obs = x)

#$minimum
#[1] 0.8995461
#
#$objective
#[1] 6.635162

This is not too far away from sample mean: 1.11, given that you only have 6 observations which is insufficient for a close estimate anyway.

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It is pretty sufficient to use optimize here, as you work with univariate optimization. If you want to use optim , set method = "Brent" . You can read Error in optim(): searching for global minimum for a univariate function for more.