Maximum Product of Spacings using R - r

Question

I would like to estimate the parameter for exponential distribution using Maximum Product of Spacings (MPS). I will have to minimize:
-(1/(n + 1))*(sum of log D[i] from i = 1 to n + 1),
where D[i] = F(x[i]) - F(x[i - 1])
And the following is my R code:
n<- 10
mydata<- rexp(n, rate=2)
x<- sort(mydata)
fnn<- function(lambda,x){
for (i in 2:n){
c<- 1-exp(-lambda*x[i])
d<- 1-exp(-lambda*x[i-1])
}
s<- (1/(n-1))*sum(log(c-d))
return(-s)
}
optim(0.8, fnn, x=x)
Can someone please verify if I am doing the right thing here?
The output I obtained is far from the true value of lambda = 2.
$`par`
[1] 0.92375
$value
[1] 0.1847188
$counts
function gradient
18 NA
$convergence
[1] 0
$message
NULL
what modifications should I include?

Answer 1

The problem with your code is that it is rewriting c and d each time through the for loop. It also had a bug in the computation of the multiplicative constant 1/(n + 1).
Here is a corrected version. The key is to reserve memory before the loop with numeric(n - 1).
I also include a simpler version, taking advantage of R's built-in pexp.
fnn <- function(lambda, x){
n <- length(x)
c <- numeric(n - 1)
d <- numeric(n - 1)
for (i in 2:n){
c[i - 1] <- 1 - exp(-lambda*x[i])
d[i - 1] <- 1 - exp(-lambda*x[i-1])
}
s <- (1/(n + 1))*sum(log(c - d))
return(-s)
}
fnn2 <- function(lambda, x){
n <- length(x)
D <- log(pexp(x[-1], rate = lambda,) - pexp(x[-n], rate = lambda))
s <- sum(D)/(n + 1)
-s
}
set.seed(1234)
n <- 10
mydata <- rexp(n, rate = 2)
x <- sort(mydata)
opt <- optim(0.8, fnn, x = x)
opt2 <- optim(0.8, fnn2, x = x)
opt$par
#[1] 2.9225
opt2$par
#[1] 2.9225
identical(opt$par, opt2$par)
#[1] TRUE