Hessian matrix in optim() in R

Question

I'm having some trouble using optim() in R to solve for a likelihood involving an integral and obtain the hessian matrix for the 2 parameters. The algorithm converged but I get an error when I use the hessian=TRUE option in optim(). The error is:

Error in integrate(integrand1, lower = s1[i] - 1, upper = s1[i]) : non-finite function value

Also had a warning message of NAs

Here's my code:

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

Thanks for your help!

Answer 1

optim minimizes the function. You could plot the likelihood function given the argument rate. It needs a bit of a fiddling to get a plot. Do it like this:

z2 <- function(rate) {
    par <- numeric(2)
    par[1] <- .68
    par[2] <- rate
    y <- LLL(par)
    y
}

z1 <- Vectorize(z2,vectorize.args="rate")

curve(z1, from=.001,to=1)

You will see that the function is minimal for the lowest value for rate. Same if you change from to .1. I cannot judge if the estimate is valid.