[R] estimate likelihood

[ProfJCNash]

Not possible, because the hessian is singular. Recoded as follows (your
code should be executable before you put it in a help request).

# asindii2.R -- Is it possible to estimate the likelihood parameter
#    and test for significant as follows:

x <- c(1.6, 1.7, 1.7, 1.7, 1.8, 1.8, 1.8, 1.8)
y <- c( 6, 13, 18, 28, 52, 53, 61, 60)
n <- c(59, 60, 62, 56, 63, 59, 62, 60)

DF <- data.frame(x, y, n)

# note: there is no need to have the choose(n, y) term in the likelihood
fn <- function(p, DF) {
   z <- p[1]+p[2]*DF$x
   sum( - (DF$y*DF$z) - DF$n*log(1+exp(DF$z*DF$x)))
}
out <- nlm(fn, p = c(-50,20), hessian = TRUE, print.level=2, DF=DF)
print(out)

eigen(out$hessian)
sqrt(diag(solve(out$hessian)))
## ----- end of snippet ---


On 16-03-03 01:30 PM, Alaa Sindi wrote:
> Hi all,
> 
> Is it possible to estimate the likelihood parameter and test for significant as follows:
> 
> x <- c(1.6, 1.7, 1.7, 1.7, 1.8, 1.8, 1.8, 1.8)
> y <- c( 6, 13, 18, 28, 52, 53, 61, 60)
> n <- c(59, 60, 62, 56, 63, 59, 62, 60)
> 
> # note: there is no need to have the choose(n, y) term in the likelihood
> fn <- function(p)
> 
> 
> z = p[1]+p[2]*x
> sum( - (y*z) - n*log(1+exp(z*x))) 
> 
> out <- nlm(fn, p = c(-50,20), hessian = TRUE, print.level=2)
> 
> out
> 
> eigen(out$hessian)
> sqrt(diag(solve(out$hessian)))
> 
> 
> Thanks
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