Dimensional error computing a Hessian (for maximum likelihood)

Question

Given that I am forcing, via the for loops, for Hessian to be an 8x8, why do i get error that it is a 2x1?

Running an 8x8 Hessian to generate maxLik on nx8 regressor dataset. Getting an error that Hessian is 2x1 instead of 8x8, even though this is forced in Hessian[i,j]<-.

x <- independent.variable.matrix.nby8
param <- c(1,1,1,1,1,1,1,1)

Hessian <- matrix(ncol=8,nrow=8)
hess.lik <- function(param) {
  loglambd <- x%*%param       
  lambd <- exp(loglambd)
  for (i in 1:8){
    for (j in 1:8){
      moment <- x[,i]*x[,j]
      return(dim(moment))
      Hessian[i,j] <- -sum(moment%*%lambd)
    } 
  }
}

The dataset, x, is nxk (4406x8).
The param (initial iteration of estimators, start= argument under maxLik) vector is 8 numeric string, so loglambda is (nxk)%*%(8x1) a nx1 vector. Right?
The moment is then nx1 * nx1 i.e. nx1 as element multiplication. Right?
and each element of the Hessian should be scalar as it's a summed kx1 vector (result from moment %*% lambda). This part of the code is non conformable.

The main qn is, even if parts are not the right dimension (likely), the Hessian[i,j <- -sum(vector) should form a scalar, which through {i,j} in 1:8 loops should force an 8x8. Why does that not work?

Can you make the problem reproducible. The code needs to define `x` and `param`. — Matt Brigida, Sep 28 '15 at 21:32
If the error you're getting is complaining about you hessian being `2 x 1`, it may be because you're doing `return(dim(moment))`- the return value of your `hess.lik` function will just be the dimensions of `moment`, when you probably want to be returning `Hessian`, after running through the two nested for loops. — Marius, Sep 29 '15 at 05:31

Dimensional error computing a Hessian (for maximum likelihood)

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