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I'm using the solnp() function in the R package Rsolnp to solve a nonlinear regression with constraints. It works well, converges with no problem. I want to use the Hessian matrix to calculate standard errors of the four parameter estimates, but the Hessian is not 4 by 4 as I had expected, but 5 by 5. I looked around on SO and didn't see anyone else with an unexpected Hessian size. All the examples I found with the Hessians printed showed them to be the expected size of p by p (e.g., 2x2, 3x3, and 4x4).

How can I get standard errors for my 4 parameters from this 5 by 5 Hessian?

df <- data.frame(
  Recruit.N = c(78.4, 79.8, 106, 57.4, 81.7, 94.4, 74.1, 42, 61.6, 47.7, 61.8, 
    28.1, 32.3, 19, 23.4, 20.1, 27), 
  Stock.5 = c(66.6, 90.3, 138.5, 79.8, 77.3, 78.4, 79.8, 106, 57.4, 81.7, 94.4, 
    74.1, 42, 61.6, 47.7, 61.8, 28.1), 
  Stock.6 = c(25.2, 66.6, 90.3, 138.5, 79.8, 77.3, 78.4, 79.8, 106, 57.4, 81.7, 
    94.4, 74.1, 42, 61.6, 47.7, 61.8), 
  Stock.7 = c(23.8, 25.2, 66.6, 90.3, 138.5, 79.8, 77.3, 78.4, 79.8, 106, 57.4, 
    81.7, 94.4, 74.1, 42, 61.6, 47.7)
)

lossfcn <- function(parz, mydat) {
  alpha <- parz[[1]]
  beta <- parz[[2]]
  p5 <- parz[[3]]
  p6 <- parz[[4]]
  p7 <- 1 - p5 - p6
  S <- with(mydat, p5*Stock.5 + p6*Stock.6 + p7*Stock.7)
  Obs <- mydat$Recruit.N
  Pred <- alpha * S * exp(-beta*S)
  Resid <- log(Obs) - log(Pred)
  sigma <- sqrt(mean(Resid^2))
  LL <- dlnorm(Obs, meanlog=log(Pred), sdlog=sigma, log=TRUE)
  -sum(LL)
}

inequal <- function(parz, mydat) {
  parz[3] + parz[4]
}

library(Rsolnp)
solnp(pars=c(1, 0.008, 1/3, 1/3), fun=lossfcn, mydat=df,
  ineqfun=inequal, ineqLB=0, ineqUB=1, 
  LB=c(0, 0, 0, 0), UB=c(1000, 1000, 1, 1), control=list(trace=0))

$pars
[1] 6.731317e-01 1.888572e-10 8.141363e-01 1.858631e-01

$convergence
[1] 0

$values
[1] 79.87150 75.50927 75.50927 75.50927

$lagrange
          [,1]
[1,] -2.028222

$hessian
           [,1]          [,2]         [,3]        [,4]         [,5]
[1,]  0.3350868 -3.359077e-01     17.84919  -0.4306057   -0.3382811
[2,] -0.3359077  1.993956e+02 -10161.63351  -7.0844295   -2.2749785
[3,] 17.8491854 -1.016163e+04 548099.69224 -85.9544831 -224.0362766
[4,] -0.4306057 -7.084429e+00    -85.95448  25.1086694    5.8817704
[5,] -0.3382811 -2.274979e+00   -224.03628   5.8817704    4.1978178

$ineqx0
[1] 0.9999995

$nfuneval
[1] 142

$outer.iter
[1] 3

$elapsed
Time difference of 0.03016496 secs

$vscale
[1] 1 1 1 1 1 1
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1 回答 1

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Unlike the 3 posts you linked, you have an inequality constraint. Check the ineqx0 in the returned values: the other posts have NULL but you have 0.9999995. With an inequality constraint, there is a slack variable, so the problem is augmented. The hessian matrix returned is for this augmented set of parameters. Just retain the first 4 x 4 submatrix of the hessian for your wanted parameters.

于 2019-01-26T11:56:55.307 回答