[R] Fitting a Weibull/NaNs

[Spencer Graves]

      If the algorithm works properly, you should get exactly the same 
answer using a linear or a log scale for the parameters. 

      The bigger question is not bias but the accuracy of a normal 
approximation for confidence intervals and regions.  I have evaluated 
this by making contour plots of the log(likelihood).  I use "outer" to 
compute this over an appropriate grid of the parameters.  Then I use 
"contour" [or "image" with "contour(..., add=TRUE)"] to see the result.  
After I get a picture, I may specify the levels, using, e.g., 
2*log(likelihood ratio) is approximately chi-square with 2 degrees of 
freedom.  The normality assumption says that the contours should be 
close to elliptical.  I've also fit log(likelihood) to a parabola in the 
parameters, possibly after deleting points beyond the 0.001 level for 
chi-square(2).  If I get a good fit, I'm happy.  If not, I try a 
different parameterization. 

      When I've done this, I've found that I tend to get more nearly 
normal contours by throwing the constraint to (-Inf) than leaving it at 
0, i.e., by

      Bates and Watts (1988) Nonlinear Regression Analysis and Its 
Applications (Wiley) explain that parameter effects curvature seems to 
be vastly greater than the "intrinsic curvature" of the nonlinear 
manifold, onto which a response vector is projected by nonlinear least 
square.  This is different from maximum likelihood, but I believe that 
this principle would still likely apply. 

      Does this make sense? 
      spencer graves

         p.s.  I don't understand what you are saying about "0.41 3.70 
1.00" below.  You are giving me a set of three numbers when you are 
trying to estimate two parameters and getting NAs, Inf's and NaNs.  I 
don't understand.  Are you printing out "x" when the log(likelihood) is 
NA, NaN or Inf?  If yes, is one component of "x" <= 0? 

Eric Rescorla wrote:

>Spencer Graves <spencer.graves at pdf.com> writes:
>  
>
>>      I have not used "nlm", but that happens routinely with function
>>      minimizers trying to test negative values for one or more
>>      component of x.  My standard approach to something like this is
>>      to parameterize "llfunc" in terms of log(shape) and log(scale),
>>      as follows: llfunc <- function (x) {
>>      -sum(dweibull(AM,shape=exp(x[1]),scale=exp(x[2]), log=TRUE))}
>>
>>      Have you tried this?  If no, I suspect the warnings will
>>      disappear when you try this.
>>    
>>
>
>This works. I've got some more questions, though:
>
>(1) Does it introduce bias to work with the logs like this?
>
>(2) My original data set had zero values. I added .5 experimentally,
>    which is how I got to this data set. This procedure doesn't work
>    on the original data set.
>
>    Instead I get (with the numbers below being the values
>    that caused problems):
>
>[1] 0.41 3.70 1.00
>[1] 0.41 3.70 1.00
>[1] 0.410001 3.700000 1.000000
>[1] 0.410000 3.700004 1.000000
>[1] 0.410000 3.700000 1.000001
>Warning messages: 
>1: NA/Inf replaced by maximum positive value 
>2: NA/Inf replaced by maximum positive value 
>3: NA/Inf replaced by maximum positive value 
>4: NA/Inf replaced by maximum positive value 
>
>Thanks,
>-Ekr
>
>  
>