On 2011-03-01 06:38, Schatzi wrote:
Here is a reply by Bart:
Yes you're right (I should have taken off my glasses and looked closer).
However, the argument is essentially the same:
Suppose you have a solution with a,b,k,l. Then for any positive c, [a+b-bc]
+ [bc] + (bc) *exp(kl')exp(-kx) is also a solution, where l'
= l - log(c)/k .
Cheers,
Bert
(Feel free to post this correction if you like)
This is from me:
The problem with dropping the "l" parameter is that it is supposed to
account for the lag component. This equation was published in the literature
and has been being solved in SAS. When I put it in excel, it solves, but not
very well as it comes to a different solution for each time that I change
the starting values. As such, I'm not sure how SAS solves for it and I'm not
sure what I should do about the equation. Maybe I should just drop the
parameter "a." Thanks for the help.
When you say 'published in the literature' you should
provide a reference; you may be misinterpreting what's
published.
If SAS provides a 'solution', then there's an added
assumption being made (perhaps 'l' is being fixed?).
What Excel does is of little interest.
'Dropping' the parameter 'a' is equivalent to setting a=0.
You could also set, say, a = -10 or l = 50, or ...
The point is that, as Bert says, the model is
nonidentifiable.
Peter Ehlers
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