On Tue, Jan 17, 2012 at 8:06 AM, Kenneth Frost <kfr...@wisc.edu> wrote:
> Sorry, that wasn't to helpful...I see that the intervals and se.fit argument 
> are currently ignored.

Yes, because the fitted values are nonlinear in the parameters, which
makes finding exact confidence regions impossible. I think the "usual"
approach (subject to correction by experts) is to use a delta method
approximation for the fitted variances from the varcov matrix of the
parameters at the converged optimum (itself an approximation) and then
a standard t-interval  based on that. However, this approximation can
be quite bad, because "degrees of freedom" don't mean much for
nonlinear models -- in fact, that's the essential (and huge!)
difference between linear and nonlinear models -- and the likelihood
surface may not be close enough to quadratic. So one may do better
with, e.g. a bootstrap approximation, although this can be
problematic, too, due to convergence and other issues.

What I think can be said with some certainty is that the idea of
approximating by a segmented regression and then using CI's for each
linear part in the "usual" way is a particularly bad one -- the CI's
will be underestimated because they don't take into account the
uncertainty in the location of the fitted breakpoints, which are
nonlinear **and** non-smooth functions of the data.

So if confidence intervals for the fitted values are really important,
I suggest that Julian work with his local statistician to come up with
the best approach for his particular situation. It's tricky.

Cheers,
Bert

>
> On 01/17/12, crimsonengineer87  <julianjonre...@gmail.com> wrote:
>> Dear Forum,
>>
>> I have been wracking my head over this problem for the past few days. I have
>> a dataset of (x,y). I have been able to obtain a nonlinear regression line
>> using nls. However, we would like to do some statistical analysis. I would
>> like to obtain a confidence interval for the curve. We thought we could
>> divide up the curve into piecewise linear regressions and compute CIs from
>> those portions. There is a package called strucchange that seems helpful,
>> but I am thoroughly confused.
>>
>> 'breakpoints' is used to calculate the number of breaks in the data for
>> linear regressions.  I have the following in my script:
>>
>> bp.pavlu <- breakpoints(Na ~ f(yield, a, b), h=0.15, breaks=3,
>> data=pavludata)
>> plot(bp.pavlu)
>> breakpoints(bp.pavlu)
>>
>> But I am confused as to how to graph the piecewise functions that make up
>> the curve. I am not even sure if I am using breakpoints correctly. Do I just
>> give it a linear relationhip (Na ~ yield), instead of what I have?
>>
>> Is there an easier way to calculate the confidence interval for a non-linear
>> regression?
>>
>> I am new to R (as I've read in many questions), but I have most certainly
>> tried many things and am just getting frustrated with the lack of examples
>> for what I'd like to do with my data... I'd appreciate any insight. I can
>> also provide more information if I am not clear. Thanks in advance.
>>
>> Julian
>>
>> --
>> View this message in context: 
>> http://r.789695.n4.nabble.com/breakpoints-and-nonlinear-regression-tp4303629p4303629.html
>> Sent from the R help mailing list archive at Nabble.com.
>>
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>>
>>
>
> ______________________________________________
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>



-- 

Bert Gunter
Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: 467-7374
Website:
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