Re: Curve Fitting to Exponential Data

mcdonnellb@xxxxxxxxx wrote in
news:1164121169.646648.138510@xxxxxxxxxxxxxxxxxxxxxxxxxxxx:

Thank you very much for your reply.

You say there is a non-zero "horizontal"
asymptote, and you haven't placed it in your model

Try y=A*e^(Bx)+ C

Even better, you might want to MEASURE your asymptotic value,
subtract it a priori from your data, and then estimate your original
model.

I've tried this and it resulted in this superior looking graph:
http://i81.photobucket.com/albums/j201/tacoflies/sample_and_curve2.png

There was a significant issue though:
The curve fitting method I linked to above involves taking ln(y).
When y is below the asymptote and I subtract the asymptote value, y is
negative and ln(y) is undefined.
To avoid this problem I ignored any values of Y where y <= 0, but this
obviously will alter my resulting fitted curve. Actually, thinking
about it some more, this only happens with the least reliable data, so
maybe it's not a big deal.

It is a concern, and can bias the B value toward an overestimate. Your
ln method is a popular way of dealing with exponentials (At least on
paper, but the practical issues are certainly real!), but Myers points
out the problem that you really can't correctly transform y=ae^bx + eps
that way, as ln(y)=ln(a)+ bx+eps, because that is not the transformation
of the error term. It works if you have a multiplicative error, though.
If y=ae^bx*eps, and then ln(y)=ln(a)+ bx+ln(eps). In any case, nonlinear
regression techniques, which do not require the log, might be preferable
(though maybe not available).




Since datapoints at lower X's are more reliable than those at higher
X's perhaps I should be doing some weighting in my curve fitting
calculations?

Off the cuff, this looks a little double exponentially to me, or

y=Ae^(Bx) + Ce^(Dx) + F

Looking at the curve fit now, do you still think think that
distribution looks double exponential...? I think I can forget about
solving it if it is!


I still think it looks a little like a double exponential, but that
certainly does not mean that a single exponential fit is without value.

--
Scott
Reverse name to reply
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