"Michael Hochster" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> In sci.math Phil Sherrod <[EMAIL PROTECTED]> wrote:
>
> : On 27-Apr-2004, Michael Hochster <[EMAIL PROTECTED]> wrote:
>
> :> : This isn't a "simple linear regression" problem.  It is a nonlinear
> :> : regression problem.  There are a number of nonlinear regression
programs
> :> : that can solve your problem for a and b.  Here is such a program that
I
> :> ran
> :> : through my NLREG program (http://www.nlreg.com)
> :>
> :> Yes, it is a simple linear regression problem: ordinary regression
> :> of y on e^-x. As the author of regression software, you should know
> :> better.
>
> : I agree, by transforming the input variables this function is easily
> : converted to a linear regression.  But it can be handled more easily and
> : properly as a nonlienar regression where no transformations are
required.
> : Remember that fitting a function to a transformed independent variable
does
> : not always yield the same fitting parameter results as fitting the
function
> : to the non-transformed input -- minimizing the sum of squared deviations
for
> : X is not the same as log(X) or sin(X). The difference can be
significant.
>
> There is a closed form for the a and b minimizing sum[y - a - b*e^(-x)]^2,
> provided by the usual linear regression formulas. Are you saying it is
easier
> and/or more proper to do a numerical search for a and b?
>
> Mike


It is not the method of solving the equations but that is being debated, but
the procedure used to estimate the parameters.
One method is linear regression which as you point out has a closed form
solution. There are alternatives, many of which make different assumptions
about the form the error takes. These lead to different estimates for the
parameters a,b. To use linear regression you are assuming that the error
terms in the equation (y=a+be^(-x)+error)  all come from the same normal
distribution (and are not correlated). This is of course possible and may be
true for this case, I haven't checked. But in general, for a model of this
type I would have made a different starting assumption.


Rod



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