[EMAIL PROTECTED] (CatharineStacimer) wrote in message news:<[EMAIL PROTECTED]>...
> I'm not quite sure how to phrase this possibly basic question: can one
> solve a regression equation for two (or more) coefficients if one
> coefficient is a function of another? i.e. There is both an intercept
> and slope coefficient, but the slope is a function of the intercept.
>
> Specifically, I've been exploring the logistic equation of:
> Pt = 1 / (1 + A exp(-Bt))
> (A is the timing variable, B is the slope, t is time)
> and reading articles using it and working through some examples to
> remember the maths and clear out the cobwebs.
>
> Now, one variation of the equation I've seen has B being a function of
> A:
> B = (1/D * LN {A/C}) where C and D are positive constants
>
> so substituting: this in, Pt
> = 1 / [1 + A * exp (-(1/D * LN {A/C}) * t)]
> = 1 / [1 + A * exp (-(t/D * LN {A/C})]
> and using the fact that for positive N, N^X = e^(X LN(N))
> =1 / [1 + (A * {A/C}^{-t/D})]
>
> So are there two coefficients one could solve for here? I'm used to
> seeing Y = B1 + B2X2. To me this varation looks somewhat like Y = B1 +
> B1X2.
With the equation relating A and B in there (assuming C and D are
known, not estimated), you have a nonlinear model of a special form.
It may perhaps be better to work in terms of B rather than A.
Depending on where the equation relating B and A comes from and how
certain you are it is appropriate, it might also be prudent to
consider the model with A and B "unrelated" and see whether fit with
the extra d.f. in the model (or lack of fit without it) indicates that
the constrained model is insufficient.
Glen
.
.
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