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I've got a conjecture about the relationship of a dataset. My intuition says it's a decaying exponential, but I want to know which a and b in $e^{-ax^{b}}$ best fits the data. What's the best way to go about it? Most standard stat program doesn't seem to have this capability. I don't think looking at the log of the data would work unless I give it a b to work with either, right?

Benjamin Lindqvist
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    Usually there's no $b$ in "decaying exponential". Does your intuition insist it may be $\ne 1$? On the other hand there is usually a multiplicative constant $ce^{-ax}$ or $e^{C-ax}$. – hmakholm left over Monica Dec 29 '13 at 21:16
  • I have to ask - would it be atypical to the point of unlikeliness for the x being of another order than one? – Benjamin Lindqvist Dec 30 '13 at 07:51
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    x @Benjamin: It wouldn't be something one would usually call a "decaying exponential". Whether it is likely in your situation depends on what your situation is. However $b\ne 1$ only makes sense if the zero point of the $x$ scale is "special" to the process the data come from. With an ordinary exponential decay ($b=1$) the zero point is just an arbitrary convention. – hmakholm left over Monica Dec 30 '13 at 10:29

2 Answers2

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You can transform the suggested model to linear model as follow: $\ln(ln(y))=ln(-a)+bln(x)$. Using linear model will give you the best ln(-a) and b, which get you a and b. So the transformation variables are: $t=ln(ln(y))$, and $w = ln(x)$.

bankrip
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The answer user117552 gave is obviously good but, be careful : the values you will so obtain for $a$ and $b$ are only estimates. From there, you must start a nonlinear regression.

A simple clarification : in your problam, you want, I suppose, to minimize the sum of the squared errors on the $y$'s. The linearization step will provide you the solution corresponding to the minimum sum of the squared errors on the $\ln(ln(y))$'s which is not the same at all

Claude Leibovici
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