2
$\begingroup$

If the data (Y and X) is to be fit by a function of the type $f(x)=e^{ax}$ and I fit the Y vs. log(X) via f, and obtain the fit parameter $a$, is this $a$ parameter different than the desired one (in f(x)) even though i fit Y vs. log(X) instead of Y vs. X?

Or do I have to transform the parameter somehow?

Improve this question
$\endgroup$
4
  • $\begingroup$ Do you mean $y=e^{ax}+\epsilon$ and $\log(y)$ vs. $x$? $\endgroup$
    – GeoMatt22
    Aug 30, 2016 at 0:05
  • 1
    $\begingroup$ Can you write out the equation you are fitting? And also the method you use to fit (e.g. least squares)? Please put these in the question, rather than a new comment. $\endgroup$
    – GeoMatt22
    Aug 30, 2016 at 0:18
  • $\begingroup$ But the equation im fitting is given in the post, it is that of f(x) $\endgroup$
    – user929304
    Aug 30, 2016 at 0:21
  • 1
    $\begingroup$ So you are fitting $f(y)\approx x$? Depending on the error model and fitting method, the estimate for $a$ could be biased, but the "units" should be OK. $\endgroup$
    – GeoMatt22
    Aug 30, 2016 at 0:30

2 Answers 2

Reset to default
2
$\begingroup$

Unless the data fit the equation perfectly, then the results will not be the same.

If you have a linear (Gaussian) model and specify

$Y_i = e^{ax_i} + \epsilon_i; \epsilon_i \sim N(0,\sigma^2)$

then this is not the same as saying

$\log(Y_i) = ax_i + \epsilon_i$.

As for which is correct, it depends on the situation. If the errors are all likely to have the same variance before taking logs then the first form is correct. If the errors are multiplicative then the second is appropriate. Alternatively, the errors might be doing something else and then neither approach works.

Improve this answer
$\endgroup$
0
$\begingroup$

If you find the best linear fit between $Y$ and $\log_e(X)$, its slope will corresponds exactly to the parameter $a$ in your function.

Note that this only works if you're taking the log base $e$, otherwise you'd have to correct for the change of base.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ @user929304 that's what this answer is saying. In fact you would have to transform your data (or use a more sophisticated nonlinear optimization tool) if you didn't log-transform $X$ $\endgroup$
    – shadowtalker
    Aug 30, 2016 at 0:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.