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I am running the following regression:

$y= bX^a + noise$

but instead of using a nonlinear regression, I use a grid approach to find best fit alpha (lowest MSE). I am getting a good fit with independent normal homoskedastic residuals. However, my question is regarding the standard error of alpha. Is there a way to estimate that without resorting to nonlinear approach?

Note: $y$ takes on both positive and negative values so can't linearize by taking logs. I will post a scatter plot shortly.

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    $\begingroup$ Do you mean your model is $y_i = b x_i^a + \epsilon_i$? Do you know anything about the distribution of $\epsilon_i$? $\endgroup$ Commented Feb 7, 2017 at 8:54
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    $\begingroup$ Is $Y$ always negative or just sometimes? How often Is $X$ negative, zero or positive? Can you list and/or plot the data? $\endgroup$ Commented Feb 7, 2017 at 10:00
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    $\begingroup$ In my view this isn't clear until we know more about the data. The data might make a power function plausible or implausible and even if it is plausible advice on how to fit it hinges on the precise nature of the data. $\endgroup$ Commented Feb 7, 2017 at 15:28
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    $\begingroup$ This question is confusing because it clearly is a nonlinear regression but asks about something in the "linear setting." What is this supposed to mean? $\endgroup$ Commented Feb 7, 2017 at 15:57
  • $\begingroup$ I have edited the question above with some details and will also post the scatter plot shortly to give a better idea about the data. $\endgroup$ Commented Feb 7, 2017 at 16:46

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Why don't you take logarithm of the right hand side and regress $y=d+a\log(X)$? So redefining the independent variable to be $\log(X)$ and running a standard linear regression.

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  • $\begingroup$ Sorry forgot to mention that y takes on negative values also so can't do that. Edited above. $\endgroup$ Commented Feb 7, 2017 at 7:44
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    $\begingroup$ Why not? As long as you don't take logarithm of $y$, you should be fine. Or does $X$ also take negative values? $\endgroup$ Commented Feb 7, 2017 at 7:45
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    $\begingroup$ That is a different model. You need to take $\log$ also of $y$ to keep the model the same. $\endgroup$ Commented Feb 7, 2017 at 7:46
  • $\begingroup$ Yes, you're right. Can you do some sort of bootstrapping to find out the standard error of $\alpha$? $\endgroup$ Commented Feb 7, 2017 at 8:08

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