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I'm learning about allometric relationships and how to derive the parameters from regression equations.

I've seen that you can fit a linear regression model by taking the log of both the X and Y variables of your data that have an allometric relationship.

Then you can use the coefficients of the slope and intercept to create a power law of the form $Y=aX^b$

However, I've read that you need to raise your intercept to whatever base you're using but not your slope. For example, if I got an intercept of -1.2 after conducting a regression on the $log_{10}$ of both X and Y I need to apply $10^{-1.2}$ before using it in my power law.

My question is why don't you have to do this for the slope coefficient?

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You have probably found estimates for $p$ and $q$ of the following model:

$$\log_{10}Y = p +q\log_{10}X$$

This is equivalent to

$$10^{\log_{10}Y} = 10^{p +q\log_{10}X} =10^{p +\log_{10}X^q} $$

$$\implies Y = 10^{p}X^q$$

Comparing with your model's original form,

$$Y=aX^b$$

you can see $a=10^p$ and $b=q$, i.e. you only take the power of your slope's estimate.

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  • $\begingroup$ Ah okay, so when I take the $log_{10}$ of my Y and X data I don't incorporate the coefficient estimate (i.e. the $\beta_1$ value which is the slope or q value in your answer). Rather it's because the slope is a coefficient of logged data in the first place that it's interpreted differently to the intercept. $\endgroup$
    – adkane
    Jan 17 '17 at 22:36
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    $\begingroup$ @Manassa Mauler, Yes, the slope you are finding is for the logged data, thats why you need the power for it to make sense for the original data. $\endgroup$ Jan 17 '17 at 22:46
  • $\begingroup$ Thanks. I just found it confusing because you use typically use the slope to find the estimate of the intercept. $\endgroup$
    – adkane
    Jan 17 '17 at 22:53

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