# Propagate errors from logarithmic scale to linear model

So I have the given exponential model: $$y = \beta_0 * x ^{\beta_1}$$

In order to estimate $$\beta_i$$ I take the $$\log_{10}$$ of the system to work with linear model and use linear regression: $$\log_{10}y = \log_{10}\beta_0 + \beta_1 \log_{10}x$$. After fitting I get the associated standard deviation error $$\delta\log_{10}\beta_0$$ and $$\delta\beta_1$$.

How can I propagate the errors back to the exponential form so that I can write the result as: $$y = (\beta_0\pm\delta\beta_0) * x ^{(\beta_1\pm\delta\beta_1)}$$

Would it simply be propagating the error of $$\beta_0$$ as $$\delta\beta_0 = \frac{\partial \log_{10}\beta_0}{\partial \beta_0}*\delta\log_{10}\beta_0$$ and leave $$\delta\beta_1$$ as it is?

• Commented Oct 11, 2022 at 18:50
• It would have been slightly less messy if you'd worked with natural logarithms.
– J.G.
Commented Oct 11, 2022 at 20:24

Use the transformation law of random variables to calculate how the uncertainty in X transforms into the uncertainty of Y $$𝑌=𝑓(𝑋) ⇒ 𝑑𝑌 = \left|\frac{∂𝑓}{∂𝑋}\right|\cdot 𝑑𝑋$$ However, be aware that this method yields different optimal fit-parameters (and uncertainty) compared to a non-linear fit. This is due to the fact that least-square fits put more weight onto "larger" deviations compared to "small" deviations. There exists a vast literature on how to optimise least-square fits by using different penalty function.