Why is $Y=\beta_0 x^{\beta_1} e$ a linear model? When we apply the transform, it becomes $lnY = ln\beta_0+\beta_1 lnx +lne$, and why is it still linear when the $\beta_0$ part is under ln?


2 Answers 2


The term "linearity" is context-dependent, so a linear regression model is not necessarily the same as a linear function.

A linear function is classified as such via the superposition principle, requiring both additivity and homogenity. We can generalize linear maps with to multivariate functions simply:

  • Additivity: $f(\vec{x}_1+\vec{x}_2)=f(\vec{x}_1)+f(\vec{x}_2)\enspace \forall\ \vec{x}_1,\vec{x}_2\in \Bbb{R}^n$
  • Homogeneity: $f(\alpha \vec{x})=\alpha f(\vec{x})\enspace \forall\ \vec{x}\in \Bbb{R}^n$ and $\alpha\in \Bbb{R}$

So the equation $f(\beta_0,\beta_1)=\beta_0 x^{\beta_1}e$ is a nonlinear function because it is not additive nor homogeneous.

However, w.r.t. linear regression, linear models are of the form $Y=\beta_0+\beta_1f_1(x_1)+\beta_2f_2(x_2)+\cdots+\beta_nf_n(x_n)+c$, regardless of whether any $f_i$ is a nonlinear map.

So, after a nonlinear transformation such as natural logarithm, you can consider the resulting regression model as the form $Y'=\beta_0'+\beta_1ln(x)+1$ where the primes denote the natural log transformed variables. This demonstrates linearity between $Y'$ and the parameters $\beta_0'$ and $\beta_1$.


It depends of your point of view: when one speaks about linear you usually refer the variables linked by the linear relationship. The model with x as input and Y as target is not linear however the model with input $\log(x)$ and target $\log(Y)$ is linear:

$$ \tilde{Y} = \tilde{\beta} + \beta_1\tilde{x} + \tilde{e} $$

with $\tilde{Y}=\log(Y)$, $\tilde{\beta_0}=\log(\beta_0)$, $\tilde{x}=\log(x)$ and $\tilde{e}=\log(e)$

However, this example shows how features transformation can help learn non-linear features within a linear structure.


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