# Why is $Y=\beta_0 x^{\beta_1} e$ a linear model?

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?

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.