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So the definition of linear regression is that the response variable is a linear function of the estimators. If we consider univariate regression (for ease of visualization), we have $$ y = \beta_1x + \beta_0 $$ But we could also have $$ y = \beta_1x^2 + \beta_0 \\ y = \beta_1 \exp(\log(x^3)) $$ which also satisfies the condition that the response variable is a linear function of the ESTIMATORs.

I find this terminology to be a bit confusing as I would expect "linear" regression to be restricted to strict lines in univariate regression.

When linear regression is introduced in courses, the examples are always straight lines, and I think some instructors even introduce linear regression as fitting a LINEAR line to a set of data, but that's not true.

So isn't it rather confusing that it's called "linear" regression? I feel like "linear" regression connotes that the fit will be a straight line (in the univariate case).

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  • $\begingroup$ I've personally thought (I'm not a statistician), that if we don't change the parameter, and by using a substitution of variables can make it linear and solve it with least squares, then it is linear. A rational would be a problem but in this if $z = x^2$ then $y\left(z\right)$ is linear in z. $\endgroup$ – EngrStudent Jun 25 at 18:47
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    $\begingroup$ Contrast $y = \beta_0 + \beta_1 x$ (linear in parameters) with $y = b_0 + x^{\beta_1}$ or $y = \beta_0 + \beta_{1}^{x}$ (nonlinear in parameters). $\endgroup$ – Alexis Jun 25 at 20:23
  • $\begingroup$ @EngrStudent Right, that's how I was thinking about it just now. If we consider something like $y = sin(x)$. We could define the transformation $z = sin(x)$. Then $y=z$ is linear in the connotated sense of the word. $\endgroup$ – anonuser01 Jun 25 at 20:27
  • $\begingroup$ I like how MathematicalMonk addresses this: youtube.com/watch?v=rVviNyIR-fI. “It’s not just lines & planes.” $\endgroup$ – Dave Jun 26 at 13:50
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    $\begingroup$ The sense in which regressing $y$ on nonlinear transformations of $x$ still is just a matter of lines, planes, and linear subspaces generally is explained and illustrated at stats.stackexchange.com/a/354256/919. $\endgroup$ – whuber Jun 26 at 15:38
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Check out these threads:

How to tell the difference between linear and non-linear regression models?

Why must linear regressions only generate linear functions that resemble "lines or planes" (*Introduction to Statistical Learning* question)?

"Linear" is talking about the calculation that happens to the unknown coefficients $\beta$. Any regression of the form

$$y \approx \beta^Tf(x)$$

is reasonably called linear as long as $f$ and $x$ are known. This is because a lot of the same theory and computation applies when you go to estimate $\beta$, regardless of what values $f$ and $x$ take.

EDIT: When might you know $f$?

  • Spline regression
  • RBF's or Nardaraya-Watson as in PRML Section 6.3 (This is like an interaction term on steroids; it's completely nonparametric. It's still linear in $\beta$.)
  • When using a wavelet basis

How do you know if you should be using a wavelet basis or a spline? Sorry, that's a whole different question.

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  • $\begingroup$ "linear" means no interaction terms. $\endgroup$ – LBogaardt Jun 25 at 18:52
  • $\begingroup$ @LBogaardt When I think of interaction, I think of interaction between $\beta_i$ and $\beta_j$ for $i \neq j$. But $\beta_i^2$ would no longer be linear. $\endgroup$ – anonuser01 Jun 25 at 19:00
  • $\begingroup$ As far as I know, $\beta_i x_i + \beta_j x_j$ is called 'linear', so is $\beta_i log(x_i) + \beta_j x_j^5$, but $\beta_{ij} x_i x_j$ is not. $\endgroup$ – LBogaardt Jun 25 at 19:35
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    $\begingroup$ @LBogaardt The latter should still be within the definition of linear regression because $y$ is still a linear function of $\beta$. You can introduce a new variable $z = x_ix_j$. It's just $z$ will have a strong degree of multicollinearity with $x_i$ and $x_j$. $\endgroup$ – anonuser01 Jun 25 at 19:51
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    $\begingroup$ When you're given an input variable, in this case, $f(x)$, which situations would you ever know what $f(x)$ is? $\endgroup$ – anonuser01 Jun 25 at 21:09

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