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I just started taking a regression analysis course. My professor was emphasizing that the class only focuses on Linear regression. The professor gave an example that:

> (1) Y = B0 + B1 * (X^2) is linear, and 
> (2) Y = B0 + (B1^2) * X is NOT linear.

The professor said that (1) is linear because the x-axis can just be changed to x^2. I don't really understand this part. This seems to be contrary to grade 12 math in which y = b + m*x^2 is considered quadratic. I argued that for (2), B1 is always constant. So B1^2 is just any other constant number, let's say B2, to which my professor said B1 is already fixed. But don't we also pretend to fix X as well - it is a controlled variable? I would really appreciate it if someone could explain this to me. Thank you in advance. Nobody else in the class raised a question about this, so maybe I am neglecting something small.

tidbits
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1 Answers1

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In regression, the model is (or not) linear or nonlinear with respect to the parameters.

That is to say that if the model is $$y=f(a,b,x)$$ you need to look at the expression of $$\frac{dy}{da}$$ If this derivative is independent of $a,b,c$, then the model is linear with respect to $a$ and so on.

Does this clarify the point ?

In your examples, the first model is $$y=b_0+b_1 x^2 \implies \frac{dy}{db_0}=1 \qquad \text{and}\qquad \frac{dy}{db_1}=x^2$$ So, it is linear with respect to both $b_0$ and $b_1$.

At the opposite, for the second model $$y=b_0+b_1^2 x \implies \frac{dy}{db_0}=1 \qquad \text{but}\qquad \frac{dy}{db_1}=2 b_1x$$So, it is linear with respect to $b_0$ but nonlinear with respect to $b_1$. So, it is nonlinear.

Claude Leibovici
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    This makes sense mathematically, thank you. But here are my follow-up questions: a) Why are b0 and b1 treated as variables? Aren't they fixed? b) What is the reasoning behind this being linear and non-linear? – tidbits Sep 10 '17 at 05:01
  • @tidbits. You are in the frame of regression and $b_0,b_1$ are the parameters you want to adjust to fit your model based on $n$ data points $(x_i,y_i)$. Is it better ? – Claude Leibovici Sep 10 '17 at 05:11