Why linear in linear regression?

In linear regression, one estimates parameters that are supposed to be linear with respect to the dependent variable, for instance

Is it not true that neither ##y(\theta_0)## nor ##y(\theta_0,...,\theta_n)## are linear functions, but rather affine functions of the parameters?

Affine linear is often abbreviated by calling it linear. It's for convenience, and it ##\epsilon=0## in your examples, then it is linear. However, it's not only convenience. E.g. if we consider the tangent line at ##x=2## at the curve ##y=x^2##, then it is an affine linear object on the surrounding ##x,y-##plane. But we also speak of the tangent space at ##x=2##. In that case we have implicitly identified the point ##(2,4)## with the origin of the tangent space and all of a sudden the affine linear line in the plane, became a linear line in the tangent space.

So, as long as not both terms are necessary in a certain context, affine linear is often just called linear.

Could you explain how the line ##x=2## is "an affine linear object on the surrounding ##x,y##-plane"? An affine transformation is a linear transformation plus a translation.

If ##\epsilon \neq 0##, are then both terms necessary?

schniefen said:

Could you explain how the line ##x=2## is "an affine linear object on the surrounding ##x,y##-plane"? An affine transformation is a linear transformation plus a translation.

If ##\epsilon \neq 0##, are then both terms necessary?

Both terms are necessary if you have a global coordinate system and you perform geometry. Then the distinction is reasonable. If you only want to say: not curved, then linear will do.

schniefen said:

TL;DR Summary: It appears to be common to say linear regression, but is this correct?

In linear regression, one estimates parameters that are supposed to be linear with respect to the dependent variable, for instance

##y=\theta_0 e^x+\epsilon \ ,##

or

##y=\theta_0+\theta_1 x_1+\theta_2x_2+...+\theta_n x_n+\epsilon \ . ##
​

Is it not true that neither ##y(\theta_0)## nor ##y(\theta_0,...,\theta_n)## are linear functions, but rather affine functions of the parameters?

​

In regression you make an assumption about the form of the functional relationship between your response and any predictor(s). LINEAR regression doesn't require the function to be itself linear, only that it be linear in the parameters to be estimated.

This expression would qualify as a functional form we'd lump into linear regression.

## y = \beta_0 + \beta_1 x_1^2 + \beta_2 \frac{x_2}{x_2^2 + 5} ##

This is not a linear function of the parameters.

## y = \beta_0 e^{\beta_1 x_1 + \beta_2 x_2} ##

since, for given values of the predictors it is a linear function of the betas. You don't need to worry about the affine stuff.