2

One standard example when introducing OLS in econometric classes is modelling the log-wage by education and experience. Often, the example models account for experience by not only by the experience itself but additionally but the squared experience. This is done because a non-linear relationship is assumed. This yields the theoretical model

$E[\log(Y)|X] = \beta_0 + \beta_1Education + \beta_2Experience + \beta_3Experience^2$.

However, one assumption in OLS is that the dependent variable is linear in all its components. How can then this serve as an example when apparently some nonlinear relationship is existing?

What confuses me is that even though the squared experience and log-wage may have a linear relationship, there is still the original variable in the model which has apparently a nonlinear relationship with log-wage. So, in this case it's linear in Education and squared Experience but it's still nonlinear in Experience itself.

random_guy
  • 2,410
  • The nature of your confusion is unclear. The title asks how OLS can model non-linear relationships among explanatory and response variables. At the end you observe that the example you provide actually is such a non-linear model. What is the problem, then? – whuber Apr 17 '15 at 15:47
  • My problem is that three explanatory variables are in the model: education, experience and squared experience. Even though with squaring experience, it accounts for nonlinearities, I don't understand why it is correct to us such a model when there is still experience in it which has a nonlinear relationship to log-wage. – random_guy Apr 17 '15 at 15:56
  • 1
    I am totally baffled by that comment. Perhaps you could elaborate on your understanding of "nonlinear relationship"? That might hold the key to the constructive interpretation of your question. Another term you might explain further is your sense of what "correct" means in the context of a modeling exercise. – whuber Apr 17 '15 at 15:59
  • With nonlinear I mean for instance that if I plotted Y against X, the relation would look rather look like $y = \beta x^2$ rather than $y = \beta x$. With correct I mean the assumptions for OLS are fulfilled such that it is BLUE. – random_guy Apr 17 '15 at 16:07
  • 2
    You might benefit from a more precise understanding of "nonlinear." This concept comprises any function that does not have a constant derivative. So, for instance, if you were to plot $y = x + x^2$, its derivative $1 + 2x$ would clearly be nonconstant. Letting $x$ be your "Experience" reproduces the example you are asking about. A phrase like "linear in $x^2$ but nonlinear in $x$" makes no sense from this point of view, because you cannot vary one without simultaneously varying the other. – whuber Apr 17 '15 at 16:13

1 Answers1

2

Your example does not violate the linearity assumption in linear regression. Linear Regression assumes linearity in the coefficients $\beta$. In your example, you could think of $experience^2$ as a brand new explanatory variable. You could arbitrarily call it $experience.version2$ and now it's just some variable in a linear combination of $\beta$ coefficients and explanatory variables.

An example of a non-linear relationship would be something like $experience^\beta$ or $\frac{e^{\beta*experience}}{1+e^{\beta*experience}}$.

TrynnaDoStat
  • 8,124
  • My problem is, however, that even though squaring experience can make the relationship linear, there is still the original variable in the model which has apparently a nonlinear relationship with log-wage. So, in this case it's may be linear in Education and squared Experience but it's still nonlinear in Experience itself. – random_guy Apr 17 '15 at 15:28
  • 1
    @random_guy You are right, it's not linear in experience but that's okay and does not violate an assumption of OLS. – TrynnaDoStat Apr 17 '15 at 16:09
  • 1
    So it is only assumed there is a parameter vector which describes a linear relationship? Meaning, Y does not need to have a linear relationship with each explanatory variable? – random_guy Apr 17 '15 at 16:20
  • Yup, you've got it. – TrynnaDoStat Apr 17 '15 at 16:37