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Isn't a linear relationship between the dependent variable and the independent variable a necessary condition? So why semi-log and reciprocal models fit with this type of assumption?

John Doe
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You've got to distinguish the condition "linear in parameters" from "linear in variables." Often it is the case that a nonlinear relationship between variables can be transformed into a linear relationship in transformed variables. For example, $$ y = a + bx + cx^2 + \epsilon $$ can be transformed into $$ y = a + bx + cz + \epsilon, $$ with $z=x^2$ as a new explanatory variable. This new relationship is linear in the parameters $(1,x,z)$. If the Gauss-Markov assumptions still hold after this transformation, then OLS will work as we normally would expect it to.

EDIT:

Consider this example taken from Wikipedia. This is a nonlinear model, $$ y = a e^{bx} U, $$ that can be transformed into a linear model,

$$ ln(y) = ln(a) + bx + u, $$ where the error term $u = ln(U)$.

The trick is to consider whether the Guass-Markov assumptions have been seriously violated. As the Wikipedia article mentions, linearization must be done with caution.

jmbejara
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    Gauss-Markov gives the conditions that make the OLS estimators "best linear unbiased", thus it's primarily about linearity of the estimators, not of the model equation. Could you please clarify? – Michael M Nov 19 '13 at 13:23
  • @MichaelMayer, but still. Linearity in the parameters just mean we have constant betas, but not necessarily a linear relationship with the dependent variable -- right? – John Doe Nov 19 '13 at 19:56
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    @MichaelMayer, Gauss-Markov says that the OLS estimator is the "best linear unbiased estimator" (BLUE). So, yes, this means that the estimator is linear. However, one of the assumptions (see classical linear regression model assumptions) of Gauss-Markov is that the model is also linear (in parameters). If you add the assumption that the disturbance term is normally distributed conditional on the regressors, then OLS achieves the Cramer-Rao lower bound and is BUE (best unbiased estimator). That is, it will be the most efficient even among nonlinear estimators – jmbejara Nov 20 '13 at 20:43
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    @jmbejara. Exactly. Nevertheless I missed the reason why Gauss-Markov still holds in the answer given above. (The focus is on the correct model structure implying mean zero errors. This might or might not hold if you transform or don't transform variables.) – Michael M Nov 20 '13 at 20:56
  • So just to keep it simple: so gauss markov holds for some semi-log models, right? Log-log it is ok (it is intuitive because, well, it is linear), I get it. If we would graph a simple linear regression, y=a+ßlogx+e, we would not see a linear relationship by any means between dependent and independent variable, correct? But if the betas are are linear and the other Gauss Markov assumptions hold, we still can estimate the betas with OLS and have them BLUE. Please tell me if I understood it correctly. – John Doe Nov 20 '13 at 23:02
  • Depends how you define 'independent variable'. If the true process is $Y_i=a + bX_i + \varepsilon_i$ and the errors are uncorrelated with mean zero and constant variance, then the OLS estimator for a and b are BLUE. Taking a log in the same setting would destroy the mean zero assumtion (and also equal variance). – Michael M Nov 21 '13 at 06:14