Suppose that I wish to estimate the parametes $\alpha$ and $\beta$ in the following regression model: $$ Y=K^{\alpha}L^{\beta}\epsilon $$ A standard procedure is to take logs and estimate $$ \text{log}\left(Y\right)=\alpha\text{log}\left(K\right)+\beta\text{log}\left(L\right)+\text{log}\epsilon $$ Now, even if $\epsilon$ is statistically mean independent of $K,L$ in the original multiplicative model, it is not necessarily the case in the second model. This is because of Jensen's inequality, and the fact that if there is heteroskedasticity in the original model, then the conditional mean of$\text{log }\epsilon$ will vary, depending on the form of heteroskedasticity, and if it is related to levels of $K$ and $L.$ In that case, OLS will produce biased estiamtes on $\alpha$ and $\beta.$ My question is as follows- in the original model: can we not redefine the minimization problem as: $$ \text{min}\sum_{i=1}^{N}\left(\frac{Y}{K^{\alpha}L^{\beta}}\right)^{2} $$ to obtain NLS estimates instead? This should obviate the endogeneity problem in my opinion..is this OK?

  • $\begingroup$ I realized this is not a well-defined minimization problem: alpha and beta would be set to infinity in this process... $\endgroup$ – ChinG Aug 8 '18 at 14:31
  • $\begingroup$ It's easy to see why: quadratic measures work best for simmetric errors around zero. your error is probably centered abkve zero $\endgroup$ – Aksakal Aug 8 '18 at 15:45

The mean independence assumption is mostly a technical assumption that guarantees some desirable properties of the OLS and other estimators.

This means that it matters to make it at the level of estimation. In other words, we first apply the logarithmic transformation and then we impose the assumption of mean-independence.

Another standard approach is to insert the error term as $e^{\epsilon}$ in the output function.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.