I'm not really sure why I couldn't take any of these models and perform OLS on them. Would appreciate help in understanding/finding the answer

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  • $\begingroup$ From memory, with OLE you are trying to minimise the sum of the square of the differences. This occurs when the derivative of this is 0. You could try this approach on each of the four equations in the question, ending up with another equation for each that gives you estimators for the parameters to take. $\endgroup$ Mar 10, 2017 at 20:39

1 Answer 1


Recall that one of the assumptions (actually, a constraint) of the OLS method is a linearity over the parametric set, i.e., check whether the first derivative w.r.t $\beta$ depends on the unknown $\beta$.

1) Simple linear regression. $\frac{\partial }{\partial \beta} \mathbb{E}[Y|X=x] = \ln (x) $. You can denote $\ln(x) = x^*$ if you find it more convenient.

2) Try $1/Y_i = \beta_1 + a \frac{1}{X_i}\to Y_i^* = \beta_0^* + \beta_1^*X_i^*$.

3) $\ln Y_i = \ln(a) + \beta_1\ln(X_i)$

4)Just denote $X_i^2 = X_i^*$.

5) $\ln(Y_i) = \ln (a) + \beta_1X_i. $

  • $\begingroup$ Awesome- thank you! Really appreciate it- I also have another question I'd really appreciate some guidance on, you're a lifesaver math.stackexchange.com/questions/2180873/… $\endgroup$ Mar 10, 2017 at 22:23
  • $\begingroup$ You're welcome. I've answered the other question too. If the answers are correct/satisfactory - you can click on "accept". $\endgroup$
    – V. Vancak
    Mar 10, 2017 at 23:11
  • $\begingroup$ For (3) I suspect $\ln Y_i = \ln(a)+\beta_1\ln (X_i)$ $\endgroup$
    – Henry
    Mar 11, 2017 at 0:12

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