I have to apply a transformation to the following to fit it to the general linear regression model $Y_{i} = \beta_{0} + \beta_{1}X_{i1} + \beta_{2}X_{i2} + ... + \beta_{p}X_{i(p-1)} + \epsilon_{i}$ With p-1 predictor variables. My problem is this:
$Y_{i} = \epsilon* exp[\beta_{0} + \beta_{1}X_{i} + \beta_{2}X_{i}^2]$
What I am thinking is to let $Y' = ln(Y)$ and to let $X_{i2} = X_{i}^2$, giving me $Y'_{i} = (\beta_{0} + \beta_{1}X_{i1} + \beta_{2}X_{i2}) *ln( \epsilon_{i})$
What do I do with that stray $ln(\epsilon_{i})$?
Or is this a case where it cannot be expressed in the standard form?
Thanks!
If I properly understand, you need to fit a model which write $$y=e^{\beta_0+\beta_1 x+\beta_2 x^2}$$ which is nonlinear with respect to the parameters and so it will require nonlinear regression which needs good estimates for starting iterations.
As you notice, you can make this problem linear writing $$\log(y)=z=\beta_0+\beta_1 x+\beta_2 x^2$$ and obtain the parameters using the classical linear lesat-square regression.
However, the values you obtain must be considered as estimates of the real parameters because what you minimized is $$\sum_{i=1}^n \Big(\beta_0+\beta_1 x_i+\beta_2 x_i^2-\log(y_i)\Big)^2$$ while the original problem requires the minimisation of $$\sum_{i=1}^n \Big(e^{\beta_0+\beta_1 x_i+\beta_2 x_i^2}-y_i\Big)^2$$ which will not correspond to the same.
So, you do not face any problem with the $\epsilon_i$'s as long as you solve the second problem using the first for generating easily the estimates of the $\beta$'s.