0
$\begingroup$

I am considering the model:

$$ y_t = \beta_0\left(\Pi_{i=0}^{K}x_{i,t}^{\beta_i}\right)\left(\Pi_{j = K+1}^{L}e^{\beta_{j}x_{j,t}}\right) $$

where we want to have multiplicative effect between variables and linear for the others. My question is the following: in this model, how can we interprate and estimate the coefficients ? I mean we cannot keep the interpretation from the simple linear regression model where the coefficient $\beta_i$ represents the impact of the explanatory variable $x_{i,t}$ on the explained variable $y_t$.

The advantage of a such model is to separate explanatory variables in two sets in order to have something realistic about the model when we should have $y_t = 0$ because in the set of fundamental explanatory variables we have one variable $x_{i,t} = 0$ so there is a kind of ''interaction" between variables.

$\endgroup$
13
  • 2
    $\begingroup$ If you take the logs of both sides, it becomes a linear model - linear in $\log x_i$ for the first $K$ terms and linear in the $x_i$ themselves for the remaining $L-K$ terms. $\endgroup$
    – jbowman
    Commented Apr 6, 2023 at 15:00
  • 1
    $\begingroup$ since you have x^beta, it is not a linear model. IN a linear model, if you want to have multiplicative effect of a variable on the outocme you can just log-transform it. $\endgroup$
    – rep_ho
    Commented Apr 6, 2023 at 15:02
  • 2
    $\begingroup$ This model is incompletely specified. How it is fit and interpreted depends on your probability model for the deviations between the left and right sides. What is that model? $\endgroup$
    – whuber
    Commented Apr 6, 2023 at 15:50
  • 2
    $\begingroup$ The coefficients are exactly the same as in the model formulation you've written, so the interpretation is too. Note also @whuber's comment. $\endgroup$
    – jbowman
    Commented Apr 6, 2023 at 16:50
  • 1
    $\begingroup$ Start with the log-log model (which is readily fit using ordinary least squares regression) and study the residuals. But before doing that, please explain what you mean about the zero values: that raises many red flags but it's unclear what you're trying to say. $\endgroup$
    – whuber
    Commented Apr 6, 2023 at 21:45

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.