# Linearization of multiplicative and exponential regression models and their OLS estimation

So far I have been under the impression that you can "linearize" multiplicative models of the form (1) $$y=\alpha * \beta_1x_1 * \beta_2x_2 * \beta_3x_3$$ and exponential models of the form (2) $$y=\alpha * x_1^{\beta_1} * x_2^{\beta_2} * x_3^{\beta_3}$$ by taking the logarithm and then doing a regular OLS estimation.

For exponential models this makes sense to me as the logarithm gives us (3) $$y=\alpha + \beta_1*log(x_1) + \beta_2*log(x_2) + \beta_3*log(x_3)$$ but for the multiplicative model we receive (4) $$y=\alpha + log(\beta_1*x_1) + log(\beta_2*x_2) + log(\beta_3*x_3)$$.

Can we estimate a regression like (4) with standard OLS? How do we tell the statistics software the difference to (3)? How would a regression estimation like this look like in R using the lm command?

• Model (1) isn’t identified. It’s indistinguishable from $y=\beta x_1 x_2 x_3$. Dec 20, 2018 at 13:15

There is no need to "linearize" the "multiplicative" form. You have $$\alpha\beta_1\beta_2\beta_3x_1x_2x_3$$. So simply multiplying all three x variables, then the coefficient of their product is $$\alpha\beta_1\beta_2\beta_3$$. In this setup, you're assuming there is no intercept. For the exponential form, everything you have in (3) is correct other than the $$\alpha$$ in the log solution should be $$\ln(\alpha)$$.

In R, you'd have the first setup as:

lm(y ~ 0 + x1:x2:x3) # 0 + takes out the intercept


You get one coefficient, it is $$\hat\alpha\hat\beta_1\hat\beta_2\hat\beta_3$$

For the second setup:

lm(log(y) ~ log(x1) + log(x2) + log(x3))


You get four coefficients: $$\{\hat\delta,\hat\beta_1,\hat\beta_2,\hat\beta_3\}$$ where $$\delta=\ln(\alpha)$$.

You can rewrite the model

$$(1) \ \ y_i = \alpha\beta_1x_1\beta_2x_2\beta_3x_3u_i$$

by taking logs

$$\log y_i = [\log \alpha + \log\beta_1 + \log \beta_2 + \log\beta_3]+ \log x_1 +\log x_2+ \log x_3 + \log u_i$$

and in estimating the equation

$$\log y_i = \lambda_0 + \lambda_1 \log x_1 +\lambda_2 \log x_2+ \lambda_3 \log x_3 + e_i$$ you would get an estimate of $$\lambda_0$$ $$\lambda_0 = [\log \alpha + \log\beta_1 + \log \beta_2 + \log\beta_3]$$

as the Laconic says $$(\alpha,\beta_1,\beta_2,\beta_3)$$ are not seperately identifiable.

You would also get estimates of $$\lambda_1,\lambda_2,\lambda_3$$ and could test if they are all equal to 1 as they should be if $$(1)$$ where the true model.