Linearization of multiplicative and exponential regression models and their OLS estimation

Question

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?

Answer 1

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 $\delta=\ln(\alpha)$.

Answer 2

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.