# Linear regression expression of a non-linear model

$$Y=\frac{x_1x_2}{β_0+β_1x_1+β_2x_2}$$

It was written on some slide of my econometrics class that such a model could be expressed in the form of a linear model, but I am struggling to derive it by myself.

Here is what I have attempted so far:

$$\frac{1}{Y} = \frac{\beta_0+\beta_1x_1+\beta_2x_2}{x_1x_2}$$

$$\frac{1}{Y} = \frac{\beta_0}{x_1x_2}+\frac{\beta_1x_1}{x_1x_2}+\frac{\beta_2x_2}{x_1x_2}$$

$$\frac{1}{Y} = \frac{\beta_0}{x_1x_2}+\frac{\beta_1}{x_2}+\frac{\beta_2}{x_1}$$

$$Y^*=0 + \beta_0X_1^*+\beta_1X_2^*+\beta_2X_3^*$$

where $$X_1^*=\frac{1}{x_1x_2}$$, $$X_2^*=\frac{1}{x_2}$$ and $$X_3^*=\frac{1}{x_1}$$

• Welcome to Cross Validated! I like your middle equation. Why do you have a problem with lacking an intercept?
– Dave
Commented Oct 21, 2022 at 2:06
• Thank you for your answer. It's a good question that you raise lol. I felt like there should be one but I could run the model without intercept... I have edited my post and corrected for the re-inversion mistake. What do you think ? Commented Oct 21, 2022 at 10:28
• Try it only with y=x_1/beta0, and see what happens, maybe try a few values. Commented Oct 22, 2022 at 21:18

$$\dfrac{1}{Y}=\beta_0\dfrac{1}{x_1x_2}+\beta_1\dfrac{1}{x_2}+\beta_2\dfrac{1}{x_1}$$

The $$\dfrac{1}{x_1}$$, $$\dfrac{1}{x_2}$$, and $$\dfrac{1}{x_1x_2}$$ are then nonlinear basis functions of the original variables.

Yes, the inversion lacks an intercept, but linear models don’t require intercepts. It’s often advisable to include an intercept in your model, but an intercept-free linear model remains a linear model.