Cindy Noble

2022-10-05

How to write an equation where both independent variables and dependent variables are log transformed in a multiple regression?

How to write the multiple regression model when both the dependent variable and independent variables are log-transformed?

I know that without any log transformation the linear regression model would be written as enter image description here

$y={\beta}_{0}+{\beta}_{1}({x}_{1})+{\beta}_{2}({x}_{2})+\dots $

But now I have transformed both my dependent variables and independent variable with log. So is correct to write as enter image description here $\mathrm{log}(y)={\beta}_{0}+{\beta}_{1}\cdot \mathrm{log}({x}_{1})+{\beta}_{2}\cdot \mathrm{log}({x}_{2})+\dots $

Or since I am transforming both sides of question so can I write it as enter image description here

$\mathrm{ln}(y)={\beta}_{0}+{\beta}_{1}({x}_{1})+{\beta}_{2}({x}_{2})+\dots $

How to write the multiple regression model when both the dependent variable and independent variables are log-transformed?

I know that without any log transformation the linear regression model would be written as enter image description here

$y={\beta}_{0}+{\beta}_{1}({x}_{1})+{\beta}_{2}({x}_{2})+\dots $

But now I have transformed both my dependent variables and independent variable with log. So is correct to write as enter image description here $\mathrm{log}(y)={\beta}_{0}+{\beta}_{1}\cdot \mathrm{log}({x}_{1})+{\beta}_{2}\cdot \mathrm{log}({x}_{2})+\dots $

Or since I am transforming both sides of question so can I write it as enter image description here

$\mathrm{ln}(y)={\beta}_{0}+{\beta}_{1}({x}_{1})+{\beta}_{2}({x}_{2})+\dots $

Reagan Tanner

Beginner2022-10-06Added 8 answers

I am afraid that all your transformed forms are false.

Starting from

$y={\beta}_{0}+{\beta}_{1}({x}_{1})+{\beta}_{2}({x}_{2})+...$

you can write :

$\mathrm{ln}(y)=\mathrm{ln}{\textstyle (}{\beta}_{0}+{\beta}_{1}({x}_{1})+{\beta}_{2}({x}_{2})+...{\textstyle )}$

This is not equivalent to ${\beta}_{0}+{\beta}_{1}(\mathrm{ln}({x}_{1}))+{\beta}_{2}(\mathrm{ln}({x}_{2}))+...$

Also you can change of variables :

${X}_{1}={e}^{{x}_{1}}\phantom{\rule{1em}{0ex}};\phantom{\rule{1em}{0ex}}{X}_{2}={e}^{{x}_{2}}\phantom{\rule{1em}{0ex}},...$

$y={\beta}_{0}+{\beta}_{1}\mathrm{ln}({X}_{1})+{\beta}_{2}\mathrm{ln}({X}_{2})+...$

And with the change of :

$Y={e}^{y}$

$\mathrm{ln}(Y)={\beta}_{0}+{\beta}_{1}\mathrm{ln}({X}_{1})+{\beta}_{2}\mathrm{ln}({X}_{2})+...$

Sorry if I misunderstood your question.

Starting from

$y={\beta}_{0}+{\beta}_{1}({x}_{1})+{\beta}_{2}({x}_{2})+...$

you can write :

$\mathrm{ln}(y)=\mathrm{ln}{\textstyle (}{\beta}_{0}+{\beta}_{1}({x}_{1})+{\beta}_{2}({x}_{2})+...{\textstyle )}$

This is not equivalent to ${\beta}_{0}+{\beta}_{1}(\mathrm{ln}({x}_{1}))+{\beta}_{2}(\mathrm{ln}({x}_{2}))+...$

Also you can change of variables :

${X}_{1}={e}^{{x}_{1}}\phantom{\rule{1em}{0ex}};\phantom{\rule{1em}{0ex}}{X}_{2}={e}^{{x}_{2}}\phantom{\rule{1em}{0ex}},...$

$y={\beta}_{0}+{\beta}_{1}\mathrm{ln}({X}_{1})+{\beta}_{2}\mathrm{ln}({X}_{2})+...$

And with the change of :

$Y={e}^{y}$

$\mathrm{ln}(Y)={\beta}_{0}+{\beta}_{1}\mathrm{ln}({X}_{1})+{\beta}_{2}\mathrm{ln}({X}_{2})+...$

Sorry if I misunderstood your question.

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