2
$\begingroup$

I have a variable X1 = (a - b) / (a + b). This variable shows a higher correlation to Y that any of (a, Y) and (b, Y).

In a multiple regression model like Y ~ X1, X2, does it make sense to use the X1 formula, or should I always use the base variables a and b?

In this post somebody pointed out that

Intuitively you'd be a lot more confident about inferences from an observed ratio of 1 (boys to girls) if it came from seeing 100 boys and 100 girls than > from seeing 2 and 2. Consequently, if you have covariates you'll have more information about their effects and potentially a better predictive model.

Fine, but can the multiple linear model rebuild the same (X1, Y) predictive relationship just by a and b least square analysis?

$\endgroup$
  • $\begingroup$ As @jbowman gets at below, I think your theory about how the world is likely to work is important for choosing how you specify your model... do you have a good theoretical reason to believe a variable Z = (a-b)/(a+b) is related to Y? $\endgroup$ – Michael Bishop Dec 5 '11 at 0:19
2
$\begingroup$

The short answer is no. The relationship $Y = \beta_1 (a-b)/(a+b) + \beta_2 X_2 + e$ is not the same as $Y = \beta_1 a + \beta_2 b + \beta_3 X_2 + e$, and you can't get from one to the other.

Having said that, there's nothing intrinsically wrong with transforming your right hand side variables before running a regression. Consider predicting the weight of a tree from its measured height and circumference at its base; you're much better off computing its approximate volume from height and circumference, and estimating a regression with weight as the dependent variable and volume as the independent variable, than using height and circumference as the independent variables.

If the transformed variable will have a lot of measurement error associated with it, which seems to be your concern, then other problems will arise, known generically as "errors in the variables." What to do about that depends in part upon how severe the problem is, whether it's localized to a small subset of the data, and other factors - but that's a different question.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.