11
$\begingroup$

If this is a duplicate question, please point to the right way, but the similar questions I've found here haven't been sufficiently similar. Suppose I estimate the model $$Y=\alpha + \beta X + u$$

and find that $\beta>0$. However, it turns out that $X=X_1+X_2$, and I suspect $\partial Y/\partial X_1 \ne \partial Y / \partial X_2$, and in particular, that $\partial Y/\partial X_1 > \partial Y / \partial X_2$. So I estimate the model $$Y=\alpha + \beta_1 X_1 + \beta_2 X_2 +u$$and find significant evidence for $\beta_1,\beta_2>0$. How can I then test whether $\beta_1> \beta_2$? I considered running another regression $$Y=\alpha +\gamma(X_1 - X_2) + u$$ And testing whether $\gamma>0$. Is this the best way?

Also, I need to generalize the answer to many variables, i.e. suppose we have $$Y=\alpha + \beta^1 X^1 + \beta ^2 X^2 + \dots + \beta^nX^n + u$$ where for each $j=1,\dots,n$, $X^j=X_1^j+X_2^j$, and I would like to test for each $j$ whether $\partial Y/\partial X_1^j \ne \partial Y / \partial X_2^j$.

By the way, I am primarily working in R.

$\endgroup$
16
$\begingroup$

Is this the best way?

No, that won't actually do what you want.

Let $\gamma = \beta_1 - \beta_2$.

$\beta_1 X_1 + \beta_2 X_2 = (\gamma+\beta_2) X_1 + \beta_2 X_2 = \gamma X_1 + \beta_2 (X_1 + X_2)$.

Hence the model $Y=\alpha + \beta_1 X_1 + \beta_2 X_2 +u$ becomes $Y=\alpha + \gamma X_1 + \beta_2 (X_1 +X_2) +u$

So you supply predictors $X_1$ and $X_3=X_1+X_2$, and then you can perform a straight hypothesis test of whether $\gamma>0$ (against the null of equality).

$\endgroup$

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.