I asked a similar question on Cross Validated, but got no answer. The following question is sufficiently different.

Consider the following deterministic relationship:$$Y_{t}=C_{t}+I_{t}+G_{t}+(X_{t}-M_{t})$$

If we run an OLS regression of Y on the covariates, of course, we get an $R^{2}$ of 1, and the vector of coefficients equal to their population counterparts, i.e. $$\hat{\beta}= \beta=1$$

Now, this means the marginal effect of each regressor is constant. However, the contribution of each regressor depends on its relative variance to y. In fact, in the univariate case, it can be shown that:$$R^{2}=\beta^{2}\frac{var(x)}{var(y)}$$ which shows that even when the marginal effect of $x$ on $y$ is large, its contribution in the partial $R^{2}$ sense is high only when it varies a lot, causing y to vary with it as well.

Now, consider the same above expression:$$Y_{t}=C_{t}+I_{t}+G_{t}+(X_{t}-M_{t})$$

Let us divide both sides by $Y_{t}$ for the entire vector of coefficients and the design matrix.

We get:$$1=\frac{C_{t}}{Y_{t}}+\frac{I_{t}}{Y_{t}}+\frac{G_{t}}{Y_{t}}+\frac{(X_{t}-M_{t})}{Y_{t}}$$

This holds for each realization of $Y_{t}$ and the design matrix by construction. Now, each regressor here is the relative contribution (which we can multiply by 100 to get percentage contribution) to the dependent variable.

Now that we have the relative contribution for each regressor for each observation, we can get the mean contribution of the regressor across observations.

I ran some simulations, and found that the mean contribution of each regressor is close, but not exactly equal to the partial $R^{2}$ of each regressor. Is there any relationship between them? Intuitively, they should be the same, right? Thanks a lot!

  • $\begingroup$ I think this question would be more appropriate if posted in cross-validated. $\endgroup$ – Fuca26 Jan 28 '16 at 8:27

Your Answer

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

Browse other questions tagged or ask your own question.