1
$\begingroup$

Suppose $(X_{1,t},X_{2,t})$ is i.i.d bivariate normally distributed with correlation $p$ and both $X_{1,t}$ and $X_{2,t}$ have mean zero and variance 1. If we consider $X_{1,t}$ and $X_{2,t}$ as regressors then the first part of the OLS estimator is given by $(X'X)^{-1}X'$ where $X=[X_1, X_2]$. If we now zoom in on column $t$ of $(X'X)^{-1}X'$ we see the two elements $$\frac{X_{1,t}\sum_{t=1}^{n}X_{2,t}^2-X_{2,t}\sum_{t=1}^{n}X_{1,t}X_{2,t}}{\sum_{t=1}^{n}X_{1,t}^2\sum_{t=1}^{n}X_{2,t}^2-(\sum_{t=1}^{n}X_{1,t}X_{2,t})^2}$$, and $$\frac{X_{2,t}\sum_{t=1}^{n}X_{1,t}^2-X_{1,t}\sum_{t=1}^{n}X_{1,t}X_{2,t}}{\sum_{t=1}^{n}X_{1,t}^2\sum_{t=1}^{n}X_{2,t}^2-(\sum_{t=1}^{n}X_{1,t}X_{2,t})^2}$$

It turns out that as I decrease the correlation, $p$, between $X_{1,t}$ and $X_{2,t}$ then the two elements above are more likely to be simultaneously above 0.

Is there any way to derive this rigorously from the two elements above or to grasp this intuitively in the context of the regression model?

$\endgroup$
3
  • $\begingroup$ What does this question have to do with $Y$ or $\epsilon$? Isn't it just asking about $X_1$ and $X_2$? $\endgroup$
    – whuber
    Commented Jun 16, 2017 at 18:53
  • $\begingroup$ You are right, it is just about $X_1$ and $X_2$. The model was meant for sketching the context from which my question arises, but maybe I should delete it if it obscures the question. $\endgroup$
    – Joogs
    Commented Jun 16, 2017 at 19:10
  • 1
    $\begingroup$ Extraneous information like that can distract readers from understanding what you really need. BTW, aren't you asking about column $t$ of $(X^\prime X)^{-1}X^\prime$ (which is a $2\times n$ matrix)? $\endgroup$
    – whuber
    Commented Jun 16, 2017 at 20:55

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.