0
$\begingroup$

I'm studying for an Econometrics exam and going over an old lecture slide (see picture). In it the lecturer is deriving \beta_1 in a basic OLS regression with one variable.

I understand everything up to equation 10 and I am able to calcualte eq.11, but I do not understand why the additions/subtractions need to be made in eq. 11. From what I can tell \beta_1 in 10 is not equal to \beta_1 in 11?

Can someone explain why the calcutations are made in this way? And is equation 10 equal to 11? This is very important for me to know because deriving \beta_1 has been a question in basically every exam.

enter image description here

EDIT: I realize now that this question should probably be posted on the statistics-SE instead. I am not eligible to migrate questions so I would appreciate the help of a moderator for doing this. :)

$\endgroup$

1 Answer 1

3
$\begingroup$

Working backwards from the numerator in 11 to the numerator in 10, we have:

$$\begin{align}\tag{11}\sum_{i=1}^n(X_i-\bar{X})(Y_i-\bar{Y}) &=\sum_{i=1}^n (X_iY_i-X_i\bar{Y}-\bar{X}Y_i+\bar{X}\bar{Y})\\ &=\sum_{i=1}^n X_iY_i- \sum_{i=1}^nX_i\bar{Y}- \sum_{i=1}^n\bar{X}Y_i+\sum_{i=1}^n\bar{X}\bar{Y}\\ &=\sum_{i=1}^n X_iY_i- \bar{Y}\sum_{i=1}^nX_i-\bar{X} \sum_{i=1}^nY_i+n\bar{X}\bar{Y}\\ &=\sum_{i=1}^n X_iY_i- \bar{Y}(n\bar{X})-\bar{X} (n\bar{Y})+n\bar{X}\bar{Y}\\ &=\sum_{i=1}^n X_iY_i- n\bar{X}\bar{Y}-n\bar{X}\bar{Y}+n\bar{X}\bar{Y}\tag{B}\\ &=\sum_{i=1}^n X_iY_i- n\bar{X}\bar{Y}\tag{A}\\ &=\sum_{i=1}^n X_iY_i- n \left(\frac{1}{n}\sum_{i=1}^nX_i\right)\left(\frac{1}{n}\sum_{i=1}^nY_i\right)\\ &=\sum_{i=1}^n X_iY_i- \frac{1}{n}\left(\sum_{i=1}^nX_i\right)\left(\sum_{i=1}^nY_i\right) \tag{10}\\ \end{align}$$

The term $\bar{X}\sum_{i=1}^nY_i=n\bar{X}\bar{Y}$ is added and subtracted when going from $(\text{A})$ to $(\text{B})$.

$\endgroup$
1
  • $\begingroup$ Thank you! It took me a while to get, even whith your step-by-step solution. Now onto the denominator! :D $\endgroup$
    – AoMRos
    Jul 20, 2023 at 17:47

Your Answer

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

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