# Regression proof for decomposition of sums of squares [duplicate]

I got as far as distributing the summation across the Left Side so that I have:

$$\sum_i y_i^2 - \sum_i 2 y_i \bar{y} + \sum_i \bar{y}^2$$

Not sure where to go from there.

## marked as duplicate by Xi'an, John, Silverfish, whuber♦Sep 26 '16 at 14:31

• Please add the [self-study] tag & read its wiki. – Silverfish Sep 24 '16 at 21:50

Now we just need to show that the rightmost term equals zero. You should be able to convince yourself that $\sum_{i=1}^{n} (y_i - \hat{y}_i) = 0$ by plugging in the formula for $\hat{y}_i$ so we only need to prove that $\sum_{i=1}^{n} (y_i - \hat{y}_i) \hat{y}_i = 0$,
and the remaining steps can be found in my answer to this question: Proof that $\hat{\sigma}^2$ is an unbiased estimator of $\sigma^2$ in simple linear regression.
• I actually had some typos in my original answer, sorry about that. Yes you should be able to show that $\sum_{i=1}^{n} (y_i - \hat{y}_i) = 0$. Anyways, I showed the rest of the steps from there. – dsaxton Sep 26 '16 at 0:57