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In the lecture notes, in order to prove that TSS=ESS+RSS.

My instructor writes that:

$$TSS = \sum_t^n (y_t -\bar{y})^2= \sum_t^n y_t^2 -n\bar{y}^2$$

I don’t understand the second equality. Instead, I would write

$$ = \sum_t^n y_t^2 - 2\sum_t^n y_t\bar{y} + n\bar{y}^2$$. But the instructor’s expression is different. How does he get it?

Tom Solid
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    This is a big FAQ, but since it's always expressed as formulas, it's almost impossible to search for answers. In the end it's pure algebra, depending on the observation that the residuals sum to zero. – whuber Dec 03 '20 at 13:00
  • I found the duplicate with this search: https://stats.stackexchange.com/search?q=TSS%3DESS%2BRSS. – whuber Dec 03 '20 at 14:23

1 Answers1

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$$\sum_t^n (y_t -\bar{y})^2= \sum_t^n ( y_t^2 -2\bar{y}y_t + \bar{y}^2) =$$ $$=\sum_t^n y_t^2 -2\bar{y}\sum_t^n y_t + \sum_t^n \bar{y}^2 = \sum_t^n y_t^2 -2\bar{y}n\bar{y} + n\bar{y}^2 $$

which equals to $$ \sum_t^n y_t^2 -n\bar{y}^2$$

QED

Tom Solid
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