In a simple linear regression

$SSR = \sum(\hat{y} - \bar{y})^2$

$SSXY = \sum(x - \bar{x})(y-\bar{y})$

$SSX = \sum(x - \bar{x})^2$

$SSR$ can be computed by dividing $SSXY^2$ by $SSX$. Namely, $SSR = SSXY^2/SSX$

Could someone illustrate mathematically how this is obtained? Thanks!!


1 Answer 1


From the answers at Is there a way to use the covariance matrix to find coefficients for multiple regression? we know the least squares equations for the estimated coefficient $\widehat\beta$ are

$$SS_X \widehat\beta = SS_{XY}$$

where the predicted values are given by


Squaring both sides and summing, recognizing that $\Sigma_i x_i^2=SS_X$, yields

$$SS_R=SS_X \widehat\beta^2.$$

Plugging in the solution $\widehat\beta^2 = \left(SS_{XY} / SS_X\right)^2$ from the first equation turns this into

$$SS_R=SS_X \left(SS_{XY} / SS_X\right)^2 =SS_{XY}^2/SS_X.$$

  • $\begingroup$ I can follow the logic of your explanation, but after the step $\widehat{y}_i=x_i\widehat{\beta}.$, I can't seem to get things right. For example, when I square $\widehat{y}_i$, what I get is the sum of the square of each predicted value, not $\sum(\hat{y}-\bar{y})$. Could you explain a little as to what might be my mistake? (I was doing it in R, so I basically did y_hat %*% y_hat) $\endgroup$
    – Alex
    Jul 30, 2014 at 0:19
  • 1
    $\begingroup$ Without any loss of generality, you may assume that the means of the $x$ and $y$ are zero (center the variables). $\endgroup$
    – whuber
    Jul 30, 2014 at 1:45

Your Answer

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

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