Explained sum of squares: Difference between revisions

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Covariance and variance doesn't make sense for realised (non-stochastic) variables. So removed the expressions with them and expanded the last term.
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:<math>\hat{y_i} = \hat{a} + \hat{b}x_i</math>
:<math>\bar{y} = \hat{a} + \hat{b}\bar{x}</math>
:<math>\hat{b} = \frac{Cov(x,y)}{Var(x)} = \frac{\sum_{i=1}^n (x_i-\bar{x})(y_i-\bar{y})}{\sum_{i=1}^n (x_i-\bar{x})^2}</math>
So,
:<math>\hat{y_i} - \bar{y} = \hat{b}(x_i - \bar{x})</math>
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Therefore,
:<math>\sum_{i=1}^n 2(\hat{y}_i-\bar{y})(y_i-\hat{y}_i) = 2\hat{b}\sum_{i=1}^n (x_i-\bar{x})(y_i-\hat{y}_i) = 2\hat{b}\sum_{i=1}^n (x_i-\bar{x})((y_i - \bar{y}) - \hat{b}(x_i - \bar{x}))</math>
:<math> 2b\sum_{j= 21}^{n}\left((x_j-\bar{x})(y_j-\hat{by}_j)-\frac{\sum_{i=1}^n (Cov(x_i-\bar{x,})(y_i-\bar{y})}{\sum_{i=1}^n (x_i- \hatbar{bx})^2}Var(x_j-\bar{x})^2\right) = 2\hat{b}n (0) = 0</math>
 
==Partitioning in the general ordinary least squares model==