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I've worked the slope all the way down to $\sum [x_i(y_i - \bar{y})] = \hat\beta_1 \sum[x_i(x_i - \bar{x})]$

But I can not figure out how to show the steps for:

$\sum[x_i(y_i - \bar{y})] = \sum(x_i - \bar{x})(y_i - \bar{y})$

and

$\hat \beta_1 \sum[x_i(x_i - \bar{x})] = \hat \beta_1 \sum(x_i - \bar{x})^2$

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  • $\begingroup$ Probably $x_i$ are centralised ( Original variable is subtracted by its mean value). If that is the case $\bar{x} =0$. $\endgroup$
    – vinux
    Mar 27 '12 at 3:51
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    $\begingroup$ @vinux, usually if some term is in the equation it is assumed that it is not zero. $\endgroup$
    – mpiktas
    Mar 27 '12 at 7:25
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Add and subtract and then look whether the unwanted term disappears:

$$\sum[x_i(y_i-\bar y)]=\sum[(x_i-\bar x+\bar x)(y_i-\bar y)]=\sum[(x_i-\bar x)(y_i-\bar y)]+\sum [\bar x(y_i-\bar y)].$$

Now

$$\sum [\bar x(y_i-\bar y)]=\bar x\sum [y_i-\bar y]=\bar x\left[\sum y_i-n\bar y\right]=0,$$

where $n$ is the number of the terms in the sum. Apply the same trick for the second equation.

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