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$\begingroup$

I can not understand how to get from one side to the other.

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

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You need to show $\sum \overline{x} (y_i - \overline{y}) = 0$. This sum is equal to $$\overline{x} \left(\sum y_i - \sum \overline{y} \right)$$ and you have $\sum \overline{y} = n \overline{y}$ since you're just summing up the same number $n$ times. I'll let you try to finish it from here. (Hint: What is the definition of $\overline{y}$ in terms of $y_1,\dots,y_n$?)

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    How am I getting x-bar out of the equation? I can see sum[x(y-ybar)] --> sum(xy - xybar) --> sum(xy) - sum(xybar) but I do not see how to get x-bar from this. I understand that sum(xi) = nxbar. So can sum(xy) = sum(x)*sum(y)?2012-03-29
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    $\overline{x}$ is just a number that does not depend on $i$, so it can be pulled out of the sum. Formally, if $\lambda, z_1, \dots, z_n$ are numbers, then $\sum \lambda z_i = \lambda \sum z_i$.2012-03-29
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    Is this a correct first step? $\sum (x_i- \bar{x})(y_i - \bar{y}) = (\sum x_i- \sum\bar{x})(\sum y_i - \sum \bar{y})$ = $(\sum x_i - n \bar{x})(\sum y_i - n \bar{y}) $2012-03-29
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    You can't split the sum up across multiplication, but what you can do is write $\sum (x_i - \overline{x})(y_i - \overline{y}) = \sum (x_i y_i - x_i \overline{y} - \overline{x} y_i + \overline{x}\overline{y}$ and then sum up each term individually using the fact that $\sum x_i = n \overline{x}$ and similarly for the $y_i$'s.2012-03-29
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    Now I got it.... $\sum (x_i- \bar{x})(y_i - \bar{y}) = \sum (x_iy_i - x_i \bar{y} - \bar{x}y_i + \bar{x} \bar{y}) $ = $\sum (x_iy_i - x_i \bar{y}) - \sum (\bar{x} y_i + \bar{x} \bar{y})$ = $ \sum[x_i (y_i - \bar{y})] + \bar{x} \sum(y_i - \bar{y}) $ so then $\bar{x} \sum(y_i - \bar{y}) = \bar{x}(n \bar{y} - n\bar{y}) = 0 $2012-03-29
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    forgot about the dang + sign at the end... took me a minute. Thanks again for the help.2012-03-29