I can't understand how did they solve the second part in this attached image. The part where (xi-(average of x))(yi-(average of y)) becomes xi (yi-(average of y)) then switches and finally we get the end general form. Can you please explain the process? I understand A.7 but I don't get A.8. I cannot find the rules they used for this. A.7 is pretty straightforward. Thank you so much!
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Very confused. Need urgent help
- Thread starter jabadaba
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HallsofIvy
Elite Member
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- Jan 27, 2012
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Clearly \(\displaystyle \sum (x_i- \overline{x})(y_i- \overline{y})= \sum x_i(y_i- \overline{y})- \sum \overline{x}(y_i- \overline{y})\). Since \(\displaystyle \overline{x}\) is a constant, that is \(\displaystyle \sum x_i(y_i- \overline{y})- \overline{x}\sum (y_i- \overline{y})\). The key point is that \(\displaystyle \sum (y_i- \overline{y})= \sum (y_i)- \overline{y}= \overline{y}- \overline{y}= 0\).
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Go back to the definition of \(\displaystyle \overline x\),I can't understand how did they solve the second part in this attached image. The part where (xi-(average of x))(yi-(average of y)) becomes xi (yi-(average of y)) then switches and finally we get the end general form. Can you please explain the process? I understand A.7 but I don't get A.8. I cannot find the rules they used for this. A.7 is pretty straightforward. Thank you so much!
\(\displaystyle \overline x = \frac{\Sigma x_i}{n}\)
so
\(\displaystyle \Sigma (x_i - \overline x) = \Sigma (x_i) - n \overline x\)
Similiarly for \(\displaystyle \Sigma (y_i - \overline y)\)
Thus, for any a,
\(\displaystyle a * \Sigma (x_i - \overline x) = a * \Sigma (y_i - \overline y) \) = ?
Thank you a lot guys. I just have one more question. I understand how the constant becomes zero. Then what is the explanation for xi (yi-y) being equal to yi (xi-x)? Does it have something to do with the deviation of the x data and the y data having some sort of relation?
Thank you a lot guys. I just have one more question. I understand how the constant becomes zero. Then what is the explanation for xi (yi-y) being equal to yi (xi-x)? Does it have something to do with the deviation of the x data and the y data having some sort of relation?
No,it is just that both are zero.
No,it is just that both are zero.
Thsnks a lot!