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The correlation coefficient of the data pairs $(x_i,y_i), i=1,\dots,n$ is given by $$r = \frac{\sum_{i=1}^n(x_i-\bar{x})(y_i - \bar{y})}{(n-1)s_x s_y}.$$

Hence $r= 0$ iff $\frac{1}{n}\sum_{i=1}^n x_i y_i = \bar{x}~\bar{y}.$

How does one interpret this? I am teaching a course in Probability and Statistics for the Master's students in Math, Engg, Economics, and Astronomy. I would like to give an intuitive explanation for this. Any help is greatly appreciated.

kjetil b halvorsen
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Ashok
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    In my opinion it is easier to intuitively explain the numerator of the original formula. If the quadrants generated by $\bar x$ and $\bar y$ are considered, the summands $(x_i - \bar x)(y_i - \bar y)$ will have positive or negative signs depending on which quadrants the data points fall in. If all/most are in the upper right or lower left quadrants (positive linear situation), $r$ will be positive. Similar argument for $r$ negative if most in upper left and lower right. Taking the magnitudes also into consideration now $r = 0$ only if the data points balance out in the four quadrants. – Just_to_Answer Aug 06 '17 at 16:53
  • @Just_to_Answer: Thank you very much for your response. I haven't thought about this before indeed. However, still, is there some intuition for $\frac{1}{n}\sum_{i=1}^n x_i y_i = \bar{x}~\bar{y}$? – Ashok Aug 07 '17 at 04:33
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    By subtracting the right hand side from the left, you obtain the covariance. This is extensively discussed and interpreted at https://stats.stackexchange.com/questions/18058. Are you looking for anything different than that? – whuber Aug 07 '17 at 16:23
  • @whuber: Thank you for the interest shown. //Are you looking for anything different than that?// Yes, really. Can we say something about product of the averages being equal to average of the products? – Ashok Aug 08 '17 at 11:35
  • The link I provided explains how to interpret the covariance, so it comes down to interpreting what it means for the covariance (or the correlation) to be zero. If you are looking for more than is in that link, then please search this site for [Anscombe's quartet](https://stats.stackexchange.com/search?q=Anscombe+quartet). – whuber Aug 08 '17 at 12:57

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