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If we normalize (scale and center) x and y, then the slope from y_norm ~ x_norm is equal to cor(x, y). Why?

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Because, by definition y_norm and x_norm are unit-variance random variables whose covariance is related to the covariance of $X$ and $Y$ by $$\operatorname{cov}\left(\frac{X-\mu_X}{\sigma_X},\frac{Y-\mu_Y}{\sigma_Y}\right) = \frac{\operatorname{cov}(X,Y)}{\sigma_X\sigma_Y}.$$ Now apply the definition of cor(x,y).

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  • $\begingroup$ My question is very similar to: $\endgroup$
    – Alireza
    Oct 25, 2016 at 5:10
  • $\begingroup$ stats.stackexchange.com/questions/183778/… $\endgroup$
    – Alireza
    Oct 25, 2016 at 5:10
  • $\begingroup$ Not sure how to do a newline here! Anyways, I'm still not sure how the slope becomes equal to the correlation. And also the answers to the above post are not very clear. I came upon this problem when I was trying both y_norm ~ x_norm and x_norm ~ y_norm, and to my surprise, the slope were the same. $\endgroup$
    – Alireza
    Oct 25, 2016 at 5:13
  • $\begingroup$ See this question and its answers. $\endgroup$ Oct 25, 2016 at 16:35
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cor(x, y)=Cov(x,y)/Sd(x)sd(y) when you standardize.... your standard deviation of x and y will equal to 1 and therefore you will have 1 on your denominator and your cor(x,y) = cov(x,y)

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  • $\begingroup$ I understand this, but my whole question is why cov(x, y) is the same as slope of y ~ x. $\endgroup$
    – Alireza
    Oct 28, 2016 at 22:07

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