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the following procedure has the scope to describe how, from the Least square method, I obtained the Pearson coefficient and the determination coefficient. At the end of the procedure I have several questions.

If I apply the least square method, finding the residual $r_{xi}$:

$r_{xi}=y_{i}-b_{x}x_{i}$ (1)

The least square method consists, firstly, in doing the sum of the square of the residual:

$\sum_{i}r_{i}^2=\sum_{i}(y_{i}-b_{x}x_{i})^2$ (2)

If I want to find the minimum of the square of the residual, so I do the partial derivation and is equal to 0:

$\frac{d\sum_{i}r_{i}^2}{db_{x}}=0$ (3)

$\frac{d\sum_{i}(y_{i}-b_{x}x_{i})^2}{db_{x}}=0$ (4)

$\sum_{i}(2b_{x} x_{i}^2-2 y_{i} x_{i})=0$ (5)

$b_{x} = \frac{\sum_{i} x_{i} y_{i}}{\sum_{i} x_{i}^2}$ (6)

Viceversa, if I start from from this residual $r_{yi}$:

$r_{yi}=x_{i}-b_{y}y_{i}$ (7)

Doing the same procedure I obtain:

$b_{y}= \frac{\sum_{i} y_{i}x_{i}}{\sum_{i} y_{i}^2}$ (8)

If I do the product of $b_{y}$ and $b_{x}$ I obtain

$ b_{y} b_{x} = \frac{(\sum_{i} y_{i}x_{i})^{2}}{\sum_{i} y_{i}^2\sum_{i} x_{i}^2} $ (9)

Finally if I do the square of this product I obtain:

$ \sqrt{b_{y} b_{x}} = \frac{\sum_{i} (y_{i}x_{i})}{\sqrt{\sum_{i} y_{i}^2\sum_{i} x_{i}^2}} $ (10)

In doing the least square method, that I posted above , I noticed that if I substitute in (10) $y_{i}$ and $x_{i}$ with $y’_{i}$ and $x’_{i}$ which are equal to :

$y’_{i}=y_{i}-\overline{y}$ (11)

$x’_{i}=x_{i}-\overline{x}$ (12)

where $\overline{y}$ and $\overline{x}$ are the mean of the samples $y_{i}$ and $x_{i}$; $ \sqrt{b_{y} b_{x}}$ is like the Pearson coefficient, whereas $ b_{y} b_{x}$ is like the determination coefficient. So my questions are:

  1. Is this the typical procedure to obtain the Pearson coefficient(PC) and the determination coefficent (DC)?
  2. Why the PC and the DC take in consideration only the fluctuation of $x$ and $y$?
  3. Theoretically, why taking the square root of the product of the angular coefficient ($ \sqrt{b_{y} b_{x}}$ ) should have the meaning of the PC, i.e. a parameter that assesses the correlation between the two parameter taken in consideration? Thank you in advance
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    $\begingroup$ numerator of $b_yb_x$ is incorrect in (9) $\endgroup$
    – gunes
    Dec 5, 2020 at 12:41
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    $\begingroup$ corrected! Thanks $\endgroup$ Dec 5, 2020 at 12:45
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    $\begingroup$ You should square the entire sum $\endgroup$
    – gunes
    Dec 5, 2020 at 13:11
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    $\begingroup$ Thanks, corrected again :), Lukily the mistake does not affect the purpose of the answer. Do you have an idea of how to answer? $\endgroup$ Dec 5, 2020 at 13:15

1 Answer 1

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The question is interesting. A possible explanation can be the fact that substituting in (10) $y_{i}$ and $x_{i}$ with $y’_{i}$ and $x’_{i}$ which are equal to :

$y’_{i}=y_{i}-\overline{y}$

$x’_{i}=x_{i}-\overline{x}$

Implicitly you are using a formula

$y=bx+a$

So the correlation coefficient take in account the possibility of an intercept a that has the formula:

$a=\overline{y}-b\overline{x}$

An other hint can be the fact that the formula (10) is a geometric mean.

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